introduction to frequency-domain analysis of continuous-time, linear...

Introduction to Frequency-Domain Analysis ofContinuous-Time, Linear and Time-Invariant Systems

• Time-domain analysis of transient response

• Fourier series of periodic Dirichlet signals

• Bode plots of system frequency-response

• Bilateral Fourier transform for zero-state response

• Unilateral Laplace transform for transient response

c�2013 George Kesidis

1

Time-domain analysis of continuous-time LTI systems

• Signals: properties, operations, construction, important signals

• Single-input, single-output systems: properties

• The zero-input response (ZIR): characteristic values and modes

• The zero (initial) state response (ZSR): the impulse response, convolution

• System stability

• The (zero state) system transfer function and eigenresponse

• Resonance and step-response of second-order tuned circuits


2

Introduction to signals - discrete and continuous time

• A function x is a mapping of elements of a domain A to a range B, i.e.,

x : A ! B,

in such a way that for all elements t of A, x(t) is a single element of B, i.e.,

8t 2 A, x(t) 2 B.

• The strict range of x is {x(t) | t 2 A} ⇢ B.

• For example x(t) = t2, t 2 R, has strict range the non-negative real numbers,R�0

:= {z 2 R | z � 0} =: [0,1)

where “a := b” means true by definition of a.

• In this course, we will use the words “function” and “signal” interchangeably.

• Also, we will consider a one-dimensional, ordered domain A for our signals, and associatethat domain with time.

• Note that in image processing, signals (images) will have two-dimensional spatial domains.

• Video signals are three-dimensional (two-dimensional images as a function of time).


3

Signal/function types - discrete and continuous time (cont)

• If the domain A has “countably” many values, then x : A ! B is said to be discrete time.

• A set is countable if there is a one-to-one mapping between it and a subset of the set ofintegers, Z.

• In this course, we will consider only continuous time-signals x whose domains are subsetsof the real line, R, typically either A = R = (�1,1) or A = R�0

= [0,1), andhence uncountably infinite in size.

• We will also consider signals that are either real or complex valued, i.e., their range B = Ror B = C.


4

A signal operation/transformation - time shift

• For all signals x : R ! R and scalars ⌧ 2 R, define a new signal �⌧ x as

8t 2 R, (�⌧ x)(t) = x(t� ⌧),

i.e., the signal x delayed by ⌧ seconds (equivalently, advanced by �⌧ seconds).

• An example pulse x is depicted in Figure (a).

• ��2

x (x delayed by -2 seconds, or advanced by 2 seconds) is depicted in Figure (b); notethat 2 = (��2

x)(�1) = x(�1� (�2)) = x(1).

• �

2

x (x delayed by 2 seconds) is depicted in Figure (c); note that 2 = (�

2

x)(3) =

x(3� 2) = x(1).


5

Periodic and aperiodic signals

• For the following introductory discussion, we will use examples where x : R ! R.

• A signal is said to be periodic if there is a fixed real number T 6= 0 such that

8t 2 R, x(t) = x(t� T ).

• The smallest such T > 0 of a periodic signal x is said to be the period of x.

• We take constant (DC) signals, i.e.,

9c 2 R such that 8t 2 R, x(t) = c,

to be periodic with “fundamental frequency” 1/T = 0 (which motivates us to choosetheir period T = 1 by definition).

• A periodic signal with period T will be called T -periodic.

• A signal that is not periodic is said to be aperiodic.


6

Periodic signals - examples

• For example, the sinusoid x(t) = sin(5t), t 2 R, has period 2⇡/5.

• The signal (a) in the following figure has period 4.

• The truncated exponential signal (b) is aperiodic.


7

Expressing periodicity with the delay operator

• Let’s recall the time-delay operator � so as to restate the definition of periodicity infunctional form.

• A signal x is said to be periodic if there is a real T 6= 0 such that

x = �T x,

recalling that this statement expresses functional equivalence, i.e.,

8t 2 R, x(t) = (�T x)(t) := x(t� T ).

• So if x is T -periodic (for T > 0) then x is an “eigenfunction” of �nT for all n 2 Z.


8

Sums of periodic signals

• A periodic signal with period T > 0 “repeats itself” every nT seconds, for any integern > 0;

• that is, the pattern of the signal is repeated every consecutive interval of duration nTseconds.

• Suppose x and y are periodic signals with periods Tx > 0 and Ty > 0, respectively.

• If there are positive integers nx > 0 and ny > 0 such that nxTx = nyTy =: T , then thesum of the signals, x+ y, will repeat itself every T seconds, where to be clear,

8t 2 R, (x+ y)(t) := x(t) + y(t);

i.e., x+ y will also be a periodic signal with period T .

• The converse statement is also true, leading to:

• If x and y are periodic, then: x+ y is periodic if and only if Tx/Ty is rational (= ny/nx

for integers nx, ny > 0).


9

Sums of periodic signals - examples


10

Sums of periodic signals - examples (cont)

• Consider the following periodic signals: for t 2 R,

x1

(t) = 5cos(7t+1), x2

(t) = 13 sin(12t� 1),

y1

(t) = 4 sin(3⇡t+ ⇡/2), y2

(t) =

p2 sin(5⇡t� ⇡/6)

• x1

+ x2

is periodic because the period of x1

is 2⇡/7 and that of x2

is 2⇡/12; so theratio of their periods, 7/12 (or 12/7), is rational.

• y1

+ y2

is periodic because the period of y1

is 2⇡/(3⇡) = 2/3 and that of y2

is2⇡/(5⇡) = 2/5; so, the ratio of their periods, 5/3 (or 3/5), is rational.

• In these cases, the period of the sum is the least common integer multiple of the components’periods, e.g., the period of x

1

+ x2

is 2⇡, while that of y1

+ y2

is 2.

• However, x1

+y2

is aperiodic because the ratio of their periods is (2⇡/7)/(2/5) = 5⇡/7which is not rational (⇡ = 3.1415... is not a rational number).

• Adding a constant to a periodic signal results in a periodic signal, i.e., if x is T -periodicand c 2 R, then x+ c is T -periodic where (x+ c)(t) := x(t) + c, t 2 R.


11

Energy signals

• A signal x : R ! R is said to be an “energy signal” if

0 < Ex :=

Z 1

�1|x(t)|2dt < 1.

• Note that if x is either the voltage or current signal associated with a resistive load, thenEx would be proportional to the energy dissipated by the load.

• For example, for parameters A, T, b 2 R with b > 0, if

x(t) =

⇢

Ae�b(t�T ) if t � T0 if t < T,

then

Ex =

Z 1

�1|x(t)|2dt = A2

Z 1

T

e�2b(t�T )dt = A2/(2b);

since Ex = A2/(2b) < 1, x is an energy signal.

• The exponential signal x(t) = �7e�5t, t 2 R, is not an energy signal because Ex = 1.


12

Power signals

• A signal x : R ! R is said to be a “power signal” if the following limit exists and

0 < Px := lim

⌧!1

1

⌧

Z ⌧/2

�⌧/2

|x(t)|2dt < 1,

i.e., the Px is average energy per unit time, or the power, of x.

• Note that if x was an energy signal, then Px = 0.

• Conversely, if Px > 0, then Ex = 1.

• If a signal x is periodic with period T , then

Px =

1

T

Z

T

|x(t)|2dt

whereR

Tdenote integration over any interval of R of length T , i.e., over any period of x.

• To see why, note that if ⌧/T is an integer, then the numerator of Px can be written as asum of ⌧/T identical, consecutive integrals over periods of length T .

• Power signals are not necessarily periodic, though all of our examples will be.


13

Power versus energy signals - examples

• Fig. (b) is an energy signal with

Eg =

Z 1

�1g2(t)dt =

Z 1

�1

4e�2(t+1)dt = �2e�2(t+1)

�

�

�

1

�1

= 2.

• Fig. (a), a wave with period T = 4, is a power signal with

Pf =

1

4

Z

2

�2

f2

(t)dt =1

4

✓

Z

0

�2

1

2dt+

Z

2

0

(1� t)2dt

◆

=

t

4

�

�

�

�

0

�2

�1

12

(1� t)3�

�

�

�

2

0

=

2

3

.

• Adding a non-zero constant c 2 R to (i.e., spatially shifting) any energy signal x gives apower signal x+ c with Px = c2.

• Also, for any T -periodic, power-signal x and non-zero constant c 2 R, Px+c = Px + c2

only if x has zero mean, i.e.,R

Tx(t)dt = 0, cf. Parseval’s theorem.


14

Power versus energy signals - examples

• Consider a real-valued sinusoid of canonical form (referenced to cosine):

x(t) = A cos(wot+ �), t 2 R,where the parameter A > 0, A 2 R (i.e., A 2 R>0) is the amplitude, period T =

2⇡/wo 2 R>0, and phase � 2 R (since cosine has period 2⇡, we can take � 2 [�⇡,⇡)).

• A sinusoid is a power signal with

Px =

1

2⇡/wo

Z

2⇡/wo

0

A2

cos

2

(wot+ �)dt

=

A2wo

2⇡

Z

2⇡/wo

0

✓

1

2

cos(2wot+2�) +1

2

◆

dt

=

A2wo

2⇡

✓

1

4wosin(2wot+2�) +

1

2

t

◆

�

�

�

�

2⇡/wo

0

=

A2wo

2⇡

✓

0+

⇡

wo

◆

=

A2

2

.

• Thus, the root mean-square (RMS) value of a sinusoid x ispPx = A/

p2.


15

Periodic extensions of aperiodic signals of finite support

• The support of a signal x is the interval of time over which it is not zero.

• A signal of finite support is sometimes called a pulse.

• In this example, the pulse x has support of duration 4.

• For T � 4, the T -periodic extension of x depicted in this figure is

xT(t) :=

1X

n=�1x(t� nT ) =

1X

n=�1(�nT x)(t), t 2 R,

i.e., superposition of time-shifted pulses x by all integer (n) multiples of T .

• If x is an energy signal then xT

will be a power signal with PxT

= Ex/T .


16

Neither all signals of finite support are energy signalsnor all periodic signals are power signals

• Consider the signal

x(t) =

⇢

1

1�t0 t < 1

0 else

• The support of this signal is of duration 1.

• But note that limt"1 x(t) = 1, i.e., the signal has finite escape time.

• This signal is not an energy signal because Ex = 1.

• Also, any periodic extension of this signal would not be a power signal.


17

Bounded signals

• A signal x is bounded if there exists a real M < 1 such that

8t 2 R, |x(t)| M.

• Obviously, the signal of the previous example is not bounded.

• The sinusoid x(t) = A cos(wot+ �), t 2 R is bounded by M = A > 0.

• The signal x(t) = c+ A cos(wot+ �), t 2 R, where the constant c 2 R, is boundedby M = max{A+ c, |�A+ c|} (note that c may be negative).

• Note that the signal x(t) = 1/pt for t � 1, and x(t) = 0 else, is a bounded signal

(M = 1) with limt!1 x(t) = 0, but it is

– not an energy signal since Ex = lim⌧!1R ⌧

1

t�1dt = lim⌧!1 log(⌧) = 1, and

– not a power signal since Px = lim⌧!1 ⌧�1

log(⌧/2) = 0.


18

Bounded signal - here 8t, |x(t)| < M


19

Signal types - discussion

• We will define other types of signals in the following as needed, e.g., causal signals, Dirichletsignals.

• Such definitions can be extended to C-valued signals by, e.g.,

– simultaneously applying the definition to the both the real and complex components ofthe signal, e.g., for a C-valued signal to be periodic, both its real and imaginary partsneed to be periodic with rational ratio of their periods; or

– using modulus instead of absolute value, i.e., for x 2 C,

|x|2 = |Re{x}|2 + |Im{x}|2.


20

Signal operations - time scaling

• Recall the time-delay operator �⌧ for any fixed ⌧ 2 R: (�⌧ x)(t) := x(t� ⌧) 8t 2 R.

• A signal x : R ! R time-dilated by factor a > 0 (equivalently, time-contracted by factor1/a) is the signal

(Sa x)(t) := x(t/a), t 2 R.

• In the example for x(t/a), the rectangular pulse portion ends at t/a = 2, i.e., at t =

2a = 4 for a = 2, and t = 2a = 1 for a = 1/2.

• Also, note how the y-intercept does not change since, obviously, 0 = 0/a for all a > 0.


21

Time inversion

• A signal x : R ! R time-inverted (or time-reflected) is the signal

(S�1

x)(t) := x(�t), t 2 R.

• Basically, the time-inverted signal of x is x’s reflection in the y-axis, as the following exampleillustrates.


22

Combinations of temporal (domain) operations

• Example: Given a signal x : R ! R, how would one plot {x(�2t+2), t 2 R}?

• Note that

x(�2t+2) = (��2

x)(�2t) = (S

1/2��2

x)(�t) = (S�1

S

1/2��2

x)(t), t 2 R.

• So, beginning with the signal x (Fig. (a) below), do the following:

(b) time delay by -2 (advance by 2) giving x(t+2) =: r(t), then

(c) time dilate by 1/2 (contract by 2), r(2t) = x(2t+2) =: v(t), then

(d) time reflect v(�t) = x(�2t+2).

• To verify one point: by Fig. (d) 1 = x(�2t+2)|t=1

= x(0) = 1 by Fig. (a).


23

Combinations of temporal (domain) operations (cont)

• It’s easy to verify that time dilation and reflection (S) operations generally commute; forexample

S�1

S

0.5 = S

0.5 S�1

= S�0.5 .

• But time-dilation/reflection S and time-shift � operations generally do not commute.

• In the previous example, Fig. (e) depicts ��2

S�0.5 x which is obviously di↵erent fromS�0.5��2

x depicted in Fig. (d); this is simply because �2(t+2) 6= �2t+2.

• We can also deal with combined temporal operations by first factoring, e.g.,

x(�2t+2) = x(�2(t� 1)) = (S�0.5 x)(t� 1) = (�

1

S�0.5 x)(t), t 2 R.

• So, we would do the time-scale/reflection operations before the time-shift, but obviouslywith di↵erent parameters than used without first factoring as above.

• In summary for this example, we’ve shown

�

1

S�0.5 = S�0.5��2

6= ��2

S�0.5 .


24

Spatial (range) operations

• For a signal x : R ! R:

– A spatial increase (upward shift) by constant b 2 R leads to the signal(x+ b)(t) = x(t) + b, t 2 R.

– A spatial amplification (dilation) by factor a 2 R leads to the signal(ax)(t) = a · x(t), t 2 R.

– A spatial reflection (in the time axis) leads to the signal�x(t), t 2 R.

• As with temporal operations, spatial shift and spatial dilation operations generally do notcommute.

• If combinations of spatial and temporal operations are in play, then any spatial operationwill commute with any temporal operation.


25

Spatial (range) operations - example

• For the example pulse x of Fig. (a), �3x(t) + 5 can be plotted by

(b) first spatially reflecting and amplifying x by 3 to get �3x(t),

(c) then shift up by 5.

• Alternatively, since �3x + 5 = �3(x � 5/3), one can first shift x down by 5/3, thenspatially reflect and amplify by 3.

• It would be a mistake to first shift by 5 then scale by -3 as this would result in�3(x+5) =

�3x� 15 6= �3x+5.


26

Even and odd signals

• With the notion of time-inversion, we can define other signal attributes.

• A signal x : R ! R is even if x(t) = x(�t) 8t 2 R.

• A signal x is odd if x(t) = �x(�t) 8t 2 R.

• Clearly, {cos(t), t 2 R} is an even signal, {sin(t), t 2 R} is an odd signal,and {3e�5t, t 2 R} is neither.

• Any signal can be written as a sum of even and odd signals through the identity

x(t) =

x(t) + x(�t)

2

+

x(t)� x(�t)

2

8t 2 R, i.e., x =

x+S�1

x

2

+

x� S�1

x

2

,

where xe

(t) :=

1

2

(x(t) + x(�t)) is even and xo

(t) :=

1

2

(x(t)� x(�t)) is odd.

• For example, we can write

3e�5t= 3

e�5t+ e5t

2

+ 3

e�5t � e5t

2

= 3cosh(5t) + (�3sinh(5t)) 8t 2 R,

where sinh (hyperbolic sine) is odd () -3sinh is odd) and cosh is even.


27

Even and odd signal decomposition - graphical example

• For the sub-domain �1 t 1:x(t) = 1 + t, x(�t) = 1 � t,xe

(t) = (1 + t + 1 � t)/2 = 1, andxo

(t) = (1+ t� (1� t))/2 = t.

• For the sub-domain 1 t 2: x(t) =

2, x(�t) = 0, xe

(t) = (2 + 0)/2 =

1, and xo

(t) = (2� 0)/2 = 1.


28

Even and odd signal - properties

• It’s easy to show that:

– If x is odd (respectively, even) then cx is odd (respectively, even) for all scalars c 2 R.

– The sum of odd functions is odd.

– The sum of even functions is even.

– The product of even functions is even.

– The product of odd functions is even.

– The product of an even and an odd function is odd.

• For example, if f and g are arbitrary odd functions, then 8t 2 R,

(fg)(�t) = f(�t)g(�t) = (�f(t))(�g(t)) = f(t)g(t) = (fg)(t),

i.e., fg is even.


29

Some important signals - polynomials

• This course will focus on certain families of signals and combinations made from them.

• x is a polynomial of degree or order n 2 {0,1,2, ...} if 8t 2 R,x(t) = ant

n+ an�1

tn�1

+ ...+ a1

t+ ao

=

nX

k=0

aktk,

where we will take the scalar coe�cients ak 2 R 8k 2 {0,1,2, ..., n}.

• By the fundamental theorem of algebra, x has n roots in C, i.e., there are n solutions tkto x(t) = 0 (allowing for multiple, identical roots) so that

x(t) = an(t� t1

)(t� t2

)...(t� tn)

= an

nY

k=1

(t� tk).

• When all coe�cients are real, all roots of x are either real or come in complex-conjugatepairs.

• Obviously, the constant signal x = a0

is a polynomial of order zero.


30

Exponential signals

• Real-valued exponential signals have the form x(t) = Aebt, t 2 R, where A, b 2 R.

• Periodic, complex-valued exponential signals have the form x(t) = Aej(wt+�), t 2 R,where j :=

p�1 and A > 0, w,� 2 R and the period is 2⇡/w.

• In electrical engineering, we don’t use i forp�1 as i typically symbolizes electrical current.

• Since the signal {ejv, v 2 R} has period 2⇡, we can take the phase � 2 [�⇡,⇡).

• By the Euler-De Moivre identity,

x(t) = Aej(wt+�)= A cos(wt+ �) + jA sin(wt+ �), t 2 R.

• Here, the constant (or “direct current” (DC)) signal obviously corresponds to w = 0

(again, infinite period).

• Also, here w is the real-valued frequency – we will consider complex-valued frequencies(s 2 C) when we study transient response by unilateral Laplace transform.


31

Exponential signals (cont)

• For the frequency s = �+jw 2 C, consider the complex exponential function x : R ! C:

x(t) = Aest+j�= Ae�tej(wt+�), t 2 R.

• Note the real part of this signal is a sinusoid with exponential envelope:

Re{x}(t) = Ae�t cos(wt+ �), t 2 R.

• In the figure, e�2tcos(10⇡t) is plotted with its envelope ±e�2t, i.e., w = 10⇡ so that

the period T = 2⇡/(10⇡) = 0.2, and � = �2 < 0.


32

The unit-step function, u

• The unit-step will be denoted by u:

u(t) =

⇢

1 if t � 0

0 else (t < 0)

• Note that u(0) = 1 is indicated by the solid period in the figure; and just before time 0,u(0�) = 0.

• The unit step obviously has a single jump discontinuity of unit height at the time-origin.


33

The unit impulse (Dirac delta function), �

• Suppose we wish to di↵erentiate the unit-step function, u, i.e., computeu = du/dt = Du, where we have defined the time-derivative operator

D :=

d

dt

• Clearly, (Du)(t) = 0 for all t 6= 0, but what happens at the jump discontinuity, t = 0?

• Consider approximating the derivative (Du)(0) by a di↵erence quotient �" = u��"u"

:

�"(t) :=

u(t)� u(t� ")

", t 2 R,


34

The unit impulse (Dirac � function), � (cont)

• As " ! 0, �" is taller and thinner but with constant area = 1, i.e.,

8" > 0,

Z 1

�1�"(t) = 1.

• That is, as " ! 0, the support of �" disappears, but its area remains constant as its valueat the origin diverges " 1.

• So as " ! 0, �" converges to the unit impulse �, graphically depicted as

"


35

The unit impulse, � (cont)

• Since the unit step is constant except at the origin,

Du = �,

i.e., the unit impulse is the derivative at a unit jump discontinuity.

• We will see later how this peculiar function is crucial to our understanding of linear systems.

• For now, we will examine some of its important properties.

• We’ve already ascertained that

8t 6= 0, �(t) = 0

8t 2 R,Z t

�1�(⌧)d⌧ = u(t)

where the second fact follows from � = Du and also implies that, for all intervals I ⇢ Rcontaining zero in their interior,

Z

I�(⌧)d⌧ = 1.


36

Unit impulse, � - ideal sampling property

• Since 8t 6= 0, �(t) = 0, if the signal x is continuous at the origin, then

(x�)(t) = x(t)�(t) = x(0)�(t)

where x(0)�(t) is an impulse but a unit impulse only if x(0) = 1.

• By definition, the impulse “↵�” for scalar ↵ 2 R satisfiesZ 1

�1↵�(t)dt = ↵.

• The ideal sampling property of the unit impulse is: For all signals x continuous at the origin,Z 1

�1x(t)�(t)dt = x(0).

• By a simple change of variables of integration: for all signals x continuous at time T 2 R,Z 1

�1x(t)�(t� T )dt = x(T ).


37

Unit impulse, � - jump discontinuity example


38

Signal construction - example

• For ⌧ > �, note the function:

(�� u��⌧ u)(t) = u(t� �)� u(t� ⌧) =

⇢

1 if � t < ⌧0 else,

• That is, �� u � �⌧ u is a rectangular pulse of unit height and support [�, ⌧); as such,it is a “window” function on this interval.

• This signal x is linear (x(t) = t + 1) just on the time-interval [�1,1], and a constant(x(t) = 2) just on [1,2].

• So, x(t) = (t+1)(u(t+1)� u(t� 1)) + 2(u(t� 1)� u(t� 2)), t 2 R.


39

Window function


40

Dirichlet signals

• In this course, we will focus on signals that are Dirichlet so as to study signals and linearsystems in a frequency domain.

• A signal x : R ! R:

– Satisfies the weak Dirichlet condition if it is absolutely integrable, i.e., ifZ 1

�1|x(t)|dt < 1,

or just absolutely integrable over any period T if x is periodic, i.e.,Z

T

|x(t)|dt < 1.

– Satisfies the strong Dirichlet conditions if over any finite interval of time: (i) it has afinite number of jump discontinuities, and (ii) it has a finite number of minima andmaxima.

• Note how the strong Dirichlet condition (ii) is not possessed by the bounded signal x(t) =

cos(1/t), t 2 R as it has infinitely many extrema in any interval containing the origin(t = 0).


41

Systems - single input, single output (SISO)

• In the figure, f is an input signal that is being transformed into an output signal, y, by thedepicted system (box).

• To emphasize this functional transformation, and clarify system properties, we will writethe output signal (i.e., system “response” to the input f) as

y = Sf,where, again, we are making a statement about functional equivalence:

8t 2 R, y(t) = (Sf)(t).

• Again, Sf is not S “multiplied by” f , rather a functional transformation of f .


42

SISO systems (cont)

• The n signals xk, k 2 {1,2, ..., n}, are the internal states of the system.

• We will see that states can be taken as “outputs of integrators,” e.g., the states of a linearcircuit can be taken as the currents through inductors and voltages across capacitors.

• In discussing the following properties of systems, we will generally consider Dirichlet, continuous-time, complex-valued input signals, f : R ! C.

• But if there is a finite origin of time (e.g., f : [0,1) ! C) the following properties pertainto systems with state variables that are initially zero (8k, xk(0) = 0).


43

Linear systems

• A linear system S has the following two properties:

– (Homogeneity/Scaling) For all input signals f and scalars a,

S(af) = aSf,where we remind that, by definition, 8t 2 R, (af)(t) = a ·f(t) and “·” is just scalarmultiplication.

– (Superposition) For all signals f1

and f2

,

S(f1

+ f2

) = Sf1

+ Sf2

,

where we remind that, by definition, 8t 2 R, (f1

+ f2

)(t) = f1

(t) + f2

(t) and“+” on the right-hand-side is just scalar summation.

• Implicitly, when defining such system properties, we are restricting our attention to signalsf for which Sf : R ! C is a well-defined (output) signal, restricting our attention toDirichlet signals in particular.


44

Linear systems - examples

• Suppose a system is defined by the input-output relationship,

8t 2 R, y(t) = 3

Z t

�1f(⌧)d⌧ = (Sf)(t).

• This system is linear owing to the linearity property of the integration operator (and com-mutativity of scalar multiplication and addition).

• Scaling: For all input signals f , scalars a, and time t 2 R,

(Saf)(t) = 3

Z t

�1af(⌧)d⌧ = a3

Z t

�1f(⌧)d⌧ = a(Sf)(t),

i.e., S(af) = aSf .

• Superposition: For all input signals f1

, f2

, and time t 2 R,

(S(f1

+ f2

))(t) = 3

Z t

�1(f

1

(⌧) + f2

(⌧))d⌧ = 3

Z t

�1f1

(⌧)d⌧ +3

Z t

�1f2

(⌧)d⌧

= (Sf1

+ Sf2

)(t),

i.e., S(f1

+ f2

) = Sf1

+Sf2

- by the absolute integrability property of Dirichlet signals,the integral of the sum (of integrands) is the sum of the integrals (of each integrand).

45

Linear systems - examples (cont)

• Suppose a system is defined by the input-output relationship,

8t 2 R, y(t) = 3f(t) + 7;

• or just y = 3f + 7 = Sf where “7” here is understood to be the constant signal = 7

for all time.

• Clearly, in general,

S(f1

+ f2

) = 3(f1

+ f2

) + 7 6= 3f1

+ 7+ 3f2

+ 7 = Sf1

+ Sf2

.

• So, this is not a linear system because it does not have the superposition property.

• One can easily check that the scaling property also fails, i.e., that:for all signals f 6= 0 and scalars a such that

S(af) 6= aSf.

• This system is said to be a�ne.

• For a signal, we will use the adjectives a�ne and linear interchangeably.


46

Disproving system linearity - discussion

• The logical negation of a universal statement is an existential one (i.e., there is a counter-example).

• Also recall De Morgan’s rule: for all statements p and q,

⇠ (p and q) = (⇠ p) or (⇠ q),

where ⇠ here represents logical negation.

• So, to disprove linearity, we either need only find a scalar and input-signal for which thescaling property does not hold or a pair of input signals for which the superposition propertydoes not hold.

• In the previous example of an a�ne system, both properties more generally do not hold,i.e., not just for a single counter-example.


47

Linear systems - examples (cont)

• Consider the system whose (zero initial-state) output y to input f are related by

3D y + y = f2

• To check whether this system is linear, consider arbitrary input signals f1

and f2

and scalars↵1

and ↵2

.

• For k 2 {1,2}, let yk be the system output to fk, i.e., 3D yk + yk = f2

k .

• Thus, by linearity of the derivative D operator,

3D(↵1

y1

+ ↵2

y2

) + (↵1

y1

+ ↵2

y2

) = ↵1

(3D y1

+ y1

) + ↵2

(3D y2

+ y2

)

= ↵1

f2

1

+ ↵2

f2

2

6= (↵1

f1

+ ↵2

f2

)

2,

i.e., ↵1

f2

1

+ ↵2

f2

2

6= (↵1

f1

+ ↵2

f2

)

2 for arbitrary signals fk and scalars ↵k.

