Chapter 7
Laplace Transforms
Applications of Laplace Transform
• Easier than solving differential equations
– Used to describe system behavior
– We assume LTI systems
– Uses S-domain instead of frequency domain
• Applications of Laplace Transforms/
– Circuit analysis
• Easier than solving differential equations
• Provides the general solution to any arbitrary wave (not just LRC)
– Transient
– Sinusoidal steady-state-response (Phasors)
– Signal processing
– Communications
• Definitely useful for Interviews!
notes
Building the Case…
http://web.cecs.pdx.edu/~ece2xx/ECE222/Slides/LaplaceTransformx4.pdf
Laplace Transform
Laplace Transform
• We use the following notations for Laplace Transform pairs
– Refer to the table!
Laplace Transform Convergence
• The Laplace transform does not converge to a finite value for all signals
and all values of s
• The values of s for which Laplace transform converges is called the
Region Of Convergence (ROC)
• Always include ROC in your solution!
• Example:
f (t )  e  at u (t );
0+ indicates greater than
zero values

F ( s )   f (t )e  st dt


1
   e  at e  st dt 
e(  s a )t
0
sa
 1 (  j  a )t

e
sa


0
1
; Re( s )  a
sa

; note : s    j
0
 1 (  a )t  jt

e
e
sa

 Re( s  a)  0
0
Remember: e^jw is
sinusoidal; Thus, only the
real part is important!
Example of Bilateral Version
f (t )  e at u (t );
Find F(s):
ROC
F (s)  
S-plane


0
  e at e  st dt 

Re(s)<a
a
Find F(s):
f (t )e  st dt

1
e(  s  a )t
sa
0


1
; Re(  s  a )  0
sa
1
1
; Re(  s )   a  
; Re( s )  a
sa
sa
f (t )  e  at u (t );

F ( s )   f (t )e  st dt

Remember
These!
0

   e  at e  st dt  


1
e (  s a )t
sa
0


1
; Re(  s  a )  0
sa
1
1
; Re(  s )  a 
; Re( s )  a
sa
sa
Note that Laplace can also be found for periodic functions
Example – RCO may not always exist!
f (t )  e 2t u (t )  e 3t u (t )

F ( s )   f (t )e  st dt

1
e u (t ) 
; Re( s )  2
s2
1
 3t
e u (t )  
; Re( s )  3
s3
1
1
F (s) 

;
s2 s3
2t
Note that there is no common ROC  Laplace Transform can not be applied!
Example – Unilateral Version
• Find F(s):
f (t )  e  at u (t ); a  0

• Find F(s):

F ( s)   f (t )e  st dt
F ( s)   f (t )e dt
 st
0
0


  e  st (t t 0 )dt  e  st0
  e  at e  st dt
0
0

• Find F(s):
1
; Re( s  a)  0  Re( s)  a
sa
f (t )  e at ; a  0
• Find F(s):

F ( s )   f (t )e  st dt
0

  e e dt
at  st
0
1
[0  1] 
sa
1

; Re( s  a)  0  Re( s )  a
sa

f (t )   (t t 0 )
 f (t )   (t )
F ( s)  1; s
f (t )  u (t )

F ( s )   f (t )e  st dt
0

1
  e  st u (t )dt   [lim t  e  st  1]
0
s
1
  [lim t  e (  j ) t  1]
s
1
 ; Re( s )  0
s
Example
f (t )  cos(bt )
f t   1 / 2e
jbt
 1 / 2e
 jbt

F ( s )   f (t )e  st dt
0
1
e  at 
; Re( s )  a
sa
1/ 2
1/ 2
1 / 2e jbt 
;1 / 2e  jbt 
; Re( s )  0
s  jb
s  jb
1/ 2
1/ 2
s
F ( s) 

 2
; Re( s )  0
s  jb s  jb s  b 2
f (t )  sin( bt )
f t   1 / 2 je jbt  1 / 2 je jbt

