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Chapter 1: Classification of Signal and System

Houshou Chen

Dept. of Electrical Engineering,National Chung Hsing University

E-mail: [email protected]

H.S. Chen Chapter1: Classification of signals and systems 1

• Siganls:

1. What is a signal

2. Classification of signals

3. Basic operations on signals

4. Elementary signals

• Systems:

1. What is a system

2. Classification of systems

3. LTI systems: circuit example

4. More example and motivation

Department of Electrical Engineering, National Chung Hsing University


Signals

• Signal: a continuous-time signal x(t) (discrete-time signal x[n]) isa function of an independent continuous variable t (discretevariable n).

• Elementary continuous-time signals:

1. x(t) = es0t, s0 = σ0 + jω0 (complex exponential)

2. x(t) = ejω0t, s0 = jω0 (periodic complex exponential)

3. x(t) = eσ0t, s0 = σ0 (real exponential)

4. x(t) = cos ω0t = Reejω0t (sinusoidal signals)

5. impulse function: δ(t)

6. unit function: u(t)

7. ramp function: r(t)



• Elementary discrete-time signals:

1. x[n] = zn0 , z0 = r0e

jω0 (complex exponential)

2. x[n] = ejΩ0n, z0 = ejΩ0 (periodic complex exponential)

3. x[n] = rn0 , z0 = r0 (real exponential)

4. x[n] = cos Ω0n = ReejΩ0n (sinusoidal signals)

5. impulse function: δ[n]

6. unit function: u[n]

7. ramp function: r[n]

• We will treat continuous-time and discrete-time signalsseparately but in parallel.



Classification of signals

1. continuous-time x(t) vs. discrete-time x[n]

• Usually a discrete-time signal x[n] is obtained from acontinuous time signal x(t) by sampling:

x[n] = x(nT ), n = 0,±1,±2... for some fixed T.

2. even vs. odd signals

• even (real): x(−t) = x(t)

• odd (real): x(−t) = −x(t)

• symmetric (complex): x(−t) = x∗(t)

• anti-symmetric (complex): x(−t) = −x∗(t)



Any signal x(t) can be decompose into the even part xe(t) andthe odd part xo(t) by:

x(t) =12[x(t) + x(−t)] +

12[x(t) − x(−t)],

where

xe(t) =12[x(t) + x(−t)] and xo(t) =

12[x(t) − x(−t)]

• It is easy to check that xe(t) = xe(−t) , xo(t) = −xo(t).



3. periodic vs. aperiodic signals

• A signal x(t) (x[n]) is called a periodic signal if there exist realnumber T (integer N) such that:

x(t + T ) = x(t) (x[n + N ] = x[n]).

• The smallest T0 (N0) such that :

x(t + T0) = x(t) (x[n + N0) = x[n])

is called the (fundamental) period of x(t) (x[n]).

• 2πT0

( 2πN0

) is called the fundamental frequency ( radsec ) of x(t) (x[n]).

• x(t) (x[n]) is called aperiodic if it is not periodic.



4. deterministic vs. random

• deterministic signal x(t)⇒ x(t0) is a number, no uncertainity

• random signal x(t)⇒ x(t0)is a random variable (with some probability specification)x(t) = random signal = random process = stochastic process

5. energy signal vs. power signal

• for a continuous signal x(t):E =

∫ ∞−∞ x2(t)dt : energy

P=limT→∞ 1T

∫ T2

−T2

x2(t)dt : power

= 1T

∫ T2

−T2

x2(t)dt if x(t) is periodic with period T



• for a discrete signal x[n]E =

∑∞n=−∞ x[n]: energy

P = limn→∞ 12N

∑N−1n=−N x2[n]: power

= 1N

∑N−1n=0 x2[n] periodic with period N

• x(t)(x[n]) is an energy signalif 0 < E < ∞or is a power signalif 0 < P < ∞

• A signal x(t) (x[n]) can not be an energy signal and a powersignal simultaneously.



Difference between x(t) and x[n]

• There are many similarities between x(t) and x[n] , but there isone important difference.

• For a continuous time x(t) = ejw0t we have:

1. ejw1t = ejw2t if w1 = w2, i.e., any two signals with twodifferent frequencies are distinct.

2. w1 > w2 ⇒ ejw1t oscillates faster than ejw2t.

3. ejw0t is periodic for any w0, T0 = 2πw0

.



• The above three properties are not true for a discrete-time signalx[n] = ejΩ0n.