• For example, if f1

⌘ 1 ⌘ f2

(the constant signal 1) and ↵1

= 1 = ↵2

, then↵1

f2

1

+ ↵2

f2

2

= 2 6= 4 = (↵1

f1

+ ↵2

f2

)

2.

• So, this system is not linear.


48

Time-invariant systems

• Recall the time-shift (delay) operator �⌧ f ,

8t 2 R, (�⌧ f)(t) = f(t� ⌧).

• A system S is time-invariant if for all ⌧ 2 R and Dirichlet f : R ! C,

S�⌧ f = �⌧Sf,i.e., if S and �⌧ commute.

• That is, if y = Sf , then the system S is time-invariant if for all signals f and delays ⌧ ,the response to �⌧ f (i.e., “f(t� ⌧)”) is �⌧ y (i.e., “y(t� ⌧)”).

• Or even more simply, a delay in the input results in the same delay in the output.


49

Time-invariant systems - examples

• For example, consider again the integrator system

8t 2 R, (Sf)(t) =

Z t

�1f(r)dr.

• For all Dirichlet f and ⌧ 2 R:

8t 2 R, (S�⌧ f)(t) =

Z t

�1(�⌧ f)(r)dr

=

Z t

�1f(r � ⌧)dr

=

Z t�⌧

�1f(r0)dr0, r0 := r � ⌧

= (Sf)(t� ⌧)

= (�⌧Sf)(t).

• Thus, this system is time invariant.


50

Time-invariant systems - examples (cont)

• Consider again the a�ne system defined by y = Sf = 3f +7.

• Clearly, for all signals f : R ! C and delays ⌧ 2 R: 8t 2 R,

S(�⌧ f)(t) = 3(�⌧ f)(t) + 7 = 3f(t� ⌧) + 7 = �⌧(Sf)(t) = y(t� ⌧),

where y := Sf .

• So, this system is time-invariant.

• For the system 8t 2 R, (Sf)(t) = 3f(t) + 7t: if ⌧ 6= 0 then

S(�⌧ f)(t) = 3(�⌧ f)(t) + 7t = 3f(t� ⌧) + 7t 6= 3f(t� ⌧) + 7(t� ⌧) = (�⌧Sf)(t).

• That is, there are cases (in fact, all but ⌧ 6= 0) where time-shift in the input does not leadto the same-time-shift in the output.

• So, this system is time varying, i.e., not time invariant.

• One can similarly show that the system 8t 2 R, (Sf)(t) = 3tf(t) + 7 is also time-varying.


51

Time-invariant systems - examples (cont)

• Consider the system whose output to input f is y(t) = (Sf)(t) = f(t2), 8t 2 R.

• To check for time-invariance, take an arbitrary input f and scalar T and let g = �Tf ,i.e., g(t) = f(t� T ), 8t 2 R.

• The output to g is (Sg)(t) = g(t2) = f(t2 � T ), 8t 2 R.

• Also, (�Ty)(t) = f((t� T )

2

), 8t 2 R.

• So, this system is not time-invariant because for arbitrary f, T ,

(S�Tf)(t) = f(t2 � T ) 6= f((t� T )

2

) = (�TSf)(t), 8t 2 R.

• For example, if f(t) = t, 8t 2 R, and T = 1, then(S�Tf)(t) = t2 � 1 6= (t� 1)

2

= (�TSf)(t), 8t 6= 1.


52

Causal signals and systems

• A signal f is causal if f(t) = 0 for all t < 0.

• Note that for any signal x : R ! R, f = xu is a causal signal.

• A system S is said to be causal if: for all causal input signals f , the output signal (response)Sf = y is also causal.

• As we will be focusing on time-invariant (and linear) systems, the choice of time origin willbe arbitrary, and we will typically choose 0.

• If the system has non-zero response at negative time, it is said to be anticipating the causalsignal which does not “arrive” till time 0.

• So, non-causal systems are said to be anticipative.


53

Causal systems - examples

• Consider again, the integrator system 8t 2 R, (Sf)(t) =

R t

�1 f(r)dr.

• Take arbitrary causal f , f(t) = 0 for all t < 0.

• Thus, for all t < 0, (Sf)(t) =

R t

�1 f(r)dr =

R t

�1 0dr = 0.

• So, this system is causal.

• Now consider the example 8t 2 R, (Sf)(t) = 3f2

(t+2).

• If f is causal, then for t = �1 note that

(Sf)(�1) = 3f2

(�1+ 2) = 3f2

(1)

• So, (Sf)(�1) = 0 only if f(1) = 0, which may not be the case for an arbitrary causalsignal.

• So, for arbitrary causal f and t < 0, (Sf)(t) 6= 0 (e.g., when f(1) 6= 0).

• Hence this system is not causal.

• One can easily show that this system is also not linear, but it is time invariant.


54

Causal systems - examples (cont)

• Consider the system 8t 2 R, (Sf)(t) = 3tf(t) + 7.

• If f is causal, then for all t < 0,

(Sf)(t) = 3tf(t) + 7 = 3t · 0+ 7 = 7 6= 0.

• So this system is also not causal.

• Exercise: Is this system linear? time-invariant?

• Exercise: Is the system (Sf)(t) = 3f(t) + 7t, t 2 R, causal? linear? time-invariant?


55

Stateless/memoryless systems

• In a memoryless (or stateless) system, 8t 2 R and input signals f , the output at time t,y(t), depends on the input f only through f(t).

• For example, the system 8t 2 R, y(t) = (Sf)(t) = 3tf(t) + 7 is obviously a memo-ryless system.

• But in the integrator system 8t 2 R, y(t) = (Sf)(t) =

R t

�1 f(r)dr, clearly the outputy(t) generally depends on f(r) for r 6= t (here, for all r t).

• So this integrator system is stateful.

• We will see that stateful systems have associated internal state signals (again, we will denotethem with x).


56

Canonical representation of a SISO LTIC system by ODE

• In this course, we focus on linear and time-invariant (LTI) SISO systems S in continuoustime.

• If y = Sf represents the response of the system y to input f , then we will represent thesystem with the following ordinary di↵erential equation (ODE) relating y and f :

dny

dtn+ an�1

dn�1y

dtn�1

...+ a1

dy

dt+ a

0

y = bmdmf

dtm+ bm�1

dm�1f

dtm�1

...+ b1

df

dt+ b

0

f,

with integers n,m � 0, and the scalar coe�cients ak, bk 2 R, 8k.

• Note that we can take the coe�cient an = 1 without loss of generality, i.e., one can dividea system equation by the coe�cient of dny/dtn if it is not 1 to get a system equation inthe above form.

• Also note that this system equation is a functional equivalence in that it holds for all time,i.e., one can stipulate “(t)” for all participating signals giving an equation that would holdfor all t 2 R.


57

Compact ODE forms of a SISO system equation

• We will write such system equations more compactly in two ways.

• First, using the time-derivative operator

D :=

d

dt,

we can write the system equation as

D

n y + an�1

D

n�1 y...+ a1

D y + a0

y = bmD

m f + bm�1

D

m�1 f...+ b1

D f + b0

f.

• Note that D0 is just the identity map: 8f, D0f = f .

• Furthermore, we can define the system polynomials (with real coe�cients, an = 1),

Q(s) :=

nX

k=0

aksk and P (s) :=

mX

k=0

bksk, 8s 2 C.

• So, the system equation can be written simply as

Q(D)y = P (D)f

where P is a polynomial of degree m and Q, the characteristic polynomial of the system,has degree n.


58

This SISO system is LTIC

• The system with input f and output y (the response to f) satisfying

D

n y + an�1

D

n�1 y...+ a1

D y + a0

y = bmD

m f + bm�1

D

m�1 f...+ b1

D f + b0

f

is Linear, Time-Invariant and Causal (LTIC) because:

– the time-derivative operators Dk are LTIC for all integers k � 0, and

– the coe�cients bk, ak of the polynomials P and Q are all assumed to be constant,i.e., no time-varying coe�cients.

• In the following, we will explicitly “solve” this system (for y in terms of f) and more clearlydemonstrate that it is LTIC.


59

SISO Canonical ODE - resistive circuit example

• Consider this simple, ideal resistive circuit.

• By Kirchho↵’s current law (KCL) at node y (i.e., zero is the sum of the currents referencedinto the node between the resistors) and Ohm’s law (current in terms of node voltages),

f � y

R1

+

0� y

R2

= 0,

where “0” on the left-hand-side is the ground voltage.

• In canonical form, the input-output relationship is the voltage divider:

y =

R2

R1

+R2

f.

• Obviously, this system is memoryless/stateless.


60

SISO Canonical ODE - passive RC circuit example

• In this passive RC circuit, the KCL nodal equation at y is

f � y

R+ CD(0� y) +

0� y

R+ CD(0� y) = 0.

• In canonical form,

D y +

1

RCy =

1

2RCf.


61

SISO Canonical ODE - passive RC circuit example (cont)

• We now solve the system D y + (RC)

�1y = (2RC)

�1f , i.e., we find the output yexplicitly in terms of the input f and initial conditions.

• Using the co-factor method, assume the output is of the form

y(t) = A(t)e�t/(RC)

and substitute it into the system equation: 8t � 0,

(DA)(t)e�t/(RC) �1

RCy +

1

RCy =

1

2RCf(t)

) (DA)(t) = f(t)1

2RCet/(RC)

) A(t) = A(0�) +

Z t

0

f(r)1

2RCer/(RC)dr

) y(t) = A(0�)e�t/(RC)

+

Z t

0

f(r)1

2RCe�(t�r)/(RC)dr

• where 0 is the origin of time, and clearly this system is stateful.


62

SISO Canonical ODE - passive RC circuit example (cont)

• If the initial condition y(0�) is given just before the origin of time, then clearly A(0�) =

y(0�), i.e.,

8t � 0�, y(t) = y(0�)e�t/(RC)

+

Z t

0�f(r)

1

2RCe�(t�r)/(RC)dr.

• We will see how this form of the solution is important for transient analysis as it separatesout

– the portion of the output {y(t) | t � 0} due only to the initial condition, here simplythe constant signal y(0�), is called the Zero-Input Response (ZIR), and

– the portion due only to the input, hereR t

0� f(r) 1

2RCe�(t�r)/(RC)dr, t � 0 - this is

called the Zero-State Response (ZSR) or zero initial-state response.

• Note for the above example, if input f ⌘ 0, then the voltage source is a short circuit andthe energy stored in the capacitor will eventually dissipate, i.e., ZIR! 0 over time.

• For the example of Fig. 1.26 of Lathi p. 78:check that the system equation is D y = RD f + f/C so that 0 is the characteristicvalue - see equ. (1.36a). Here if f ⌘ 0, then the current source is an open circuit so thecapacitor’s energy does not dissipate, i.e., ZIR is constant - see (1.42).


63

SISO Canonical ODE - active RC circuit example

• Recall that with an op-amp (having a

power supply not shown), a relatively low-

powered input signal can control a sepa-

rately high-powered output signal.

• Recall that an ideal op-amp

– draws zero current from its two (±)

input leads, and

– has zero di↵erential input voltage,

– i.e., has infinite input impedance.


64

SISO Canonical ODE - active RC circuit example (cont)

• So, for this circuit, the voltage at the negative input terminal of the op-amp is also f .

• Thus, KCL nodal equation at the “�” input terminal is

vA � f

R1

+ C1

D(y � f) = 0

) vA = f +R1

C1

D f �R1

C1

D y.

• Also, KCL at node A is

f � vA

R1

+

y � vA

R2

+ C2

D(0� vA) = 0.

• Substituting vA into KCL at A and then rewriting in canonical form gives:

D

2 y +

1

C2

✓

1

R1

+

1

R2

◆

D y +

1

R2

C2

R1

C1

y

= D2f +

✓

1

R1

C1

+

1

C2

✓

1

R1

+

1

R2

◆◆

D f +

1

R2

C2

R1

C1

f.


65

Time-Domain Analysis of LTIC SISO Systems - Introduction

• Consider a LTIC SISO system with input f : [0,1) ! R and output y : [0,1) ! R,i.e., with time origin 0.

• We will now study solutions of general LTIC SISO systems of the form

Q(D)y = P (D)f,

with initial conditions (Dn�1 y)(0�), (Dn�2 y)(0�), ..., (D y)(0�), y(0�), where

– D is the time derivative operator;

– Q is the system’s characteristic polynomial of degree n � 1 with real coe�cients:Q(D) =

Pnk=0

akDk, an = 1;

– P is a polynomial of degree m with real coe�cients: P (D) =

Pmk=0

bkDk;

• By “solution” we mean to find the entire signal y : [0,1) ! R given:

– a causal input signal f with zero initial conditions, (Dk f)(0�) = 0 for all integersk � 0 (so f typically specified with a step-function u factor);

– the n initial conditions; and

– the model of the system itself, P,Q.


66

Approach to solution: ZIR and ZSR

• The total solution y to P (D)f = Q(D)y and the given initial conditions is a sum of twoparts:

– the ZSR, yZS

, which solves

P (D)f = Q(D)yZS

with zero initial conditions, i.e., (Dk y)(0�) = 0 8 integers k � 0; and

– the ZIR, yZI

, which solves

0 = Q(D)yZI

with the given initial conditions.

• The total response y of the system to f and the given initial conditions is, by linearity,

y = yZI

+ yZS

.

• We will determine the ZIR by finding the characteristic modes of the system.

• We will determine the ZSR by convolution of the input with the impulse response.


67

Total response - example

• Recall that this RC circuit with input f and output y is described by

D y +

1

RCy =

1

2RCf

i.e., Q(D) = D+1/(RC) with degree n = 1, and P (D) = 1/(2RC) with degreem = 0.


68

Total response - example

• Also recall that the total-response is

y(t) = y(0�)e�t/(RC)

+

Z t

0�f(r)

1

2RCe�(t�r)/(RC)dr 8t � 0,

i.e., 8t � 0,

yZI

(t) = y(0�)e�t/(RC),

yZS

(t) =

Z t

0�f(r)

1

2RCe�(t�r)/(RC)dr.


69

Comment: Solution by homogeneous and particular components

• You may have learned how to find the total response by summing

– any “particular” solution yp

satisfying P (D)f = Q(D)yp

irrespective of initial con-ditions and possibly including characteristic modes, and

– a generic “homogeneous” solution yh

satisfying 0 = Q(D)yh

consisting only of ncharacteristic modes, with n free parameters/coe�cients.

• The n free parameters of the homogeneous solution are determined by applying the n initialconditions to the total response, y = y

p

+ yh

.

• The di↵erence of this approach is simply that

– yp

may be the ZSR together with a portion of the ZIR, and

– yh

is the remaining portion of the ZIR.

• That is, the ZSR is the special “particular” solution with zero initial conditions, and theZIR is the special “homogeneous” solution that accounts for the initial conditions by itself.


70

ZIR - the characteristic values

• To solve for the ZIR, i.e., solve

Q(D)y = 0 given (D

n�1 y)(0), (Dn�2 y)(0), ..., (D y)(0), y(0),

recall that exponential functions, {est | t 2 R} for frequency s 2 C, are eigenfunctions ofdi↵erential operators of the form Q(D) for a polynomial Q.

• That is, for scalar s 2 C, if we substitute y = {est | t 2 R}, we get8t 2 R, (Q(D)y)(t) = Q(D)est = Q(s)est.

• So, to solve Q(s)est = 0 for all time t � 0, where est is not always zero over this timeperiod, we require

Q(s) = 0, the characteristic equation of the system.

• If s is a root of the characteristic polynomial Q of the system,

– s would be a characteristic value of the system, and

– the signal {est | t � 0} is a characteristic mode of the system, i.e., Q(D)est = 0

8t � 0.

• Since Q has degree n, there are n roots of Q in C, each a system characteristic value.


71

ZIR - the characteristic values (cont)

• Though there may be some repeated roots of the characteristic polynomial Q, there willalways be n di↵erent, linearly independent characteristic modes, µk, i.e.,

8t � 0,

nX

k=1

�kµk(t) = 0 , 8k, scalars �k = 0.

• So, by system linearity, we will be able to write

8t � 0, yZI

(t) =

nX

k=1

ckµk(t),

for scalars ck 2 C that are found by considering the given initial conditions

(D

l y)(0�) = (D

lnX

k=1

ckµk)(0�) =

nX

k=1

ck(Dl µk)(0�), for 0 l n� 1,

i.e., n equations in n unknowns (ck).

• The linear independence of the modes implies linear independence of these n equations inck, and so they have a unique solution.


72

ZIR - the case of n di↵erent and real characteristic values

• If there are n di↵erent roots of Q in R, s1

, s2

, ..., sn, then there are n characteristic modes:for k 2 {1,2, ...n},

µk(t) = eskt, t � 0.

• Therefore,

8t � 0, yZI

(t) =

nX

k=1

ckeskt.

• Since 8k 2 {1,2, ..., n}, l � 0,

Dlµk = slkµk and µk(0) = 1,

the n unknown scalars ck 2 R can be solved using the n equations:

(D

l y)(0�) =

nX

k=1

ckslk, for 0 l n� 1.


73

ZIR - the case of n di↵erent and real characteristic values - example

• For this series RLC circuit with input f and output y, KVL gives

8t � 0, Ri(t) + L(D i)(t) + vc(0) +1

C

Z t

0

i(⌧)d⌧ = f(t).

• We can di↵erentiate this equation and substitute i = y/R to get

D y + LR�1

D

2 y + (RC)

�1y = D f

) D

2 y +RL�1

D y + (LC)

�1y = RL�1

D f,

i.e., P (s) = RL�1s and the characteristic polynomial of this second-order (n = 2)circuit is

Q(s) = s2 +RL�1s+ (LC)

�1.


74

ZIR - the case of n di↵erent and real char. values - example (cont)

• For example, if R = 700⌦, L = 0.1H, and C = 1µF, then

Q(s) = s2 + 7000s+10

7

= (s+5000)(s+2000).

• So, the characteristic values are -5000, -2000.

• The characteristic modes are exp(�5000t), exp(�2000t) 8t � 0.

• The ZIR is of the form

8t � 0�, yZI

(t) = c1

e�5000t+ c

2

e�2000t.

• If the given initial conditions are

(D y)(0�) = �2000V/s, y(0�) = 10V,

then

�2000 = �5000c1

� 2000c2

= (D yZI

)(0�)

10 = c1

+ c2

= yZI

(0�)

• Thus, c1

= �6 and c2

= 16, i.e.,

8t � 0�, yZI

(t) = �6e�5000t+16e�2000t


75

ZIR - the case of not-real characteristic values

• The characteristic polynomial Q may have non-real roots, but such roots come in complex-conjugate pairs because Q’s coe�cients ak are all real.

• For example, if the characteristic polynomial is

Q(s) = s3 + 5s2 + 11s+15 = (s+3)(s2 + 2s+5)

then the characteristic values (Q’s roots) are

�3, � 1± 2j recalling j =

p�1.

• Because we have three di↵erent characteristic values 2 C, we can specify three correspond-ing characteristic modes,

e�3t, e(�1+2j)t, e(�1�2j)t

and construct the ZIR as

yZI

(t) = c1

e�3t+ c

2

e(�1+2j)t+ c

3

e(�1�2j)t 8t � 0� .


76

ZIR - not-real characteristic values with real characteristic modes

• By the Euler-De Moivre identity,

8t � 0, yZI

(t) = c1

e�3t+2(c

2

+ c3

)e�tcos(2t) + 2j(c

2

� c3

)e�tsin(2t)

• Because all initial conditions are real and yZI

is real valued, c2

and c3

will be complexconjugates, i.e.,

c2

= c3

2 C.

• So, we can write

8t � 0�, yZI

(t) = c1

e�3t+ c

2

e�tcos(2t) + c

3

e�tsin(2t), where

c2

= (c2

+ c3

) = 2Re{c2

} 2 Rc3

= 2j(c2

� c3

) = �2Im{c2

} 2 Ri.e., just two “degrees of freedom” with these two coe�cients to accommodate the initialconditions, so not an “underspecified” scenario.

• Thus, instead of e(�1±2j)t, the following real-valued characteristic modes can be used

e�tcos(2t), e�t

sin(2t)


77

ZIR - not-real char. values with real char. modes (cont)

• So for t � 0�, if the ZIR is of the form

yZI

(t) = c1

e�3t+ c

2

e�tcos(2t) + c

3

e�tsin(2t),

) (D yZI

)(t) = �3c1

e

�3t+ c

2

[�e

�tcos(2t)� 2e

�tsin(2t)] + c

3

[�e

�tsin(2t) + 2e

�tcos(2t)],

) (D

2 yZI

)(t) = 9c1

e

�3t+ c

2

[�3e

�tcos(2t) + 4e

�tsin(2t)] + c

3

[�3e

�tsin(2t)� 4e

�tcos(2t)],

and the initial conditions are

(D

2 y)(0�) = 3, (D y)(0�) = 10, y(0�) = 5,

then the three equations to solve for the three unknown constants c1

, c2

, c3

are

5 = c1

+ c2

= yZI

(0�)

10 = �3c1

� c2

+ 2c3

= (D yZI

)(0�)

3 = 9c1

� 3c2

� 4c3

= (D

2 yZI

)(0�)

whose solution is

c1

= 6, c2

= �1, c3

= 13.5.


78

ZIR - not-real char. values with real char. modes (cont)

• In summary, for the case of a complex-conjugate pair of characteristic values ↵± j�, with↵,� 2 R, we can use the real-valued modes (with two di↵erent associated real coe�cients),

e↵t cos(�t), e↵t sin(�t) 8t � 0,

instead of complex-valued modes (with associated complex-conjugate coe�cients)

e(↵±j�)t 8t � 0,

in either case, the pair of coe�cients give two degrees of freedom to meet the given initialconditions.

• Restricting the span to real-valued signals, we can use these two real-valued modes because

span{e(↵+j�)t, e(↵�j�)t} = span{e↵t cos(�t), e↵t sin(�t)}, t 2 [0,1).

• Generally, linear combinations of characteristic modes span all possible real-valued ZIRsignals.


79

ZIR - the case of repeated characteristic values

• Consider the case where at least one characteristic value is of order > 1, i.e., repeatedroots of the characteristic polynomial, Q.

• For example, Q(D) = (D+7)

3

(D�2) has a triple (twice repeated) root at �7 and asingle root at 2.

• Again, {e�7t, t � 0} is a characteristic mode because Q(D)e�7t ⌘ 0 follows from

(D+7)e�7t= D e�7t

+7e�7t

= �7e�7t+7e�7t

= 0 8t � 0.

• Also, {te�7t, t � 0} is a characteristic mode because Q(D)te�7t ⌘ 0 follows from

(D+7)

2te�7t= (D

2

+14D+49)te�7t

= D

2 te�7t+14D te�7t

+49te�7t

= (�14e�7t+49te�7t

) + 14(e�7t � 7te�7t) + 49te�7t

= (�14+ 14)e�7t+ (49� 98+ 49)te�7t

= 0 8t � 0.


80

ZIR - the case of repeated characteristic values (cont)

• Similarly, t2e�7t is also a characteristic mode because (D+7)

3t2e�7t ⌘ 0.

• Note that without three such linearly independent characteristic modes {e�7t, te�7t, t2e�7t;

t � 0} for the twice-repeated (triple) characteristic value -7, the initial conditions will createan “overspecified” set of n equations involving fewer than n “unknown” coe�cients (ck)of the linear combination of modes forming the ZIR.

• For this example,

yZI

(t) = c0

e�7t+ c

1

te�7t+ c

2

t2e�7t+ c

3

e2t, t � 0� .

• If the given initial conditions are, say,

(D

3 y)(0�) = 10, (D

2 y)(0�) = �5, (D y)(0�) = 6, y(0�) = 12,

the four equations to solve for the four unknown coe�cients ck are:

yZI

(0�) = c0

+ c3

= 12

(D yZI

)(0�) = � 7c0

+ c1

+ 2c3

= 6

(D

2 yZI

)(0�) = 49c0

� 14c1

+ 2c2

+ 4c3

= �5

(D

3 yZI

)(0�) = � 343c0

+ 147c1

� 42c2

+ 8c3

= 10


81

ZIR - general case of repeated characteristic values

• In general, a set of k linearly independent modes corresponding to a characteristic values 2 C repeated k � 1 times (i.e., of multiplicity k) are,

tk�1est, tk�2est, ..., test, est, for t � 0.

• Also, if s = a ± jw are characteristic values repeated k � 1 times, with a,w 2 R andw 6= 0, we can use the 2k real-valued modes

treat cos(wt), treat sin(wt), for r 2 {0,1,2, ..., k � 1}.


82

Zero State Response - the impulse response

• Let h be a LTI system’s ZSR to the unit impulse, �.

• At this point, it’s di�cult to see why h should be a well-defined signal, let alone a crucialone to the study of LTI systems.

• In the following, we will derive h by inverse Fourier or inverse Laplace transform of thesystem transfer function H = P/Q.

• For a procedure to compute h directly from P (D)� = Q(D)h see, for example, Sec. 2.3of Lathi.

• For now, we will assume that h is known for a given LTI system and demonstrate how itcan be used to derive the ZSR to an arbitrary (Dirichlet) input signal f by the process ofconvolution.


83

ZSR - LTI property and convolution with impulse response

• Consider a LTI system with impulse response h and recall the approximate impulse function,

�" := "�1

(u��"u),

where 0 < " ⌧ 1.

• For arbitrary integer k, consider a short interval of time [k", (k + 1)") and supposeover this interval the input signal f , i.e., {f(t), k" t < (k +1)"}, is smooth enoughto be well approximated by a constant f(k"), i.e., for k" t < (k +1)",

f(t) ⇡ f(k")(u(t� k")� u(t� (k +1)")) = f(k")�"(t� k") ",

where "f(k")�"(t� k") is zero outside of this short interval of time.


84

ZSR - LTI property and convolution with impulse response (cont)

• Piecing together these non-overlapping approximations leads to a Riemann approximationof f over R:

f(t) ⇡1X

k=�1

f(k")�"(t� k")", 8t 2 R

• Note that as " # 0 (i.e., " # d⌧ , a di↵erential), this approximation becomes more precisetending to a Riemann integral that is the sampling property of the unit impulse:

f(t) =

Z 1

�1f(⌧)�(t� ⌧)d⌧, 8t 2 R.


85



86


• Again, we have constructed a Riemann approximation of the input signal as a linear com-bination of delayed approximations of the impulse:

f ⇡1X

k=�1

f(k") �k"�" "

• For the (zero initial states) LTI system under consideration, let the ZSR of �" be h".

• So, by time-invariance, the zero-state response to �k"�" is �k"h".

• Thus, by linearity of the system, the approximate ZSR to f is

yZS

⇡1X

k=�1

f(k") �k"h" "

i.e., 8t, yZS

(t) ⇡1X

k=�1

f(k") (�k"h")(t) "


87


• As " # 0, this Riemann approximate integral converges to the precise ZSR to f is 8t,

yZS

(t) =

Z 1

�1f(⌧)(�⌧h)(t)d⌧

=

Z 1

�1f(⌧)h(t� ⌧)d⌧

=: (f ⇤ h)(t), 8t 2 R,i.e., the convolution of the input f and (zero state) impulse response

h = lim

"#0h".

• Note how the convolution is a consequence of the LTI property (of the zero-state system).

• Also note that the convolution operator (⇤) is a bilinear mapping of two signals (here, fand h) to a third signal (y = f ⇤ h).


