Transcript Laplace Transform (1) Definition of Bilateral Laplace Transform (b for bilateral or two-sided transform) Let s=σ+jω Consider the two sided Laplace transform as the Fourier.

Laplace Transform (1)
Definition of Bilateral Laplace
Transform
(b for bilateral or two-sided transform)
Let s=σ+jω
Consider the two sided Laplace transform as the
Fourier transform of f(t)e-σt. That is the Fourier transform
of an exponentially windowed signal.
Note also that if you set the evaluate the Laplace
transform F(s) at s= jω, you have the Fourier transform
(F(ω))
Unilateral Laplace Transform
(Implemented in Mathematica)
Difference Between the
Unilateral Laplace Transform
and Bilateral Laplace transform
• Unilateral transform is used when we
choose t=0 as the time during which
significant event occurs, such as
switching in an electrical circuit.
• The bilateral Laplace transform are
needed for negative time as well as for
positive time.
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
0

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 Unilateral Laplace
Bilateral Laplace
f (t )  e  at u (t );

F ( s )   f (t )e  st dt

0

   e  at e  st dt 


1 ( s  a )t
e
sa
0


1
; Re( s  a )  0
sa
1
1
; Re( s )  a 
; Re( s )  a
sa
sa
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!
Laplace Transform & Fourier
Transform
• Laplace transform is more general
than Fourier Transform
– Fourier Transform: F(ω). (t→ ω)
– Laplace Transform: F(s=σ+jω) (t→ σ+jω, a
complex plane)
How is Laplace Transform Used
(Building block of a
negative feedback system)
This system becomes unstable if βH(s) is -1. If you subsittuted
s by jω, you can use Bode plot to evaluate the stability of
the negative feedback system.
Understand Stability of a system
using Fourier Transform (Bode Plot)
(unstable)
Understand Stability of a System
Using Laplace Transform
Look at the roots of Y(s)/X(s)
Laplace Transform
• We use the following notations for Laplace
Transform pairs – Refer to the table!
Table 7.1
Table 7.1 (Cont.)
Laplace Transform Properties (1)
Laplace Transform Properties (2)
Model an Inductor in the SDomain
• To model an inductor in the S-domain, we need to
determine the S-domain equivalent of derivative
(next slide)
Differentiation Property
Model a Capacitor in the SDomain
If initial voltage is 0, V=I/sC
1/(sC) is what we call the impedance of a capacitor.
Integration Property (1)
Integration Property (2)
Application
• i=CdV/dt (assume initial voltage is 0)
• Integrate i/C with respect to t, will get
you I/(sC), which is the voltage in
Laplace domain
• V=Ldi/dt (assume initial condition is 0)
• Integrate V/L with respect to t, get you
V/(sL), which is current in Laplace
domain.
Next time
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  st 0
  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   [limt  e  st  1]
0
s
1
  [limt  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 
f (t )  e
 at
cos(bt)
Example
f t   1 / 2e jbt e  at  1 / 2e  jbt e  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
Extra Slides
Building the Case…
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!
Example of Bilateral Version
Find F(s):
f (t )  e at u (t );
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 );

Remembe
r
These!
F ( s)   f (t )e  st dt

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