1. For a discrete-time signal, we havex[n] = ej(Ω0+2π)n = ejΩ0n × ej2πn = ejωon

i.e., the signal x[n] at frequency (Ω0 + 2π) is the same as that atfrequency Ω0, that is unlike the continuous case:ejw1t = ejw2t if w1 = w2

2. I.e., for continuous-time signal, ejw0t are all distinct for distinctw0. On the other hand, in discrete-time, the signalx[n] = ejΩ0n = ej(Ω0+2mπ)n for any m ∈ Z.=⇒ we only need to consider a frequency interval of length 2π,usually −π ≤ Ω < π or 0 ≤ Ω < 2π.



3. Ω0 is larger ⇒ ejΩ0n oscillate faster is not true in discrete-timecase

• In discrete-time, since we only need to consider a frequencyinterval of length 2π, say −π ≤ Ω < π or 0 ≤ Ω < 2π. Wehave: frequencies close to 0, 2π are termed as low frequenciesand frequencies close to π, or −π are termed as highfrequencies.

• I.e., As Ω → 0, 2π, ejΩ0n oscillates slower, and as Ω → π,−π,ejΩ0n oscillates faster.

• cos(0n) = 1, cos(πn8 ), cos(πn

4 ), cos(πn2 ), cos(πn

1 ), Ω from 0 toπ, ejΩn oscillates slower to faster

• cos( 3πn2 ), cos( 3πn

4 ), cos( 8πn7 )cos(2πn) = 1, Ω from π to 2π,

ejΩn oscillates faster to slower



4. The period of discrete-time signal ejΩ0n

• ejΩ0(n+N) = ejΩ0n ∗ ejΩ0N = ejΩ0n( need ejΩ0N = 1)⇒ Ω0N = 2πm ⇒ Ω0

2π = mN

i.e., a discrete-time signal ejΩ0n is not necessary periodic for anyΩ0. For a periodic ejΩ0n, we must have Ω0 = s2π, where s ∈ Q.

• ej nπ4 (Ωo = π

4 = 182π, N = 8) periodic

• ej3n (Ωo = 3 = mN 2π) not periodic



Figure 1: (−1)n



Operations on signals

• operation on

⎧⎨⎩

t − axis

x − axisof x(t)

• On dependent variable x(t)i.e., given x(t), =⇒ want to find y(t) = Ax(t) + B⎧⎨⎩

y1(t) = Ax(t) scaling first

y2(t) = y1(t) + B shift next⇒ y2(t) = y(t) = Ax(t) + B.



• y(t) = Ax(t) + B

– Remark:

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

|A| > 1 expand(A < 0 reverse)

|A| < 1 compress

B > 0 shift up

B < 0 shift down



• y(t) = 3x(t) + 4

Figure 2:



If we do⎧⎨⎩

y1(t) = x(t) + B shift next

y2(t) = Ay1(t) scaling first=⇒ y2(t) = A(x(t) + B)

• Conclusion:

– y(t) = Ax(t) + B then A first ⇒ B next

– y(t) = A(x(t) + B) then B first ⇒ A next



• On independent variable t

i.e., given x(t) ⇒ y(t) = x(at + b)⎧⎨⎩

y1(t) = x(t + b) shift first

y2(t) = y1(at) scaling next⇒ y2(t) = y(t) = y1(at + b)

• Remark:

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

|a| > 1 compress(a < 0 reverse)

|a| < 1 expand

b > 0 shift left (advance version)

b < 0 shift right (delayed version)



Figure 3:



If we do

⎧⎨⎩

y1(t) = x(at) scaling first

y2(t) = y1(t + b) shift next⇒ y2(t) = y1(t + b) = x(a(t + b)) = x(at + ab)

Conclusion:y(t) = x(at + b) b first⇒ a nexty(t) = x(a(t + b)) a first⇒ b next



Why need this: convolutional sum, integral

• x(t) ⇒ Ax(t) + B A first,B next

• x(t) ⇒ x(at + b) b > 0 shift leftb first, a next b < 0 shift rightor equivalent x(t) ⇒ x(at − b) b > 0 shift right

x(t) ⇒ x(at − b) b < 0 shift left

by changing variable∫ ∞−∞ h(t − τ)x(τ)dτ ⇒ t − τ = λ ⇒ τ = t − λ ⇒ dτ = −dλ

=∫ −∞∞ h(λ)x(t − λ)(−dλ) =

∫ ∞−∞ x(λ)h(t − λ)(dλ) = x(t)