88

ZSR - impulse response - example

• Recall that the total response y to this (LTIC) circuit with causal input f and initialcondition y(0�) is 8t � 0,

y(t) = y(0�)e�t/(RC)

+

Z t

0�f(r)

1

2RCe�(t�r)/(RC)dr

= y(0�)e�t/(RC)

+

Z 1

0�f(r)

1

2RCe�(t�r)/(RC)u(t� r)dr

• Clearly, the ZSR portion is yZS

= f⇤h with impulse response h(t) = (2RC)

�1e�t/(RC)u(t),t 2 R, and f(t) = 0 for t < 0 (i.e., f = fu) since f is causal.

89

Total response - example - circuit with switches thrown at 0

• The circuit is operational for time t 2 R, i.e., including negative time when the switch onthe right is closed and the one on the left is open.

• At time zero, the two switches are simultaneously thrown.

• Prior to time zero, the circuit was in DC steady state (DC battery of A volts).

• So, y(0�) = 0 and vc(0�) = A (capacitor has infinite DC impedance).

• Since vc is a state of the circuit (it’s the output of an integrator (of y)), it’s continuous at0 so that vc(0�) = vc(0) = A.


90

Total response - example - circuit with switches thrown at 0 (cont)

• But y is not a state variable and at time 0, y(0) = (f(0)� vc(0))/R; so y will not becontinuous at 0, i.e., y(0�) 6= y(0), unless f(0) = A.

• By KVL, 8t � 0, f(t) = vc(t) +Ry(t) = vc(0) + C�1

R t

0� y(⌧)dt+Ry(t).

• Thus, R�1

D f = D y + (RC)

�1y.

• Exercise: Show by the co-factor method and then IBP that the total response is:

8t � 0, y(t) = y(0�)e�t/(RC)

+

Z t

0�R�1

(D f)(⌧)e�(t�⌧)/(RC)d⌧

= y(0�)e�t/(RC)

+

Z t

0�f(⌧)

✓

1

R�(t� ⌧)�

1

R2Ce�(t�⌧)/(RC)

◆

d⌧

= y(0�)e�t/(RC)

+

Z 1

0�f(⌧)h(t� ⌧)d⌧,

where the impulse response h(t) =

1

R�(t)� 1

R2Ce�t/(RC)u(t).

• Note: The total response will be discontinuous at time 0 when f(0) 6= A, i.e., y(0) =

y(0�) + f(0)/R, because h has an impulse component (again, f(0�) := 0).

91

Convolution of causal signals

• Generally, if

– the system is causal, so h is a causal signal, and

– the origin of time is zero, so that we can also assume the input f(t) = 0 for t < 0

(i.e., f is a causal signal too),

then

(f ⇤ h)(t) =

Z 1

�1f(⌧)h(t� ⌧)d⌧ =

Z t

0�f(⌧)h(t� ⌧)d⌧, 8t � 0.

• Here, since h is causal, the integrand is zero for ⌧ > t.

• And since f is causal, is zero for ⌧ < 0.

• Here, we integrate from 0� so that integrals involving impulses � at the origin are unam-biguous, i.e., we’re not “splitting hairs”.


92

Discussion: the ZSR has zero initial conditions

• We’ve identified the ZSR yZS

to a causal input f of a LTIC system with (causal) impulseresponse h as the convolution,

yZS

(t) =

Z t

0�f(⌧)h(t� ⌧)d⌧, t � 0.

• Here, we assume that the input f has zero “initial conditions,”

(D

k f)(0�) = 0 for all integers k � 0.

• So, clearly, yZS

(0�) = 0 and by di↵erentiating the above display, we get that (Dk yZS

)(0�) =

0 for all integers k > 0 as well.

• But yZS may not be continuous at time 0.


93

ZSR and ZIR - op-amp circuit example

• For another example, consider the above circuit input f , output y, and vc(0) = 2V.

• By KCL at “-” and vc, respectively,

y � f

1

+

vc � f

2

= 0 andf � vc

2

+ 0.5D(0� vc) = 0. (1)

• So, vc = 3f � 2y and by substitution,

Q(D)y = D y + y =

3

2

D f + f = P (D)f. (2)


94

ZSR and ZIR - op-amp circuit example (cont)

• The system characteristic polynomial is Q(D) = D+1, so that the system characteristic-value (system pole, root of Q) is -1, and characteristic mode is e�t.

• Thus, yZI

(t) = y(0�)e�t, t � 0�, where (by KCL at “-” op-amp input at time 0�)

y(0�) =

3

2

f(0�)�1

2

vc(0�) =

3

2

0�1

2

2 = �1;

noting vc(0�) = vc(0) since vc is a continuous state variable.

• For the ZSR, if the forced response yF

is known, one might be tempted to simply write

yZS

(t) = yF

(t) + ce�tu(t),

and solve for the coe�cient c of the characteristic mode by using yZS

(0) = 0.

• However, we’ll see later that when the degree of P equals that of Q (as in this example),and the input f(0) 6= 0, then y

ZS

will be discontinuous at time 0 too.


95

Op-amp circuit example: ZSR discontinuous at time 0

• The “transfer function” for this op-amp circuit,

H(s) =

P (s)

Q(s)=

1.5s+1

s+1

= 1.5�0.5

s+1

,

is proper but not strictly proper, where b1

= 1.5 is the leading coe�cient of P .

• We’ll see later that the impulse response h of this system has an impulse component �(indicating direct coupling between input and output),

h(t) = 1.5�(t)� 0.5e�tu(t).

• Thus, yZS = h ⇤ f = bnf + (h� bn�) ⇤ f ,

– where h�bn� corresponds to a “strictly proper” system ˜P/Q = P/Q�bn = H�bn,

– and so (h� bn�) ⇤ f is continuous at time 0 and so zero initial (at time 0) conditionsapply,

– but yZS

is discontinuous at time zero if f(0) 6= 0.


96

Op-amp circuit example: ZSR discontinuous at time 0 (cont)

• For example, if the input f(t) = e�2tu(t) then f is discontinuous at time 0 and theforced response (or “eigenresponse” since f is exponential) of this system is

yF

(t) = H(�2)f(t) = 2f(t).

• If we take yZS

(t) = 2f(t) + ce�tu(t) and find c using yZS

(t) = 0, we get c = �2.

• However, c = �0.5 for the true ZSR.

• One can verify this by finding the true ZSR by convolution, yZS

= h ⇤ f (or by using theunilateral Laplace transform which we will later discuss in detail).

• Indeed by linearity, for all t � 0,

yZS

(t) = (f ⇤ h)(t)

= e�2tu(t) ⇤ (1.5�(t)� 0.5e�tu(t))

= 1.5e�2t � 0.5

Z t

0

e�2⌧e�(t�⌧)d⌧ = 1.5e�2t � 0.5e�t(1� e�t

)

= 2e�2t � 0.5e�t.


97

Graphical convolution - example 1

• We now describe how to compute convolutions with the help of graphical illustrations forthe di↵erent cases of integration.

• For example, let’s compute f ⇤ h where

f(t) = 2e�tu(t) and h(t) = u(t)� u(t� 3), 8t 2 R.

• Again, (f ⇤ h)(t) :=

R1�1 f(⌧)h(t� ⌧)d⌧ .

• For this example, we will see graphically that there are three cases for time t (which is fixedwhen integrating over the dummy variable of integration ⌧ ).


98

Graphical convolution - example 1 - case 1

• Case 1: When t < 0,

f(⌧)h(t� ⌧) = 0, 8⌧ 2 R.

• So, simply because the integrand is constantly zero in this case,

(f ⇤ h)(t) =

Z 1

�1f(⌧)h(t� ⌧)d⌧ = 0.


99


• Case 2: When t � 0 and t� 3 < 0 (i.e., 0 t < 3),

(f ⇤ h)(t) =

Z 1

�1f(⌧)h(t� ⌧)d⌧

=

Z t

0

f(⌧)h(t� ⌧)d⌧

=

Z t

0

2e�⌧ · 1d⌧ = 2� 2e�t.


100


• Case 3: When t� 3 � 0 (i.e., t � 3),

(f ⇤ h)(t) =

Z 1

�1f(⌧)h(t� ⌧)d⌧

=

Z t

t�3

f(⌧)h(t� ⌧)d⌧

=

Z t

t�3

2e�⌧ · 1d⌧ = 2e�(t�3) � 2e�t.


101

Graphical convolution - example 1 - summary

• In summary,

(f ⇤ h)(t) =

8

<

:

0 t < 0

2� 2e�t0 t < 3

2e�(t�3) � 2e�t3 t

• Note that f ⇤ h is continuous even though both f and h have/For jump discontinuities.


102

Convolution is commutative

• Convolution is commutative, i.e.,

8f, h, h ⇤ f = f ⇤ h,

• This is proved by simple change of variable of integration (⌧ 0= t� ⌧ ).

• To illustrate this, let’s recompute the previous example instead as

(h ⇤ f)(t) =

Z 1

�1h(⌧)f(t� ⌧)d⌧, 8t 2 R.

• Again, we have the same three cases in t.


103

Graphical convolution - example 1b - case 1

• Case 1: When t < 0,

h(⌧)f(t� ⌧) = 0, 8⌧ 2 R.

• Thus, (h ⇤ f)(t) = 0.


104


• Case 2: When 0 t < 3,

(h ⇤ f)(t) =

Z 1

�1h(⌧)f(t� ⌧)d⌧

=

Z t

0

h(⌧)f(t� ⌧)d⌧

=

Z t

0

1 · 2e�(t�⌧)d⌧

= 2e�t

Z t

0

e⌧d⌧ = 2e�t(et � 1) = 2� 2e�t.


105


• Case 3: When t � 3,

(h ⇤ f)(t) =

Z 1

�1h(⌧)f(t� ⌧)d⌧

=

Z

3

0

h(⌧)f(t� ⌧)d⌧

=

Z

3

0

1 · 2e�(t�⌧)d⌧ = 2e�t

Z

3

0

e⌧d⌧ = 2e3�t � 2e�t

• In summary, we confirm that f ⇤ h = h ⇤ f for this example.


106

Graphical convolution - example 2

• Now consider a graphical convolution example where each signal is of finite span.

• Suppose we wish to compute f ⇤ h where

f(t) = (2� t)(u(t)� u(t� 2)) and h(t) = u(t+1)� u(t� 2), 8t 2 R.

• Note that h is not causal.

• Graphically, we will see that there are five cases for t.


107


• Case 1: When t+1 < 0 (i.e., t < �1), f(⌧)h(t� ⌧) = 0, 8⌧ 2 R.

• Thus,

(f ⇤ h)(t) =

Z 1

�1f(⌧)h(t� ⌧)d⌧

= 0.


108


• Case 2: When 0 t+1 < 2 (i.e., �1 t < 1),

(f ⇤ h)(t) =

Z 1

�1f(⌧)h(t� ⌧)d⌧

=

Z t+1

0

f(⌧)h(t� ⌧)d⌧

=

Z t+1

0

(2� ⌧) · 1d⌧ = 2� 0.5(1� t)2


109


• Case 3: When t+1 � 2 and t� 2 < 0 (i.e., 1 t < 2),

(f ⇤ h)(t) =

Z

2

0

f(⌧)h(t� ⌧)d⌧

=

Z

2

0

(2� ⌧) · 1d⌧

= 2


110


• Case 4: When 0 t� 2 < 2 (i.e., 2 t < 4),

(f ⇤ h)(t) =

Z

2

t�2

f(⌧)h(t� ⌧)d⌧

=

Z

2

t�2

(2� ⌧) · 1d⌧

= 0.5(4� t)2


111


• Case 5: When t� 2 � 2 (i.e., t � 4), f(⌧)h(t� ⌧) = 0 8⌧ 2 R.

• Thus,

(f ⇤ h)(t) = 0.


112

Convolution - example 2 - summary

• In summary for this example, we again have a continuous f ⇤ h:

(f ⇤ h)(t) =

8

>

>

>

<

>

>

>

:

0 t < �1

2� 0.5(1� t)2 �1 t < 1

2 1 t < 2

0.5(4� t)2 2 t < 4

0 4 t


113

Convolution - additive finite support-lengths property

• Also note that in the previous example,

– the support of f ⇤ h (time interval over which f ⇤ h is non-zero) is of length 5,

– while that of f is 2 and that of h is 3.

• Convolution has a general property that, 8f, h, the length of support of f ⇤ h is the sumof that of f and that of h.


114

Convolution - other important properties

• Recall ZSR of LTI systems by convolution with an impulse response.

• One can directly show that convolution is a commutative bi-linear mapping of pairs of(Dirichlet) signals to another, i.e., 8 signals f, g, h and scalars ↵,� 2 C,

(↵f + �g) ⇤ h = ↵(f ⇤ h) + �(g ⇤ h)

• By changing order of integration (Fubini’s theorem), one can easily show that convolutionis associative, i.e., 8 signals f, g, h,

(f ⇤ g) ⇤ h = f ⇤ (g ⇤ h), cf. next slide.

• We’ll use these properties when composing more complex systems from simpler ones.

• By the ideal sampling property, the identity signal for convolution is the impulse �,i.e., 8 signals f ,

f ⇤ � = � ⇤ f = f

• By just changing variables of integration, we can show how to exchange time-shift withconvolution (time-invariance), i.e., 8 signals f, h : R ! C and times ⌧ 2 R,(�⌧f) ⇤ h = �⌧(f ⇤ h) (or just associativity & commutativity with g = �⌧�).


115

Convolution - proof of associative property

• 8 signals f, g, h and 8t 2 R,

((f ⇤ g) ⇤ h)(t) =

Z 1

�1(f ⇤ g)(⌧)h(t� ⌧)d⌧

=

Z 1

�1

Z 1

�1f(�)g(⌧ � �)d�h(t� ⌧)d⌧

=

Z 1

�1

Z 1

�1g(⌧ � �)h(t� ⌧)d⌧ f(�)d�, by Fubini’s theorem

=

Z 1

�1

Z 1

�1g(⌧ 0

)h((t� �)� ⌧ 0)d⌧ 0 f(�)d�, ⌧ 0

= ⌧ � �

=

Z 1

�1(g ⇤ h)(t� �)f(�)d�

= (f ⇤ (g ⇤ h))(t).

• Q.E.D.


116

Total response - second-order, series RLC circuit example

• Recall that for this series RLC circuit with R = 700⌦, L = 0.1H, and C = 1µF:

D

2 y +RL�1

D y + (LC)

�1y = RL�1

D f

(D+2000)(D+5000)y = 7000D f

Q(D)y = P (D)f


117

Total response - second-order, series RLC circuit example

• We’ll show later that the impulse response for this circuit is

h(t) =

✓

�14000

3

e�2000t+

35000

3

e�5000t

◆

u(t), t 2 R.


118

Total response - second-order, series RLC circuit example (cont)

• If the input is, say, f(t) = 5e�1000tu(t), t 2 R, then the ZSR yZS

= f ⇤ h is:

8t < 0, yZS

(t) = 0

8t � 0, yZS

(t) =

Z t

0�h(⌧)f(t� ⌧)d⌧

=

Z t

0

(�14000

3

e�2000⌧+

35000

3

e�5000⌧)5e�1000(t�⌧)d⌧

= �7000

3

5e�1000t

Z t

0

(2e�1000⌧ � 5e�4000⌧)d⌧

= �7000

3

5e�1000t

✓

2

1000

(1� e�1000t)�

5

4000

(1� e�4000t)

◆

= �7

4

· 5e�1000t+

70

3

e�2000t �175

12

e�5000t

• Note how the ZSR consists of a forced response plus characteristic modes.

• Re. the forced (eigen-) response, note that the transfer function

H(�1000) = �7

4

where H(s) =

P (s)

Q(s)=

7000s

(s+2000)(s+5000)

.

119

Total response - second-order, series RLC circuit example (cont)

• Recall if the initial conditions are (D y)(0�) = �2000V/s and y(0�) = 10V, thenthe ZIR is

8t � 0�, yZI

(t) = 16e�2000t � 6e�5000t.

• Combining with the the ZSR just derived, the total response of the system (to both initialconditions and input) is: 8t � 0�,

y(t) = yZI

(t) + yZS

(t)

=

nX

k=1

ckµk(t) + (h ⇤ f)(t)

= 16e�2000t � 6e�5000t+

✓

�7

4

· 5e�1000t+

70

3

e�2000t �175

12

e�5000t

◆

u(t)


120

Total response - discussion of existence and uniqueness

• The existence of the ZIR follows from the linear independence of characteristic modes.

• Similarly, the existence of the ZSR follows from the assumption that the input and impulseresponse are Dirichlet signals, so that the convolution is well defined.

• When the initial conditions are (respectively, input signal is) zero, the unique solution forthe ZIR (respectively, ZSR) is the zero signal.

• Given this, it’s easy to prove by contradiction the uniqueness of the ZIR and ZSR in general.

• For example, assume that yZI

and yZI

are two di↵erent ZIRs for the same system (Q) andthe same initial conditions:

– Note that the signal yZI

� yZI

is the ZIR for Q with zero initial conditions.

– That is, yZI

� yZI

= 0, a contradiction.

– So, the original assumption of the existence of two di↵erent ZIRs is false.


121

System stability - ZIR - asymptotically stable

• Consider a SISO system with input f and output y.

• Recall that the ZIR yZI

is a linear combination of the system’s characteristic modes, wherethe coe�cients depend on the initial conditions.

• A system is said to be asymptotically stable if for all initial conditions,

lim

t!1yZI

(t) = 0.

• So, a system is asymptotically stable if and only if all of its characteristic values have strictlynegative real part.

• For example, if the characteristic polynomial Q(s) = (s+5)(s+7), then

– the system’s characteristic values (roots of Q) are �7,�5 (strictly negative and real),

– the ZIR is of the form yZI

(t) = c1

e�7t+ c

2

e�5t, t � 0�,

– So, yZI

(t) ! 0 as t ! 1 for all c1

, c2

, and

– hence is asymptotically stable.

• If instead Q(s) = (s+5)

2, then the characteristic modes would be e�5t, te�5t, both ofwhich ! 0 as t ! 1, so this system is still asymptotically stable.

122

System stability - bounded signals

• Recall that a signal x is said to be bounded if

9 M < 1 s.t. 8t 2 R, |x(t)| M ;

otherwise x is said to be unbounded.

• For example, x(t) = 3e�5tu(t), t 2 R is bounded (can use M = 3).

• Also, 3cos(5t), t 2 R, is bounded (again, can use M = 3).

• But both 3e2t cos(5t) and 3e2t are unbounded; for the latter example, obviously

lim

t!13e2t = 1.


123

System stability - ZIR - marginally stable

• A system is said to be marginally stable if it is not asymptotically stable but yZI

is always(for all initial conditions) bounded.

• A system is marginally stable if and only if

– it has no characteristic values with strictly positive real part,

– it has at least one purely imaginary characteristic value, and

– all purely imaginary characteristic values are not repeated.

• That is, a marginally stable system

– has some characteristic modes of the form cos(wt) or sin(wt),

– while the rest are of the form tke�↵tcos(wt) or tke�↵t

sin(wt),

– for real ↵ > 0, frequency w � 0 and integer degree k � 0 (and time t � 0).

• For example, if the characteristic polynomial is Q(s) = s(s2 + 2)(s + 3), with char-acteristic values 0, ±j

p2 and �3, then the system is marginally stable with modes 1,

cos(

p2t), sin(

p2t) and e�3t, the first three of which are bounded but do not tend to

zero as time t ! 1.


124

System stability - ZIR - unstable

• A system that is neither asymptotically nor marginally stable (i.e., a system with unboundedmodes) is said to be unstable.

• For example, if the characteristic polynomial is Q(s) = (s2+2)

2

(s+3) then the purelyimaginary characteristic values ±j

p2 are repeated, and hence the two additional modes

t sin(p2t), t cos(

p2t) are unbounded - so this system is unstable.

• Similarly, the system with Q(s) = s2(s2 + 2)(s+3) is unstable owing to the repeatedcharacteristic value at 0 with associated modes 1 and t, the latter unbounded.

• For another example, the system with Q(s) = (s2 + 2)(s� 3) is unstable owing to the(unstable) characteristic value at 3 with associated unbounded mode e3t.


125

System stability - ZIR - series RLC circuit example

• If the output signal is redefined as the capacitor voltage, then P (s) = (LC)

�1

= w2

obut the characteristic polynomial remains

Q(s) = s2 + 2⇣wos+ w2

o , where

– resonant frequency wo = 1/pLC, and

– damping factor ⇣ = (R/2)p

C/L.

– The characteristic values (roots of Q) of the system are

�wo

✓

⇣ ±q

⇣2 � 1

◆


126

System stability - ZIR - series RLC circuit example (cont)

Cases regarding the discriminant ⇣2 � 1 (assuming fixed L,C > 0):

• ⇣ = 0 (R = 0): two purely imaginary characteristic values ±jwo, i.e., marginally stable.

• 0 < ⇣ < 1 (0 < R < 2

p

L/C): two complex conjugate characteristic values of strictly

negative real part and magnitude wo, c± := �wo(⇣ ± jp

1� ⇣2), i.e., asymptoticallystable, underdamped.

• ⇣ = 1 (R = 2

p

L/C): a repeated (double) real characteristic value at �wo < 0,i.e., asymptotically stable, critically damped.

• 1 < ⇣ < 1 (R > 2

p

L/C): two di↵erent real strictly negative char. values, r± :=

�wo(⇣ ±p

⇣2 � 1), i.e., asymptotically stable, overdamped.


127

System stability - ZSR - BIBO stable

• A SISO system is said to be Bounded Input, Bounded Output (BIBO) stable if8 bounded input signals f , the ZSR y

ZS

is bounded.

• A necessary condition for BIBO stability is absolute integrability of the impulse response,Z 1

0

|h(⌧)|d⌧ < 1.

• To see why: For arbitrary bounded input f (by Mf with 0 Mf < 1): 8t � 0,

|yZS

(t)| = |(f ⇤ h)(t)|

=

�

�

�

�

Z t

0

f(t� ⌧)h(⌧)d⌧

�

�

�

�

Z t

0

|f(t� ⌧)h(⌧)|d⌧ (by the triangle inequality)

Z t

0

Mf |h(⌧)|d⌧

Mf

Z 1

0

|h(⌧)|d⌧ =: My < 1,


128

System stability - ZSR - BIBO stable

• The condition of absolute integrability of the impulse response,Z 1

0

|h(⌧)|d⌧ < 1,

is also necessary for, and hence equivalent to, BIBO stability.

• When we study LTI systems in the frequency domain and derive the impulse response, wewill readily see how asymptotic stability (of ZIR) is equivalent to BIBO stability (of ZSR).

• Example: If the system has char. values ±jwh of multiplicity 1, and the input is f(t) =

cf cos(wft)u(t), then yZS

(t) has the termR t

0

ch cos(wh⌧)cf cos(wf(t� ⌧))d⌧

=

chcf

2

Z t

0

�

cos((wh � wf)⌧ + wft) + cos((wh + wf)⌧ � wft)�

d⌧ ;

so if wf 6= wh, then yZS

is bounded, else yZS

(t) has the term 0.5chcft cos(wft) whichis unbounded - i.e., marginally stable system is not BIBO.

• For another example, see Ex. 2.6-4 of Lathi.


129

ZSR - the transfer function, H

• Recall that for any polynomial Q and frequency s 2 C (including s = jw, w 2 R),

Q(D)est = Q(s)est, 8t � 0.

• So, if we guess that yp

(t) = H(s)est, t � 0, is a “particular” solution ofQ(D)y = P (D)f in response to input f(t) = estu(t), we get by substitution that

8t � 0, s 2 C, H(s)Q(s)est = Q(D)y = P (D)f = P (s)est

) H(s) = P (s)/Q(s).

• The “rational polynomial” H = P/Q is known as the system’s transfer function and willfigure prominently in our study of frequency-domain analysis.

• So, the ZSR (forced response + characteristic modes) would be of the form:

yZS

(t) = H(s)est + linear combination of char. modes, t � 0,

again with (D

k yZS

)(0�) := 0 8 integers k � 0.

• Recall for the previous series RLC circuit example with input f(t) = 5e�1000t, t � 0:

H(s) =

P (s)

Q(s)=

7000

(s+2000)(s+5000)

, s 2 C

yZS

(t) = H(�1000)f(t) + linear combination of char. modes, t � 0.

130

ZSR - impulse response h, transfer function H, and eigenresponse

• yZS

(t) = H(s)Aest+j�+ linear combination of char. modes, with t � 0 and A >

0,� 2 R, is the ZSR to input f(t) = Aest+j�u(t), with (D

k yZS

)(0�) := 0 for allintegers k � 0.

• The eigenresponse is a special case of the forced response for exponential inputs of stable,LTI systems.

• If s = jw for w 2 R (i.e., input f is sinusoidal) and the system is asymptotically stable,then the ZSR tends to the steady-state eigenresponse of the system:

y(t) ⇠ H(jw)Aej(wt+�) as t ! 1.

• Since y = h ⇤ f , we get that as t ! 1 for a causal and asymptotically stable system,

yZS

(t) =

Z t

0

h(⌧)Aej(w(t�⌧)+�)d⌧ =

✓

Z t

0

h(⌧)e�jw⌧d⌧

◆

Aej(wt+�) ⇠ H(jw)Aej(wt+�),

)Z 1

0�h(⌧)e�jw⌧d⌧ = H(jw), 8w 2 R.

131


ZSR for single-pole systems using eigenresponse

• We will see that yZS

will be continuous at time 0 when H = P/Q is strictly proper(i.e., bn = 0 so that m := the degree of P < n := degree of Q); thus, h has no bn�term and there is no “direct coupling” between input and output in the system.

• So given a strictly proper system with a single pole p (characteristic value), i.e., n = 1,the ZSR to a sinusoidal input f = Aesot+j� will be of the form

yZS

(t) = H(so)f(t) + ce�pt= H(so)Aesot+j�

+ ce�pt, 8t � 0.

• So, if H is strictly proper, we can solve for the coe�cient c of the system mode e�pt usingyZS

(0) = 0, i.e.,

0 = H(so)Aej� + c ) c = �H(so)Aej�.


132

Remark: time origin at �1 for an asymptotically stable system

• If we take the origin of time to be �1 for an asymptotically stable LTI system, then thesystem is in steady-state in “finite time” when the characteristic-modes component of theresponse will be negligible.

• So, for frequency w 2 R, the response to the persistent sinusoidal input f(t) = Aej(wt+�)

is, 8t 2 R,Z 1

�1h(⌧)Aejw(t�⌧)+j�d⌧ =

✓

Z 1

�1h(⌧)e�jw⌧d⌧

◆

Aej(wt+�)= H(jw)Aej(wt+�),

where the frequency w 2 R and the integral is finite since h is assumed Dirichlet.

• Thus, H is the bilateral Fourier transform of the impulse response h,Z 1

�1h(⌧)e�jw⌧d⌧ = H(jw), 8w 2 R.

• For a causal system (and as in the previous slide),Z 1

0

h(⌧)e�jw⌧d⌧ = H(jw), 8w 2 R.


133

Resonance of the marginally stable series RLC circuit

• Recall that the series RLC circuit with capacitor-voltage output is marginally stable whenR = 0 (damping factor ⇣ = 0) with two purely imaginary characteristic values at±j/

pLC =: ±jwo.

• All sinusoidal inputs are, of course, bounded.

• Consider an input of the form

f(t) = ejwt= cos(wt) + j sin(wt).

• When w 6= wo, the steady-state total response to this complex-valued, bounded sinu-soidal input, consisting of the eigenresponse H(jw)f(t) plus a linear combination of the(persistent) characteristic modes e±jwot, is bounded

• However, if w = wo then Q(jwo) = 0 and P (jwo) = w2

o 6= 0; so |H(jw)| divergesas w ! wo.


