S. Awad, Ph.D. M. Corless, M.S.E.E. E.C.E. Department University of Michigan-Dearborn Laplace Transform Math Review with Matlab: Application: Linear Time.

S. Awad, Ph.D.

M. Corless, M.S.E.E.

E.C.E. Department

University of Michigan-Dearborn

LaplaceTransform

Math Review with Matlab:

Application:Linear Time Invariant (LTI) Systems

U of M-Dearborn ECE DepartmentMath Review with Matlab

2

Laplace Transform:X(s) Linear Time Invariant Systems

Linear Time Invariant (LTI)Systems

Definition of a Linear Time Invariant System

Impulse Response

Transfer Function Simple Systems Simple System Example Pulse Response Example Transient and Steady State Example


3


System Definition A system can be thought of as a black box with an input

and an output

The signal connected to the input is called the Excitation

The system performs a Transformation, T, (function) on the input

Given an input excitation, the output signal is called the Response

Excitation Response

Output Signal

y(t)

Input Signal

x(t)

System

Transformation

T


4


Differential Equations

Time domain systems are often described using a Differential Equation

N

N

N

M

M

M

dt

tyd

dt

tdy

dt

tdyty

dt

txd

dt

tdx

dt

tdxtx

)(...

)()()(

)(...

)()()(

2

2

210

2

2

210

Recall that time domain Differentiation corresponds to Laplace Transform domain Multiplication by s with subtraction of Initial Conditions

Output Signal

y(t)

Input Signal

x(t)System


5


Linear Systems A system is Linear if it satisfies the Superposition Principle

( where and are constants ):

This can be restated given the excitation and response relationships:

)()()( 22113 txtxtx

)()()( 22113 tytyty

)()( 1111 txTty )()( 2222 txTty

)()()()( 22112211 txTtxTtxtxT

Then an Excitation of:

Results in a Response of:


6


Time Invariance A system is time-invariant if its input-output relationship

does not change as time evolves

0 t0 t

)(tx )(ty

)()( tytxT

0 t 0 t)( ty)( tx


7


Impulse Response The Impulse Response signal, h(t), of a linear system is

determined by applying an Impulse to the Input, x(t), and determining the output response, y(t)

Due to the properties of a Linear Time Invariant System, the Impulse Response Completely Characterizes the relationship between x and y for all x such that:

Where * denotes the Convolution operation

dthxthtxty )()()(*)()(


8


Laplace Transform Since Convolution may be Mathematically Intensive,

the Laplace Transform is often used as an aid to analyze the Linear Time Invariant Systems.

Recall the relationship between Convolution in the Time-Domain and Multiplication in the Laplace Transform-Domain

)(*)()( thtxty

)()()( sHsXsY

LT LT LT


9


Transfer Function The Transfer Function, H(s), of a system is the Laplace

Transform of the Impulse Response, h(t)

)(

)()()(

sX

sYthLTsH

)(*)()( thtxty

)()()( sHsXsY

The Transfer Function completely specifies the relationship between the excitation (input) and response (output) in the Laplace Transform-Domain


10


Simple Systems Most systems can be created by combining the

following simple system building blocks:

Linear Operations: Multiplication by a Constant Addition of Signals

Time-Domain Differentiation Time-Domain Integration Time-Domain Delay


11


Linear Operations

Linear operations have a direct correlation between the Time-Domain and Laplace Transform-Domain (s-domain) counterparts

Time-Domain

)(tx )(1 txk1k

Laplace Transform-Domain

)(sX )(1 sXk1k

Time-Domain)(1 tx

)(2 tx

)()( 21 txtx

Laplace Transform-

Domain)(1 sX

)(2 sX

)()( 21 sXsX


12


Time-Domain Differentiation Time-Domain Differentiation Operation

Equivalent Laplace Transform-Domain Operation

)(ssX)(sXMultiplication

s

dt

tdx )()(txDifferentiation

dt

d


13


Time-Domain Integration Time-Domain Integration Operation (no initial conditions)


dttx )()(txIntegration

dt

s

sX )()(sX

Division

s1


14


Time-Domain Delay Time-Domain Delay Operation


)( 0ttx )(tx Delay by t0

Operation

)(0 sXe st)(sXMultiplication

0ste


15


Automatic Gain Controls for a radio

Car Mufflers (mechanical filter) Suspension Systems (mechanical low pass filter) Cruise Control (motor speed control)