F ( s )   f (t )e  st dt
0
1
; Re( s )  a
sa
1/ 2 j
1/ 2 j
1 / 2 je jbt 
;1 / 2e  jbt 
; Re( s )  0
s  jb
s  jb
1/ 2 j 1/ 2 j
b
F ( s) 

 2
; Re( s )  0
2
s  jb s  jb s  b
e  at 
Example
f (t )  e  at cos(bt )
f t   1 / 2e jbte  at  1 / 2e  jbte  at

F ( s )   f (t )e  st dt
0
1
e 
; Re( s )  a
sa

1
1
1
F (s)  

2  s  (a  jb) s  (a  jb) 
sa

; Re( s  a )  0
2
2
(s  a)  b
 at
Properties
• The Laplace Transform has many difference properties
• Refer to the table for these properties
Linearity
Scaling & Time Translation
Scaling
Do the time translation first!
e  sb / a
f (at  b)u(at  b) 
F (s / a)
a
Time Translation
b=0
Shifting and Time Differentiation
Shifting in s-domain
Differentiation in t
Read the rest of
properties on your
own!
Examples
f (t )  5e 0.3t
1
e at 
; Re( s)  a
sa
1
F (s)  5
; Re( s)  0.3
s  0.3
f (t )  5e 0.3( t 2 )u(t  2)
1
; with _ time _ shift
sa
1
F ( s)  5
e 2 s ; Re( s )  0.3
s  0.3
e at 
Note the ROC did not change!
f (t )  5e 0.3(t )u (t  2)
1
e  at 
; with _ time _ shift
sa
f (t )  5e 0.3(t )u (t  2){e  2 ( 0.3)  e 2( 0.3) }
 5e  2 ( 0.3) {e 0.3(t ) e 2( 0.3)u (t  2)}
 5e  2 ( 0.3) {e 0.3(t  2 )u (t  2)}
e  2( 0.3)  2 s 2.744  2 s
F ( s)  5
e 
e ; Re( s )  0.3
s  0.3
s  0.3
Example – Application of Differentiation
g (t )  tf (t )
{tf (t )}  ?

G ( s )   tf (t )e  st dt
0
 
   f (t )e  st dt
ds 0

G (s)   F (s)
ds
g (t )  t cos(bt )
{t cos(bt )}  ?

G ( s)   F ( s)
ds

s
 { 2
}; Re( s )  0
2
ds s  b
s 2  b2

; Re( s )  0
2
2 2
s b
Read Section 7.4
Read about Symbolic Mathematics:
http://www.math.duke.edu/education/ccp/materials/diffeq/mlabtutor/mlabtut7.html
And
http://www.mathworks.de/access/helpdesk/help/toolbox/symbolic/ilaplace.html


Matlab Code:
Example
e  sb / a
f (at  b)u(at  b) 
F (s / a)
a
• What is Laplace of t^3?
– From the table: 3!/s^4 Re(s)>0
Time transformation
• Find the Laplace Transform:
g (t )  sin( 12t   / 2)u (4t   / 6)
g (t )  sin{ 3(4t   / 6)}u (4t   / 6)  sin( 3 )u ( )   4t  / 6
Time _ Translatio n :
a  4; b   / 6
e  s / 24
G( s) 
F ( s / a);
4
f ( )  sin( 3 )u ( )  F ( s ) 
3
s2  9
Note that without u(.) there will
be no time translation and thus,
the result will be different:
e  s / 24
3
G( s) 
;
4 ( s / 4) 2  9
Assume t>0
Given Laplace find f(t)!
A little about Polynomials
• Consider a polynomial function:
• A rational function is the ratio of two polynomials:
Has roots and zeros; distinct roots, repeated
roots, complex roots, etc.
• A rational function can be expressed as partial fractions
• A rational function can be expressed using polynomials
presented in product-of-sums
Finding Partial Fraction Expansion
• Given a polynomial
• Find the POS
(product-of-sums) for the denominator: G ( s)  N ( s) 
D( s)
• Write the
partial fraction expression
for the polynomial
G( s) 
N ( s)
( s  p1 )( s  p2 )( s  p3 )...
k1
k2
k3