⊗h(t)

x[n]⊗

h[n] =∑−∞

k=∞ h[n − k]x[k]=

∑−∞m=∞ h[m]x[n − m] = x[n]

⊗h[n]



Recall h(τ) ⇒ h(t − τ) = h(−τ + t)(h(−τ − (−t)) = h(−(τ − t)))

1. y(t) =∫ ∞−∞ h(t − τ)x(τ) = h(t)

⊗x(t)

2. y[n] =∑

k h[n − k]x[k] = h[n]⊗

x[n]

3. X(D)h(D)

Figure 4:



Figure 5:



Other Elementary signals

1. ramp function:

r(t) =

⎧⎨⎩

0 t ≤ 0

t t ≥ 0

r[n] =

⎧⎨⎩

0 n ≤ 0

n n ≥ 0

Figure 6:



2. unit function

u(t) =

⎧⎨⎩

0 t ≤ 0

1 t ≥ 1 step function

u[n] =

⎧⎨⎩

0 n = −1,−2, . . .

1 n = 0, 1, . . .

Figure 7:



Remark: Many functions x(t) can be written in term of stepfunction. This will be very useful since we can deal with thetransform of x(t) by the transform of u(t), e.g., r(t) = tu(t).



• u(t) − u(t − 1)

• u(t − a) − u(t − b)

Figure 8:



• t · (u(t) − u(t − 1))

• t · (u(t) − u(t − 1)) + (u(t − 1) − u(t − 2))

Figure 9:



In general, if we have x(t) in the form as follows.

Figure 10:



We can always partition x(t) into:x(t) = g1(t)[u(t − a1) − u(t − a2)]

+ g2(t)[u(t − a2) − u(t − a3)]

+...

+ gn(t)[u(t − an) − u(t − an+1)] as follows.

Figure 11:



3. impulse function

δ(t) =

⎧⎨⎩

0 t = 0

1 · ∞ t = 0 impulse function

δ[n] =

⎧⎨⎩

0 n = 0

1 n = 0 delta function

In general, δ(t) is not a function, it is a generalized function.(but δ[n] is a function).



For example, δ(t) can be defined as the limit of some function.

• We can think of the continuous-time impulse function with theproperty ∫ ∞

−∞δ(t)dt = 1

and δ(t) =

⎧⎨⎩

0 (t = 0)

∞ (t = 0)

• In other words, continuous-time impulse δ(t) has the property:δ(t) = 0 for all t except at t = 0 and the total area under δ(t) is 1.



Figure 12:



Properties of impulse function

There are many property of δ(t)

1. sampling property:

x(t) ∗ δ(t − t0) = x(t0) ∗ δ(t − t0)

2. sifting property:∫ ∞−∞ x(t)δ(t − t0)dt = x(t0)

∫ b

ax(t)δ(t − t0)dt =

⎧⎪⎪⎨⎪⎪⎩

x(t0) if t0 ∈ [a, b]

0 else



sampling and sifting property

Figure 13:



3. δ(at) = 1|a|δ(t)

Figure 14:



4. δ(at + b) = δ(a(t + ba )) = 1

|a|δ(t + ba )

Figure 15:



• All of these properties can be proved by thinking δ(t) as ageneralized function.

• From the above properties, we have

x(t0) =∫ ∞−∞ x(t)δ(t − t0)dt ( by 1)

=∫ ∞−∞ x(τ)δ(τ − t0)dτ ( replace t by τ)

=∫ ∞−∞ x(τ)δ(t0 − τ)dτ ( by 3)

Since this is true for ∀t0 ∈ (−∞,∞), we can replace t0 by t.

• Finally, we have ∫ ∞

−∞x(τ)δ(t − τ)dτ = x(t), ∀t

⇒ x(t) = x(t) ⊗ δ(t)



From this property, δ(t) (or δ[n]) is the identity of convolutionalintegral (convolutional sum)

• x(t) =∫ ∞−∞ x(τ)δ(t − τ)dτ

or =∫ ∞−∞ x(t − τ)δ(τ)dτ (continuous-time)

• x[n] = Σ∞k=−∞x[k]δ[n − k]

or = Σ∞k=−∞δ[k]x[n − k] (discret-time)

We see that any signal x(t) (x[n]) can be written as the ”linearcombination” of δ(t) (δ[n]) and it’s shift version δ(t − τ) (δ[n − k]),i.e., the linear integral for continuous-time, and linear sum fordiscrete-time.



