Laplace Transform •  Review

•  Slides borrowed from Evans, U. of Texas @ Austin

Zero-State Response •  Linear constant coefficient differential equation

Input x(t) and output Zero-state response: all initial conditions are zero Laplace transform both sides of differential equation with

all initial conditions being zero and solve for Y(s)/X(s)

)()()( tytyty statezeroinputzero −− +=

( ) ( )sYty ↔

( ) ( )sYstydtd rr

r

↔

0)0()()()('

=

=+−y

txtyty

( ) ( )sXtx ↔

( ) ( )sXstxdtd kk

k

↔

11

)()()(

)()()(

+==

=+

ssXsYsH

sXsYssY

Transfer Function •  H(s) is called the transfer function because it

describes how input is transferred to the output in a transform domain (s-domain in this case) Y(s) = H(s) X(s) y(t) = L-1{H(s) X(s)} = h(t) * x(t) ⇒ H(s) = L{h(t)}

•  Transfer function is Laplace transform of impulse response

( ) ( )( ) ( )

( ) ( )( )

Ts

Ts

esXsYsH

esXsYTtxty

−

−

==

=

−=

Transfer Function Examples •  Laplace transform

•  Assume input x(t) and output y(t) are causal •  Ideal delay of T seconds

Initial conditions (initial voltages in delay buffer) are zero

( ) ( )∫∞ −−

=0

dtetxsX ts

Tx(t) y(t)

( )

( ) ( ) ( )( ) ( )

( )s

sXsYsH

sXsxsXssY

txdtdty

==

=−=

=

− )(0

)(( ) ( )

( ) ( ) ( )

( )ssX

sYsH

ys

sXs

sY

dxtyt

1)()(

0110

==

+=

=

−

∫ −ττ

Transfer Function Examples •  Ideal integrator with

y(0-) = 0 •  Ideal differentiator

with x(0-) = 0

( )•dtdx(t) y(t) ( )dt

t

0∫ −•

x(t) y(t)

Cascaded Systems •  Assume input x(t) and output y(t) are causal

•  Integrator first, then differentiator

•  Differentiator first, then integrator

•  Common transfer functions A constant (finite impulse response) A polynomial (finite impulse response) Ratio of two polynomials (infinite impulse response)

1/s s X(s)/s

( )∫ −

tdx

0ττx(t)

X(s)

x(t)

X(s)

s 1/s s X(s)

)(txdtd

x(t)

X(s)

x(t)

X(s)

Block Diagrams H(s) X(s) Y(s)

H1(s) + H2(s) X(s) Y(s) H1(s)

X(s) Y(s) H2(s)

Σ =

H1(s) X(s) Y(s) H2(s) H1(s)H2(s) X(s) Y(s) = W(s)

G(s) 1 + G(s)H(s) X(s) Y(s) G(s) X(s) Y(s)

H(s)

Σ - =

E(s)

19 - 8

Cascade and Parallel Connections •  Cascade

W(s) = H1(s) X(s) Y(s) = H2(s)W(s) Y(s) = H1(s) H2(s) X(s) ⇒ Y(s)/X(s) = H1(s)H2(s)

One can switch the order of the cascade of two LTI systems if both LTI systems compute to exact precision

•  Parallel Combination Y(s) = H1(s)X(s) + H2(s)X(s) èY(s)/X(s) = H1(s) + H2(s)

H1(s) X(s) Y(s) H2(s) H2(s) X(s) Y(s) H1(s) ⇔

H1(s) + H2(s) X(s) Y(s) H1(s)

X(s) Y(s) H2(s)

Σ =

19 - 9

Feedback Connection •  Governing equations

•  Combining equations

( ) ( ) ( ) ( )( ) ( ) ( )sEsGsY

sYsHsFsE

=

−=

( ) ( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )

( ) ( )( ) ( )

( )sFsHsG

sGsY

sFsGsYsHsGsY

sYsHsFsGsY

1

+=

=+

−=

•  What happens if H(s) is a constant K? Choice of K controls all

poles in transfer function

G(s) 1 + G(s)H(s) F(s) Y(s) G(s) F(s) Y(s)

H(s)

Σ - =

E(s)

External Stability Conditions •  Bounded-input bounded-output stability

Zero-state response given by h(t) * x(t) Two choices: BIBO stable or BIBO unstable

•  Remove common factors in transfer function H(s) •  If all poles of H(s) in left-hand plane,

All terms in h(t) are decaying exponentials h(t) is absolutely integrable and system is BIBO stable

•  Example: BIBO stable but asymptotically unstable

⎟⎠

⎞⎜⎝

⎛+

=⎟⎠

⎞⎜⎝

⎛+−

⎟⎠

⎞⎜⎝

⎛−

=⎟⎠

⎞⎜⎝

⎛

−

−=

11

11

11

11)( 2 ss

sss

ssH

Based on slide by Prof. Adnan Kavak

Internal Stability Conditions •  Stability based on zero-input solution •  Asymptotically stable if and only if