134

Resonance of the marginally stable series RLC circuit (cont)

• That is, though f(t) = ejwot is a bounded input, the output (of the marginally stablecircuit with resonant frequency wo) is not bounded.

• This phenomenon is called resonance.

• When we study transient response of resonant circuits in the (complex) frequency domain,we will see that the total response of this system to f(t) = ejwot has an unbounded outputcomponent tejwot.

• In the following, we will refer to the characteristic values of the system (roots of thecharacteristic polynomial Q, asymptotes of transfer function H), as system poles.


135

Zero-state step response of causal LTI systems

• The zero-state step response of a causal LTI system with impulse response h is yu

= u⇤h.

• That is, 8t < 0, yu

(t) = 0, and 8t � 0,

yu

(t) =

Z 1

�1h(⌧)u(t� ⌧)d⌧

=

Z t

0

h(⌧)d⌧

= H(0) · 1 + linear combo. of char. modes

where the (steady-state, zero-frequency/DC) eigenresponse is

H(0) =

Z 1

0

h(⌧)d⌧.

• So, if the system is asymptotically stable, then

lim

t!1yu

(t) = H(0).


136

Step response of a first-order circuit - time constant, 90% rise time

• Recall that this circuit has system equation D y +(RC)

�1y = (2RC)

�1f and impulseresponse h(t) =

1

2RCe�t/(RC)u(t).

• Also, H(0) = P (0)/Q(0) = 0.5 (consistent with the infinite DC impedance of acapacitor so that the circuit is an even voltage divider in DC steady-state).

• The (ZS) step response is yu

(t) = (h ⇤u)(t) =

R t

0

h(⌧)d⌧ = 0.5(1� e�t/(RC)

)u(t),t � 0.

• The 90% rise-time of a system is the first time t90

at which it attains 90% of the terminal(steady-state) value of its step response,

yu

(t90

) = 0.9 · 0.5 ) t90

= RC ln(10)

where RC is the time constant of this circuit; i.e., t90

is ln(10) ⇡ 2.3 time-constants.

137

Step response of series RLC circuit - preliminaries

• In this series RLC circuit, the capacitor voltage has been designated the output signal.

• Thus, the system equation is D

2 y +RL�1

D y + (LC)

�1y = (LC)

�1f .

• Again, P (s) = (LC)

�1, but the characteristic polynomial remains

Q(s) = s2 +RL�1s+ (LC)

�1

= s2 + 2wo⇣s+ w2

o ,

where again the resonant frequency and damping factor are, respectively,

wo = 1/pLC and ⇣ = (R/2)

p

C/L.

• So the characteristic values / poles again are �wo(⇣ ±p

⇣2 � 1).


138

Step response of series RLC circuit - overdamped case, ⇣ > 1

• In the overdamped case (⇣ > 1), there are two di↵erent, negative real poles, r±.

• Again, we’ll see later how the impulse response is

h(t) =

wo

2

p

⇣2 � 1

�

er+t � er�t�

u(t), t 2 R.

• Thus, the ZS step response is yu

= h ⇤ u,

yu

(t) =

✓

1+

w2

o

r+

(r+

� r�)er+t +

w2

o

r�(r� � r+

)

er�t◆

u(t), t 2 R.

• Again, note how the zero-state step response consists of

– the eigenresponseH(0)u(t) = u(t), withH(s) = P (s)/Q(s) = (LC)

�1/(s2+RL�1s+ (LC)

�1

) so that H(0) = 1,

– and a linear combination of characteristic modes, whose coe�cients are such that theinitial conditions of y

u

are zero consistent with a zero-state step response, yu

(0) = 0

and (D y)(0) = 0.


139

Step response of series RLC circuit - overdamped case, ⇣ > 1 (cont)

• One can directly check that the step response monotonically increases:

yu

(0) = 0

(D yu

)(t) � 0 8t � 0

lim

t!1yu

(t) = 1 = H(0) = P (0)/Q(0).

• To find an approximate 90% rise time in the special case when ⇣ � 1 (so thatexp(r�t90) ⌧ exp(r

+

t90

) < 1),

0.9 = 0.9yu

(1) = yu

(t90

) ⇡ 1+

w2

o

r+

(r+

� r�)er+t90

) t90

⇡1

�r+

ln

✓

10w2

o

�r+

(r+

� r�)

◆


140

Step response of series RLC circuit - overdamped case (cont)

• This is the step response yu

(t) for the series RLC circuit example with R = 700⌦,L = 0.1H, and C = 1µF, so that

wo = 3162rad/s and ⇣ = 1.107.

• Since ⇣ > 1, this is the overdamped case.

• From the plot, t90

⇡ 0.0015s.


141

Step response of series RLC circuit - critically damped case, ⇣ = 1

• In the critically damped case (⇣ = 1), there is a repeated (double), negative real pole at�wo.

• The impulse response is now

h(t) = w2

o te�wotu(t), t 2 R.

• Thus, the ZS step response h ⇤ u is (after integration by parts),

yu

(t) =

�

1� wote�wot � e�wot�

u(t), t 2 R,

noting the characteristic modes e�wot and te�wot.

• Again, the step response monotonically increases:

yu

(0) = 0

(D yu

)(t) � 0 8t � 0

lim

t!1yu

(t) = 1 = H(0).


142

Step response of series RLC circuit - underdamped case, 0 < ⇣ < 1

• In the underdamped case (0 < ⇣ < 1), there are two complex-conjugate poles of magnitudewo and negative real part,

�wo(⇣ ± j

q

1� ⇣2).

• Here the impulse response is

h(t) =

wop

1� ⇣2e�wo⇣t

sin(wo

q

1� ⇣2 t)u(t), 8t 2 R.

• Thus, the ZS step response h ⇤ u is (after integration by parts): 8t 2 R,

yu

(t) =

1� e�wo⇣tcos(wo

q

1� ⇣2 t)�⇣

p

1� ⇣2e�wo⇣t

sin(wo

q

1� ⇣2 t)

!

u(t).


143

Step response of series RLC circuit - underdamped case (cont)

• One can directly check that

yu

(0) = 0 and lim

t!1yu

(t) = 1,

but yu

is not monotonically increasing in the underdamped case.

• Indeed, yu

overshoots its terminal value yu

(1) = 1.

• By solving the first-order condition, D yu

= 0, we get that local extrema of yu

occur attimes k⇡/(wo

p

1� ⇣2) for integers k > 0.

• In particular, k = 1 corresponds to the maximum overshoot,

yu

⇡

wo

p

1� ⇣2

!

� yu

(1) = e�⇡⇣/p

1�⇣2

> 0,

see the following figure.

• Note that the overshoot decreases with damping factor ⇣.


144

Step response of series RLC circuit - underdamped case (cont)

• This is the previous numerical example but with the resistance reduced to R = 100⌦ sothat ⇣ = 0.16 < 1, i.e., the underdamped case, and unchanged w

0

= 3162rad/s.

• Here, we see maximum overshoot of about 0.6 ⇡ exp(�⇡⇣/p

1� ⇣2) at about 0.001 ⇡⇡/(wo

p

1� ⇣2) seconds.


145

Toward frequency domain methods

• We now begin our journey into the frequency domain.

• We first study Fourier-series representation of periodic signals in the time domain.

• Our origin of time will be �1 so that a steady-state eigenresponse of an asymptoticallystable system is reached in finite time.

• Subsequently, we will study the full Zero-State Response (ZSR, including modes) using thebilateral Fourier transform on the real frequency domain, w 2 R.

• The Fourier transform can accommodate aperiodic input signals and not asymptoticallystable and noncausal systems.

• Finally, as we have just done in the time domain, we will study the total system response(ZSR + ZIR) using the unilateral Laplace transform in the complex frequency domain,s 2 C.


146

Fourier-series representation of periodic signals - Outline

• Approximating Dirichlet signals f : I ! R by minimizing square error over a finite time-interval (period) I ⇢ R

– Linear vector space of signals

– Norm, inner product, and LLSE estimation

– Orthogonal signal-sets

– LLSE given (projection onto the span of) orthogonal signal-sets

• Fourier series representation of periodic, Dirichlet signals f : R ! R

– Trigonometric Fourier series: DC, fundamental, and harmonic components.

– Compact trigonometric Fourier series.

– Exponential Fourier series.

• Parseval’s theorem for signal power.

• Eigenresponse to a periodic input, i.e., steady-state response of an asymptotically stableLTI system.


147

Dirichlet signals on a finite time-interval

• We begin our study of Fourier series by reviewing basics of linear algebra

– for continuous-time signals f := {f(t) | t 2 I} on a finite time-interval I ⇢ R (latera period of the assumed-periodic f),

– instead of vectors v 2 Rk, which can be thought of as discrete-time signals on a finitetime-interval, i.e., v := {vl | l 2 {1,2, ..., k}} for positive integer k � 1.

• Again, we will restrict consideration to Dirichlet signals f : I ! R, i.e., satisfying

– the weak Dirichlet condition of absolute integrability,R

I |f(t)|dt < 1,

– the strong Dirichlet conditions: over any finite interval ⇢ I: (i) f has a finite numberof jump discontinuities, and (ii) f has a finite number of minima and maxima.

• The set of all such signals forms a linear vector space, e.g., it is closed under scalar multi-plication and vector addition: 8 such signals f, g and scalars ↵,� 2 R, ↵f +�g : I ! Ris also Dirichlet.


148

Square error between a signal and its estimator

• Suppose the signal ˆf is an approximation of f over the time-interval I.

• The square error between f and ˆf is

||f � ˆf ||2 :=

Z

I|f(t)� ˆf(t)|2dt,

i.e., the squared L2

norm of the error signal " := f � ˆf .

• This definition of norm has the properties of the Euclidean norm of vectors of a finite-dimensional vector space, i.e.,

– 8 (Dirichlet) signals x, ||x|| :=q

R

I |x(t)|2dt = 0 , x = 0, i.e., x is the zero

signal (x(t) = 0 for “almost all” t 2 I),

– 8 signals x and scalars c 2 R, and ||cx|| = |c| · ||x||.

– 8 signals x, y, ||x+ y|| ||x||+ ||y|| (triangle inequality).


149

LLSE estimator - projection and inner product

• Consider signals f, g : I ! R with g 6= 0.

• We wish to approximate f “given g”, i.e., using ˆf = ↵g by minimizing over the scalar↵ 2 R the square error

||f � ˆf ||2 = ||f � ↵g||2 =: E(↵).

• By definition of norm and linearity of the integral,

E(↵) =

Z

I|f(t)� ↵g(t)|2dt

=

Z

I|f(t)|2dt� 2↵

Z

If(t)g(t)dt+ ↵2

Z

I|g(t)|2dt,

i.e., E is quadratic.

• Since E is convex, it is minimized at ↵ satisfying the first-order optimality condition,

0 = E0(↵) = �2

Z

If(t)g(t)dt+2↵

Z

I|g(t)|2dt

) ↵ =

R

I f(t)g(t)dtR

I |g(t)|2dt


150

LLSE estimator - projection and inner product (cont)

• That is, the Linear, Least Square-Error (LLSE) estimator, of f given g 6= 0 is✓

R

I f(t)g(t)dtR

I |g(t)|2dt

◆

g =

< f, g >

||g||2g,

where we have identified the inner product operator mapping two signals to a scalar,

< f, g > :=

Z

If(t)g(t)dt 2 R.

• The LLSE estimator of the signal f given the signal g 6= 0 is the projection of f onto theone-dimensional vector space (of signals)

{↵g | ↵ 2 R} = span{g}.


151

LLSE estimator - LLSE signal and given signals are orthogonal

• The error signal for this LLSE estimator is

" := f �< f, g >

||g||2g = f � ˆf.

• Note that the inner product of real-valued signals is obviously linear in both arguments andcommutative.

• Also, for all signals g,

< g, g > = ||g||2.

• Thus, if g 6= 0 then

< ", g > = < f �< f, g >

||g||2g, g > = < f, g > �

< f, g >

||g||2< g, g >

= 0,

i.e., the error " = f � ↵g of the LLSE estimator of f given g is orthogonal to g.


152

LLSE estimator given orthogonal signal-sets

• Consider a group of k � 1 non-zero signals

g1

, g2

, ..., gk : I ! Rthat are orthogonal to each other, i.e.,

8p 6= q 2 {1,2, ..., k},Z

Igp(⌧)gq(⌧)d⌧ =: < gp, gq > = 0.

• For any signal f : I ! R, we can easily generalize the LLSE estimator given one signalto find the LLSE that is a linear combination of these k � 1 given signals: for scalars↵1

,↵2

, ...↵k 2 R,ˆf = ↵

1

g1

+ ↵2

g2

+ ...+ ↵kgk,

i.e., the LLSE estimator.

• To this end, define the square-error function,

E(↵1

,↵2

, ...↵k) = ||f � ˆf ||2 = ||"||2 = < ", " > .


153

LLSE estimator given orthogonal signal-sets (cont)

• By the linearity and commutativity property of inner product,

E(↵1

,↵2

, ...↵k) =

Z

ˆI

|f(⌧)� ↵1

g1

(⌧)� ↵2

g2

(⌧)� ...� ↵kgk(⌧)|2d⌧

= < f �kX

l=1

↵lgl, f �kX

l=1

↵lgl >

= ||f ||2 + 2

X

p<q

↵p↵q < gp, gq > +

X

l

�

�2↵l < f, gl > +↵2

l ||gl||2�

• By the orthogonality assumption, i.e., 8p 6= q, < gp, gq >= 0,

E(↵1

,↵2

, ...↵k) = ||f ||2 +

X

l

�

�2↵l < f, gl > +↵2

l ||gl||2�

• Again, the coe�cients ↵l that minimize E are obtained by the k first-order conditions

0 =

@E

@↵l

= �2↵l < f, gl > +2↵l||gl||2, l 2 {1,2, ..., k}.


154

LLSE estimator given orthogonal signal-sets (cont)

• Thus, the LLSE coe�cients are

↵l =

< f, gl >

||gl||2=

R

I f(⌧)gl(⌧)d⌧R

I |gl(⌧)|2d⌧

• So for an orthogonal signal-set, the LLSE coe�cients are the same as those of LLSEestimators given just the individual signals.

• Note: If the given signal-set {g1

, g2

, ..., gk} is not orthogonal, the Gram-Schmidt procedurecan be used to find an orthogonal signal-set that spans the linear vector space of the givenone, where

span{g1

, g2

, ..., gk} = {↵1

g1

+ ↵2

g2

+ ...+ ↵kgk | ↵1

,↵2

, ...,↵k 2 R}.

• Actually, Gram-Schmidt will work on a given signal-set that is not even linearly independentto produce an orthonormal basis of its span.


155

LLSE estimators with signals illustrated as vectors


156

LLSE estimator on a finite time-interval - example

• Consider the exponential signal f = 5e2t for t 2 I = [�2,2].

• To find the LLSE estimator of f given the signals g1

(t) = 1 and g2

(t) = t, t 2 I,ˆf = ↵

1

g1

+ ↵2

g2

,

first note that

< g1

, g2

> =

Z

2

�2

g1

(t)g2

(t)dt =

Z

2

�2

1 · t dt =

1

2

t2�

�

�

�

2

�2

= 0.

• Thus, g1

and g2

are orthogonal.

• Recall that the product of an even and odd signal is odd, and that zero is the integral (oraverage value) of an odd signal over an interval of time I that equally straddles the origin.


157

LLSE estimator on a finite time-interval - example (cont)

• Since g1

and g2

are orthogonal, the LLSE coe�cients are

↵1

=

< f, g1

>

||g1

||2=

R

2

�2

f(t)g1

(t)dtR

2

�2

|g1

(t)|2dt=

R

2

�2

5e2t · 1dtR

2

�2

1

2dt=

(5/2)e2t�

�

2

�2

4

= (5/8)(e4 � e�4

) = (5/4) sinh(4).

↵2

=

< f, g2

>

||g2

||2=

R

2

�2

f(t)g2

(t)dtR

2

�2

|g2

(t)|2dt=

R

2

�2

5e2ttdtR

2

�2

t2dt=

(5/2)te2t � (5/4)e2t�

�

2

�2

(1/3)t3|2�2

= (45/64)e4 + (75/64)e�4

• So, the LLSE of f given {g1

, g2

} is

ˆf = (5/4) sinh(4)g1

+ ((45/64)e4 + (75/64)e�4

)g2

.


158

Fourier series representation - Basic theorem

• Now consider periodic signals on R.

• Recall that if the ratio of the periods of g1

and g2

is rational, or if one is periodic and theother is constant, then all linear combinations ↵

1

g1

+ ↵2

g2

are periodic, i.e., 8 scalars↵1

,↵2

2 R.

• The following theorem is the basic to representation periodic signals by Fourier series.

• Theorem: The span of the set of signals

{cos(kwot), t 2 R}, k 2 {0,1,2,3, ...} and {sin(kwot), t 2 R}, k 2 {1,2,3, ...}includes all piecewise-continuous, periodic Dirichlet signals with period T := 2⇡/wo > 0

that are valued at the midpoint of their jump discontinuities.

• Note that the constant signal is included in this signal set:if k = 0 then 8t 2 R, 1 = cos(kwot).

• Also for k > 0, the period of cos(kwot) and sin(kwot) is 2⇡/(kwo) = T/k.

• So, linear combinations of these signals will be periodic with period T/k for somek 2 {1,2, ...} or constant (corresponding to k = 0).


159

Fourier series coe�cients (coordinates w.r.t. trigonometric basis)

• Suppose f : R ! R is Dirichlet and periodic with period

T =:

2⇡

wo> 0,

where wo is the fundamental frequency of f .

• Define the infinitely large, trigonometric signal-set:

gc,k := {cos(kwot), t 2 R}, k 2 Z�0 and g

s,k := {sin(kwot), t 2 R}, k 2 Z�1

• Note that all of these g have the property that 8t 2 R, g(t) = g(t+T ), consistent withg either being constant (k = 0), or having period T/k for some integer k > 0 so that

8t 2 R, g(t) = g(t+ T/k)

= g(t+ T/k + T/k) = g(t+2T/k) ...

= g(t+ (k � 1)T/k + T/k)

= g(t+ T ).

• The LLSE of f given this signal set over an arbitrary interval I of length T is

ˆf =

1X

k=0

↵kgc,k +

1X

k=1

�kgs,k = ↵0

+

1X

k=1

(↵kgc,k + �kgs,k),

where by the previous theorem, the {↵k,�k} are the coordinates of f w.r.t. the trig basis,i.e., the LLSE has zero square error, ||"||2 = ||f � ˆf ||2 = 0.

160

Orthogonality of the given trigonometric signals - di↵erent cosines

• Again, wo := 2⇡/T is the fundamental frequency of T -periodic f .

• LetR

T...dt represents integration over any interval of length T (i.e., any period of f).

• 8 integers k 6= l � 0,

< gc,k, gc,l > =

Z

T

cos(kwot) cos(lwot)dt

=

Z T

0

0.5cos((k + l)wot) + 0.5cos((k � l)wot) dt

=

0.5

(k + l)wosin((k + l)wot) +

0.5

(k � l)wosin((k � l)wot)

�

�

�

�

T

0

=

0.5

(k + l)wosin((k + l)woT ) +

0.5

(k � l)wosin((k � l)woT )

=

0.5

(k + l)wosin((k + l)2⇡) +

0.5

(k � l)wosin((k � l)2⇡)

= 0.


161

Orthogonality of the given trigonometric signals - di↵erent sines

• 8 integers k 6= l � 1,

< gs,k, gs,l > =

Z

T

sin(kwot) sin(lwot)dt

=

Z T

0

0.5cos((k � l)wot)� 0.5cos((k + l)wot) dt

=

0.5

(k � l)wosin((k � l)wot)�

0.5

(k + l)wosin((k + l)wot)

�

�

�

�

T

0

= 0


162

Orthogonality of the given trigonometric signals - a sine and cosine

• 8 integers k � 0, l � 1,

< gc,k, gs,l > =

Z

T

cos(kwot) sin(lwot)dt

=

Z T

0

0.5 sin((l+ k)wot) + 0.5 sin((l � k)wot) dt

= �0.5

(l+ k)wocos((l+ k)wot)�

0.5

(l � k)wocos((l � k)wot)

�

�

�

�

T

0

= �0.5

(l+ k)wo(cos((l+ k)2⇡)� 1)�

0.5

(l � k)wo(cos((l � k)2⇡)� 1)

= 0


163

Fourier coe�cients/coordinates

• By the orthogonality of the given trigonometric signal-set, the LLSE of f over a periodT = 2⇡/wo has coe�cients

8k � 0, ↵k =

< f, gc,k >

||gc,k||2

=

R

Tf(t) cos(kwot)dt

R

T| cos(kwot)|2dt

=

8

>

>

<

>

>

:

2

T

Z

T

f(t) cos(kwot)dt if k > 0

1

T

Z

T

f(t)dt if k = 0

8k � 1, �k =

< f, gs,k >

||gs,k||2

=

R

Tf(t) sin(kwot)dt

R

T| sin(kwot)|2dt

=

2

T

Z

T

f(t) sin(kwot)dt.

• Note that 8k > 0, T�1||gc,k||2 = T�1||g

s,k||2 = 1/2, consistent with our previouscalculation of the power of a sinusoid with unit amplitude.


164

Fourier coe�cients/coordinates (cont)

• So, at all points t of continuity of f (when f =

ˆf),

f(t) = ↵0

+

1X

k=1

(↵k cos(kwot) + �k sin(kwot)) =:

ˆf(t).

• Note that ↵0

= T�1

R

Tf(t)dt is just the mean/average value (DC component) of f .

• The T -periodic (k = 1) component is the fundamental of T -periodic f .

• For k � 2, the T/k-periodic component is the (k � 1)

th harmonic of f .


165

Trigonometric Fourier series - square-wave example

• The Fourier coe�cients of this square wave f of period 7 and duty cycle 4 are

↵0

=

1

T

Z T/2

�T/2

f(t)dt =

1

7

Z

2

�2

5dt =

5

7

t

�

�

�

�

2

�2

=

20

7

.

8k � 1, ↵k =

2

T

Z T/2

�T/2

f(t) cos(kwot)dt =

2

7

Z

2

�2

5 cos(k(2⇡/7)t)dt

=

5

k⇡sin(k(2⇡/7)t)

�

�

�

�

2

�2

=

10

k⇡sin(4k⇡/7).

8k � 1, �k =

2

T

Z T/2

�T/2

f(t) sin(kwot)dt = 0.


166

Trigonometric Fourier series - square-wave example (cont)

• So,

f(t) =

20

7

+

1X

k=1

10

k⇡sin(4k⇡/7) cos(2k⇡t/7) =:

ˆf(t), 8t 2 R,

except possibly at points t of jump discontinuity of f where ˆf(t) = 5/2, e.g., at t = ±2.

• Note �k = 0 8k � 1 simply because the integrand is odd on the interval of integration[�3.5,3.5] - recall that the product of an even (f) and odd (sine) function is odd.

• Also note the interesting identity that ensues at time t = 0 (a point of continuity of f):

�2⇡

7

=

1X

k=1

sin(4k⇡/7)

k.


167

Trigonometric Fourier series - triangle-wave example

↵0

=

1

T

Z T/2

�T/2

f(t)dt =

1

10

Z

5

�5

3t dt =

3

20

t2�

�

�

�

5

�5

= 0.

8k � 1, ↵k =

2

T

Z T/2

�T/2

f(t) cos(kwot)dt = 0.

8k � 1, �k =

2

T

Z T/2

�T/2

f(t) sin(kwot)dt =

1

5

Z

5

�5

3t sin(k(2⇡/10)t)dt

=

3

5

✓

�5

k⇡t cos(k⇡t/5) +

5

2

(k⇡)2sin(k⇡t/5)

◆

�

�

�

�

5

�5

after integration by parts

= �30

k⇡cos(k⇡) =

30

k⇡(�1)

k+1


168

Trigonometric Fourier series - even and odd signals

• Thus, for this triangle wave f ,

f(t) =

1X

k=1

30

k⇡(�1)

k+1

sin(k⇡t/5)

8t 2 R except possibly at t = 5k for odd k 2 Z (i.e., points of jump discontinuity in fwhere the RHS ˆf(5k) = 0).

• Note ↵k = 0 8k � 0 in the previous triangle-wave example simply because the integrandis odd on the interval of integration [�5,5] - recall that the product of an odd (f) andeven (cos) function is odd.

• In general, if T -periodic f(t) is even on [�T/2, T/2], then 8k � 1, �k = 0, i.e., f hasa cosine (even) trigonometric Fourier series.

• Alternatively, if f(t) is odd on [�T/2, T/2] then 8k � 0, ↵k = 0, i.e., f has a sine(odd) trigonometric Fourier series.


169

Partial Fourier series approximation

• Consider the partial Fourier series ˆfK of T -periodic, Dirichlet f up to the (K � 1)

th

harmonic, i.e., 8K � 0, t 2 R,

ˆfK(t) := ↵0

+

KX

k=1

↵k cos(kwot) + �k sin(kwot),

• For Dirichlet f (in L2

),

lim

K!1

Z

T

|ˆfK(t)� f(t)|2dt = lim

K!1||ˆfK(t)� f(t)||2 = 0.

• So, at points t of continuity of f ,

lim

K!1ˆfK(t) :=

ˆf(t) = f(t).

• In the following we will no longer point out possible di↵erences between f and its (complete)Fourier series ˆf at points of jump discontinuity of f .


170

Partial Fourier series approximation - square-wave example

• In the following, we plot ˆf0

= 20/7, ˆf1

, ˆf10

, and ˆf40

for the previous square-wave example.

• The non-uniform convergence near the jump discontinuity is called the Gibbs phenomenon.


171

Parseval’s theorem for signal power

• Recall that the average power of a sinusoid A cos(wt+ �), t 2 R, is A2/2.

• Discrepancies only at points of jump discontinuity will not a↵ect mean power calculations.

• So, the Fourier series representation of a T -periodic, Dirichlet signal f can be used tocompute its mean power, i.e.,

Pf = T�1

Z

T

|ˆf(t)|2dt

= T�1

Z

T

�

�

�

�

�

↵0

+

1X

k=1

(↵k cos(kwot) + �k sin(kwot)

�

�

�

�

�

2

dt

= T�1

Z

T

↵2

0

+

1X

k=1

↵2

k cos2

(kwot) + �2

k sin2

(kwot) dt by orthogonality

= ↵2

0

+

1X

k=1

✓

↵2

k

2

+

�2

k

2

◆


172

Parseval’s theorem for signal power (cont)

• Suppose we want to compute the fraction ' of T -periodic f ’s power in the frequencyband [w

1

, w2

] where w2

> w1

� 0 are frequencies in radians/s and wo = 2⇡/T is thefundamental of f .

• First identify which harmonics of f lie in the band, i.e., for what integers k � 0 iskwo 2 [w

1

, w2

], equivalently

k1

:= dw1

/woe = dw1

T/(2⇡)e k k2

:= bw2

/woc = bw2

T/(2⇡)c ,where for z 2 R, dze is the smallest integer � z and bzc is the greatest integer z.

• So,

' =

1

Pf

k2

X

k=k1

↵2

k + �2

k

2

,

where we’ve defined �0

= ↵0

to simplify this expression when k1

= 0,and

Pk2

k=k1

... := 0 when k1

> k2

.

• A complementary problem is to find the smallest frequency interval [w1

, w2

] for which' � 0.95.


173

Parseval’s theorem for signal power - example

• To find the fraction of the power of the previous square wave (with period T = 7) in thefrequency band [20.5Hz,100.5Hz], recall its Fourier series:

20

7

+

1X

k=1

10

k⇡sin(4k⇡/7) cos(2k⇡t/7).