 ....
( s  p1 ) ( s  p2 ) ( s  p3 )
k1  [( s  p1 ) I ( s )]
• Find the constants
– If the rational polynomial has
distinct poles then we can use the
following to find the constants:
s  p1
k2  [( s  p2 ) I ( s )]
s  p2
k2  [( s  p3 ) I ( s )]
s  p3
......
http://cnx.org/content/m2111/latest/
Application of Laplace
• Consider an RL circuit with R=4, L=1/2. Find i(t) if
v(t)=12u(t).
di (t )
 Ri (t )  v (t )
dt
di (t )
0.5
 4i (t )  v (t )
dt
U sin g _ Laplace :
(0.5s  4) I ( s )  V ( s )
L
1
 I ( s )  H ( s ).V ( s )
0 .5 s  4
Given
v (t )  12u (t )  V ( s )  12 / s
Partial fraction expression
24
k
k
I ( s )  H ( s ).V ( s ) 
 1 2
s( s  8) s s  8
H ( s)  I ( s) / V ( s) 
k1  [( s  p1 ) I ( s )]
k2  [( s  p2 ) I ( s )]
 I ( s) 
p1  0
3
p 2  8
 3
3 3

 i (t )  3u (t )  3e 8t ; t  0
s s8
Matlab
Code
Application of Laplace
• What are the initial [i(0)] and final
values:
i (0  )  lim s  sI ( s )  lim s 
24
0
s 8
lim s  i (t )  lim s 0 sI ( s )  lim s 0
24
3
s 8
– Using initial-value property:
– Using the final-value property
Note: using Laplace Properties
Initial _ value : f (0 )  lim f (t )  lim sF ( s )
t 0
t 
Final _ value : f ()  lim f (t )  lim sF ( s )
t 
t 0
Note that
Initial Value: t=0, then, i(t) 3-3=0
Final Value: t INF then, i(t) 3
i(t )  3u(t )  3e8t ; t  0
Using Simulink
v(t
)
H(s)
i(t)
Actual Experimentation
• Note how the voltage looks like:
di (t )
 Ri (t )  v(t )
dt
di (t )
0.5
 4i (t )  v(t )
dt
i (t )  3u (t )  3e 8t ; t  0
L
v(t )  0.5
di (t )
 12e 8t ; t  0
dt
Output Voltage:
Input Voltage:
Partial Fraction Expansion
(no repeated Poles/Roots) – Example
• Using Matlab:
• Matlab code:
b=[8 3 -21];
a=[1 0 -7 -6];
[r,p,k]=residue(b,a)
We can also use
ilaplace (F); but the
result may not be
simplified!
Finding Poles and Zeros
• Express the rational function as
the ratio of two polynomials each
represented by product-of-sums
• Example:
F ( s) 
4s  8
2( s  2)

2 s 2  8s  6
( s  1)( s  3)
Pole
S-plane
zero
H(s) Replacing the Impulse Response
x(t)
h(t)
convolution
y(t)
X(s)
H(s)
multiplication
Y(s)
H(s) Replacing the Impulse Response
x(t)
h(t)
y(t)
X(s)
multiplication
convolution
Example: Find the output
X(t)=u(t); h(t)
H(s)
Y(s)
h(t)
1
h(t )  u (t )  u (t  1)
0
1
y(t)
1
0
1
1 es
H ( s)  
s s
1
X (s) 
s
e^-sF(s)
1  es 1 es
Y ( s )  H ( s ). X ( s ) 
 2 2
s2
s
s
 y (t )  tu(t )  (t  1)u (t  1)
This is commonly used in D/A converters!
Dealing with Complex Poles
• Given a polynomial
• Find the POS (product-of-sums)
for the denominator:
• Write the partial fraction
expression for the polynomial
• Find the constants
– The pole will have a real and
imaginary part: P=|k|f
• When we have complex poles
{|k|f} then we can use the
following expression to find the
time domain expression:
http://cnx.org/content/m2111/latest/
a  Re( P); b  Im( P); angle _ of _ P  f
f (t )  2 | k | e at cos(bt  f )
Laplace Transform Characteristics
• Assumptions: Linear Continuous Time
Invariant Systems
• Causality
– No future dependency
– If unilateral: No value for t<0; h(t)=0
• Stability
– System mode: stable or unstable
– We can tell by finding the system
characteristic equation (denominator)
We need to add control
mechanism to make the
overall system stable
H ( s) 
1
1