• Remark:

⎧⎪⎪⎨⎪⎪⎩

r′(t) = u(t)

u′(t) = δ(t)⎧⎪⎪⎨⎪⎪⎩

∫ t

−∞ δ(τ)dτ = u(t)

∫ t

−∞ u(τ)dτ = r(t)



Also

⎧⎪⎪⎨⎪⎪⎩

r[n] − r[n − 1] = u[n]

u[n] − u[n − 1] = δ[n]⎧⎪⎪⎨⎪⎪⎩

Σnk=−∞δ[k] = u[n]

Σnk=−∞u[k] = r[n]



• The relationship between u[n] and δ[n]

• From the identity of convolutional sum, we have

u(t) =∫ ∞

−∞u(t − τ)δ(τ)dτ

=∫ t

−∞δ(τ)dτ

• Similarly, we have

u(t) =∫ ∞

−∞u(τ)δ(t − τ)dτ

=∫ ∞

0

δ(t − τ)dτ



• The relationship between u[n] and δ[n]

• From the identity of convolutional sum, we have

u[n] =∞∑

k=−∞u[n − k]δ[k]

=n∑

k=−∞δ[k]

• Similarly, we have

u[n] =∞∑

k=∞δ[n − k]u[k]

=∞∑

k=0

δ[n − k]



System

A continuous-time (discrete-time) system H is an operator thattransfer the input x(t) (x[n]) into the output y(t) (y[n]). We denotethe process by

Figure 16:



Example: the RLC circuit

Figure 17:

How to describe the relationship between the input vi(t) and theoutput v0(t)?



Classification of system

1. linear vs. nonlinearH is called linear if H has the superposition property:⎧⎨

⎩Hx1(t) + x2(t) = Hx1(t) + Hx2(t)Hcx(t) = cHx(t)

⇔ Hc1x1(t) + c2x2(t) = c1Hx1(t) + c2Hx2(t)

⇔ H∑ni=1 cixi(t) =

∑ni=1 ciHxi(t)



Figure 18:



2. time-invariant vs. time-variant

• H is called time-invariant if the following is trueHx(t) = y(t) =⇒ Hx(t − t0) = y(t − t0)

• I.e., a time-shift to in the input x(t) results in an identicaltime-shift to in the output



Figure 19:



3. memory vs. memoryless

• A system H is memoryless if the value y(t0) (i.e.,y(t = t0)) onlydepends on the value x(t0) for any t0.

• example: y(t) = x2(t) is memoryless since y(t0) = x2(t0) for ∀t0.

• example: y(t) = x(t − 1) is a system with memory sincey(t0) = x(t0 − 1), e.g., y(0) = x(−1). y(t0) depends on x(t) att = t0 − 1, not at t0.

• In other words, output y(t) at current time t = t0 is only affectedby input x(t) at current time t = t0



4. causal vs. noncausal

• A system H is causal if the value y(t0) only depends onx(t) : t ≤ t0.

• I.e., current output is produce by current input and past input,not future input.

• the system y[n] = x[n − 1] is causal (y[0] = x[−1])

• the system y[n] = x[n + 1] is noncausal (y[0] = x[1])

• the system y(t) = x(t + a) is causal if a ≤ 0 and is noncausal ifa > 0



5. stable vs. nonstable

• H is stable if | x(t) |≤ Mx < ∞ ∀t

then | y(t) |≤ My < ∞ ∀t

• I.e., bounded input x(t) produces bounded output y(t)



We will focus on a linear time-invariant system (LTI system) H.If H is a LTI system, x(t) and y(t) are usually described by

• impulse response h(t)

• transfer function H(s)

• differential equation

• block diagram



Preview and Review: t and s domain

1. t-domain: impulse response h(t)

x(t) =∫ ∞−∞ x(τ)δ(t − τ)dτ

⇒ y(t) = Hx(t) = H∫ ∞−∞ x(τ)δ(t − τ)dτ

=∫ ∞−∞ x(τ)Hδ(t − τ)dτ =

∫ ∞−∞ x(τ)h(t − τ)dτ

2. s-domain: transfer function H(s)

x(t) =∫ ∞−∞ X(w)ejwtdw ⇒ y(t) = Hx(t)