Characteristic roots are in left-hand plane (LHP) Roots may be repeated or non-repeated

•  Unstable if and only if (i) at least characteristic root in right-hand plane and/or (ii) repeated characteristic roots are on imaginary axis

•  Marginally stable if and only if There are no characteristic roots in right-hand plane and Some non-repeated roots are on imaginary axis

Based on slide by Prof. Adnan Kavak

Frequency-Domain Interpretation

•  y(t) = H(s) e s t for a particular value of s

•  Recall definition of frequency response:

( ) ( )

( ) ( )

( )( )

sH

sts

ts

ts

dehe

deh

ethty

∫∫

∞

∞−

−

∞

∞−

−

=

=

∗=

ττ

ττ

τ

τ

h(t) y(t) est

h(t) y(t) ej 2π f t

( ) ( )

( ) ( )

( )( )

fH

fjtfj

tfj

tfj

dehe

deh

ethty

∫∫

∞

∞−

∞

∞−

−

=

=

∗=

ττ

ττ

τππ

τπ

π

2 2

2

2

Frequency-Domain Interpretation •  Generalized frequency: s = σ + j 2 π f •  We may convert transfer function into

frequency response by if and only if region of convergence of H(s) includes the imaginary axis

•  What about h(t) = u(t)? We cannot convert H(s) to a frequency response However, this system has a frequency response

•  What about h(t) = δ(t)?

( ) { } 0Refor 1>= s

ssH

( ) ( ) 1 allfor 1 freq =⇒= fHssH

( ) ( )fjs

sHfHπ2freq

==

Frequency Selectivity in Filters •  Lowpass filter

•  Highpass filter

•  Bandpass filter

•  Bandstop filter

Linear time-invariant filters are BIBO stable

f

|Hfreq(f)|

f

|Hfreq(f)|

f

|Hfreq(f)| 1

f

|Hfreq(f)|

1

Passive Circuit Elements •  Laplace transforms

with zero-valued initial conditions

•  Capacitor

•  Inductor

•  Resistor

( )

( ) ( )

( ) ( )( )

sLsIsVsH

sIsLsVdtdiLtv

==

=

=

( )

( ) ( )

( ) ( )

( ) ( )( ) sCsIsVsH

sIsC

sV

sVsCsIdtdvCti

1

1

==

=

=

=

( ) ( )( ) ( )

( ) ( )( )

RsIsVsH

sIRsVtiRtv

==

=

=

+

– v(t)

+

– v(t)

+

– v(t)

First-Order RC Lowpass Filter

x(t) y(t) ++

C

R

X(s) Y(s) ++

R

sC 1

Time domain

Laplace domain CRs

CRsXsY

sX

sCR

sCsY

sIsC

sY

sCR

sXsI

1

1

)()(

)(

1

1

)(

)(

1)(

1)()(

+=

+=

=

+=

i(t)

I(s)

Plot Bode plot – Amplitude & Phase?

First-Order RC Highpass Filter

x(t) y(t) ++

R

C

X(s) Y(s) ++

sC 1

Time domain

Laplace domain

CRs

ssXsY

sX

sCR

RsY

sIRsYsC

R

sXsI

1)(

)(

)(

1)(

)( )(

1)()(

+=

+=

=

+=

i(t)

RI(s) Frequency response is

also an example of a notch filter Plot Bode plot – Amplitude & Phase?

Passive Circuit Elements •  Laplace transforms

with non-zero initial conditions

•  Capacitor

( )

( ) ( ) ( )[ ]( ) ( )

( ) ( )⎥⎦

⎤⎢⎣

⎡−=

−=

−=

=

−

−

−

sisIsL

iLsIsL

isIsLsVdtdiLtV

0

0

0

•  Inductor

( )

( ) ( ) ( )[ ]

( ) ( ) ( )

( ) ( )[ ]−

−

−

+=

+=

−=

=

0

1

0

1

0

vCsIsC

svsI

sCsV

vsVsCsIdtdvCti

Operational Amplifier •  Ideal case: model this nonlinear circuit as

linear and time-invariant Input impedance is extremely high (considered infinite) vx(t) is very small (considered zero)

+ _

y(t) +

_

+ _ vx(t)

Operational Amplifier Circuit •  Assuming that Vx(s) = 0,

•  How to realize a gain of –1? •  How to realize a gain of 10?

