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Eel2010 C29

This document discusses the unilateral Laplace transform and provides proofs of two properties. It defines the unilateral Laplace transform of a continuous-time signal x(t) from 0- to infinity. It then proves that the Laplace transform of the derivative of x(t) is equal to sX(s) - x(0-), and that the Laplace transform of the second derivative of x(t) is equal to s^2X(s) - sX(0-) - x'(0-). The document concludes with an example problem that uses these properties to find the time domain response y(t) given a differential equation and initial conditions.

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Dikshant Gupta (B21CI014)
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15 views4 pages

Eel2010 C29

This document discusses the unilateral Laplace transform and provides proofs of two properties. It defines the unilateral Laplace transform of a continuous-time signal x(t) from 0- to infinity. It then proves that the Laplace transform of the derivative of x(t) is equal to sX(s) - x(0-), and that the Laplace transform of the second derivative of x(t) is equal to s^2X(s) - sX(0-) - x'(0-). The document concludes with an example problem that uses these properties to find the time domain response y(t) given a differential equation and initial conditions.

Uploaded by

Dikshant Gupta (B21CI014)
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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EEL2010: Signals and Systems

IIT Jodhpur

Institute Core, Semester II, 2022-2023

March 30, 2023

IIT Jodhpur (Lecture 29) EEL2010: Signals and Systems March 30, 2023 1/4
Unilateral Laplace Transform
The unilateral transform of a continuous-time signal x(t) is defined as
Z ∞
X (s) = x(t)e−st dt
0−

Prove the following.


 
L dxdt
= sX (s) − x(0− ).
 2 
dx(t)
L ddt2x = s2 X (s) − sx(0− ) − dt t=0−
.

IIT Jodhpur (Lecture 29) EEL2010: Signals and Systems March 30, 2023 2/4
dx
= sX (s) − x(0− )

L dt


dx
 Z ∞ dx −st
L = e dt
dt 0− dt
∞ Z ∞
= x(t)e−st +s x(t)e−st dt
0− 0−
= sX (s) − x(0− ).

 
d2 x dx(t)
L dt2
= s2 X (s) − sx(0− ) − dt t=0−

d2 x
   
d dx
L = L
dt2 dt dt
 
dx
= sL − x′ (0− )
dt
= s(sX (s) − x(0− )) − x′ (0− )
= s2 X (s) − sx(0− ) − x′ (0− ).

IIT Jodhpur (Lecture 29) EEL2010: Signals and Systems March 30, 2023 3/4
Example
d2 y(t) dy(t)
Consider a system described with differential equation dt2 + 3 dt + 2y(t) = x(t) with
initial conditions y(0− ) = β and y ′ (0− ) = γ. Let x(t) = αu(t). Find y(t).

d2 y(t) dy(t)
+3 + 2y(t) = x(t)
dt2 dt
2 α
s Y(s) − βs − γ + 3sY(s) − 3β + 2Y(s) =
s
β(s + 3) γ
Y(s) = +
(s + 1)(s + 2) (s + 1)(s + 2)
α
+
s(s + 1)(s + 2)
1 1 3
Y(s) = − + (for α = 2, β = 3, γ = −5)
s s+1 s+2
y(t) = [1 − e−t + 3e−2t ]u(t)

IIT Jodhpur (Lecture 29) EEL2010: Signals and Systems March 30, 2023 4/4

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