• Thus, its total mean power is

Pf :=

✓

20

7

◆

2

+

1X

k=1

1

2

✓

10

k⇡sin(4k⇡/7)

◆

2

.

• Since the frequency band in radians/s is [41⇡,201⇡] and the signal’s period is 7,

– the lower harmonic index in the band is k1

= d41⇡/(2⇡/7)e = d143.5e = 144,

– the upper one is k2

= b201⇡/(2⇡/7)c = b703.5c = 703.

• So, the fraction of f ’s mean power in [20.5Hz,100.5Hz] is

P�1

f

703

X

k=144

1

2

✓

10

k⇡sin(4k⇡/7)

◆

2

.


174

Parseval’s theorem for signal power - example (cont)

• Note that Pf is easily computed directly as

Pf =

1

T

Z

T

|f(t)|2dt =

1

7

5

2 · 4 =

100

7

⇡ 14.

• Also note that the f ’s DC power is✓

1

T

Z

T

f(t)dt

◆

2

= ↵2

0

=

✓

20

7

◆

2

⇡ 8.


175

The compact trigonometric Fourier series

• Using the trigonometric identity

8↵,�, w 2 R, ↵ cos(wt) + � sin(wt) =

q

↵2

+ �2

cos(wt� tan

�1

(�/↵)).

we obtain the compact trigonometric Fourier series (referenced to cosine) of any T -periodic,Dirichlet f from its trigonometric one:

ˆf(t) = A0

cos(�0

) +

1X

k=1

Ak cos(kwot + �k), t 2 R, where

Ak =

q

↵2

k + �2

k

�k = � tan

�1

(�k/↵k)

with �0

:= 0.

• Note that the amplitudes Ak � 0, 8k � 0.

• Also, if ↵0

< 0 then A0

= |↵0

| and �0

= ⇡.


176

Trigonometric and compact trigonometric Fourier series

• Note that f(t) = 5cos(6t+3)+7sin(13t+5) is periodic since the ratio, 13/6 or6/13, of the periods, 2⇡/6 and 2⇡/13, of its components is rational.

• The period of f is the least common integer multiple of its components’ periods, i.e.,T = 2⇡ = 6(2⇡/6) = 13(2⇡/13) seconds

• So f ’s fundamental is wo = 2⇡/T = 1 radian/s = the greatest common divisor of 6 and13.

• The above f is given in compact trigonometric form and has only 5th and 12

th harmonics(no DC or fundamental components in particular):f(t) = 5cos(6t+3)+ 7cos(13t+5� ⇡/2)

• We can write the above f as a trigonometric Fourier series,

f(t) = [5 cos(3) cos(6t)� 5 sin(3) sin(6t)] + [7 sin(5) cos(13t) + 7cos(5) sin(13t)]

• The derived trigonometric Fourier series of the previous: (a) even square-wave happened tobe in compact form, and (b) odd triangle-wave can be transformed to compact form againby using the identity 8v 2 R, sin(v) = cos(v � ⇡/2).


177

Compact trigonometric Fourier series - square-wave example

• For this (not even) square-wave example with period T = 7, the DC trigonometric Fouriercoe�cient is

↵0

=

1

T

Z �2.5+T

�2.5

f(t)dt =

1

7

Z

3

�1

5dt =

5

7

t

�

�

�

�

3

�1

=

20

7

.


178

Compact trigonometric Fourier series - square-wave example (cont)

• The other trigonometric Fourier coe�cients are, 8k � 1:

↵k =

2

T

Z �2.5+T

�2.5

f(t) cos(kwot)dt =

2

7

Z

3

�1

5 cos(2k⇡t/7)dt

=

5

k⇡sin(2k⇡t/7)

�

�

�

�

3

�1

=

5

k⇡(sin(6k⇡/7)� sin(�2k⇡/7))

=

10

k⇡cos(2k⇡/7) sin(4k⇡/7).

�k =

2

T

Z �2.5+T

�2.5

f(t) sin(kwot)dt =

2

7

Z

3

�1

5 sin(2k⇡t/7)dt

= �5

k⇡cos(2k⇡t/7)

�

�

�

�

3

�1

= �5

k⇡(cos(6k⇡/7)� cos(�2k⇡/7))

=

10

k⇡sin(2k⇡/7) sin(4k⇡/7).


179

Compact trigonometric Fourier series - square-wave example (cont)

• So, the compact trigonometric Fourier coe�cients are A0

= 20/7,�0

= 0, and 8k � 1,

Ak =

q

↵2

k + �2

k =

10

k⇡| sin(

4k⇡

7

)|r

cos

2

(

2k⇡

7

) + sin

2

(

2k⇡

7

) =

10

k⇡| sin(

4k⇡

7

)|

�k = � tan

�1

(�k/↵k) = � tan

�1

✓

sin(4k⇡/7) sin(2k⇡/7)

sin(4k⇡/7) cos(2k⇡/7)

◆

=

⇢

� tan

�1

(tan(2k⇡/7)) = � 2k⇡/7 if sin(4k⇡/7) > 0

� tan

�1

(tan(2k⇡/7)) + ⇡ = � 2k⇡/7+ ⇡ if sin(4k⇡/7) < 0

) ˆf(t) =

20

7

+

1X

k=1

10

k⇡| sin(4k⇡/7)| cos(2k⇡t/7� 2k⇡/7+ ⇡1{sin(4k⇡/7) < 0})

=

20

7

+

1X

k=1

10

k⇡sin(4k⇡/7) cos(2k⇡(t� 1)/7),

consistent with this square wave being the previous example even square-wave delayed byone second, where

• the indicator function 1B = 1 if B is true, else 1B = 0.


180

The complex-exponential Fourier series

• Recall the Euler-De Moivre identities:

cos(wt) =

ejwt+ e�jwt

2

sin(wt) =

ejwt � e�jwt

2j=

�jejwt+ je�jwt

2

) ↵ cos(wt) + � sin(wt) =

↵k � j�k

2

ejwt+

↵k + j�k

2

e�jwt

• We obtain the complex-exponential Fourier series of any T -periodic, Dirichlet signal f bysubstituting this identity into its trigonometric one:

ˆf(t) =

1X

k=�1

Dk ejkwot,

where, again, wo = 2⇡/T ,

↵0

= D0

and 8k � 1,

Dk =

↵k � j�k

2

and D�k =

↵k + j�k

2

.


181

The complex-exponential Fourier series (cont)

• By substituting the expressions for the trigonometric Fourier coe�cients, we get that

8k 2 Z, Dk =

1

T

Z

T

f(t)e�jkwotdt.

• Note that if f is real valued, then 8k � 0, ↵k,�k 2 R

) 8k 2 Z, D�k = Dk, the complex conjugate of Dk 2 C.

• In turn, 8k � 1, Dk ejkwot= D�k e�jkwot, consistent with f being real valued.

• Similarly, using Euler’s identity ejwt= cos(wt) + j sin(wt), we can transform the

complex-exponential to the trigonometric Fourier coe�cients: 8k � 1,

↵k = Dk +D�k and �k = j(Dk �D�k).


182

Complex-exponential Fourier series - example

• Define 8t 2 R,

f(t) =

1X

k=�1

7e�2(t�5k)(u(t� 5k)� u(t� 3� 5k)),

i.e., a Dirichlet signal with period T = 5 and duty cycle 3 (within each period).

• The complex-exponential Fourier coe�cients are, 8k 2 Z,

Dk =

1

T

Z

T

f(t)e�jkwotdt =

1

T

Z T

0

f(t)e�jkwotdt

=

1

5

Z

3

0

7e�2te�jk2⇡t/5dt =

1

5

·7

�(2 + jk2⇡/5)e�(2+jk2⇡/5)t

�

�

�

�

3

0

=

7

10+ jk2⇡

⇣

1� e�3(2+jk2⇡/5)⌘

.

183

Complex-exponential Fourier series - orthogonality

• For any period T > 0, the C-valued signal-setn

ge,k(t) := ejk2⇡t/T = ejkwot, t 2 R

o

k2Z

is orthogonal (and a basis for the set of all periodic Dirichlet functions with fundamentalfrequency wo or some whole multiple, kwo).

• Orthogonality is easily verified: 8k 6= l 2 Z

< ge,k, ge,l > :=

Z

T

ge,k(t)ge,l(t)dt =

Z

T

ejkwote�jlwotdt

=

1

j(k � l)woej(k�l)wot

�

�

�

�

T

0

= 0

• Moreover, 8k 2 Z,

||ge,k||2 =< g

e,k, ge,k > =

Z

T

ge,k(t)ge,k(t)dt =

Z

T

|ge,k(t)|2dt =

Z

T

1dt = T.


184

Complex-exponential Fourier series - orthogonality (cont)

• Note that inner product for C-valued signals is linear in the first argument and conjugatelinear in the second argument.

• So, for inner product thus defined, we have again that the complex-exponential Fouriercoe�cients for T -periodic, Dirichlet f are, 8k 2 Z,

Dk =

< f, ge,k >

||ge,k||2

=

R

Tf(t)g

e,k(t)dt

T

=

1

T

Z

T

f(t)e�jkwotdt.


185

Spectrum (frequency-domain representation) of a signal

• Recall the previous 10-periodic, odd, triangle-wave example signal, f .

• The trigonometric Fourier coe�cients of f are ↵k = 0 for k � 0 and �k = (�1)

k30/(k⇡)

for k � 1.

• Thus, the exponential Fourier coe�cients are

Dk =

8

<

:

0 if k = 0

(�1)

k+1j15/(k⇡) if k > 0

(�1)

kj15/(k⇡) if k < 0

• We can plot magnitude and phase spectrum of f , respectively:

|Dk| =

15

k⇡=

3

kwofor k 6= 0, and

\Dk =

8

<

:

don’t care if k = 0

⇡/2 if k > 0 and odd or k < 0 and even�⇡/2 if k < 0 and odd or k > 0 and even

versus kwo = k2⇡/10 = k⇡/5 for k 2 Z.


186

Spectrum (frequency-domain representation) of a signal (cont)


187

Complex-exponential Fourier series - Parseval’s theorem

• We can argue as before that the mean power of T -periodic, Dirichlet f is

Pf = T�1

Z

T

|f(t)|2dt = T�1

Z

T

f(t)f(t)dt

= T�1

Z

T

1X

k=�1

Dkejkwot

1X

`=�1

D`ej`wotdt

= T�1

1X

k=�1

1X

`=�1

DkD`

Z

T

ejkwot ej`wot dt

= T�1

1X

k=�1

DkDk

Z

T

ejkwot ejkwotdt by orthogonality

=

1X

k=�1

|Dk|2T�1

Z

T

1dt

=

1X

k=�1

|Dk|2 = |D0

|2 + 2

1X

k=1

|Dk|2.

where the last equality is because f is R-valued ) 8k 2 Z, D�k = Dk ) |Dk| = |D�k|.


188

Complex-exponential Fourier series - Parseval’s theorem - example

• Recall that 5-periodic f(t) =

P1k=�1 7e�2(t�5k)

(u(t�5k)�u(t�3�5k)), t 2 R,had complex-exponential Fourier series coe�cients: 8k 2 Z,

Dk =

7

10+ jk2⇡

⇣

1� e�3(2+jk2⇡/5)⌘

=

7

10+ jk2⇡(1� e�6

cos(6k⇡/5) + je�6

sin(6k⇡/5)).

• By Parseval’s theorem, the mean power of f is

Pf =

1X

k=�1

�

�

�

�

7

10+ jk2⇡

⇣

1� e�3(2+jk2⇡/5)⌘

�

�

�

�

2

=

1X

k=�1

7

p

100+ (k2⇡)2

q

(1� e�6

cos(6k⇡/5))2 + (e�6

sin(6k⇡/5))2

=

1

T

Z

T

|f(t)|2dt =

1

5

Z

3

0

49e�4tdt =

49

20

(1� e�12

).


189

Complex-exponential Fourier series - power in a frequency band

• Consider the frequency band [w1

, w2

] with w2

> w1

� 0 radians/s.

• To compute the amount of T -periodic f ’s mean power in this band, recall that we need toidentify which of its harmonics k lie in it:

k1

:= dw1

/woe k bw2

/woc =: k2

,

where again the fundamental frequency of f is wo = 2⇡/T .

• Using the complex-exponential Fourier series of f , note that we need to account for bothpositive (kwo) and negative (�kwo) frequencies for this index set k.

• That is, if {Dk}k2Z are the complex-exponential Fourier coe�cients of f and k1

> 0,then the amount of f ’s power in this band is

k2

X

k=k1

(|D�k|2 + |Dk|2) =

�k1

X

k=�k2

|Dk|2 +

k2

X

k=k1

|Dk|2 = 2

k2

X

k=k1

|Dk|2.

• Note that if k1

= 0 (i.e., w1

= 0), then we should include the term |D0

|2 (representingf ’s DC power) only once in the expression above.


190

Eigenresponse

• Recall that for an asymptotically stable system with transfer function H, the steady-stateresponse to the (persistent) sinusoidal input f(t) = Aej(wt+�), t 2 R, with amplitudeA > 0, frequency w, and phase �, is the eigenresponse,

y(t) = H(jw)Aej(wt+�)= |H(jw)|Aej(wt+�+\H(jw)), t 2 R.

• That is,

– themagnitude response or system gain of the system is |H(jw)|, or 20 log

10

|H(jw)|in decibels (dB):

input magn. 20 log

10

A dB ! output magn. 20 log

10

A+20 log

10

|H(jw)| dB

– the phase response of the system is \H(jw):

input phase � radians ! output phase �+ \H(jw) radians

• So, the eigenresponse to real-valued f(t) = A cos(wt+ �), t 2 R, is real-valued

y(t) = A|H(jw)| cos(wt+ �+ \H(jw)), t 2 R.


191

Eigenresponse and Fourier series

• Given a complex-exponential Fourier series of an arbitrary (Dirichlet) periodic input,

f(t) =

1X

k=�1

Dkejkwot

the eigenresponse of an asymptotically stable, LTI system H is, by linearity,

y(t) =

1X

k=�1

DkH(jkwo)ejkwot, t 2 R.

• Equivalently, the eigenresponse to f with the compact Fourier-series representation

f(t) =

1X

k=0

Ak cos(kwot+ �k)

is

y(t) =

1X

k=0

|H(jkwo)|Ak cos(kwot+ �k + \H(jkwo)), t 2 R.


192

Eigenresponse and Fourier series - circuit example

• For this example LTIC circuit, the transfer function is

H(jw) =

R

R+ (jwL||R)

=

R

R+ ((jwL)�1

+R�1

)

�1

=

jw/2+R/(2L)

jw +R/(2L)

=

P (jw)

Q(jw)


193

Eigenresponse and Fourier series - circuit example (cont)

• The transfer function can be verified by the system equation Q(D)y = P (D)f andH = P/Q, where by KCL

0 =

f � y

R+

0� y

R+ i where LD i = f � y

) 0 =

D f �D y

R+

0�D y

R+D i =

D f �D y

R+

0�D y

R+

f � y

L

) Q(D)y := D y +

R

2Ly =

1

2

D f +

R

2Lf =: P (D)f.


194


• So, the magnitude response in dB at frequency w is

20 log

10

|H(jw)| = 20 log

10

p

(w/2)2 + (R/(2L))2p

w2

+ (R/(2L))2= 10 log

10

(w/2)2 + (R/(2L))2

w2

+ (R/(2L))2

and the phase response is

\H(jw) = tan

�1

✓

w/2

R/(2L)

◆

� tan

�1

✓

w

R/(2L)

◆

.

• Note that for very low frequencies limw!0

|H(jw)| = H(0) = 1 (= 0 dB), and forhigh frequencies limw!1 |H(jw)| = 0.5 (⇡ �6 dB).

• So, higher frequencies are attenuated by about 6 dB, while low frequencies are not signif-icantly attenuated, i.e., they pass through the system (because the DC impedance of theinductor is zero).


195


• Again recall that f(t) =

P1k=�1 7e�2(t�5k)

(u(t� 5k)� u(t� 3� 5k)), t 2 R, hadcomplex-exponential Fourier series coe�cients,

Dk =

7

10+ jk2⇡

⇣

1� e�3(2+jk2⇡/5)⌘

f(t) =

1X

k=�1

Dkejk2⇡t/5

• So, if R = 10k⌦ and L = 0.1H (i.e., R/L = 10

5), then the eigenresponse of thiscircuit to this input is, 8t 2 R,

y(t) =

1X

k=�1

H(jk2⇡/5)Dkejk2⇡t/5

=

1X

k=�1

jk⇡/5+ 10

5/2

jk2⇡/5+ 10

5/2·

7

10+ jk2⇡

⇣

1� e�3(2+jk2⇡/5)⌘

ejk2⇡t/5


196

LTIC system magnitude and phase responses

• Recall that a SISO LTIC system relating input signal f to output y by P (D)f = Q(D)yhas (zero state) transfer function H(s) = P (s)/Q(s).

• The roots of Q are called system characteristic values or poles, while the roots of P aresystem zeros.

• Recall that for an asymptotically stable system with transfer function H, the steady-state(eigen-) response to real-valued, persistent, sinusoidal input

f(t) = A cos(wt+ �), t 2 Ris real-valued sinusoidal output at the same frequency

y(t) = A|H(jw)| cos(wt+ �+ \H(jw)), t 2 R,with input parameters: amplitude A > 0, frequency w � 0, and phase �.

• the magnitude response or system gain of the system is |H(jw)|, or 20 log

10

|H(jw)|in decibels (dB):

input magnitude 20 log

10

A dB ! output magnitude 20 log

10

A+20 log

10

|H(jw)| dB

• the phase response of the system is \H(jw):

input phase � radians ! output phase �+ \H(jw) radians


197

Bode plots of the system

• Themagnitude Bode plot of the system depicts the amplitude gain in decibels 20 log

10

|H(jw)|as a function of the log-frequency log

10

w for w � 0, i.e., a “log-log” plot.

• The phase Bode plot of the system depicts the phase change \H(jw) as a function oflog

10

w for w � 0, i.e., a semi-log plot.

• In the following discussion of Bode plots, logarithm is assumed base 10.

• Note that DC (w = 0) gain/phase-change is depicted at logw = �1, i.e., the originon a Bode plot is a low frequency logw = 0 ) w = 1 radians/s, but not DC.

• A change in frequency by a factor of 10 is called a decade (by a factor of 8 is called anoctave).


198

Example of a second-order transfer function

• Recall that for this second-order, series RLC circuit, the transfer function is

H(s) =

LC�1

s2 +RL�1s+ (LC)

�1

.

• Consider the overdamped case where there are two real poles r± < 0, and define thepositive frequencies w

1

:= �r+

< w2

:= �r�.


199

Example of a second-order transfer function (cont)

• Thus, the transfer function can be written as

H(jw) =

H(0)

(1 + jw/w1

)(1 + jw/w2

)

=

H(0)

(1� jw/r+

)(1� jw/r�),

where here the DC gain is

H(0) =

LC�1

w1

w2

=

LC�1

r+

r�= 1.

• For example, if R = 1010⌦, L = 10mH and C = 1µF, then w1

= 10

3 radians/s andw

2

= 10

5 radians/s.


200

“Standard form” transfer function to explain Bode plots

• Consider a transfer function that is a proper, rational polynomial

H(jw) =

P (jw)

Q(jw)

, for frequencies w � 0,

where by “proper” we mean that the degree of P that of Q.

• In polar form (not indicating dependence on real frequency w � 0),

|H|ej\H =

|P |ej\P

|Q|ej\Q

• Thus,

|H| =

|P ||Q|

and \H = \P � \Q.


201

“Standard form” transfer function to explain Bode plots (cont)

• So, for the above second-order RLC circuit, we can write

20 log |H(jw)| = 20 log |H(0)|� 20 log |1+ jw/w1

|� 20 log |1+ jw/w2

|\H(jw) = \H(0)� \(1 + jw/w

1

)� \(1 + jw/w2

)

• More generally, assume that P and Q have only real roots and no common roots, i.e., 8k, l,wz,k 6= wp,l.

• Factoring P and Q we get for the case of no DC poles or zeroes (i.e., 8k, �wz,k,�wp,k 6=0, so that DC gain in decibels is finite):

H(jw) = H(0)

Qmk=1

(1 + jw/wz,k)

Qnk=1

(1 + jw/wp,k)

(i)

• If H has r m DC zeroes (i.e., 0 DC gain or �1 in decibels) or r n DC poles (i.e., 1DC gain), then for some scalar ⌘ 2 R, respectively:

H(jw) = ⌘(jw)

rQm�r

k=1

(1 + jw/wz,k)

Qnk=1

(1 + jw/wp,k)

(ii)

or H(jw) = ⌘

Qmk=1

(1 + jw/wz,k)

(jw)

rQn�r

k=1

(1 + jw/wp,k)

(iii)


202

“Standard form” transfer function to explain Bode plots (cont)

• Since 8w > 0, |jw| = w and \jw = ⇡/2, the phase and magnitude for the above threecases (i)-(iii) are, for w � 0:

20 log |H(jw)| = 20 log |H(0)|+mX

k=1

20 log |1+ jw/wz,k|�

nX

k=1

20 log |1+ jw/wp,k| (i)

\H(jw) = \H(0) +

mX

k=1

\(1 + jw/wz,k)�

nX

k=1

\(1 + jw/wp,k) (i)

20 log |H(jw)| = 20 log |⌘|+ r20 logw +

m�rX

k=1


nX

k=1

20 log |1+ jw/wp,k| (ii)

\H(jw) = \⌘ + r⇡

2

+

m�rX

k=1

\(1 + jw/wz,k)�

nX

k=1

\(1 + jw/wp,k) (ii)

20 log |H(jw)| = 20 log |⌘|� r20 logw +

mX

k=1


n�rX

k=1

20 log |1+ jw/wp,k| (iii)

\H(jw) = \⌘ � r⇡

2

+

mX

k=1

\(1 + jw/wz,k)�

n�rX

k=1

\(1 + jw/wp,k) (iii)


203

Magnitude Bode plots with DC zeros or poles

• If there is a DC pole or zero, the magnitude Bode plot in decibels is obviously not finite atDC.

• Note the terms

20 log |⌘| ± r20 logw

for the above magnitude Bode plots with DC poles or zeros (of degree r).

• As a function of logw, these terms have

– slope ±20r and

– y-intercept 20 log |⌘|, where

– the y-axis is logw = 0, i.e., w = 1 radians/s (not DC where logw = �1 orw = 0).


204

Magnitude Bode plot: log |1+ jw/w1

| vs. logw with w1

> 0

• So, to construct the magnitude Bode plot, we first need to construct that of its componentterms of the form

20 log |1+ jw/w1

| = 10 log(1 + (w/w1

)

2

) versus logw, w � 0,

for an arbitrary frequency w1

> 0.

• Case 1: If w < w1

/10 (i.e., logw < �1+ logw1

), then 1.1 > |1+ jw/w1

| > 1 sothat |1+ jw/w

1

| ⇡ 1 and

20 log |1+ jw/w1

| ⇡ 0.

• Case 2: If w = w1

, then

20 log |1+ jw/w1

| = 20 log |1+ j| = 20 log

p2 ⇡ 3.

• Case 3: If w > 10w1

(i.e., logw > 1+ logw1

), then |1+ jw/w1

| ⇡ w/w1

and

20 log |1+ jw/w1

| ⇡ 20 logw � 20 logw1

;

as a function of logw, this line is of slope 20dB/decade with x-intercept logw1

.


205


| vs. logw with w1

> 0 (cont)

• These three cases are summarized in the following figure.

• Note that if w1

is real but negative, the magnitude Bode plot above does not change; theabove analysis applies to �w

1

= |w1

|.

• If w1

is complex-valued,

– the same three cases above apply using its modulus |w1

| (instead of w1

) leading tothe same asymptotic magnitude Bode plot,

– however at log |w1

| in Case 2, the true magnitude Bode plot will not di↵er by 3dBfrom (the corner of) the asymptotic plot, cf. the resonant circuit example.


206


| vs. logw (cont)


207

Magnitude Bode plots - rules of thumb

• So, to sketch the asymptotic magnitude Bode plot of a transfer function H:

1. Mark o↵ on the x-axis the modulus of the non-zero zeroes and poles of H.

2. Each zero, respectively pole, contributes an additional 20 dB/decade slope only forfrequencies larger than (i.e., to the right of) its modulus.

3. If the DC (from �1 = log0) gain is finite, then mark it o↵ on the y-axis; else

4. If there are r DC zeroes (Case (ii)), then the initial (again from �1 = log0) slopeis 20r dB/decade.

5. If there are r DC poles (Case (iii)) then the initial slope is -20r dB/decade.

• If the kth repeated zero (resp., pole) is real, then the true Bode plot will be 3k dB larger(resp., smaller) than the asymptotic Bode plot at the zero (resp., pole).


208

Magnitude Bode plot - second-order system example

• Suppose two poles at �w1

= �10

3 radians/s and �w2

= �10

5 radians/s and DC gainH(0) = 1 (20 log1 = 0 dB), consider again the example second-order transfer function

H(jw) =

H(0)

(1 + jw/w1

)(1 + jw/w2

)

) 20 log |H(jw)| = 0� 20 log |1+ jw/103|� 20 log |1+ jw/105|.


209

Phase Bode plot - \(1 + jw/w1

) vs. logw for w1

> 0

• To plot

\(1 + jw/w1

) = tan

�1

(w/w1

), for w1

> 0,

we again take three cases.

• Case 1: If w < w1

/10 (i.e., logw < �1+ logw1

), then

0 tan

�1

(w/w1

) < tan

�1

(1/10) ⇡ 0.

• Case 2: If w = w1

, then tan

�1

(w/w1

) = tan

�1

(1) = ⇡/4 radians (=45 degrees).

• Case 3: If w > 10w1

(i.e., logw > 1+ logw1

), then

⇡/2 � tan

�1

(w/w1

) > tan

�1

(10) ⇡ ⇡/2 radians (=90 degrees).

• These three cases are summarized in the following figure.


210

Phase Bode plot - \(1 + jw/w1

) vs. logw for w1

> 0

• So, the change in phase due to the term

\(1 + jw/w1

) = tan

�1

(w/w1

)

is ⇡/2 radians over a two decade interval [w1

/10,10w1

], i.e., a slope of 45 degrees/decade.


211

Phase Bode plot - discussion

• Unlike the magnitude response, for the phase response:

– the pole or zero has an e↵ect on the one decade before w1

,

– the e↵ect of the pole or zero “saturates” to a total change in phase of ±⇡/2 onedecade after w

1

, and

– we do not annotate deviations between the true and asymptotic plots (as 3 dB gaps).

• If �1 < H(0) < 0, then the initial phase \H(0) = ±⇡.

• Note that if w1

is real but negative, the phase change is instead �⇡/2 over two decadesat log(�w

1

) = log |w1

|, i.e., -45 degrees/decade.

• Also, if w1

is complex-valued, the phase change remains ⇡/2 over two decades, but thephase Bode plot departs from the asymptotic and is steeper at log |w

1

|, cf. the resonantcircuit example.


212

Phase Bode plot - second-order system example

• For two poles at �w1

= 10

3rad/s and �w2

= 10

5rad/s and DC gain H(0) = 1,consider again the example second-order transfer function

H(jw) =

H(0)

(1 + jw/w1

)(1 + jw/w2

)

) \H(jw) = 0� tan

�1

(w/103

)� tan

�1

(w/105

)


213

Bode plots - second-order system example 2

H(s) =

10s+10

7

s2 + 10100s+10

6

=

10(s+10

6

)

(s+10

4

)(s+10

2

)

=

10(1 + s/106

)

(1 + s/104

)(1 + s/102

)


214

Bode plots - example with DC zero

H(s) =

10s(s+100)

s+10

4

) |H(j1)| ⇡10 · 100

10

4

= 10

�1

= �20dB.