+ _

Y(s) +

_

+ _ Vx(s)

Zf(s)

Z(s)

+ _ F(s)

I(s) ( ) ( ) ( )

( ) ( )( )

( )( )( )

( )

( ) ( )( )

( )( )sZsZ

sFsYsH

sFsZsZ

sY

sZsFsI

sZsIsY

f

f

f

−==

−=

=

−= H(s)

Differentiator •  A differentiator amplifies high frequencies, e.g.

high-frequency components of noise: H(s) = s for all values of s (see next slide) Frequency response is H(f) = j 2 π f ⇒ | H( f ) |= 2 π | f |

•  Noise has equal amounts of low and high frequencies up to a physical limit

•  A differentiator may amplify noise to drown out a signal of interest

•  In analog circuit design, one would generally use integrators instead of differentiators

DECIBEL SCALE

210

1

10log is defined as Decibel GaindBPGP

=

22 2

2 2 2 110 10 10 102

11 1 2

1

2 2 210 10 10 2 1

1 1 1

10log 10log 10log 10log

20log 10log 20log if

dB

VP R V RG

VP V RR

V R V R RV R V

⎛ ⎞= = = +⎜ ⎟

⎝ ⎠

= − = =

2 210 10

1 1

10log 20logdBP VGP V

= =

DECIBEL SCALE

210 10

1

20log 20logdBVH HV

= =

Magnitude H Decibel Value HdB

0.001 -60 0.01 -40 0.1 -20 0.5 -6

1/√2 -3 1 0 √2 3 2 6 10 20 20 26 100 40

1 2 1 2

11 2

2

log log log

Some Properties of Logar

log log log

log loglog

ith s

1 0

m

n

PP P P

P P PP

P n P

= +

⎛ ⎞= −⎜ ⎟

⎝ ⎠

=

=

Ø  The DECIBEL value is a logarithmic measurement of the ratio of one variable to another of the same type. Ø  Decibel value has no dimension.

Ø  It is used for voltage, current and power gains.

Typical Sound Levels and Their Decibel Levels.

•  Express transfer function in Standard form.

•  Express the magnitude and phase responses.

•  Two corner frequencies at ω=2, 10 and a zero at the origin ω=0.

•  Sketch each term and add to find the total response.

EXAMPLE 1 Construct Bode plots for

( )( )200( )2 10jH

j jω

ωω ω

=+ +

10STANDARD FORM ( )(1 2)(1 10)

jHj j

ωω

ω ω=

+ +

10

-1 -1

10 10 1020log 10 20log 20log 1 2 20log 1

90 tan ( 2)

10

tan ( 10)dbH j j j

φ

ω

ω

ω

ω

ω

= ° −

= +

−

− + − +

EXAMPLE 1 Construct Bode plots for ( )( )

200( )2 10jH

j jω

ωω ω

=+ +

10 10 10 1020log 10 20log 20log 1 2 20log 1 10dbH j j jω ω ω= + − + − +

-1 -190 tan ( 2) tan ( 10)φ ω ω= °− −

1020log 2 6dB=

26 dB

X X

XX

EXAMPLE 2 Continued: Let us calculate |H| and φ at ω=50 rad/sec graphically.

26 dB

10(50) (10) 20log (50 /10) 26 20 0.7 26 14 12H H dB= − = − × = − =

10 10 10(50) 90 45 log (1/ 0.2) 90 log (20 /1) 45 log (50 / 20)90 45 0.7 90 1.3 45 0.4 90 31.5 117 18 76.5

φ = °− °× − °× − °×

= °− °× − °× − °× = °− °− °− ° = − °

ω=50

ω=50

EXAMPLE 1 Construct Bode plots for ( )210( )5

jHj j

ωω

ω ω

+=

+

2

0.4(1STANDARD F

10)( )(1 5)

ORMjH

j jω

ωω ω

+=

+

10 10 10 1020log 0.4 20log 1 10 20log 40log 1 5dbH j j jω ω ω= + + − − +

-1 -10 tan ( 10) 90 2 tan ( 5)φ ω ω= °+ − °−

•  Express transfer function in Standard form.

•  Express the magnitude and phase responses.

•  Two corner frequencies at ω=5, 10 and a zero at ω=10.

•  The pole at ω=5 is a double pole. The slope of the magnitude is -40 dB/decade and phase has slope -90 degree/decade.

•  Sketch each term and add to find the total response.

EXAMPLE 3 Construct Bode plots for ( )210( )5

jHj j

ωω

ω ω

+=

+

10

10

10

10

20 log 0.420log 1 10

20log

40log 1 5

dbHj

j

j

ω

ω

ω

= +

+ −

−

+

-1

-1

0 tan ( 10)90 2 tan ( 5)

φ ω

ω

= °+

− °−

X O

X O

PRACTICE PROBLEM 4 Obtain the transfer function for the Bode plot given.

2 2

2

2

A zero at 0.5,

A pole at 1,

Two poles at 10,

1 0.5 (1 0.5)(0.5 ) 200( 0.5)( )(1 1)(1 10) (1 100)(1 )(10 ) ( 1)( 10)

1 0.51

1 11

(1 10)

j

j

j j sHj j j j s s

j

ω

ω

ω

ω ωω

ω

ω

ω

ω ω

ω

ω

=

=

=

+ + += = =

+ + + + + +

+

+

+