Examples of LTI Systems

The building blocks described previously can be used to model and analyze real world systems such as:

Audio Equalizers (band pass filters)


16


System Example Create a system to implement

the differential equation:)(

)()( ty

dt

tdytx

1) Determine the Transfer Function directly from the

Differential Equation

2) Draw the system in the Time-Domain

3) Draw the system in the Laplace Transform-Domain

4) Write the Transfer Function from the System Diagram

5) Determine the Impulse Response


17


Directly Determine H(s) The Transfer Function H(s) can be directly determined by taking the Laplace

Transform of the differential equation and manipulating terms

)()(

)( tydt

tdytx )()()( sYssYsX

)()1()( sYssX

1

1

)(

)()(

ssX

sYsH

LT

By definition,

H(s) = Y(s) / X(s)


18


Time-Domain-System Draw time-domain system representation for: )(

)()( ty

dt

tdytx

dt

tdytxty

)()()(

)(tx )(ty

dt

tdy )(

2) Start by drawing Input and Output at far ends

+

-

3) Draw Differentiation Block connected to y(t)

4) Draw Summation Block and its connections

1) Reorder terms to create a Function for y(t)


19


Laplace Transform-Domain The Laplace Transform-Domain System can be drawn by

leaving the linear summation block and replacing the differentiating block with a multiplication by s

)(tx )(ty

dt

tdy )(

+

-Time-Domain

s

+

-

)(sX )(sY

Laplace Transform

Domain


20


Verify H(s) The Transfer Function H(s) can also be determined by writing

an expression from the Laplace Transform-Domain System

)(sX )(sY

s

+

-

)()()( ssYsXsY )()()1(

)()()(

sXsYs

sXssYsY

1

1

)(

)()(

ssX

sYsH

Reordering terms gives the same result as taking the Laplace Transform of the Differential Equation

System directly yields:


21


Impulse Response The Impulse Response of the system, h(t), is simply the

Inverse Laplace Transform of the Transfer Function, H(s)

1

1

)()(

1

1

sLT

sHLTth

)()( tueth t


22


Pulse Response Example Given a system with an Impulse Response, h(t)=e-2tu(t)

1) Find the Transfer Function for the system, H(s)

2) Find the General Pulse Response,y(t)

3) Plot the Pulse Response for T=1 sec and T=2 sec

)(

)(2 tue

tht

Impulse Response

Pulse Response Output

)()(

)(

txth

ty

0 tT

Input Pulse

)()(

)(

Ttutu

tx


23


Transfer Function The transfer function of the system is simply the Laplace

Transform of the Impulse Response:

)()( 2 tueth t

The Transfer Function can be used to find the Laplace Transform of the pulse response, Y(s), using:

)()()( sXsHsY

LT2

1)(

s

sH


24


Laplace Transform of Input Given the equation

for a General Pulse of period T

)()()( Ttututx

sTess

sX 11

)(

s

esX

sT1

)(

The general Laplace Transform is thus:

Combining Terms:

LT


25


Determine Y(s) Y(s) is found using:

Substituting for H(s) and X(s)

Distributing terms

Rewrite in terms of a new Y1(t)

)()()( 11 sYesYsY sT)2(

1)(1

sssYwhere

)()()( sXsHsY

s

e

ssY

sT1

2

1)(

)2()2(

1)(

ss

e

sssY

sT


26


Partial Fraction Expansion The Matlab function residue can be used to

perform Partial Fraction Expansion on Y1(s)

[R,P,K] = RESIDUE(B,A) B = Numerator polynomial Coefficient VectorA = Denominator Polynomial Coefficient VectorR = Residues VectorP = Poles VectorK = Direct Term Constant