215

Op-amp circuit example - noninverting configuration

• KCL at the negative input terminal (with node voltage f) gives

0 =

0� f

R1

+

y � f

RF

+ CD(y � f)


216

Op-amp circuit example (cont)

• Thus, the system equation and transfer function are:

Q(D)y = D y +

1

CRF

y = D f +

1

C

✓

1

RF

+

1

R1

◆

f = P (D)f

) H(s) =

P (s)

Q(s)=

s+ C�1

(R�1

1

+R�1

F

)

s+ (CRF

)

�1

=

s+1010

s+10

,

the last equality when C = 0.01µF, R1

= 100k⌦, RF

= 10M⌦.

• So, this system has

– a pole at �10,

– a zero at �1010 ⇡ �10

3, and

– DC gain H(0) ⇡ 10

2

= 40dB.

• The zero being larger than the pole will restore the slope to 0 dB/decade in the magnitudeplot, and the phase to 0 degrees in the phase plot.


217

Op-amp circuit example - asymptotic Bode plots


218

Op-amp circuit example 2 - inverting configuration

• Here, y = vc so KCL at the negative input terminal (with node voltage 0) gives

0 =

f � 0

R1

+

y � 0

RF

+ CD(y � 0) ) H(s) =

�(CR1

)

�1

s+ (CRF

)

�1


219

Op-amp circuit example 2 - inverting configuration (cont)

• So when C = 0.01µF, R1

= 100k⌦, and RF

= 10M⌦,

H(s) =

�10

3

s+10

• The DC gain is again 20dB, but the DC phase is ±⇡, i.e., H(0) = �100 < 0.

• At high-frequencies, |H(jw)| ⇠ 10

3/w ! 0(= �1 dB) and \H(jw) ⇠ \(�1/j) =

⇡/2.

• Note that the previous noninverting op-amp at high frequencies, |H(jw)| ⇠ 1(= 0 dB)and \H(jw) ⇠ \1 = 0.


220

Op-amp circuit example 2 - asymptotic Bode plots


221

Resonant/tuned circuits - complex-conjugate poles (0 ⇣ < 1)

• Recall the series RLC circuit transfer function with output signal the capacitor voltage,

H(s) =

(LC)

�1

s2 +RL�1s+ (LC)

�1

=

w2

o

s2 + 2⇣wos+ w2

o

• H has a complex-conjugate pair of poles when damping factor ⇣ < 1 each with magnitudewo:

�wo(⇣ ± j

q

1� ⇣2)

• H is marginally stable (divergent eigenresponse at resonant frequencies ±wo, i.e., poles ats = ±jwo) when ⇣ = 0 (R = 0).


222

Resonant/tuned circuits - complex-conjugate poles (cont)

• Consider H(jwo) = �j/(2⇣) as a function of ⇣ 1.

• For the magnitude response:

– If ⇣ = 1, then |H(jwo)| = 1/2 = �6 dB consistent with a double pole at �wo

and a DC gain |H(0)| = 1(= 0 dB).

– If ⇣ = 0.5, then |H(jwo)| = 1(= 0 dB), i.e., the 6dB gap disappears and the trueBode plot meets the asymptotic one at its corner.

– lim⇣!0

|H(jwo)| = 1; this resonance e↵ect is why this system is called a “tunedcircuit”: it dramatically amplifies only in a narrow band around wo.


223


• The phase

\H(jw) = � tan

�1

✓

2⇣

wo/w � w/wo

◆

.

• The change in phase due to the pair of poles (each with magnitude wo) is �⇡ as frequencyw increases past wo; note that the denominator is positive for w < wo and negative forw > wo.

• As ⇣ ! 0, this change in phase becomes more abrupt, not gradual at �90 degrees/decadeover two decades.

• Recall that for L = 0.1H, and C = 1µF, the resonant frequency is wo = (LC)

�0.5=

3162 radians/s.

• Furthermore, if R = 100⌦ then ⇣ = (R/2)p

C/L = 0.16 < 1 (the underdampedcase).


224


• The following Bode plots are for the underdamped cases corresponding toR = 100⌦ (⇣ =

0.16) and R = 10⌦ (⇣ = 0.016), as well as the critically damped case (⇣ = 1) with arepeated (double) pole at �wo (note the -6dB gap and more gradual -90 degrees/decadeslope at logwo).

• The overdamped case (⇣ > 1) would have two real poles straddling logwo and Bode plotssimilar to the previous two-pole example of a series RLC transfer function.


225

Resonant/tuned circuits - magnitude and radian-phase Bode plots


226

Resonant/tuned circuits - DC zero

• Recall that if the output of the RLC circuit y is instead the voltage across the resistor, thenH has a DC zero:

H(s) =

RL�1s

s2 +RL�1s+ (LC)

�1

=

2⇣wos

s2 + 2⇣wos+ w2

o

.

• The following asymptotic Bode plots are for the underdamped case case when wo = 3162

radians/s.

• The DC gain is 0(= �1 dB) and that the gain increases from DC at 20 dB/decade,consistent with a DC zero.

• The DC phase is ⇡/2 radians since \H(jw) ⇡ \j for very small frequencies w # 0.

• At frequency wo the system response is H(jwo) = 1, i.e., 0dB and 0 phase.


227

Resonant/tuned circuits - DC zero - asymptotic plots for 0 ⇣ 1


228

Bilateral Fourier transform

• We now study in the frequency domain the ZSR of a LTI system to a not necessarily periodicinput.

• We will employ the bilateral Fourier transform spanning positive and negative frequencies(as the complex-exponential Fourier series for periodic signals).

• Outline of topics to be covered:

– Derivation of bilateral Fourier transform (FT) from Fourier series of periodically ex-tended pulse of finite support.

– Basic FT pairs and properties.

– Parseval’s theorem for signal energy.

– Ideal filters and causality issues.

– Modulation and demodulation

– Poisson’s identity and Nyquist sampling.


229

Periodic extensions of aperiodic signals of finite support

• Recall periodic extensions of pulses x of finite support.

• In this example, the pulse x has support of duration 4 and for T � 4, the T -periodicextension of x depicted in this figure is

xT(t) :=

1X

k=�1

x(t� kT ) =

1X

k=�1

(�kT x)(t), t 2 R,

i.e., superposition of time-shifted pulses x by all integer (n) multiples of T .

• Consider the Fourier series of xT

: 8t 2 R,

xT

(t) =

1X

k=�1

Dk,T ejk2⇡t/T , where Dk,T =

1

T

Z T/2

�T/2

x(⌧)e�jk2⇡⌧/Td⌧.


230

From Fourier series to Fourier transform - periodic extension (cont)

• Define the function,

X(k2⇡/T ) :=

Z 1

�1x(⌧)e�jk2⇡⌧/Td⌧.

• Thus, for large T , Dk,T ⇡ T�1X(k2⇡/T ).

• Also, 8t 2 R, x(t) = limT!1 xT

(t).

• So, substituting this approximation of the Fourier coe�cientsD into the complex-exponentialFourier series of x

T

gives, by Riemann integration, 8t 2 R,

x(t) = lim

T!1

1X

k=�1

X(k2⇡/T )ejk2⇡t/T T�1

=

Z 1

w=�1X(w)ejwtdw/(2⇡)

where w = 2⇡/T so that T�1 converges to the di↵erential dw/(2⇡) as T ! 1.


231

Fourier transform and inverse Fourier transform

• In summary, we defined the Fourier transform of time-domain signal x : R ! R as

X(w) :=

Z 1

�1x(t)e�jwtdt 8 frequencies w 2 R.

• Using the complex-exponential Fourier series of the periodic extension of x, we heuristicallyderived the inverse Fourier transform of the frequency-domain signal X : R ! R as

x(t) :=

1

2⇡

Z 1

�1X(w)ejwtdw 8 time t 2 R.

• We write X = Fx (or F{x}) and x = F�1X, again emphasizing functional transfor-mation.


232

Fourier transform properties - even magnitude and odd phase

• If x : R ! R, i.e., x is a real-valued signal, then X = Fx has

– even magnitude, i.e., |X(�w)| = |X(w)| 8w 2 R, and

– odd phase, i.e., \X(w) = �\X(�w) 8w 2 R,

• That is, X(�w) = X(w) 8w 2 R:

X(�w) =

Z 1

�1x(t)e�j(�w)tdt

=

Z 1

�1x(t)ejwtdt

=

Z 1

�1x(t)e�jwtdt (since x is R-valued)

= X(w) 8w 2 R.


233

Fourier transform properties - linearity & di↵erentiation

• The Fourier transform is a linear mapping simply because of the linearity of integration.

• For all di↵erentiable (Dirichlet) signals x, the Fourier transform of a time derivative is

8w 2 R, (F Dx)(w) = jw(Fx)(w).

• Recalling D :=

ddt, this property is proved by simple integration by parts: 8w 2 R,

(F Dx)(w) =

Z 1

�1(Dx)(t)e�jwtdt

=

Z 1

�1e�jwtdx(t)

= x(t)e�jwt�

�

1�1 �

Z 1

�1x(t)de�jwt

= 0+ jw

Z 1

�1x(t)e�jwtdt (⇤)

= jwX(w),

where (*) is by the (weak) Dirichlet condition of absolute integrability which implies thatlimt!±1 x(t) = 0.


234

Fourier transforms applied to SISO systems given by ODE

• We can use the Fourier transform (FT) to analyze the zero-state response of the system

P (D)f = Q(D)y.

• Let the FT of the input and output are, respectively, F = Ff and Y = Fy.

• By the linearity and di↵erentiation properties of the FT: 8w 2 R,

P (jw)F (w) = Q(jw)Y (w).

• Thus, the transfer function,

H(jw) =

P (jw)

Q(jw)

=

Y (w)

F (w)

, 8w 2 R.

• Change of notation: In the following discussion of Fourier transforms, we will use“H(w)” instead of “H(jw)” for the transfer function, i.e.,

H(w) =

P (jw)

Q(jw)

=

Y (w)

F (w)

, 8w 2 R.


235

Fourier transform of the unit impulse, �, and H = Fh

• By the ideal sampling property, the Fourier transform of the impulse is, 8w 2 R,

(F�)(w) =

Z 1

�1�(t)e�jwtdt = e�jwt

�

�

t=0

= 1.

• So, F� ⌘ 1.

• Note that by inverse Fourier transform, this means

2⇡�(t) =

Z 1

�1ejwtdw 8t 2 R!

• Thus, the LTI SISO system with input f = � has ZSR y = h (impulse response), so thatin the frequency domain

Y = HF = = H1 = H,

and so,

Fh = H,

i.e., the Fourier transform of the impulse response is the transfer function.


236

Transfer function is Fourier transform of (ZS) impulse response

• In summary, given the input f , to solve for the ZSR y the LTI SISO system described asP (D)f = Q(D)y (or as y = f ⇤ h), we will employ the Fourier transform:

y = F�1{HF},where H = Fh and F = Ff .

• From this, we can conclude the convolution property:

F{f ⇤ h} = F{f}F{h}.

• Sometimes the inverse Fourier transform is easily computed by integration.

• Otherwise, certain Fourier-transform properties and known Fourier-transform pairs can sim-plify computing F�1.


237

Direct proof of the convolution property, F{f ⇤ h} = F{f}F{h}.

F{f ⇤ h}(w) =

Z 1

�1(f ⇤ h)(t)e�jwtdt

=

Z 1

�1

Z 1

�1f(⌧)h(t� ⌧)d⌧e�jwtdt

=

Z 1

�1

Z 1

�1f(⌧)h(t� ⌧)e�jw(t�⌧)�jw⌧d⌧dt

=

Z 1

�1

Z 1

�1f(⌧)h(t0)e�jwt0�jw⌧d⌧dt0 (t0 = t� ⌧)

=

Z 1

�1f(⌧)e�jw⌧d⌧

Z 1

�1h(t0)e�jwt0dt0

= F{f}(w)F{h}(w)

= F (w)H(w) 8w 2 R.


238

Fourier transform pairs - exponential functions of time

• Again, let X = Fx.

• If the causal signal x(t) = esotu(t), t 2 R, with so 2 C and Re{so} < 0, then

X(w) =

Z 1

�1x(t)e�jwtdt =

Z 1

0

esote�jwtdt =

1

jw � so.

• Note that if so 2 R (so that x is R-valued), then |X(w)| = (w2

+ s2o)�0.5 is even,

\X(w) = � tan

�1

(w/so) is odd, and, again, X is the frequency domain representationof an aperiodic signal x.

• Also note that the “anti-causal” signal {�esotu(�t), t 2 R} has the same Fourier trans-form expression but requires Re{so} > 0 for convergence.

• For example, 8w 2 R,

F{e�3tu(t)}(w) =

1

jw +3

and F{e2tu(�t)}(w) =

1

2� jw.

• Neither the causal signal e3tu(t) nor the anti-causal signal e�2tu(�t) are Dirichlet becausethey do not have a finite integral.


239

Fourier analysis of RL circuit with exponential input

• Recall that for this circuit, the transfer function is

H(w) =

jw/2+R/(2L)

jw +R/(2L)=

1

2

+

R/(4L)

jw +R/(2L)8w 2 R.

• So, by linearity and the Fourier-transform pairs we have already discussed, the impulseresponse of this system is

h(t) =

1

2

�(t) +R

4Le�Rt/(2L)u(t) 8t 2 R.


240

Inverse Fourier transforms by known pairs

• Again, note that if we try to use the integral definition of F�1 on the frequency domainsignal c (constant) or b/(jw+a) we do not get an integrand corresponding to an indefiniteintegral, i.e., we get

1

2⇡

Z 1

�1cejwtdw or

1

2⇡

Z 1

�1

b

jw + aejwtdw.

• We found the inverse Fourier transforms by remembering the easily computed Fourier trans-forms F{c�} for constant c 2 R and F{e�atu(t)} for constant a > 0.


241

Fourier analysis of RL circuit with exponential input (cont)

• Let R = 2⌦ and L = 1H so that H(w) = (jw/2+ 1)/(jw +1).

• So, if the input of this circuit is f(t) = 4e�3tu(t), t 2 R, then the Fourier transform ofthe ZSR y is, 8w 2 R,

Y (w) = H(w)F (w) =

jw/2+ 1

jw +1

·4

jw +3

=

2jw +4

(jw +1)(jw +3)

=

1

jw +1

+

1

jw +3

where the last equality is a partial-fraction expansion (PFE) of Y .

• Thus, the ZSR y(t) = (e�t+ e�3t

)u(t).

• Note that y is the sum of a system characteristic mode e�t and the forced response,H|jw=�3

f(t) = (1/4)4e�3t= e�3t.

• We will subsequently study in detail the PFE of rational polynomials in s = jw for purposesof computing the inverse Laplace transform.


242

Fourier transform properties - time shifts

• The time-shift property is: 8to, w 2 R,

(F{�tox})(w) =

Z 1

�1(�tox)(t)e

�jwtdt

=

Z 1

�1x(t� to)e�jwtdt

=

Z 1

�1x(t0)e�jw(t0+to)dt0 (t0 := t� to)

= e�jwto

Z 1

�1x(t0)e�jwt0dt0

= e�jwtoX(w)

• So a time shift by to corresponds to multiplication in the frequency domain by a complex-exponential sinusoid with period 2⇡/to.


243

Fourier transform properties - frequency shifts

• The frequency-shift (modulation) property is: 8wo, t 2 R,

(F�1

�woX)(t) =

1

2⇡

Z 1

�1(�wo

X)(w)ejwtdw

=

1

2⇡

Z 1

�1X(w � wo)ejwtdw

=

1

2⇡

Z 1

�1X(w0

)ejt(w0+wo)dw0

(w0:= w � wo)

= ejtwo1

2⇡

Z 1

�1X(w0

)ejtw0dw0

= ejtwox(t).

• A baseband signal x multiplied by a carrier signal at frequency wo radians/s leads to afrequency shifted (modulated) signal ejtwox(t).


244

Fourier transform properties - time and frequency scaling

• If X = Fx and scalar ↵ 6= 0, the time/frequency scaling property is easily shown bychange-of-variables:

(F S

1/↵ x)(w) := (F{x(↵t)})(w) =

1

|↵|X⇣w

↵

⌘

.

• For example,

jw

jw6

+ 2

F�1

! D6x(6t) = D6e�12tu(t) = �72e�12tu(t) + 6�(t)

where x(t) = e�2tu(t) and u(6t) = u(t); equivalently,

jw

jw6

+ 2

= 6

jw

jw +12

F�1

! 6D e�12tu(t) = �72e�12tu(t) + 6�(t)

jw

jw6

+ 2

= 36 ·1

6

·jw6

jw6

+ 2

F�1

! 36(S

1/6 Dx)(t) = 36S

1/6(�2e�2tu(t) + �(t))

= �72e�12tu(6t) + 36�(6t) = �72e�12tu(t) + 6�(t)jw

jw6

+ 2

= �72

jw +12

+ 6

F�1

! �72e�12tu(t) + 6�(t)


245

Fourier transform properties - composition order

• The order of composed frequency-shift and time-shift operations matters, i.e., these oper-ations are not commutative.

• Given X = Fx and scalars to, wo.

• If frequency shift (modulation) before time shift:

x(t� to)ejwo(t�to)= �to{x(t)ejwot} F! X(w � wo)e�jwto

• On the other hand, if time shift before frequency shift:

x(t� to)ejwot= (�tox)(t)e

jwot F! X(w � wo)e�j(w�wo)to

• Similarly, other operation pairs (e.g., involving di↵erentiation, time/frequency scaling) arenot commutative.


246

Fourier transform pairs - complex-exponentials and impulses

• Note that by the time shift property, the Fourier transform of an impulse delayed by to is acomplex-exponential sinusoid in the frequency domain with period 2⇡/to:

F{�to�}(w) = e�jtow w 2 R.

• Similarly, an impulse located at wo in the frequency domain corresponds to a sinusoid inthe time-domain with frequency wo:

F�1{�wo�}(t) =

1

2⇡ejwot t 2 R.

• The presence of impulses in the frequency domain indicates discrete frequency components,e.g., of periodic signals.

• Again note that a complex exponential (over a time or frequency domain) cannot be directlyintegrated.


247

Fourier transform pairs - sine and cosine

• By direct integration we can verify that

F�1{⇡�wo� + ⇡��wo

�}(t) = cos(wot) t 2 RF�1{�j⇡�wo

� + j⇡��wo�}(t) = sin(wot) t 2 R

• By the Euler-De Moivre identity, the aperiodic, one-sided functions (real ↵ > 0)

e�↵tcos(wot)u(t) =

1

2

e(�↵+jwo)tu(t) +1

2

e(�↵�jwo)tu(t),

e�↵tsin(wot)u(t) =

1

2je(�↵+jwo)tu(t)�

1

2je(�↵�jwo)tu(t)

• Recalling the FT for such one-sided exponential functions, we can show by linearity that for↵ > 0, w 2 R,

F{e�↵tcos(wot)}(w) =

1

2

·1

↵+ j(w � wo)+

1

2

·1

↵+ j(w + wo)=

↵+ jw

(↵+ jw)

2

+ w2

o

F{e�↵tsin(wot)}(w) =

1

2j·

1

↵+ j(w � wo)�

1

2j·

1

↵+ j(w + wo)=

wo

(↵+ jw)

2

+ w2

o


248

Fourier transform of f(t) = cos(wot), t 2 R


249

Fourier transform of tkesotu(t) is k!/(jw � so)k+1

• Recall that SISO LTI systems have characteristic modes of the form tkesotu(t) where k � 0

is an integer and so 2 C a characteristic value of the system.

• As we are considering only asymptotically stable systems in our discussion of Fourier trans-forms, we can assume Re{so} < 0.

• To evaluate the Fourier transform of such a characteristic mode, we need to integrate byparts: for k > 0 and Re{so} < 0:

F{tkesotu(t)}(w) =

Z 1

0

tkesote�jwtdt =

1

so � jw

Z 1

0

tkde(so�jw)t

=

1

so � jwtke(so�jw)t

�

�

�

�

1

0

�1

so � jw

Z 1

0

e(so�jw)tdtk

=

k

(jw � so)

Z 1

0

tk�1e(so�jw)tdt

=

k

jw � soF{tk�1esotu(t)}(w) w 2 R.

• So by induction, we can show that for integers k > 0 and Re{so} < 0:

F{tkesotu(t)}(w) =

k!

(jw � so)k+1

8w 2 R.

250

Fourier transform pairs - sinc and rectangle pulse

• Consider the even rectangular pulse of height 1 and width 2wo in the frequency domain:

R(w) = u(w + wo)� u(w � wo)

• Its inverse Fourier transform is

r(t) = F�1{R}(w) =

1

2⇡

Z 1

�1R(w)ejwtdw

=

1

2⇡

Z wo

�wo

ejwtdw

=

1

2⇡·1

jt(ejwot � e�jwot

)

=

wo

⇡·ejwot � e�jwot

2jwot

=

wo

⇡· sinc(wot).

where

sinc(v) := sin(v)/v.

• Note that by L’Hopital’s rule, limv!0

sinc(v) = 1.


251

Fourier transform pairs - sinc and rectangle pulse (cont)

• The rectangular pulse has finite support but its FT pair, the sinc, has infinite support.

• Note that as the frequency wo increases,

– the rectangle pulse R in the frequency domain is broader, but

– the sinc function in the time domain is narrower; indeed, the width between the zerocrossings of the central lobe is ⇡/wo � (�⇡/wo) = 2⇡/wo.

• The two plotted sinc functions have wo 2 {1,5}.

• In the limit as wo ! 1 this approaches the � $ 1 Fourier-transform pair.


252

Duality

• We have noted several “dual” FT properties and pairs, e.g., time and frequency shifts,sinusoids and impulses.

• Duality is a consequence of the similarity of the Fourier and inverse Fourier transforms.

• If X = Fx, i.e.,

X(w) =

Z 1

�1x(t)e�jwtdt 8w 2 R,

then we can consider the Fourier transform (not inverse FT) of X:

(FX)(w) =

Z 1

�1X(t)e�jwtdt = 2⇡ ·

1

2⇡

Z 1

�1X(t)ej(�w)tdt

= 2⇡(F�1X)(�w) = 2⇡x(�w) 8w 2 R,noting how we have used the definition of inverse FT for the last equality and have switchedthe roles of frequency (w) and time (t).

• Duality of Fourier transforms can be stated more compactly as:

X = Fx , FX = 2⇡ S�1

x,

recalling S�1

is the domain-inversion (reflection in y-axis) operation.


253

Duality - example

• For example, recall that

r(t) =

wo

⇡sinc(wot), t 2 R !F R(w) = u(w � wo)� u(w + wo), w 2 R.

• By duality, we expect (replacing parameter wo with to now representing a time value):

u(t� to)� u(t+ to), t 2 R !F 2⇡to

⇡sinc(�wto) = 2tosinc(wto), w 2 R,

where we note that the sinc function is also even.

• Exercise: Verify this by directly integrating to compute FR.

• Exercise: Suppose the “procedure” or “function” FT(⇣, v) returnsR1�1 ⇣(a)e�javda

when passed a signal ⇣ : R ! R and real number v. Show how to use this procedure tocompute (i) (Fx)(w) and (ii) (F�1X)(t).


254

Signal energy by Parseval’s theorem

• Consider an energy signal x : R ! R with Fourier transform X = Fx,

Ex =

Z 1

�1|x(t)|2dt =

Z 1

�1x(t)x(t)dt.

• Substituting x = F�1X, using Fubini’s theorem, and then F�, we get Parseval’s theorem:

Ex =

Z 1

�1

1

2⇡

Z 1

�1X(v)ejvtdv

1

2⇡

Z 1

�1X(w)ejwtdw dt

=

Z 1

�1

1

2⇡

Z 1

�1X(v)ejvtdv

1

2⇡

Z 1

�1X(w)e�jwtdwdt

=

1

2⇡

Z 1

�1X(w)

Z 1

�1X(v)

✓

Z 1

�1

ejvt

2⇡e�jwtdt

◆

dvdw

=

1

2⇡

Z 1

�1X(w)

Z 1

�1X(v)�(w � v)dvdw =

1

2⇡

Z 1

�1X(w)X(w)dw

=

1

2⇡

Z 1

�1|X(w)|2dw


255

Parseval’s theorem - example

• Recall that if X(w) = u(w+wo)� u(w�wo), w 2 R (i.e., an even rectangle pulse),then its inverse Fourier transform is x(t) =

wo

⇡sinc(wot), t 2 R.

• The energy of this signal in the time domain cannot be directly computed, but by usingParseval’s theorem,

Ex =

1

2⇡

Z 1

�1|X(w)|2dw

=

1

2⇡

Z wo

�wo

dw

=

wo

⇡

• Equating toR1�1 |x(t)|2dt and substituting ⌧ = wot, we get an interesting result:

Z 1

�1sinc2(⌧)d⌧ = ⇡.


256

Parseval’s theorem - energy in a frequency band

• Suppose we want to find the energy of the response y of an ideal band-pass filter to a giveninput signal f , with filter pass-band [w

1

, w2

] such that 0 < w1

< w2

radians/s.

• Y = Fy satisfies

|Y (w)| =

⇢

|F (w)| if w 2 [w1

, w2

] or w 2 [�w2

,�w1

]

0 else

• So, by Parseval’s theorem,

Ey =

1

2⇡

Z �w1

�w2

|F (w)|2dw +

1

2⇡

Z w2

w1

|F (w)|2dw

=

1

⇡

Z w2

w1

|F (w)|2dw,

where the last equality is because f is R-valued (and hence F = Ff is an even function).

• Exercise: Find the transfer functionH and impulse response h = F�1H of this band-passfilter.


257

Ideal Amplifier

• An ideal amplifier has impulse response h = A�⌧� for delay ⌧ > 0, and hence transferfunction H(w) = Aej⌧w, w 2 R.

• Thus, the ZSR of an ideal amplifier to any (Dirichlet) input signal f is, for some ⌧ 2 R,

y(t) = (f ⇤ h)(t) = Af(t� ⌧), t 2 R,i.e., an amplified (by A) and undistorted (though potentially delayed by ⌧ ) version of theinput signal; equivalently,

Y (w) = F (w)H(w) = AF (w)e�jw⌧ , w 2 R.

• Such a constant response (A = |H(w)| 8w 2 R) across the whole frequency band is notfeasible.

• Typically, only an approximately distortionless amplification is possible over a limited fre-quency band.


258

Ideal Filters

• Similarly, we may be interested in “filtering out” all frequencies w (H(w) = 0) of an inputsignal f except for a certain frequency band, i.e., except for the pass-band of frequenciesw where H(w) = 1 (or H(w) = e�jw⌧ for delay ⌧ ).

• Thus, the ZSR

Y (w) = F (w)H(w) =

⇢

F (w) if w is in the pass-band0 else

• If the pass-band is of the form:

– [0, w0

] for w0

> 0 (more precisely [�w0

, w0

] in the Fourier domain), then the filteris a low-pass filter, and w

0

(or sometimes 2w0

) is called the bandwidth of the filter;

– (�1,�w0

]

S

[w0

,1) for w0

> 0, then the filter is a high-pass filter;

– [�w1

,�w0

]

S

[w0

, w1

] for w1

> w0

> 0, then the filter is a band-pass filter.

• Feasibility also requires causality, equivalently the Paley-Wiener criterion in the frequencydomain which precludes H(w) = 0 on whole intervals of the real line.


259

Low-pass, band-pass and high-pass filters


260

Feasible low-pass filter (LPF) or amplifier

• We therefore see that an ideal low-pass filter (without delay) will have transfer function inthe form of an even rectangle pulse,

H(w) = A(u(w + w0

)� u(w � w0

)) 8w 2 R,where A > 1 if it’s an ideal amplifier over the “low” frequency band [�w

0

, w0

].