s 2  4 ( s  2)( s  2)
h(t )  Ae2t u (t )  Be  2t u (t )
• Stable if all the poles are on the left
plane
– Bounded-input-bounded-output (BIBO)
• Invertability
– H(s).Hi(s)=1
• Frequency Response
– H(w)=H(s);sjw=H(s=jw)
3s  1
s 2  2s  5
3 j  1
H ( ) 
  2  2 j  5
H (s) 
Frequency Response – Matlab Code
3s  1
s 2  2s  5
3 j  1
H ( ) 
  2  2 j  5
H (s) 
Inverse Laplace Transform
Example of Inverse Laplace Transform
Bilateral Transforms
• Laplace Transform of two
different signals can be the
same, however, their ROC can
1
x(t )  e  at u (t ) 
;a  Re( s )  Right  sided
be different:
sa
•  Very important to know the
1
x(t )  e  at u (t ) 
;a  Re( s )  Left  sided
ROC.
sa
• Signals can be
– Right-sided  Use the bilateral
Laplace Transform Table
– Left-sides
– Have finite duration
• How to find the transform of
signals that are bilateral!
See notes
How to Find Bilateral Transforms
• If right-sided use the table for unilateral Laplace Transform
• Given f(t) left-sided; find F(s):
– Find the unilateral Laplace transform for f(-t) laplace{f(-t)}; Re(s)>a
– Then, find F(-s) with Re(-s)>a
• Given Fb(s) find f(t) left-sided :
– Find the unilateral Inverse Laplace transform for F(s)=fb(t)
– The result will be f(t)=–fb(t)u(-t)
• Example
x(t )  2e 5t u (t )  e 4t u (t )
2
;5  Re( s )  Right  sided
s5
To _ Find :e  4t u (t )
Assume : f (t )
2e 5t u (t ) 
1
;4  Re( s )  Right  sided
s4
1
F ( s) 
;4  Re(  s )  4  Re( s )  Left  sided
s4
2
1
 F (s) 

s5 s4
e 4t u (t ) 
Examples of Bilateral Laplace Transform
x(t )  2e 5t u (t )  e  4t u (t ) 
2
1

;5  Re( s )  4
s5 s4
x(t )  2e 5t u (t )  e  4t u (t ) 
2
1

;4  Re( s ) & 5  Re( s )
s5 s4
x(t )  2e 5t u (t )  e  4t u (t ) 
2
1

;4  Re( s );5  Re( s )
s5 s4
Find the unilateral Laplace transform for f(-t) laplace{f(-t)}; Re(s)>a
Then find F(-s) with Re(-s)>a
Alternatively: Find the unilateral Laplace transform for f(t)u(-t)
(-1)laplace{f(t)}; then, change the inequality for ROC.
Feedback System
Find the system function for the
following feedback system:
X ( s)  R( s)  E ( s)  Y ( s) / F ( s)
R( s)  Y ( s).G ( s)
X(t)
+
Sum
e(t)
y(t)
F(s)
+
r(t)
 X ( s)  (Y ( s).G ( s))  Y ( s) / F ( s)
G(s)
F (s)
 Y ( s) / X ( s)  H ( s) 
1  F ( s).G ( s)
Equivalent System
X(t)
Feedback Applet:
http://physioweb.uvm.edu/homeostasis/simple.htm
H(s)
y(t)
Practices Problems
• Schaum’s Outlines Chapter 3
–
–
–
–
3.1, 3.3, 3.5, 3.6, 3.7-3.16,  For Quiz!
3.17-3.23
Read section 7.8
Read examples 7.15 and 7.16
Useful Applet: http://jhu.edu/signals/explore/index.html