=∫ ∞−∞ X(w)Hejwtdw =

∫ ∞−∞ X(w)H(w)ejwtdw



• est is an eigenfunction of a continuous-time LTL systemy(t) =

∫ ∞−∞ h(τ)x(t − τ)dτ =

∫ ∞−∞ h(τ)es(t−τ)dτ

= (∫ ∞−∞ h(τ)e−sτdτ)est = H(s)est (= H(s)x(t))

• zn is an eigenfunction of a discrete-time LTL systemy[n] =

∑∞−∞ h[k]x[n − k] =

∑∞−∞ h[k]zn−k

= (∑∞

−∞ h[k]z−k)zn = H(z)zn (= H(z)x[n])



Change of basis: two domains

A vector x in terms of one basis e1, e2 · · · , enx = (x1, x2, · · ·xn) = x1(100 · · · 0) + x2(010 · · · 0) + · · · + xn(000 · · · 1)= x1e1 + x2e2 + · · ·xnen (∈ 〈e1, e2 · · · en〉)The same vector x in terms of another basis v1, v2 · · · vn

x = x1e1 + x2e2 + · · ·xnen = x′1v1 + x′

2v2 + · · · + x′nvn = x′

• A vector x has two representations in terms of two basesx = (x1, x2, · · · , xn) = (x′

1, x′2, · · · , x′

n)

• We can change from eini=1 to vin

i=1 and vice versa; if vini=1

are eigenvectors, we can simplify operation y = Ax in eini=1

domain to y = Dx in vini=1 domain.



• The reason is as follows. In eini=1 domain, we have y = Ax.

• If Avi = λvi for all i

v1, v2 · · · vn = eigenvectors with eigenvalues λ1, λ2 · · ·λn

x = x1e1 + x2e2 + · · ·xnen = x′1v1 + x′

2v2 + · · · + x′nvn = x′

x′ = (x′1x

′2 · · ·x′

n) = x′1v1 + x′

2v2 + · · · + x′nvn

y = Ax = A(x′1v1 + x′

2v2 + · · · + x′nvn)

= x′1λ1v1 + x′

2λ2v2 + · · · + x′nλnvn=y′

1v1 + · · · y′nvn

where y′i = λix

′i



Or equivalently,A(v1v2 · · · vn) = (Av1, Av2, · · ·Avn) = (λ1v1, λ2v2, · · ·λnvn)

= (v1v2 · · · vn)

⎡⎢⎢⎢⎣

λ1 0 0

0. . . 0

0 0 λn

⎤⎥⎥⎥⎦ ⇒ AV = V D ⇒ A = V DV −1

y = V DV −1x ⇒ y′ = Dx′ ⇒ V −1y = DV −1x



Motivation of LTI system

• Motivation I: O.D.E and Circuit ⇔ signal and system

A RLC circuit

Figure 20:



Or block diagram

Figure 21:



From the circuit theory, we have⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

VR(t) = R · i(t)

VL(t) = Ldi(t)dt

i(t) = C dVC(t)dt ⇒ di(t)

dt = C d2VC(t)dt2

Therefore, by KVL, we have :VC(t) + VL(t) + VR(t) = Vs(t)

⇒ VR(t) = R · i(t) = RC dVC(t)dt

VL(t) = L · C d2VC(t)dt2



Finally, we have

L · C d2VC(t)dt2 + RC dVC(t)

dt + VC(t) = Vs(t)

⇒ V ′′c (t) + R

L V ′c (t) + 1

LcVc(t) = 1LC Vs(t)

Input signal: x(t) = Vs(t)

output signal: y(t) = Vc(t)

⇒ The differential equation describing the relationship betweeninput x(t) & output y(t) is as follows.

y′′(t) + RL y′(t) + 1

LC y(t) = 1LC x(t)

This is a 2nd order constant coefficient linear ODE.



A complete solution y(t) is given by:

y(t) = yh(t) + yp(t) (O.D.E.)

