# Inverse Laplace Transform-Theory of Signals And Systems-Lecture Slides, Slides for Signals and Systems. Birla Institute of Technology and Science

## Signals and Systems

Description: This lecture is part of lecture series for Signals and Systems course. Dr. Aishwarya Vyasa delivered this lecture at Birla Institute of Technology and Science. It includes: Inverse, Laplace, Transform, Shift, Property, Rational, Exponential, partial, Fraction, Expansion
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How Do We Perform Inverse Laplace Transform?
In 6.003, we will only deal with Laplace transform that are:
1) Rational, i.e. X(s) = N(s)/D(s) ;
And/or:
2) exponential, i.e. X(s) = e-sT .
For case 2), use shift property
For case 1), use PFE.
For case 1) & 2), i.e.
x(t!T)" # \$ e!sT X(s) (Similar to the FT property
x(t!T)" # \$ e!j
%
TX(j
%
)
X(s)=e!sT N(s) / D(s)
( )
X1(s)"x1(t)
1 2 4 4 3 4 4
Then b
x(t)=x1(t)t!t"T=x1(t"T)
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Inverse Laplace Transforms Via Partial Fraction
Expansion and Properties
Example:
Three possible ROC’s — corresponding to three different signals
Recall
X(s)=s+3
s+1
( )
s!2
( )
=A
s+1+B
s!2
A=!2
3,B=5
3
1
s+a, Re(s)<!a" # \$ !e!at u(!t) left - sided
and 1
s+a, Re(s)>!a" # \$ e!at u(t) right - sided
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Example (cont.)
ROC I: Left-sided signal.
x(t) =
ROC II: Two-sided signal, has Fourier Transform.
x(t) =
ROC III: Right-sided signal.
x(t) =
!2
3
e!tu(t)+5
3
e2tu(!t)
"
#
\$ %
&
' (0 as t(±)
Ae!tu(t)+Be 2tu(t)
=2
3
e!t!5
3
e2t"
#
\$ %
&
' u(!t) Diverges as t( ! )
=!2
3
e!t+5
3
e2t"
#
\$ %
&
' u(t) Diverges as t(+)
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Properties of Laplace Transforms
Many parallel properties of the CTFT, but for Laplace
transforms we need to determine implications for the ROC
For example:
Linearity
ROC at least the intersection of ROC’s of X1(s) and X2(s)
ROC can be bigger (due to pole-zero cancellation)
E.g. x1(t) = x2(t) and a = -b
Then a x1(t) + b x2(t) = 0 X(s) = 0
ROC entire s-plane
ax1(t)+bx2(t)! " # aX1(s)+bX2(s)
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Time Shift
x(t!T)" # \$ e!sT X(s) , same ROC as X(s)
e!sT
s+2, Re(s)>!2" # \$ e!2tu(t)t#t!T
!T="3
e3s
s+2, Re(s)>"2# \$ % e"2(t+3)u(t+3)
E.g.,
e3s
s+2, Re(s)>!2" # \$ ?
Causal or not?
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Time-Domain Differentiation
x(t)=1
2
!
jX(s)est ds
"
#j\$
"
+j\$
%,dx(t)
dt =1
2
!
jsX(s)estds
"
#j\$
"
+j\$
%
ROC could be bigger than the ROC of X(s), if there is pole-zero
cancellation. E.g.
x(t)=u(t)! " # 1
s, Re(s)>0
s-Domain Differentiation
!tx(t)" # \$ dX(s)
ds , with same ROC as X(s)
(Derivation is
similar to d
dt
!s)
!
dx(t)
dt " # \$ sX(s), with ROC containing the ROC of X(s)
dx(t)
dt =
!
(t)" # \$ 1=s%1
sROC =entire s&plane
E.g.te!atu(t)" # \$ !d
ds
1
s+a
%
&
' (
)
* =1
s+a
( )
2, Re(s)>!a
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Convolution Property
x(t)h(t)y(t)=h(t)!x(t)
For x(t)! " # X(s), y(t)! " # Y(s), h(t)! " # H(s)
This is the main reason why Laplace transform is useful
in solving problems involving LTI systems, just like the
Fourier transform. However, because Laplace transform
and its inverse transform are not symmetric, the reverse
relation
x(t)h(t) X(s)H(s)
is not true, different from Fourier transform. Thus,
L-transform is not useful in modulation and sampling.
×
Then Y(s)=H(s)!X(s)
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Convolution Property (cont.)
ROC of Y(s) = H(s)X(s), at least the overlap of the ROC’s
of H(s) & X(s)
a) ROC could be empty if there is no overlap between the
two ROCs
E.g. x(t) = etu(t) Re(s) > 1 and h(t) = -e-tu(-t) Re(s) < -1
b) ROC could be larger than the overlap of the two.
E.g. x(t)h(t) =
δ
(t) ROC entire s-plane.
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