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.
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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
sa
1 ( j a ) t
e
sa
0
1
; Re( s ) a
sa
; note : s j
0
1 ( a ) t j t
e
e
sa
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
sa
0
1
; Re( s a ) 0
sa
1
1
; Re( s ) a
; Re( s ) a
sa
sa
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
s2
1
3t
e u (t )
; Re(s ) 3
s3
1
1
F (s)
;
s2 s3
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
sa
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]
sa
1
; Re(s a) 0 Re(s ) a
sa
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
sa
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
sa
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
sa
1
1
1
F (s)
2 s (a jb) s (a jb)
sa
; 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
sa
0
1
; Re( s a ) 0
sa
1
1
; Re( s ) a
; Re(s ) a
sa
sa
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
sa
0
1
; Re( s a) 0
sa
1
1
; Re( s ) a
; Re(s ) a
sa
sa
Note that Laplace can also be found for periodic functions