= yZ.I.R(t) + yZ.S.R.(t) (circuit)

= ynatural(t) + yforced(t) (circuit)

In general, y(t) for t t0 depends on both the initial state s(t0) andthe input function x(τ), t t0 we write:y(t) = F (s(t0); x(τ), τ t0)then ZIR(t) = f(s(t0); 0); ZSR(t) = f(0; x(τ), τ t0)

1. For a linear-system, Complete system response=ZIR+ZSR

2. We will assume s(t0) = 0 from now on and turn attentionto ZIR when we discuss the Laplace Transform.



1. Solving 2nd oreder O.D.E.:yh(t): solving λ2 + R

L λ + 1LC = 0 (two roots λ1 & λ2)

⇒ yh(t) = c1eλ1t + c2e

λ2t (λ1 = λ2 distinct roots )

or yh(t) = c1eλ1t + c2te

λ2t (λ1 = λ2 repeat roots )

or yn(t) = eα1t(c1 cos β1t + c2 sin β1t)

where λ1 = α1 + ıβ1,λ2 = λ1 = α1 − ıβ1



2. Solving 1st order differential system:states:i(t) & vC(t)

⇒ V ′C(t) = 1

c i(t)

i′(t) = 1LVL(t)= 1

L (Vs(t) − VC(t) − VR(t))= 1

L (−VC(t) − Ri(t) + Vs(t))

⇒⎡⎣ VC(t)

i(t)

⎤⎦′

=

⎡⎣ 0 1

C

−1L

−RL

⎤⎦

⎡⎣ VC(t)

i(t)

⎤⎦ +

⎡⎣ 0

1LVs(t)

⎤⎦

⇒⎡⎣ x1(t)

x2(t)

⎤⎦′

= A

⎡⎣ x1(t)

x2(t)

⎤⎦ + F (t)



in general,we have X ′(t) = Ax(t) + F (t)and we have the solution of the first order differential system:X(t) = Xh(t) + Xp(t) where Xh(t) is obtained bydiagonalizing the matrix A

A[

V1 V2

]=

[λ1V1 λ2V2

]=

[V1 V2

] ⎡⎣ λ1 0

0 λ2

⎤⎦

⇒ A = V DV −1

x′(t) = V DV −1x(t)

⇒ V −1x−1(t)︸︷︷︸Y ′(t)

= D V x(t)︸︷︷︸Y (t)



⇒⎧⎨⎩

y1(t) = c1eλ1t

y2(t) = c2eλ2t

⇒⎡⎣ x1(t)

x2(t)

⎤⎦ =

[v1 v2

] ⎡⎣ y1(t)

y2(t)

⎤⎦

⇒⎡⎣ x1(t)

x2(t)

⎤⎦ =

[v1(t) v2(t)

]⎡⎣ c1e

λ1t

c2eλ2t

⎤⎦

= c1eλ1tv1 + c2e

λ2tv2



Summary

From the above example, we can see that there are several ways todescribe the relationship between the input x(t) and the output y(t)for a LIT sytem x(t) ↔ y(t). These are:

1. Block diagram

Figure 22:



2. differential equation (SISO system)

y′′ + RL y′(t) + 1

LC y(t) = 1LC x(t) (λ2 + R

L λ + 1LC = 0 two roots)

⇒ y(t) = yh(t) + yp(t) = yZIR

(t) + yZSR

(t)

where

yh(t) =

⎧⎪⎪⎨⎪⎪⎩

c1eλt + c2e

λt (λ1 = λ2 real)

c1eλt + c2te

λt (λ1 = λ2 real)

yh(t) = c1eα1t(cos βt + sinβt) (λ1 = λ2 = α + ıβ)



3. differential system (MIMO system)

y′(t) =

⎡⎣ y1(t)

y2(t)

⎤⎦′

= A

⎡⎣ y1(t)

y2(t)

⎤⎦ + F (t)

⇒ y(t) = yh(t) + yp(t)

& yh(t) =[

v1 v2

] ⎡⎣ c1e

λ1t

c2eλ2t

⎤⎦

where Av1 = λ1v1, Av2 = λ2v2(λ1 = λ2)



4. our focus

⎧⎨⎩

time domain h(t): impulse response

frequency domain H(s): transfer function

We can find the transfer function H(s), or the frequencyresponse H(jw) (H(ejΩ)) directly from the circuit diagram orfrom the differential equation (system). After that, we can getthe impulse response h(t) from H(s). The idea is connectingwith phasors in circuit theory.



