Transcript Laplace Transform - University of Michigan

Math Review with Matlab:
Laplace
Transform
Application:
Linear Time Invariant (LTI) Systems
S. Awad, Ph.D.
M. Corless, M.S.E.E.
E.C.E. Department
University of Michigan-Dearborn
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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
2
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
System Definition
 A system can be thought of as a black box with an input
and an output
Excitation
Input Signal
x(t)
System
Transformation
T
Response
Output Signal
y(t)
 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
3
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Differential Equations
Input Signal
x(t)
System
Output Signal
y(t)
dx(t )
dx2 (t )
d M x(t )
0 x(t )  1
 2 2  ...  M

M
dt
dt
dt
dy(t )
dy2 (t )
d N y(t )
0 y(t )  1
 2 2  ...  N
dt
dt
dt N
 Time domain systems are often described using a Differential
Equation
 Recall that time domain Differentiation corresponds to Laplace
Transform domain Multiplication by s with subtraction of Initial
Conditions
4
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Linear Systems
 A system is Linear if it satisfies the Superposition Principle
( where  and  are constants ):
T1x1(t) 2 x2 (t)  1Tx1(t)2Tx2 (t)
 This can be restated given the excitation and response
relationships:
1 y1(t)  T1x1(t) 2 y2 (t)  T2 x2 (t)
 Then an Excitation of:
x3 (t )  1x1 (t )  2 x2 (t )
 Results in a Response of:
y3 (t )  1 y1 (t )  2 y2 (t )
5
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Time Invariance
 A system is time-invariant if its input-output relationship
does not change as time evolves
Tx(t  )  y(t  )
x(t )
0
0
y(t )
0
t

x(t   )
t
y(t  )
0
t

t
6
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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:

y(t )  x(t ) * h(t )   x( )h(t  )d

 Where * denotes the Convolution operation
7
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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 TimeDomain and Multiplication in the Laplace TransformDomain
y(t )  x(t ) * h(t )
LT
LT
LT
Y (s)  X (s)H (s)
8
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Transfer Function
 The Transfer Function, H(s), of a system is the Laplace
Transform of the Impulse Response, h(t)
y(t )  x(t ) * h(t )
Y (s)  X (s)H (s)
Y (s)
H (s)  LTh(t ) 
X (s)
 The Transfer Function completely specifies the relationship
between the excitation (input) and response (output) in the
Laplace Transform-Domain
9
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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
10
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Linear Operations
Time-Domain
x(t )
k1
x1(t)
k1x(t )
Time-Domain

x1(t)  x2 (t)
x2 (t )
Laplace Transform-Domain
Laplace TransformDomain
X1 (s)
X (s)
k1

k1 X (s)
X1(s)  X 2 (s)
X 2 (s)
 Linear operations have a direct correlation between the Time-Domain
and Laplace Transform-Domain (s-domain) counterparts
11
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Time-Domain Differentiation
 Time-Domain Differentiation Operation
Differentiation
x(t )
d
dt
dx(t )
dt
 Equivalent Laplace Transform-Domain Operation
Multiplication
X (s)
s
sX (s)
12
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Time-Domain Integration
 Time-Domain Integration Operation (no initial conditions)
Integration
x(t )
 dt
 x(t)dt
 Equivalent Laplace Transform-Domain Operation
Division
X (s)
1
s
X (s)
s
13
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Time-Domain Delay
 Time-Domain Delay Operation
x(t )
Delay by t0
Operation
x(t  t0 )
 Equivalent Laplace Transform-Domain Operation
Multiplication
X (s)
e
 st 0
est0 X (s)
14
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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)
 Automatic Gain Controls for a radio
 Car Mufflers (mechanical filter)
 Suspension Systems (mechanical low pass filter)
 Cruise Control (motor speed control)
15
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
System Example
 Create a system to implement
the differential equation:
dy(t )
x(t ) 
 y(t )
dt
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
16
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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
dy(t )
x(t ) 
 y(t )
dt
LT
X (s)  sY (s)  Y (s)
X (s)  (s 1)Y (s)
 By definition,
H(s) = Y(s) / X(s)
Y (s)
1
H (s) 

X (s) s 1
17
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Time-Domain-System
 Draw time-domain system representation for:
1) Reorder terms to create a Function for y(t)
2) Start by drawing Input and Output at far ends
3) Draw Differentiation Block connected to y(t)
dy(t )
x(t ) 
 y(t )
dt
dy(t )
y(t )  x(t ) 
dt
4) Draw Summation Block and its connections
x(t )
+

y(t )
-
dy(t )
dt
18
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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
x(t )
+

y(t )
-
Time-Domain
X (s)
+
Laplace Transform
Domain
dy(t )
dt

Y (s)
-
s
19
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Verify H(s)
 The Transfer Function H(s) can also be determined by writing
an expression from the Laplace Transform-Domain System
X (s)
+