• Thus, the impulse response is

h(t) =

Awo

⇡sinc(wot), t 2 R.

• Note that a non-causal impulse response h that is not anti-causal can be delayed to a causalimpulse response �⌧h, where h(t) = 0 8t < ⌧ ;

• thus, the ZSR to f of the causal LTI system is f ⇤�⌧h = �⌧(f ⇤ h).

• But the impulse response of an ideal LPF is anti-causal (meaning that the support of hextends to �1).

• Hence this ideal filter is anticausal and infeasible.


261

Approximating an ideal LPF H with a causal LPF ˜H

• Consider again the ideal LPF H(w) = u(w + wo)� u(w � wo), w 2 R.

• Suppose the anticausal (sinc) impulse response h = F�1H is delayed by ⌧ > ⇡/wo > 0

then truncated to yield a causal impulse response

˜h = u�⌧h ) ˜H = (2⇡)�1U ⇤ {H(w)e�jw⌧} where U = Fu.

• The idea is that ⌧ > 0 is su�ciently large so that the square errorZ 1

�1|h(t� ⌧)� ˜h(t)|2dt =

Z �⌧

�1|h(t)|2dt

is small.

• If this is the case, the transfer function of the causal LPF ˜H = F˜h will be a slightly distortedversion of H (both in the pass and non-pass bands) with a roughly linear phase of negativeslope due to the time delay.

• Note that just the delayed version of the impulse response �⌧h still corresponds to anideal, anticausal LPF with transfer function H(w)e�j⌧w, w 2 R.

• In a subsequent course on signal processing, you will see how to feasibly trade-o↵ lowerdistortion for complexity (e.g., degree n of the causal filter’s characteristic polynomial)through the use of special polynomials, e.g., from the Butterworth or Chebyshev families.

262

Example causal LPF ˜h = u�⌧h with wo = 2 and delay ⌧ = 20


263

Modulation of a baseband signal

• A signal f = F�1F is said to be a baseband signal if it is band limited, i.e., if there isfrequency w

0

> 0 such that F (w) = 0 8|w| > w0

.

• For example, w0

⇡ 20kHz = 20000 · 2⇡ radians/s for an audio signal discernible to thehuman ear.

• Suppose the audio signal is to be (wireless) transmitted over the air in a frequency bandassigned to a radio station,

[�wc

� w0

, � wc

+ w0

]

[

[wc

� w0

, wc

+ w0

]

where typically wc

� w0

.

• So, the audio signal needs to be frequency shifted to the carrier frequency wc

prior totransmission over the air.

• Again, frequency shifting is accomplished by multiplication by a sinusoid in the time domain,

f(t) cos(wc

t), t 2 R.which is the modulated signal traveling over the air.


264

Modulation and demodulation of a baseband signal - example


265

Demodulation of a baseband signal

• The radio receiver needs to demodulate the signal f(t) cos(wc

t) after it has been receivedover the air.

• Demodulation is accomplished by first applying the same frequency shift as modulation,

f(t) cos(wc

t) · cos(wc

t) = f(t) cos2(wc

t)

= f(t)

✓

1

2

+

1

2

cos(2wc

t)

◆

, t 2 R.

and then applying a low-pass filter to eliminate the high-frequency components (at ±2wc

)to recover the baseband signal,

1

2

f.

• In later courses on communications, you will learn of di↵erent modulation techniques, includ-ing those that involve first sampling and quantizing the continuous-time baseband signal.


266

Sampling - preliminaries

• Sampling a continuous-time, band-limited signal f is a first step to processing it digitally(i.e., applying a digital filter to its sample sequence).

• Recall the ideal sampling property of the unit impulse: if f is continuous at ⌧ thenZ 1

�1�(t� ⌧)f(t)dt = f(⌧),

i.e., the sample at time ⌧ , f(⌧), is extracted from the signal f by multiplying the signalwith an impulse at ⌧ and integrating.

• Here sampling is “ideal” because f is sampled with infinite precision and exactly at time t.

• Suppose the continuous-time signal f is being sampled every Ts

= 2⇡/ws

seconds toobtain the sequence {f(kT

s

)}k2Z.


267

Sampling - picket-fence function

• To capture such periodic sampling at frequency ws

, we multiply f by the “picket-fence”function,

pTs

(t) =

1X

k=�1

�(t� kTs

) 8t 2 R,

to obtain

pTs

(t)f(t) =

1X

k=�1

f(t)�(t� kTs

) =

1X

k=�1

f(kTs

)�(t� kTs

) 8t 2 R,

i.e., the impulse �(t� kTs

) is weighted by the f -sample f(kTs

).


268

The picket-fence function (T = Ts)


269

Impact of sampling in the frequency domain - Poisson’s identity

• To visualize the impact of sampling on the signal, we now derive F{pTs

f}.

• Obviously, pTs

is Ts

-periodic.

• The complex-exponential Fourier series of pTs

has coe�cients Dk = T�1

s

8k 2 Z; thisleads to Poisson’s identity:

pTs

(t) :=

1X

k=�1

�(t� kTs

) =

1

Ts

1X

k=�1

ejkws

t, 8t 2 R!

• Thus, if F = Ff then by the linearity and frequency-shift properties,

F{pTs

f}(w) =

1

Ts

1X

k=�1

F�

f(t)ejkws

t

(w) =

1

Ts

1X

k=�1

F (w � kws

), 8w 2 R.

• So, sampling has the e↵ect of superposing copies of F frequency-shifted by all integermultiples of the sampling frequency, w

s

= 2⇡/Ts

.


270

Nyquist sampling rate for band-limited signals

• Suppose the signal f is band-limited, i.e., there is a finite wb

> 0 such that F (w) = 0

8|w| > wb.

• The frequency-shifted copies of F that constitute FpTs

f will not overlap if

ws

� wb

> wb

) ws

> 2wb

.

• If this is the case, note that we can recover the original signal f from pTs

f by applying alow-pass filter (with filter bandwidth between wb and wc � wb).

• Thus, if ws

> 2wb

, then the complete “information” in f is preserved in its sample-sequence pT

s

f , equivalently (in terms of information) the samples {f(kTs

)}1k=�1.

• 2wb

is called the Nyquist sampling frequency of f .

• Sampling at less than the Nyquist sampling frequency may prevent recovery of f from pTs

f ,e.g., aliasing.


271

Sampling band-limited signals - example and exercise

• Consider the signal x(t) = Asinc(wot) with

• X(w) := (Fx)(w) = (⇡A/wo)(u(w + wo)� u(w � wo)) (check F�1X = x).

• So, the Nyquist simpling frequency of x is 2wo radians/s, i.e., the Nyquist sampling periodis 2⇡/(2wo) = ⇡/wo seconds.

• Consider again the ideal sampling (picket fence) function p(t) =

P1k=�1 �(t� kT ) for

sampling period T > 0.

• Recall that (Fxp)(w) =

P1r=�1X(w � k2⇡/T )/T .

• Plot Fxp and show that, in particular, if T = ⇡/wo (the Nyquist sampling period) then(Fxp)(w) = (⇡A/wo)/(⇡/wo) = A (a constant).

• So by inverse Fourier transform, x(t)p(t) = A�(t).

• Does this make sense in light of x(t)p(t) =

P1k=�1 x(kT )�(t� kT )?


272

Nyquist sampling (ws

> 2wb

) and aliasing


273

Unilateral Laplace transform for transient response - Outline

• Laplace transform definition and region of convergence.

• Basic LT pairs and properties.

• Inverse LT of rational polynomials by partial fraction expansion.

• Total response of SISO LTIC systems Q(D)y = P (D)f .

• The steady-state eigenresponse revisited.

• Complex-frequency(s)-domain equivalent circuits.

• System composition and canonical realizations.


274

Unilateral Laplace transform

• We define the unilateral Laplace transform (LT) of time-domain signal x : [0,1) ! C asX = Lx = L{x} where

X(s) :=

Z 1

0�x(t)e�stdt

and:

– the LT is defined for complex frequencies s 2 C for which this integral converges,

– 0� is just before the chosen origin of time, 0. .

• The lower limit of integration is taken as “0�” so that:

– the LT of the impulse is well defined, i.e., we’re not “splitting hairs”,

1 = (L�)(s), 8s 2 C, and

– the LT can accommodate “initial conditions.”

• Input functions of the form f = gu (u the unit step), will have zero initial conditions bydefinition, i.e.,

(D

k f)(0�) = 0, for all integers k � 0.


275

Region of convergence for Laplace transform

• Suppose that the signal x : [0,1) ! C is bounded by an exponential:

9 A > 0,↵ 2 R such that 8t � 0�, |x(t)| Ae↵t,i.e., 8t � 0�, �Ae↵t |x(t)| Ae↵t.

• If x is bounded by an exponential Ae↵t, t � 0�, then the region of convergence (RoC) ofLx contains a portion of C that is an open right-half plane.

• To see why, recall absolute integrability ) integrability; and:Z 1

0�|x(t)e�st|dt =

Z 1

0�|x(t)|e�Re{s}tdt

Z 1

0�Ae(↵�Re{s})tdt by exp. bdd.

=

Ae(↵�Re{s})t

↵� Re{s}

�

�

�

�

�

1

0�

=

A

Re{s}� ↵< 1.

• Thus, if ↵� Re{s} < 0 then X(s) exists and |X(s)| < 1 by the triangle inequality.

• Note that ↵�Re{s} < 0 , Re{s} > ↵ 2 R, i.e., s is an element of an open RHP ⇢ C.

276

Laplace transforms of exponential functions of time and their RoCs

• Consider the exponential signal x(t) = esotu(t), t 2 R, with a particular so 2 C (which

is bounded by 1eRe{so}t, i.e., ↵ = Re{so} ).

• The Laplace transform of the above exponential signal is

X(s) =

Z 1

0�x(t)e�stdt =

Z 1

0�e�(s�so)tdt =

1

s� soe�(s�so)t

�

�

�

�

1

0�

=

1

s� so, for Re{s}>Re{so}.

• Note: x(t) has no �(t) term, so it doesn’t matter whether we integrate from 0� or 0.


277

Laplace transforms pairs - (Lu)(s) = 1/s when Re{s} > 0.

• The Laplace transform of the unit step is just a special case of that of an exponentialfunction.

(Lu)(s) =

1

swhen Re{s} > 0.

• If x(t) = 1 for t � 0� then x 6= u since u(0�) = 0 but

(Lx)(s) =

1

swhen Re{s} > 0,

i.e., Lx = Lu.


278

Ambiguity of inverse Laplace transform, L�1

• For Dirichlet signals x that do not have an impulse � component at the origin 0,

X = Lx = Lxu

• For example, if x(t) = esot for t � 0�, then (Lx)(s) = (Lxu)(s) = 1/(s � so) forRe{s}>Re{so}, but note that

x(0�) = 1 while (xu)(0�) = 0,

i.e., x and xu di↵er in their initial values.

• So, there is some ambiguity when inverting L�1X to find the corresponding time-domainsignal.

• This ambiguity is important as, generally, we want

– zero initial conditions when inverting s-domain terms X of the ZSR (i.e., L�1X =

xu), and

– possibly non-zero initial conditions when inverting s-domain terms of the ZIR (i.e.,L�1X =

x without the step u factor so that (Dk x)(0�) 6= 0 is possible).

• As a default, we will use the form L�1X = xu in the following.


279

Laplace transform properties - linearity

• Just as the Fourier transform, the Laplace transform is linear.

• With the Laplace transform, we must be sure to stipulate the RoC for the linear combinationas the intersection (smallest open RHP) of those of the component signals.

• For example, if x(t) = (↵0

es0

t+ ↵

1

es1

t)u(t) for ↵

0

,↵1

, s0

, s1

2 C and t � 0�, thenX = Lx is

X(s) = ↵0

·1

s� s0

+ ↵1

·1

s� s1

, when Re{s} > max{Re{s0

},Re{s1

}}.


280

Laplace transform properties - time shifting

• Again, as the Fourier transform, the Laplace transform has a time-shift property.

• If we do not delay the negative-time support of a signal x into positive time, or there is nonegative-time support of x (x is causal), then we can simply write

(L�⌧x)(s) = e�⌧s(Lx)(s) if ⌧ > 0,

where the RoC is unchanged by time shifting.

• For example, if x(t) = esotu(t) for t � 0� and fixed ⌧ > 0 then

(L�⌧xu)(s) =

Z 1

0�x(t� ⌧)e�stdt =

Z 1

0�eso(t�⌧)u(t� ⌧)e�stdt

=

Z 1

�⌧�esot0u(t0)e�s(t0+⌧)dt0 (t0 := t� ⌧)

= e�s⌧

Z 1

0

e�(s�so)t0dt0 (u and lower limit of integration)

=

e�s⌧

s� so, when Re{s}>Re{so}.


281

Laplace transform properties - time-shifting (cont)

• However, if we delay the negative-time support of a signal to positive time, then we mustsubtract it out in order to relate to (unilateral) Lx = X.

• That is, if ⌧ > 0 then

(L�⌧x)(s) = (L{x(t� ⌧); t � 0�})(s) =

Z 1

0�x(t� ⌧)e�stdt

= e�⌧s

Z 1

�⌧�x(t0)e�st0dt0 (no unit-step u factor)

= e�⌧s

Z 1

0�x(t0)e�st0dt0 + e�⌧s

Z

0�

�⌧�x(t0)e�st0dt0

= e�⌧sX(s) + e�⌧s

Z

0�

�⌧�x(t0)e�st0dt0.

• E.g., if x(t) = esot (no unit-step u factor) for t 2 R and ⌧ > 0, then

(L�⌧x)(s) =

e�⌧s

s� so�

e�⌧s � e�⌧so

s� so=

e�⌧so

s� so, when Re{s}>Re{so}.

• Similarly, if we advance (⌧ < 0) a signal x’s positive time support into negative time, weneed to add it back to relate the result to X = Lx.

282

Laplace transform pairs - sinusoids

• In the special case of an exponential function that is a sinusoid, x(t) = ejwotu(t) forwo 2 R,

X(s) =

1

s� jwo, when Re{s}> 0.

• By Euler-De Moivre identity and linearity, we get that

L{cos(wot)u(t)}(s) =

s

s2 + w2

o

, when Re{s}> 0

L{sin(wot)u(t)}(s) =

wo

s2 + w2

o

, when Re{s}> 0


283

Laplace transform properties - frequency shift

• As the for the Fourier transform, multiplication by an exponential function in time resultsin a frequency shift.

• That is, if X = Lx and so 2 C, then

L{esotx(t)}(s) = X(s� so),

for all s 2 C such that s� so is in the RoC of X.

• For example, for all ↵, wo 2 R,

L{e↵t cos(wot)u(t)}(s) =

s� ↵

(s� ↵)2 + w2

o

, when Re{s}> ↵

L{e↵t sin(wot)u(t)}(s) =

wo

(s� ↵)2 + w2

o

, when Re{s}> ↵


284

LT pairs - L{tkesotu(t)}(s) = k!/(s� so)k+1 when Re{s}> Re{so}

• Again recall that SISO LTI systems have characteristic modes of the form tkesotu(t) wherek � 0 is an integer and so 2 C a characteristic value of the system.

• As for the Fourier transform, to evaluate the Laplace transform of such a characteristicmode, we need to integrate by parts: for k > 0 and Re{s} > Re{so}:

L{tkesotu(t)}(s) =

Z 1

0�tkesote�stdt =

1

so � s

Z 1

0�tkde(so�s)t

=

1

so � stke(so�s)t

�

�

�

�

1

0��

1

so � s

Z 1

0�e(so�s)tdtk

=

k

s� so

Z 1

0�tk�1e(so�s)tdt

=

k

s� soL{tk�1esotu(t)}(s)

• So by induction, we can show that for integers k > 0 and Re{s} > Re{so}:

L{tkesotu(t)}(s) =

k!

(s� so)k+1

.


285

Laplace transform properties - di↵erentiation

• For all di↵erentiable (Dirichlet) signals x, with X := L{x}, the Laplace transform of a

time derivative is (recall D :=

ddt),

(LDx)(s) = sX(s)� x(0�).

• This property is proved by simple integration by parts: 8s in the RoC of X = Lx,

(LDx)(s) =

Z 1

0�(Dx)(t)e�stdt

=

Z 1

0�e�stdx(t)

= x(t)e�st�

�

10� �

Z 1

0�x(t)de�st

= �x(0�) + s

Z 1

0�x(t)e�stdt (⇤)

= �x(0�) + sX(s),

where (*) because limt!1 x(t)e�st= 0 is a necessary condition for the convergence of

X (i.e., for s to be in the RoC of x).


286

Laplace transform properties - di↵erentiation (cont)

• For two derivatives:

(LD

2 x)(s) = (LD(Dx))(s)

= s(LDx)(s)� (Dx)(0�)

= s(sX(s)� x(0�))� (Dx)(0�)

= s2X(s)� sx(0�)� (Dx)(0�).

• So, the derivative property can be inductively generalized to: 8s 2 RoC of X = Lx andintegers k � 1,

(LD

k x)(s) = skX(s)�k�1

X

l=0

sl(Dk�1�l x)(0�)


287

Laplace transform properties - Initial Value Theorem

• Suppose Y = Ly where y is Dirichlet and continuous at the time 0.

• The initial value theorem states that

y(0) = lim

s!1sY (s).

• To see why:

sY (s) =

Z 1

0�y(t)se�stdt and lim

s!1se�st

= �.

• So, the IVT follows from the ideal sampling property of the impulse.

• If Y =

˜P/Q is a rational polynomial with the degree of ˜P < � 1 + degree of Q,

– then sY (s) is still a strictly proper rational polynomial with degree of s˜P (s) < degreeof Q(s),

– so y(0) = 0 by L’Hopital’s rule on sY (s) and the IVT.


288

Laplace transform properties - Initial Value Theorem - example

• For example, if y(t) = 2e�2t � 2e�3t then y(0) = 0 and

Y (s) =

2

s+2

�2

s+3

=

2

(s+2)(s+3)

,

i.e., 0 = the degree of ˜P < �1+ degree of Q = �1+ 2 so that

lim

s!1sY (s) = 0 = y(0).

• For another example, if y(t) = (3+ e�2t)u(t) then y(0) = 4 and

Y (s) = (4s+6)/(s2 + 2s); thus

lim

s!1sY (s) = lim

s!1

4s+6

s+2

= 4 = y(0).


289

Laplace transform properties - Final Value Theorem

• The final value theorem states that if limt!1 y(t) exists, then

lim

t!1y(t) = lim

s!0

sY (s).

• To see why, recall that

sY (s)� y(0�) = (LD y)(s) =

Z 1

0�(D y)(t)e�stdt =

Z 1

0�e�stdy(t)

) lim

s!0

sY (s)� y(0�) =

Z 1

0�dy(t) = y(1)� y(0�)

• For example, if y(t) = (3+ e�2t)u(t) then y(1) = 3 and

lim

s!0

sY (s) = lim

s!0

s

✓

3

s+

1

s+2

◆

= lim

s!0

3 +

s

s+2

= 3 = y(1).


290

Laplace transforms applied to SISO systems given by ODE

• We can use the Laplace transform to analyze the total transient response of the SISO, LTICsystem with input and output signals, respectively, f : [0,1) ! R, y : [0�,1) ! R:

P (D)f = Q(D)y.

• We assume Dirichlet input functions of the form f = gu, u the unit step, with zero initialconditions:

(D

k f)(0�) = 0, for all integers k � 0;

so, by the derivative property, (LDkf)(s) = sk(Lf)(s).

• Recall Q(D) =

Pnk=0

ak Dk and P (D) =

Pmk=0

bmD

m, where 8k, ak, bk 2 R andan = 1.

• Again, let Y = Ly and F = Lf .


291

Laplace transforms applied to SISO systems given by ODE (cont)

• By the derivative property for the Laplace transform of the ODE,

P (s)F (s) = Q(s)Y (s)�nX

k=1

ak

k�1

X

l=0

sl(Dk�1�l y)(0�).

for s in the intersection of the RoC of f and the system characteristic modes.

• Thus, the total response,

Y (s) =

P (s)

Q(s)F (s) +

Pnk=1

akPk�1

l=0

sl(Dk�1�l y)(0�)

Q(s)


292

Laplace transforms applied to SISO systems given by ODE (cont)

• Thus, the total response

Y (s) = YZS

(s) + YZI

(s)

where

YZS

(s) =

P (s)

Q(s)F (s) = H(s)F (s),

H(s) = P (s)/Q(s) the transfer function of the system, and

YZI

(s) =

Pnk=1

akPk�1

l=0

sl(Dk�1�l y)(0�)

Q(s)


293

Laplace analysis of RL circuit with exponential input

• The nodal equation at y for this circuit is, by KCL,

f(t)� y(t)

R+ i(0�) +

Z t

0�

f(⌧)� y(⌧)

Ld⌧ +

0� y(t)

R= 0, t � 0.

• Di↵erentiating this equation, multiplying by R/2, and rearranging gives the standard form:

D y +

R

2Ly =

1

2

D f +

R

2Lf.


294

Laplace analysis of RL circuit with exponential input (cont)

• Taking Laplace transform gives (with f(0�) = 0):

sY (s)� y(0�) +

R

2LY (s) =

1

2

sF (s) +R

2LF (s)

) Y (s) = H(s)F (s) +y(0�)

Q(s), where

H(s) =

YZS

(s)

F (s)=

P (s)

Q(s)=

s/2+R/(2L)

s+R/(2L)=

1

2

+

R/(4L)

s+R/(2L).

• Note that by linearity, the impulse response of this system is

h(t) = (L�1H)(t) =

1

2

�(t) +R

4Le�Rt/(2L)u(t) 8t � 0� .


295

Laplace analysis of RL circuit with exponential input (cont)

• Suppose R = 4⌦, L = 1H, the initial condition y(0�) = 2V , and the input signalf(t) = 2e�3tu(t) for t � 0�, i.e., F (s) = 2/(s+3) for Re{s} > �3.

• Thus, the total response

Y (s) = H(s)F (s) +y(0�)

Q(s)

=

s/2+ 2

s+2

·2

s+3

+

2

s+2

=

s+4

(s+2)(s+3)

+

2

s+2

=

✓

�1

s+3

+

2

s+2

◆

+

2

s+2

where we have “expanded the fraction” for the last equality, and the RoC for Y is Re{s} >max{�3,�2} = �2.

• Thus, by linearity, the total response

y(t) = (�e�3t+2e�2t

)u(t) + 2e�2t 8t � 0�,

yZS

(t) = (�e�3t+2e�2t

)u(t) and yZI

(t) = 2e�2t 8t � 0� .

• Note that yZS

(0�) = 0 while yZI

(0�) = 2 = y(0�), and the forced/eigen-responseis H(�3)f(t) = �e�3tu(t).

296

Laplace analysis of a circuit - approach

• In the following, we will solve transient total response of circuits by

– first finding a complex-frequency(s)-domain equivalent circuit that includes its initialconditions,

– then solving the circuit for the total transient response Y (s),

– then performing partial fraction expansion separately on the ZIR and ZSR,

– finally inverting the Laplace transform separately for the ZSR and ZIR using knownLaplace-transform pairs.

• Regarding Laplace-transform pairs, we will

– use the unit-step u factor in the time-domain for ZSR terms (so that the ZSR has zeroinitial conditions at 0�),

– otherwise not for ZIR terms (so that the ZIR is continuous at time 0 and meets thegiven initial conditions at 0�).


297

Partial Fraction Expansion (PFE)

• In the previous example we inverted the Laplace transform of a strictly proper rationalpolynomial of the form M(s)/N(s) by using a PFE.

• A rational polynomial M(s)/N(s) is strictly proper if the degree of M < that of N .

• Use long division to obtain a strictly proper rational polynomial from one that is not.

• For example,

3s+2

s+7

= 3�19

s+7

) L�1

✓

3s+2

s+7

◆

(t) = 3�(t)� 19e�7tu(t), t � 0.

• Here, (3s+2)/(s+7) is not a strictly proper rational polynomial while 19/(s+7) is.

• If a transfer function H is proper but not strictly so, then: h = L�1H will have animpulse component (bn�), the ZSR may not be continuous at the origin, and there is“direct coupling” between input and output.


298

PFE (cont)

• In the following, we will consider the strictly proper rational polynomial M(s)/N(s) withall coe�cients 2 R and K poles (roots of the polynomial N).

• That is, K := degree of N > degree of M , and so we can write

N(s) =

KY

k=1

(s� pk),

i.e., we can also assume N ’s coe�cient of sK is one without loss of generality.

• We also assume M and N have no common roots, i.e., no “pole-zero cancellation” issueto consider.

• Note that the RoC for M(s)/N(s) is the open RHP {s 2 C | Re{s} > maxk Re{pk}}.


299

PFE - the case of no repeated poles

• Suppose there are no repeated “poles” for M/N (roots of N), i.e.,

8k 6= `, pk 6= p`.

• In this case, we can write the PFE of M/N as

M(s)

N(s)=

KX

`=1

c`

s� p`

where the scalars (Heaviside coe�cients) c` 2 C are

c` =

M(s)Q

k 6=`(s� pk)

�

�

�

�

�

s=p`

= lim

s!p`

M(s)

N(s)(s� p`) =

M(s)

N(s)(s� p`)

�

�

�

�

s=p`

.

• That is, to find the Heaviside coe�cient ck over the term s � pk in the PFE, we haveremoved (covered up) the term s � pk from the denominator N(s) and evaluated theremaining rational polynomial at s = pk.

• This approach, called the Heaviside cover-up method, works even when p is C-valued.

• Given the PFE of M/N , (L�1M/N)(t) =

PK`=1

c`ep`tu(t).


300

PFE - proof of Heaviside cover-up method

• To prove that the above formula for the Heaviside coe�cient c` is correct, note that theclaimed PFE of M/N is

KX

`=1

c`

s� p`=

PK`=1

c`Q

k 6=`(s� pk)

N(s)

• Thus, the PFE equals M/N if and only if the numerator polynomials are equal, i.e., i↵

M(s) =

KX

`=1

c`Y

k 6=`

(s� pk) =:

˜M(s).

• Again, two polynomials are equal if their degrees, L, are equal and either:

– their coe�cients are the same, or

– they agree at L + 1 (or more) di↵erent points, e.g., two lines (L = 1) are equal ifthey agree at 2 (= L+1) points.

• Since M is a polynomial of degree < K, it su�ces to check that whether M =

˜M forall s = pk, k 2 {1,2, ...,K}, i.e., this would create K equations in < K unknowns (thecoe�cients of M).


301

PFE - proof of Heaviside cover-up method (cont)

• But note that any pole pr of M/N is a root of all but the rth term in ˜M , so that

˜M(pr) = crY

k 6=r

(pr � pk)

=

0

@

M(s0)Q

k0 6=r(s0 � pk0

)

�

�

�

�

�

s0=pr

1

A

Y

k 6=r

(pr � pk)

=

M(pr)Q

k0 6=r(pr � pk0)

Y

k 6=r

(pr � pk)

= M(pr).

• Q.E.D.


302

PFE - the case of no repeated poles - example

• To find the inverse Laplace transform of a strictly proper rational polynomial X = M/N ,first factor its denominator N , e.g.,

X(s) =

s2 + 5s

s3 + 9s2 + 26s+24

=

s2 + 5s

(s+4)(s+3)(s+2)

, for Re{s} > �2.