Phasors

R: VR(t) = Ri(t)

i(t) = ejwt ⇒ VR(t) =

ZR︷︸︸︷R ejwt ⇒ ZR = R( independce)

L: VL(t) = Ldi(t)dt

i(t) = ejwt ⇒ VL(t) =

ZL︷︸︸︷L · jw ejwt ⇒ ZL = jwL (SL)

C: i(t) = C dVC(t)dt

VC(t) = ejwt ⇒ i(t) =

ZC︷︸︸︷jwC ejwt ⇒ ZC = 1

jwC ( 1sC )



Figure 23:

We can replace R by R, C by 1/jwC, and L by jwL; then by KVLor KCL we can solve the transfer function H(s) directly from thecircuit diagram.



therefore by voltage divider, we have

Vc =1

jwc1

jwc + R + jwL︸︷︷︸H(jw)

Vs, (∗jw 1L on top and bottom)

i.e. H(jw) =1

LC

(jw)2+ RL jw+ 1

LC

or H(s) =1

LC

S2+ RL S+ 1

LC

Note: O.D.E. V ′′c (t) + R

L V ′c (t) 1

LC + Vc(t) = 1LC Vs(t)

• It seems that we can find H(s) aslo from the ODE.



Figure 24:

• x(t) = ejwt (or in general x(t) = est) is an eigenfunction of acontinuous-time LTI system.

• x[n] = ejΩn (or in general x[n] = zn) is an eigenfunction of adiscrete-time LTI system.

• Let x(t) = ejwt then y(t) = H(jw)ejwt

• Let x[n] = ejΩn then y[n] = H(ejΩn)ejωn



Figure 25:

I.e., mathematically, for a LTI system H, we have

1. h(t) = Hδ(t), h[n] = Hδ[n]2. H(jw) = Hejwt

ejwt , H(ejΩ) = HejΩnejΩn



e.g. the ODE for RLC circuit is:

Let x(t) = ejwt, then y(t) = H(jw)ejwt y′′(t)+ RL y′+ 1

LC y(t) = 1LC x(t)

then y′(t) = (jw)H(jw)ejwt, y′′(t) = (jw)2H(jw)ejwt

⇒ ((jw)2 + RL jw + 1

RL )H(jw)ejwt = 1RLejwt

⇒ H(jw) =1

LC

(jw)2+ RL jw+ 1

LC



This is always true for any nth order linear constant coefficient ODE.That is, given a differential equation for a LTI system

any(n)(t) + an−1y(n−1)(t) + · · · + a1y

′(t) + a0y(t)

= bmx(m)(t) + bm−1x(m−1)(t) + · · · + b1x

1(t) + b0x(t)

i.e.,∑n

i=1 aiy(i)(t) =

∑mj=1 bjx

(j)(t)



Substitute: x(t) = ejwt & y(t) = H(jw)ejwt

into the ODE & use the fact di

dti ejw = (jwt)iejwt we have

(an(jw)n + an−1(jw)n−1 + · · · + a1(jw) + a0)H(jw)ejwt

= (bm(jw)m + bm−1(jw)m−1 + · · · + b1(jw) + b0)ejwt

↔ H(jw) = bm(jw)m+···+b1(jw)+b0an(jw)n+···+a1(jw)+a0

H(s) = bmsm+···+b1sm+b0ansn+···+a1s+a0



Figure 26:



Usually H(s) = N(s)D(s) (degD(s) = n)

= A1s+p1

+ A2s+p2

+ · · · + An

s+pn(assume D(s) has n distinct toots)

(by P.E.F. partial fraction Expansion)⇒ h(t) = L−1H(s)⇒ h(t) = A1e

−p1tu(t) + A2e−p2tu(t) + · · · + Ane−pntu(t)



In general,block diagram > differential system

> O.D.E> h(t)(H(S))

where > means providing more information.

In signal & system,we study the zero-state response

Figure 27:

in particular,the system H will be a L.I.T. system.(linear & timeinvariant)



Motivation II: (linear algebra⇔ signal & system)

A =

⎡⎢⎢⎢⎢⎢⎣

a d c b

b a c d

c b a d

d c b a

⎤⎥⎥⎥⎥⎥⎦

(circulant matrix)

• How to find the eigenvectors and eigenvalues for the circulantmatrix A?

• We can use the fact that A represents a discrete-time LTI systemto find the eigenvectors and eigenvalues.



The matrix A represents a LTI system for a discrete-time withperiodic input x[n]. That is, if x is a periodic input, then y = Ax isthe periodic output with the fact that y[n] is obtained by the circularconvolution between x[n] and h[n]:

y[n] =N∑

k=1

x[k]h[n − k]

In this example, we have h[n] = (a, b, c, d).



Find the eigenvalues and eigenvectors for A. First,we can findeigenvalues of A by

Av = λv ⇒ (λI − A)v = 0

with v = 0

Therefore we must have λI − A is a singular matrix, i.e.det(λI − A) = 0 (characteristic polynomial).

This is a poly of degree n if A is a n × n matrix.