011

1

011

1

2

2

1

1

asasasa

bsbsbsb

ps

r

ps

r

ps

rk

NN

NN

NN

NN

N

N


27


» B=[0 0 1];A=[1 2 0];

Expand Y1(s) Use residue to perform partial fraction expansion

A

B

sssssY

02

1

)2(

1)(

21

ss

Ps

R

Ps

RKsY

5.0

2

5.00

)(2

2

1

11

)2(2

1

2

1)(1

ss

sY

» [R,P,K]=residue(B,A)R = -0.5000 0.5000P = -2 0K = []


28


General Solution y(t) Find y(t) by taking Inverse Laplace Transforms and

substituting y1(t) back into y(t)

)(1)(1)( 22

122

1 Ttuetuety tt

)2(2

1

2

1)(1

ss

sY

)(2

1

2

1)( 2

1 tuety t

LT-1

)()()( 11 sYesYsY sT

)()()( 11 Ttytyty

LT-1


29


Matlab Declarations The General Pulse Response can be verified using Matlab Variables must be carefully declared using proper syntax

» syms h H t s» h=exp(-2*t)

Assuming the system to be causal, T must be explicitly declared as a positive number

The Heaviside function is equivalent to the unit-step

)()()( Ttututx

)()( 2 tueth t

» T=sym('T','positive')» x=sym('Heaviside(t)-Heaviside(t-T)')


30


Matlab Verification» H=laplace(h) H = 1/(s+2)

)(1)(1)( 22

122

1 Ttuetuety tt

» X=laplace(x)X =1/s-exp(-T*s)/s» Y=H*XY =1/(s+2)*(1/s-exp(-T*s)/s)

» y=ilaplace(Y)y = -1/2*exp(-2*t)+1/2+ 1/2*Heaviside(t-T)*exp(-2*t+2*T) -1/2*Heaviside(t-T)


31


The following code recreates the Pulse Response as vectors for T=1 sec and T=2 sec

Matlab Vector Code NOTE as of Matlab 6, ezplot cannot plot functions containing

declarations of Heaviside or Dirac (Impulse)

t=[0:0.01:4]; % Time Vectortmax=size(t,2); % Index to last Time ValueT1=find(t==1); % Index to 1 secondT2=find(t==2); % Index to 2 secondsyexp=0.5*(1-exp(-2*t)); % Base exponential vectory1T=[zeros(1,T1),yexp(1:tmax-T1)];y1=yexp-y1T; % Pulse Response T=1y2T=[zeros(1,T2),yexp(1:tmax-T2)];y2=yexp-y2T; % Pulse Response T=2


32


Matlab Plots The response for T=1

and T=2 is plotted

subplot(2,1,1);plot(t,y1);title('Pulse Response T=1');grid on;subplot(2,1,2);plot(t,y2);title('Pulse Response T=2');xlabel('Time in seconds');grid on;

)(1)(1)( 22

122

1 Ttuetuety tt


33


Transient and Steady State Example

Determine an equation for the output of a system, y(t), described by the transfer function H(s) and input x(t)

From the output y(t):1. Identify the Transient Response, ytrans(t), of the system

(portion that goes to zero as t increases)

2. Identify the Steady State Response , yss(t), of the system (portion that repeats for all t)

22

2)(

2

sssH

)()2sin(

)(

tut

tx

)()(

)(

tyty

ty

sstrans


34


Laplace Transform of Input

Recall the Laplace Transform of a general sine signal with an angular frequency 0

2

2)(

)()2sin()(

2

ssX

tuttxLTLT

Find the Laplace Transform of the input signal x(t)

22

)()sin(o

oo s

tutLT


35


22

2

2

2)()()(

22 ssssHsXsY

Roots of Y(s) Determine an expression for output signal Y(s)

Determine general form for roots (poles) of denominator of Y(s)