Y (s)
-
 System directly yields:
Y (s)  X (s)  sY (s)
 Reordering terms gives the same
result as taking the Laplace Transform
of the Differential Equation
s
Y (s)  sY (s)  X (s)
(s  1)Y (s)  X (s)
Y (s)
1
H (s) 

X (s) s 1
20
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Impulse Response
 The Impulse Response of the system, h(t), is simply the
Inverse Laplace Transform of the Transfer Function, H(s)
h(t )  LT 1H (s)
1  1 
 LT 

 s  1
t
h(t )  e u(t )
21
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Pulse Response Example
 Given a system with an Impulse Response, h(t)=e-2tu(t)
Input Pulse
0
T
Impulse
Response
t
x(t )
 u(t )  u(t  T )
h(t )
 e2t u(t )
Pulse Response
Output
y(t )
 h(t )  x(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
22
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Transfer Function
 The transfer function of the system is simply the Laplace
Transform of the Impulse Response:
2t
h(t )  e u(t )
LT
1
H (s) 
s2
 The Transfer Function can be used to find the Laplace
Transform of the pulse response, Y(s), using:
Y (s)  H (s) X (s)
23
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Laplace Transform of Input
 Given the equation
for a General Pulse
of period T
 The general Laplace
Transform is thus:
 Combining Terms:
x(t )  u(t )  u(t  T )
LT
 
1 1 sT
X (s)   e
s s
1 e
X (s) 
s
 sT
24
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Determine Y(s)
 Y(s) is found using:
Y (s)  H (s) X (s)
 Substituting for H(s)
and X(s)
 1  1  e sT 
Y ( s)  

 s  2  s 
1
esT
Y (s) 

s(s  2) s(s  2)
 Distributing terms
 Rewrite in terms of a new Y1(t)
 sT
Y (s)  Y1 (s)  e Y1 (s)
where
1
Y1(s) 
s(s  2)
25
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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 Vector
A = Denominator Polynomial Coefficient Vector
R = Residues Vector
P = Poles Vector
K = Direct Term Constant
rN
bN s N  bN 1s N 1   b1s  b0
r1
r2
k



s  p1 s  p2
s  pN aN s N  aN 1s N 1   a1s  a0
26
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Expand Y1(s)
 Use residue to perform partial fraction expansion
» B=[0 0 1];A=[1 2 0];
» [R,P,K]=residue(B,A)
R =
-0.5000
0.5000
P =
-2
0
K =
[]
1
1
B
Y1 (s) 
 2

s(s  2) s  2s  0 A
R1
R
 2
s  P1 s  P2
0.5 0.5
 0

s2 s
Y1 (s)  K 
1
1
Y1 (s)  
2s 2(s  2)
27
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
General Solution y(t)
 Find y(t) by taking Inverse Laplace Transforms and
substituting y1(t) back into y(t)
1
1
Y1 (s)  
2s 2(s  2)
Y (s)  Y1 (s)  esTY1 (s)
LT-1
LT-1
 1 1  2t 
y1 (t )    e u(t )
2 2

2t 


y(t)  y1(t)  y1(t T )
2t 


y(t )  1 e u(t )  1 e u(t  T )
1

2
1

2
28
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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)
2t
h(t )  e u(t )
x(t )  u(t )  u(t  T )
» T=sym('T','positive')
» x=sym('Heaviside(t)-Heaviside(t-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
29
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Matlab Verification
» H=laplace(h)
H =
1/(s+2)
2t 


2t 


y(t )  1 e u(t )  1 e u(t  T )
1

2
» X=laplace(x)
X =
1/s-exp(-T*s)/s
» Y=H*X
Y =
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)
1

2
30
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Matlab Vector Code
 NOTE as of Matlab 6, ezplot cannot plot functions containing
declarations of Heaviside or Dirac (Impulse)
 The following code recreates the Pulse Response as vectors
for T=1 sec and T=2 sec
t=[0:0.01:4];
% Time Vector
tmax=size(t,2);
% Index to last Time Value
T1=find(t==1);
% Index to 1 second
T2=find(t==2);
% Index to 2 seconds
yexp=0.5*(1-exp(-2*t)); % Base exponential vector
y1T=[zeros(1,T1),yexp(1:tmax-T1)];
y1=yexp-y1T;
% Pulse Response T=1
y2T=[zeros(1,T2),yexp(1:tmax-T2)];
y2=yexp-y2T;
% Pulse Response T=2
31
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Matlab Plots
 The response for T=1
and T=2 is plotted
y(t )  12 1 e2t u(t )  12 1 e2t u(t  T )
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;
32
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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)
x(t ) 
sin( 2t )u(t )
2
H (s)  2
s  2s  2
y(t ) 
ytrans(t )  yss (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)
33
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Laplace Transform of Input
 Recall the Laplace
Transform of a general
sine signal with an
angular frequency w0
 Find the Laplace
Transform of the
input signal x(t)
wo
LTsin(wot )u(t )  2
2
s  wo
 sin( 2t )u(t )
LT
2
X ( s) 
2
s 2
x(t )
LT
34
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Roots of Y(s)
 Determine an expression for output signal Y(s)
 2  2