In general, it is not easy to find the eigenvalues for a given n × n

matrix A.



For this circulant matrix,we can show that

v1 =

⎡⎢⎢⎢⎢⎢⎣

1

1

1

1

⎤⎥⎥⎥⎥⎥⎦

= (ej 2π4 ·0·n)n=0,1,2,3 = (i0)n=0,1,2,3

v2 =

⎡⎢⎢⎢⎢⎢⎣

1

i

−1

−i

⎤⎥⎥⎥⎥⎥⎦

= (ej 2π4 ·1·n)0≤n≤3 = (i1·n)0≤n≤3



v3 =

⎡⎢⎢⎢⎢⎢⎣

1

−1

1

−1

⎤⎥⎥⎥⎥⎥⎦

= (ej 2π4 ·2·n)0≤n≤3 = (i2·n)0≤n≤3

v4 =

⎡⎢⎢⎢⎢⎢⎣

1

−i

−1

i

⎤⎥⎥⎥⎥⎥⎦

= (ej 2π4 ·3·n)0≤n≤3 = (i3·n)0≤n≤3

are eigenvectors of A.



eigenvalue of v1

= a + d + c + b

eigenvalue of v2

(a − c) + i(d − b)

eigenvalue of v3

(a + c) − (d + b)

eigenvalue of v4

(a − c) − i(d − b)



Moveover, v1, v2, v3, v4 are orthogonal vectors, i.e.,

(vi, vj) = 0 for any i = j

.

Let ei = 1√4vi ⇒ e1, e2, e3, e4 are orthonormal eigenvector for A.

In other words,we have A [e1, e2, e3, e4]︸︷︷︸V

= [e1, e2, e3, e4]︸︷︷︸V

D

where D =

⎡⎢⎢⎢⎢⎢⎣

λ1 0

λ2

λ3

0 λ4

⎤⎥⎥⎥⎥⎥⎦

,

and λi is an eigenvalue of ei.



we can define a = h(0), b = h(1), c = h(2), d = h(3) then

A =

⎡⎢⎢⎢⎢⎢⎣

h(0) h(3) h(2) h(1)

h(1) h(0) h(3) h(2)

h(2) h(1) h(0) h(3)

h(3) h(2) h(1) h(0)

⎤⎥⎥⎥⎥⎥⎦

=[

h((n − k))4]

where h(3)4 = h(−1), h(2)4 = h(−2), · · ·



Then

Ax =3∑

k=0

h[n − k]x[k] =3∑

k=0

h[k]x[n − k]

• This is just the discrete-time convolution sum.

• If we let x[n] = ej 2π4 nk0 (k0 = 0, 1, 2, 3)

⇒ Ax =∑3

k=0 h[k]ej 2π4 (n−k)k0

=3∑

k=0

h[k]e−j 2π4 kk0

︸︷︷︸λ

· ej 2π4 nk0︸︷︷︸

x[n]

.



• i.e., ej 2π4 nk0 , (0 ≤ k0 ≤ 3) is an eigenvector of A with eigenvalue∑

k h[k]e−j 2π4 kk0 .

In matrix language, we have⇒ y = Ax -time domainy = V DV −1x (since V −1 = V t)= V DV T x

⇒ V T y = DV T x

⇒ y′ = Dx′ -frequency domain

• If V is an orthonormal matrix,then V T = V −1



In general,we can show

A =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

h(0) h(N − 1) h(1)

h(1) h(0) h(2)

h(2) h(1) h(3)...

...

h(N − 1) h(N − 2) h(0)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

always has eigenvectors

1√N

(ej 2πN ·0n)0≤n≤N−1 = e1

1√N

(ej 2πN ·1n)0≤n≤N−1 = e2

...1√N

(ej 2πN ·(N−1)n)0≤n≤N−1 = eN , (N eigenvectors)



Such e1, . . . , eN are orthonormal eigenvector for A and

Ax =∑

k

h[n − k]x[k] =∑

k

h[k]x[n − k]

Similarly, if we let x[n] = ej 2πN nk0 , 0 ≤ k0 ≤ N − 1

⇒ Ax =∑

k

h[k]ej 2πN (n−k)k0

=∑

k

h[k]ej 2πN kk0

︸︷︷︸λ

· ej 2πN nk0︸︷︷︸x[n]

Also y = Ax = V DV −1x ⇒ V −1y = DV −1x

⇒ y = DV T x

⇒ y′ = Dx′


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