Purely Imaginary Roots

21 jp

Complex Roots

jp 12

))((

2

))((

2)(

*2211 pspspsps

sY


36


Verify Poles in Matlab

» poles=roots( conv( [1 0 2], [1 2 2]) )

poles = -0.0000 + 1.4142i -0.0000 - 1.4142i -1.0000 + 1.0000i -1.0000 - 1.0000i

))()()((

22

22

2

2

2)(

*2211

22 pspspspsssssY

jp

jp

jp

jp

1

1

2

2

2

2

1

1


37


4

3

1 21

jejp

2

2

2

2

1

1

1

1*2211 ))()()((

22)(

ps

c

ps

c

ps

c

ps

c

pspspspssY

1

*221

1

111

))()((

22

)()(

pspspsps

c

pssYpsc

2

*211

2

222

))()((

22

)()(

pspspsps

c

pssYpsc

Partial Fraction Expansion Note that since poles are complex conjugates, coefficients

will also be complex conjugates

21 jp jp 12


38


Find Coefficients in Matlab

» syms s t» p1=j*2^0.5; p1c=conj(p1); p2=(-1+j); p2c=conj(p2);» c1=(2*2^0.5)/(s-p1c)/(s-p2)/(s-p2c);» c1=subs(c1,'s',p1)c1 = 0.3536 + 0.0000i

» c2=(2*2^0.5)/(s-p1)/(s-p1c)/(s-p2c);» c2=subs(c2,'s',p2)c2 = 0.3536 - 0.3536i

4

23536.01 c

3536.03535.02 jc

1

*221

1 ))()((

22

pspspsps

c

2

*211

2 ))()((

22

pspspsps

c


39


Inverse Laplace Take Inverse Laplace Transform of Y(s)

2

2

2

2

1

1

1

111 )()(ps

c

ps

c

ps

c

ps

cLTsYLTty

)()()(*22

*11 *

22*11 tuecectuececty tptptptp

)(Re2)(Re2)( 2121 tuectuecty tptp

Reduce terms by combining complex conjugates


40


)(2

1Re2)(

4

2Re2

)(

)1(42 tueetuee

ty

tjjtjj

)(Re2)(Re2)( 2121 tuectuecty tptp

Substitute Values When substituting coefficients, it is useful to use the

polar representation to simplify cosine conversions

42 2

13536.03535.0

j

ejc

21 jp jp 12

jec4

23536.01


41


Steady State and Transient Responses

The complex signal can be converted into a function of cosines

)(2

1Re2)(

4

2Re2)( )1(42 tueetueety tjjtjj

)(4

cos)(2cos2

2)( tutetutty t

)(2cos2

2)( tuttyss )(

4cos tutey t

trans

Transient Response(Goes to 0 at t increases)

Steady State Response(Repeats as t increases)


42


Matlab Verification Matlab can be used to determine Inverse Laplace Transform Result will have transient and steady state component Result will appear different but be mathematically equivalent

» X=(2^0.5)/(s^2+2); H=2/(s^2+2*s+2);» Y=X*H; y=ilaplace(Y);» y=simplify(y); pretty(y)

1/2 1/2 - 1/2 2 cos(2 t) + 1/2 1/2 1/2 2 exp(-t) cos(t) + 1/2 2 exp(-t) sin(t)

22

2

2

2)(

22 ssssY

Steady State

Transient


43


Verify Equivalence The Hand and Matlab steady state results are equivalent

because a phase shift of is the same as negating the cosine

The Hand and Matlab transient results are equivalent by applying the relationship: 4cos2)sin()cos( xxx

)(2cos2

2)( tuttyHss

)(4

cos tutey tHtrans

)(2cos2

2)( tuttyMss

)()sin()cos(2

2tuttey t

Mtrans


44


Summary Laplace Transform is a useful technique for analyzing

Linear Time Invariant Systems

Impulse Response and its Laplace Transform, the Transfer Function, are used to describe system characteristics

Simple System Blocks for multiplication, addition, differentiation, integration, and time shifting can be used to describe many real world systems

Matlab can be used to determine the Transient and Steady-State Responses of a complex system

S. Awad, Ph.D. M. Corless, M.S.E.E. E.C.E. Department University of Michigan-Dearborn Laplace Transform Math Review with Matlab: Application: Linear Time.

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