Y (s)  X (s)H (s)   2  2
 s  2  s  2s  2 
 Determine general form for roots (poles) of denominator of Y(s)



2
2


Y (s)  
 
* 
 (s  p1 )(s  p1 )  (s  p2 )(s  p2 ) 
Purely Imaginary Roots Complex Roots
p1  j 2
p2  1 j
35
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Verify Poles in Matlab
 2  2

2 2


 
Y (s)   2  2

*
s

2
s

2
s

2
(
s

p
)(
s

p
)(
s

p
)(
s

p

1
1
2
2)


» poles=roots( conv( [1 0 2], [1 2 2]) )
poles =
-0.0000
-0.0000
-1.0000
-1.0000
p1  j 2
+
+
-
1.4142i
1.4142i
1.0000i
1.0000i
p1   j 2
p2  1  j
p2  1  j
36
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Partial Fraction Expansion
 Note that since poles are complex conjugates,pcoefficients
1  1  j  2e
will also be complex conjugates
j
3
4
2 2
c1
c1
c2
c2
Y (s) 





*

(s  p1)(s  p1 )(s  p2 )(s  p2 ) s  p1 s  p1 s  p2 s  p2
p1  j 2
p2  1 j
c1  (s  p1 )Y (s) s  p
c2  (s  p2 )Y (s) s  p
2 2
c1 
(s  p1 )(s  p2 )(s  p2* )
2 2
c2 
(s  p1 )(s  p1 )(s  p2* )
1
2
s  p1
s  p2
37
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Find Coefficients in Matlab
2 2
c1 
(s  p1 )(s  p2 )(s  p2* )
s  p1
2 2
c2 
(s  p1 )(s  p1 )(s  p2* )
s  p2
» 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 =
2
0.3536 + 0.0000i
c1  0.3536  
4
» c2=(2*2^0.5)/(s-p1)/(s-p1c)/(s-p2c);
» c2=subs(c2,'s',p2)
c2 =
2
0.3536 - 0.3536i
c  0.3535 j0.3536
38
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Inverse Laplace
 Take Inverse Laplace Transform of Y(s)




c
c
c
c
1
1
1
1
2
2
y(t )  LT Y (s)  LT 





 s  p1 s  p1 s  p2 s  p2 

 

y(t )  c1e  c e u(t )  c2e  c e u(t )
* p1*t
1
p1t
* p2*t
2
p2t
 Reduce terms by combining complex conjugates

y(t )  2 Re c1e
p1t
u(t)  2 Rec e u(t)
p2t
2
39
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Substitute Values
 When substituting coefficients, it is useful to use the
polar representation to simplify cosine conversions

y(t )  2 Re c1e
p1t
u(t)  2 Rec e u(t)
p2t
2
2 j
c1  0.3536 
e
4
1  j 4
c2  0.3535 j0.3536  e
2
p1  j 2
p2  1 j
y(t ) 
 2 j j
2 Re
e e
 4


j


1
2t
4 ( 1 j ) t
u(t )  2 Re e e
u(t )
2


40
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Steady State and Transient
Responses
 The complex signal can be converted into a function of cosines
 2 j j
y(t )  2 Re
e e
 4


j


1
2t
4 ( 1 j )t
u(t )  2 Re e e
u(t )
2




2
 
t
y(t ) 
cos 2t   u(t )  e cos t  u(t )
2
 4
Steady State Response
(Repeats as t increases)


2
yss (t ) 
cos 2t   u(t )
2
Transient Response
(Goes to 0 at t increases)
 
ytrans  e cos t  u(t )
 4
t
41
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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);
 2  2

» Y=X*H; y=ilaplace(Y);




Y (s)   2  2
» y=simplify(y); pretty(y)
 s  2  s  2s  2 
Steady State
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)
Transient
42
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Verify Equivalence
 The Hand and Matlab steady state results are equivalent
because a phase shift of  is the same as negating the cosine


2
yHss (t ) 
cos 2t   u(t )
2
 
2
yMss (t )   cos 2t u(t )
2
 The Hand and Matlab transient results are equivalent by
applying the relationship:

cos(x)  sin( x)  2 cos x   4
 
yHtrans  e cos t  u(t )
 4
t

2 t
yMtrans 
e cos(t )  sin(t )u(t )
2
43
X(s) Laplace Transform: Linear Time Invariant Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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
44