• So, by PFE

X(s) =

c4

s+4

+

c3

s+3

+

c2

s+2

=

c4

(s+3)(s+2)+ c3

(s+4)(s+2)+ c2

(s+4)(s+3)

(s+4)(s+3)(s+2)

) M(s) = 1s2 + 5s+0 = c4

(s+3)(s+2)+ c3

(s+4)(s+2)+ c2

(s+4)(s+3) =

˜M(s).

• We can solve for the 3 constants ck by comparing the 3 coe�cients of quadratic M and˜M .

• The Heaviside cover-up method suggests we try s = �2,�3,�4 to solve for c2

, c3

, c4

:

c4

=

s2 + 5s

(s+3)(s+2)

�

�

�

�

s=�4

= �2, c3

=

s2 + 5s

(s+4)(s+2)

�

�

�

�

s=�3

= 6, c2

=

s2 + 5s

(s+4)(s+3)

�

�

�

�

s=�2

= �3

• Thus, x(t) = (L�1X)(t) = (�2e�4t+6e�3t � 3e�2t

)u(t).


303

PFE - the case of a non-repeated, complex-conjugate pair of poles

• Again, recall that for polynomials with all coe�cients 2 R, all complex poles will come incomplex-conjugate pairs, p

1

= p2

.

• The case of non-repeated poles p1

, p2

= ↵±j� (↵,� 2 R, j :=

p�1) that are complex-

conjugate pairs can be handled as above, leading to corresponding complex-conjugate Heav-iside coe�cients c

1

, c2

, i.e., c1

= c2

.

• In the PFE, we can alternatively combine the terms

c1

s� p1

+

c2

s� p2

=

r1

s+ r0

(s� ↵)2 + �2

where by equating the two numerator polynomials’ coe�cients,

r0

= � c1

p2

� c2

p1

= � 2Re{c1

p2

} 2 R and r1

= c1

+ c2

= 2Re{c1

} 2 R.

• Note that

L�1

⇢

r1

s+ r0

(s� ↵)2 + �2

�

(t) = L�1

⇢

r1

s� ↵

(s� ↵)2 + �2

+

r0

+ r1

↵

�·

�

(s� ↵)2 + �2

�

(t)

= r1

e�↵tcos(�t)u(t) +

r0

+ r1

↵

�e�↵t

sin(�t)u(t), t � 0� .


304

PFE - non-repeated complex-conjugate pair of poles - example

• To find the inverse Laplace transform of

X(s) =

3s+2

s3 + 5s2 + 10s+12

,

first factor the denominator to get

X(s) =

3s+2

(s2 + 2s+4)(s+3)

.

• Note that the poles of X are �3 and �1± jp3 (so X’s RoC is Re{s} > �1).

• So, we can expand X to

X(s) =

r1

s+ r0

s2 + 2s+4

+

c3

s+3

,

where by the Heaviside cover-up method,

c3

=

3s+2

s2 + 2s+4

�

�

�

�

s=�3

= � 1.


305

PFE - non-repeated complex-conjugate pair of poles - example (cont)

• To find r1

, r0

, we will compare coe�cients of the numerator polynomials of X and its PFE,i.e.,

0s2 + 3s+2 = (r1

s+ r0

)(s+3)+ c3

(s2 + 2s+4) (⇤)= (r

1

� 1)s2 + (3r1

+ r0

� 2)s+3r0

� 4.

• Thus, by comparing coe�cients

0 = r1

� 1, 3 = 3r1

+ r0

� 2, 2 = 3r0

� 4

we get

r0

= 2 and r1

= 1.

• Note how s = �3 in (*) gives c3

= �1 as Heaviside cover-up did.


306

PFE - non-repeated complex-conjugate pair of poles - example (cont)

• Thus, by substituting, then completing the square in the denominator and rearranging thenumerator,

X(s) =

s+2

s2 + 2s+4

+

�1

s+3

=

s+1

(s+1)

2

+ (

p3)

2

+

1p3

·p3

(s+1)

2

+ (

p3)

2

�1

s+3

.

• So,

x(t) = (L�1X)(t)

=

✓

e�tcos(

p3t) +

1p3

e�tsin(

p3t)� e�3t

◆

u(t).


307

PFE - the general case of repeated poles

• If a particular pole p of M(s)/N(s) is of order r � 1, i.e., N(s) has a factor (s� p)r,then the PFE of M(s)/N(s) has the terms

c1

s� p+

c2

(s� p)2+ ...+

cr

(s� p)r=

rX

k=1

ck

(s� p)k=

M(s)

N(s)��(s)

with ck 2 C 8k 2 {1,2, ..., r}, where�(s) represents the other PFE terms ofM(s)/N(s)(i.e., corresponding to poles 6= p).

• Note that equating M/N to its PFE and multiplying by (s� p)r gives

M(s)

N(s)(s� p)r = cr +

r�1

X

k=1

ck(s� p)r�k+�(s)(s� p)r

)M(s)

N(s)(s� p)r

�

�

�

�

s=p

= cr,

i.e., Heaviside cover-up (of the entire term (s� p)r) works for cr.


308

PFE - the general case of repeated poles (cont)

• To find cr�1

, we di↵erentiate the previous display to get

d

ds

M(s)

N(s)(s� p)r =

r�1

X

k=1

ck(r � k)(s� p)r�1�k+

d

ds�(s)(s� p)r

= cr�1

+

r�2

X

k=1

ck(r � k)(s� p)r�1�k+

d

ds�(s)(s� p)r

) cr�1

=

✓

d

ds

M(s)

N(s)(s� p)r

◆

�

�

�

�

s=p

• If we di↵erentiate the original display k 2 {0,1,2, ..., r � 1} times and then substitutes = p, we get (with 0! := 1)

✓

dk

dskM(s)

N(s)(s� p)r

◆

�

�

�

�

s=p

= k!cr�k

) cr�k =

1

k!

✓

dk

dskM(s)

N(s)(s� p)r

◆

�

�

�

�

s=p

.


309

PFE - the general case of repeated poles - example

• To find the inverse Laplace transform of

X(s) =

3s+2

(s+1)(s+2)

3

,

write the PFE of X as

X(s) =

c1

s+1

+

c2,1

s+2

+

c2,2

(s+2)

2

+

c2,3

(s+2)

3

,

so clearly the RoC of X is Re{s} > �1.

• By Heaviside cover-up

c1

=

3s+2

(s+2)

3

�

�

�

�

s=�1

= �1 and c2,3 =

3s+2

s+1

�

�

�

�

s=�2

= 4.


310

PFE - the general case of repeated poles - example (cont)

• Also,

c2,2 =

1

1!

✓

d

ds

3s+2

s+1

◆

�

�

�

�

s=�2

=

1

1!

1

(s+1)

2

�

�

�

�

s=�2

= 1

c2,1 =

1

2!

✓

d2

ds23s+2

s+1

◆

�

�

�

�

s=�2

=

1

2!

�2

(s+1)

3

�

�

�

�

s=�2

= 1

• Thus,

X(s) =

�1

s+1

+

1

s+2

+

1

(s+2)

2

+

4

(s+2)

3

8Re{s} > �1

) x(t) = (L�1X)(t) =

�

�e�t+ e�2t

+ te�2t+2t2e�2t

�

u(t).

• Exercise: Find the ZSR y to input f(t) = 2ej3tu(t) of the marginally stable systemH(s) = 4/(s2 + 9).


311

PFE and eigenresponse for asymptotically stable systems

• The total response of a SISO LTI system to input f is of the form

Y (s) = H(s)F (s) +P1

(s)

Q(s)=

P (s)

Q(s)F (s) +

P1

(s)

Q(s)= Y

ZS

(s) + YZI

(s).

where P1

depends on the initial conditions and the RoC is the intersection of that of inputF = Lf and the system characteristic modes.

• Suppose the system is asymptotically stable (equivalently BIBO stable) and the input is asinusoid at frequency wo, f(t) = Aej(wot+�)u(t) with A > 0 ) F (s) = Aej�/(s �jwo).

• Since jwo cannot be a system pole (owing to asymptotic stability all poles have strictlynegative real part), we can use Heaviside cover-up on

YZS

(s) = H(s)F (s) =

P (s)

Q(s)(s� jwo)Aej�

to get that the total response is of the form

YZS

(s) =

H(jwo)Aej�

s� jwo+ modes = H(jwo)F (s) + modes,

where Aej� is the phasor of f .

312

PFE and eigenresponse for asymptotically stable systems (cont)

• Thus, the total response of an asymptotically stable system to a sinusoidal input f atfrequency wo � 0 is

y(t) = H(jwo)f(t) + linear combination of characteristic modes.

• So by asymptotic stability, the steady-state response is the eigenresponse, i.e., as t ! 1,

y(t) ⇠ H(jwo)f(t) = H(jwo)Aej(wot+�). = |H(jwo)|Aej(wot+�+\H(jwo)).


313

Complex-frequency(s)-domain equivalent circuits - introduction

• To solve for the transient total response of SISO LTIC circuits by Laplace transform, wecan use Kirchho↵’s laws and current-voltage characteristics of the individual devices to find

– the ODE P (D)f = Q(D)y relating input f to output y,

– and the initial conditions (Dk y)(0�) from those given -typically the initial inductorcurrents/fluxes and capacitor voltages/charges, i.e., initial conditions of circuit “state”variables.

• Alternatively, we can find the s-domain equivalent circuit from the s-domain equivalents ofthe individual devices, and apply KVL/KCL to solve for transient total response.

• Independent and dependent sources in the s-domain simply involve the Laplace transformof the associated signals.

• Similarly, an ideal op-amp has the same characteristics in the s-domain (infinite inputimpedance and zero di↵erential input voltage).

• By linearity of the Laplace transform, a resistor with resistance R⌦ relating a current i tovoltage v by Ohm’s law v = Ri, in the s-domain is simply V = RI where V = Lv andI = Li.


314

s-domain equivalent capacitors and inductors

• The current-voltage characteristic of an inductor is v = LD i, so in the s-domain,

V (s) = sLI(s)� Li(0�) or I(s) =

1

sLV (s) +

1

si(0�).

• The current-voltage characteristic of a capacitor is i = CD v, so in the s-domain,

I(s) = sCV (s)� Cv(0�) or V (s) =

1

sCI(s) +

1

sv(0�).

• Recall

L�1

⇢

1

s

�

= u, the unit step,

but when considering initial conditions (ZIR terms), we take

L�1

⇢

1

s

�

(t) = 1 8t � 0� .


315

s-domain equivalent capacitors and inductors (cont)

• The Thevinin or Norton equivalent may be more convenient depending on the circuit.


316

s-domain equivalent circuits - example

• The s-domain equivalent of a first-order op-amp circuit is depicted above for vC(0�) = 2Vand the Norton equivalent s-domain capacitor - so that �CvC(0�) = �0.5 · 2 = �1.

• By KCL at the negative terminal of the op-amp and at vc

:

0+

Y � F

1

+

Vc

� F

2

= 0

F � Vc

2

+ 0.5s(0� Vc

)� (�1) = 0 (impedance of cap. is 1/(0.5s))

• Eliminating Vc

= (F +2)/(s+1) we get

Y (s) =

3s+2

2(s+1)

F (s)�1

s+1

= H(s)F (s) + YZI

(s).

317

s-domain equivalent circuits - example (cont)

• So, if the input f(t) = 2e�3tu(t), t � 0�, then F (s) = 2/(s + 3) and the totalresponse,

Y (s) =

3s+2

(s+1)(s+3)

�1

s+1

= YZS

(s) + YZI

(s)

=

c1

s+1

+

c3

s+3

�1

s+1

where by Heaviside cover-up for the PFE,

c1

= [(3s+2)/(s+3)]s=�1

= �0.5 and c3

= [(3s+2)/(s+1)]s=�3

= 3.5.

• Thus,

y(t) = (�0.5e�t+3.5e�3t

)u(t)� e�t

= (�0.5e�tu(t) +H(�3)f(t))� e�t

= yZS

(t) + yZI

(t), t � 0�

• Regarding initial conditions, note that

– by KCL at the negative terminal in the time-domain,

0 = y(0�)� f(0�) + (vc

(0�)� f(0�))/2 = y(0�) + vc

(0�)/2 = y(0�) + 1,

so that, consistent with the above solution,

y(0�) = �1 = yZI

(0�).

318

Circuit example with switch instantaneously thrown at time t = 0

• The switch is thrown at time t = 0 after a long time (i.e., when t < 0).

• The input is DC so that at time 0�, the circuit (consisting of just the battery, the 2Hinductor and 1⌦ resistor) is in DC steady-state; check that for t < 0, this first-ordercircuit is asymptotically stable with transfer function 1/(2s + 1), i.e., with system pole(characteristic value) �1/2.

• An inductor has zero DC impedance, so the output y(0�) = 10V/1⌦ = 10A by Ohm’slaw.

• For time t � 0�, the input f(t) = 10u(t) so that f(0�) = 0; the negative-timeevolution of the circuit is accounted for by the initial system states, e.g., y(0�) = 10.

• Because the circuit is closed for time t < 0, the initial (time 0�) current through the 1Hinductor is zero and the initial voltage of the 1F capacity is zero.

319

Circuit example with switch thrown at time t = 0 (cont)

• Writing KVL in the time domain for time t � 0, we get

f = 10u = 2D y + vc +D y + y, wherey = D vc and vc is the capacitor voltage.

• Instead of simply di↵erentiating KVL to eliminate vc (and using D f = 10�), we will takeLaplace transform of these two equations taking care to including the initial values of the2H inductor but to use zero initial values for the 1H inductor and 1F capacitor, i.e.,

F (s) =

10

s= 2(sY (s)� y(0�)) + Vc(s) + sY (s) + Y (s) and Y (s) = sVc(s)

) Y (s) =

10

3s2 + s+1

+

20s

3s2 + s+1

= YZS

(s) + YZI

(s) where

YZS

(s) =

10

3s2 + s+1

=

s

3s2 + s+1

·10

s= H(s)F (s) and

YZI

(s) =

20s

3s2 + s+1

=

2y(0�)s

3s2 + s+1

.


320

Circuit example involving a switch thrown at time t = 0 - ZSR

• Note that the poles of YZS

(roots of 3s2 + s+1) are complex:

p =

�1+ jp11

6

and p =

�1� jp11

6

.

• To obtain the total response in the time domain, we can either complete the square (3s2+s+1 = 3((s+1/6)2 +35/108)) and then use the known Laplace transforms of sineand cosine, or just use Heaviside cover-up on the complex-conjugate poles:

YZS

(s) =

10

3s2 + s+1

=

10/3

(s� p)(s� p)=

c

s� p+

c

s� p, where

c =

10/3

s� p

�

�

�

�

s=p

=

10/3

p� p= �j

10

3

p11

.

• Thus, c = j10/(3p11) and

yZS

(t) = ceptu(t) + ceptu(t), 8t � 0� .

• Note that there is no forced response (eigenresponse) component H(0)f(t) for the DCinput f since the transfer function H has a DC zero, i.e., H(0) = 0.


321

Circuit example with switch thrown at time t = 0 - ZIR and total resp.

• Similarly, we can perform PFE on YZI

to get

YZI

(s) =

20s

3s2 + s+1

=

(20/3)s

(s� p)(s� p)=

a

s� p+

a

s� p, where

a =

(20/3)s

s� p

�

�

�

�

s=p

=

20p/3

p� p.

• So, again noting no step-function u factors,

yZI

(t) = aept + aept, 8t � 0� .

• The total response is clearly

y(t) = yZS

(t) + yZI

(t) = ceptu(t) + ceptu(t) + aept + aept, 8t � 0� .

• Exercise: check the computed y(0�) = yZI

(0�) = 2Re{a} = 10.

• This example is Lathi Problem 6.4-5, p. 465.


322

Circuit example with switch thrown at time t = 0 - s-domain equiv.

• The s-domain equivalent circuit would have input voltage F (s) = 10/s, output currentY (s), inductive impedences 2s and s, capacitive impedance 1/s, and resistance 1.

• To preserve the single loop, we use Thevinin equivalent s-domain (reactive) devices for timet � 0.

• Since the initial conditions are zero for the 1H inductor and 1F capacitor, their s-domainrepresentations (of s and 1/s impedences, respectively) will not have (i.e., have zero)associated independent voltage sources.

• The 2H inductor will have a �Ly(0�) = �20V battery referenced positive counterclock-wise.

• Thus, by KVL in the s-domain

F (s) =

10

s= 2sY (s)� 20+ s�1Y (s) + sY (s) + Y (s)

) Y (s) =

10

3s2 + s+1

+

20s

3s2 + s+1

= YZS

(s) + YZI

(s),

as computed above by applying Laplace transform to time-domain KVL.


323

s-domain equivalent circuits - RLC circuit example (w/o switch)

• For the series RLC circuit, we’ve used Thevinin equivalent devices for the s-domain equiv-alent circuit so that there remains a single loop.

• By KVL (voltage increases counterclockwise),

0 =

vc

(0�)

s+

1

sCI(s)� Li(0�) + sLI(s) + Y (s)� F (s)

I(s) = Y (s)/R

) 0 =

vc

(0�)

s+

1

sRCY (s)� Li(0�) +

sL

RY (s) + Y (s)� F (s) (eliminating I)


324

s-domain equivalent circuits - RLC circuit example (cont)

• Thus, the total response is

Y (s) =

F (s) + Li(0�)� vc

(0�)s�1

(sRC)

�1

+ sLR�1

+ 1

=

sRL�1

s2 + sRL�1

+ (LC)

�1

F (s) +sRi(0�)� v

c

(0�)RL�1

s2 + sRL�1

+ (LC)

�1

= YZS

(s) + YZI

(s), where

YZS

(s) =

sRL�1

s2 + sRL�1

+ (LC)

�1

F (s) =

P (s)

Q(s)F (s) = H(s)F (s), and

YZI

(s) =

sRi(0�)� vc

(0�)RL�1

s2 + sRL�1

+ (LC)

�1

=

P1

(s)

Q(s).

• In the following, we will take L = 1H, C = 0.25F, vc

(0�) = 2V, i(0�) = 1A.


325


• For an overdamped case, take R = 5⌦, so that Q(s) = s2+5s+4 = (s+1)(s+4).

• If the input f(t) = 3e2tu(t) () F (s) = 3/(s� 2)), then the total response is

Y (s) =

15s

(s+1)(s+4)(s� 2)

+

5s� 10

(s+1)(s+4)

=

✓

5/3

s� 2

+

5/3

s+1

�10/3

s+4

◆

+

✓

�5

s+1

+

10

s+4

◆

by PFE

) y(t) =

✓

5

3

e2t +5

3

e�t �10

3

e�4t

◆

u(t) +�5e�t+10e�4t

= yZS

(t) + yZI

(t), 8t � 0� .

• To check this solution is satisfies the initial conditions:

– By Ohm’s law, y(0�) = Ri(0�) = 5 · 1 = 5 = yZI

(0�).

– By KVL in the time-domain,

0 = vc

(0�) + LR�1

(D y)(0�) + y(0�)� f(0�) = 2+ 0.2(D y)(0�) + 5� 0

) (D y)(0�) = �35 = (D yZI

)(0�).


326


• Now consider the marginally stable case with R = 0 and the output redefined to beY := V

c

so that KVL is

0 = Y + sLI � Li(0�)� F where I = sCY � Cy(0�)

) Y (s) =

(LC)

�1

s2 + (LC)

�1

F (s) +sy(0�) + i(0�)/C

s2 + (LC)

�1

=

4

s2 + 4

F (s) +2s+4

s2 + 4


327


• So, the total response to the sinusoidal input f(t) = 3e2jtu(t) is

Y (s) =

12

(s+2j)(s� 2j)2+

2s+4

s2 + 4

=

✓

�3/4

s+2j+

3/4

s� 2j+

�3j

(s� 2j)2

◆

+

✓

2

s

s2 + 4

+ 2

2

s2 + 4

◆

) y(t) =

✓

�3

4

e�2jt+

3

4

e2jt � 3jte2jt◆

u(t) + 2cos(2t) + 2 sin(2t)

= yZS

(t) + yZI

(t), t � 0� .

• Note that resonance phenomenon for the input f is a sinusoid at the system resonantfrequency of 2 = (LC)

�0.5 radians/s, leading to an unbounded response (te2jt) to abounded input f .

• Also note that yZI

(0�) = 2 = y(0�) and i(0�) = 1 = C(D y)(0�) = 0.25(D y)(0�)

) (D y)(0�) = 4 = yZI

(0�).


328

System composition - for the ZSR of a LTIC system

• SISO system with input f and ZSR y, where F = Lf , Y = Ly:

• An ideal amplifier for scalar a (recall inverting and noninverting op-amp circuit configura-tions):

• An integrator system - recall (Lu)(s) = s�1 and (f ⇤ u)(t) =

R t

0

f(⌧)d⌧, t � 0:

• Exercise: show how a integrator can be implemented using a simple op-amp circuit with aoutput-feedback capacitor and an input-side resistor.

329

System composition - summer

• One can realize a summer with inverting and/or non-inverting op-amp circuit configurations.


330

System composition - non-inverting summer implementation

• The voltage at the positive input terminal is v+

= v� =

Rb

Rb

+Rf

y.

• With system inputs v1

, v2

, KCL at the positive input terminal is

0 =

0� v+

Ra

+

2

X

k=1

vk � v+

Rk

) y =

2

X

k=1

˜R

Rk

vk, where ˜R =

Rb

Rb

+Rf

1

Ra

+

2

X

k=1

1

Rk

!!�1

.

331

Canonical system realizations - single pole system

• Note that for the system with transfer function

H(s) =

a

s+ b=

Y (s)

F (s),

we can write

sY (s) = aF (s)� bY (s).


332

Canonical system realizations - by PFE using single-pole systems

• Given n distinct real poles, we have by PFE of strictly proper H = P/Q,

H(s) =

nX

k=1

ck

s� pk.

• So, one can realize this system by summing the outputs of n single-pole systems.


333

System composition - tandem systems

• Recall how an op-amp has a separate power supply, decoupling input from output so thatthe transfer function H

1

does not depend on its load (system H2

).


334

System composition - tandem systems - realizing repeated poles

• So, one can realize part of a transfer function H with, say, twice-repeated (triple) pole pleading to PFE terms

c1

s� p+

c2

(s� p)2+

c3

(s� p)3,

by using three tandem systems:

• Letting the input of this subsystem be F and the output be Y , note thatY (s) = c

1

F (s)s�p

+ c2

F (s)(s�p)2

+ c3

F (s)(s�p)3

.


335

Canonical system-realizations - direct form

• Consider the proper transfer function H = P/Q, i.e., m n.

• The direct-form realization employs the interior system state X := F/Q,i.e., F = QX and Y = PX where the former implies

F (s) =

nX

k=0

akskX(s) with an = 1

) snX(s) = F (s)�n�1

X

k=0

akskX(s).

• For n = 2, there are two “system states” (outputs of integrators), X and sX (x andDx):

336

Canonical system-realizations - direct form (cont)

• Now adding Y = PX, we finally get the direct-form canonical system-realization of H:

• Again, state variables taken as outputs of integrators, here: x, Dx, ... Dn�1 x.

• If bn = b2

6= 0, there is direct coupling of input and output, H is proper but not strictlyso, h = L�1H has a impulse component b

2

�, and the ZSR may be discontinuous at time0.


337

Canonical system realizations - direct form (cont)

• Note that this n = 2 example above can be used to implement a pair of complex-conjugatepoles as part of a larger PFE-based implementation (with otherwise di↵erent states); e.g., forn = 2, H(s) = P (s)/Q(s) where

Q(s) = s2 + a1

s+ a0

= (s� ↵)2 + �2

for ↵,� 2 R, so the poles are ↵± j�.


338

Canonical system realizations - by PFE

• In the general case of a proper transfer function, we can use partial-fraction expansion (afterlong division to get a strictly proper rational polynomial)

– grouping the terms corresponding to a complex-conjugate pair of poles, i.e., a second-order denominator, and

– using a direct-form realization for these terms.

• Besides the PFE-based and direct-form realizations, there are other (zero-state) systemrealizations, e.g., “observer” canonical.

• For proper rational-polynomial transfer functionsH = P/Q, all of these realizations involven (= degree of Q) integrators, the output of each being a di↵erent interior state variableof the system.


339

Signal tracking by a Proportional-Integral (PI) system

• “Open-loop” system is PI: G(s) = Kp

+Ki

/s.

• The goal is that the error signal e = f � y ! 0, i.e., the “closed-loop” output y tracksthe input f .

• Y = GE = G(Y � F ) ) Y = FG/(1 +G).

• Thus, the closed-loop transfer function (F to Y , with negative feedback) is

H(s) =

G(s)

1 +G(s)=

1

1+Kp

✓

Kp

+

↵

s+ ↵

◆

, where ↵ :=

Ki

Ki

+Kp

.


340

Signal tracking by a PI system (cont)

• Now assume that ↵ > 0 (H is stable).

• Suppose we want to see how the PI parameters Kp

,Ki

a↵ect tracking of a step f = Au,i.e., we want the (ZS) step response:

Y (s) = H(s)F (s) = H(s)A

s=

A

1+Kp

✓

Kp

s+

↵

s(s+ ↵)

◆

=

A

1+Kp

✓

Kp

+1

s�

1

s+ ↵

◆

by PFE

) y(t) = Au(t)�A

1+Kp

e�↵tu(t)

• Exercise: Find an expression for the time t > 0 at which y(t) = 0.9A.

• Exercise: Derive how this closed-loop PI system would track a ramp, f(t) = Atu(t).


341

Feedback control for stabilization by pole placement

• Consider this simple feedback control system where the fixed open-loop system (plant)G = P/Qo and constant-gain (proportional) K output-feedback combine to give theclosed-loop system H:

X = F �KY and Y = GX ) Y = GF �KGY =

G

1+KGF = HF.

• Thus, the closed-loop characteristic polynomial Q = Qo +KP because

H =

G

1+KG=

P

Qo +KP=

P

Q.

• E.g., if P (s) = 1,Qo(s) = s�5 andK > 5, thenG is unstable, butQ(s) = s�5+Khas a root at 5�K < 0 and so the closed-loop system is stable.

• So, proportional feedback control may be used to stabilize a system by pole placement.

342

Minimum-phase systems

• A LTI system H,h is said to be minimum phase if both it and its inverse 1/H,L�1{1/H}are stable and causal.

• So, a minimum-phase H has all poles and zeros with negative real part.

• E.g., for real p, z > 0, consider the following two transfer functions

Hnmp

(s) =

1+ s/(�z)

1 + s/pand H

mp

(s) =

1+ s/z

1+ s/p

• Note that the magnitude responses for these two transfer functions is the same,

|Hnmp

(jw)| = |Hmp

(jw)| =

s

1+ (w/z)2

1 + (w/p)2.

• But for all (real) frequencies w � 0, the phase response of Hmp

is smaller:

\Hmp

(jw) = tan

�1

w

z� tan

�1

w

p

tan

�1

w

�z� tan

�1

w

p= ⇡ � tan

�1

w

z� tan

�1

w

p= \H

nmp

(jw)


343

Undershoot of non-minimum-phase systems

• Consider the unit-step response of the non-minimum-phase system,

Y (s) := L{hnmp

⇤ u}(s) = Hnmp

(s)1

s=

�1/p� 1/z

1+ s/p+

1

s

) y(t) = (hnmp

⇤ u)(t) = � (1 + p/z)e�tpu(t) + u(t)

• Note that if r > p then y(0+) < 0, though limr!1 y(t) = 1;

• i.e., the non-minimum-phase system always first moves away from 1 = H(0) (here be-comes negative) before it converges to 1 = H(0) (follows the steady-state unit-stepresponse), i.e., it always undershoots.

• Exercise: Show that the minimum-phase system Hmp

never first undershoots.


344

introduction to frequency-domain analysis of continuous-time, linear...

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