Answered: determine the Laplace transform of the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

determine the Laplace transform of the given function.

Q. f(t)=sint,0 ≤t <π , f(t +π)= f(t).

Expert Solution

Step 1

Given: $f (t) = \sin t, 0 \leq t < π$
To find: The Laplace transform of the given periodic function.

Step 2

We have,
$f (t) = \sin t, 0 \leq t < π$
The Laplace transform of the periodic function is given by-
$L \{f (t)\} = \frac{1}{1 - e^{- s T}} \int_{0}^{T} f (t) e^{- s t} d t = \frac{1}{1 - e^{- s π}} \int_{0}^{π} f (t) e^{- s t} d t = \frac{1}{1 - e^{- s π}} \int_{0}^{π} (\sin t) e^{- s t} d t Let y = \int_{0}^{π} (\sin t) e^{- s t} d t \Rightarrow y = {[\sin t \int e^{- s t} d t - \int \{[\int e^{- s t} d t] \times [\frac{d}{d t} (\sin t)]\} d t]}_{0}^{π} = {[\sin t (\frac{e^{- s t}}{- s}) - \int \{(\frac{e^{- s t}}{- s}) \times (c o s t)\} d t]}_{0}^{π} = {[(\frac{- e^{- s t}}{s}) \sin t + \frac{1}{s} \int e^{- s t} c o s t d t]}_{0}^{π} = {[(\frac{- e^{- s t}}{s}) \sin t + \frac{1}{s} \{\cos t \int e^{- s t} d t - \int \{[\int e^{- s t} d t] \times [\frac{d}{d t} (\cos t)]\} d t\}]}_{0}^{π} = {[(\frac{- e^{- s t}}{s}) \sin t + \frac{1}{s} \{(\frac{- e^{- s t}}{s}) \cos t - \int \{(\frac{- e^{- s t}}{s}) \times [- \sin t]\} d t\}]}_{0}^{π} = {[(\frac{- e^{- s t}}{s}) \sin t + \frac{1}{s} \{(\frac{- e^{- s t}}{s}) \cos t - \frac{1}{s} \int e^{- s t} \sin t d t\}]}_{0}^{π} = {[(\frac{- e^{- s t}}{s}) \sin t - \frac{e^{- s t}}{s^{2}} \cos t - \frac{1}{s^{2}} \int e^{- s t} \sin t d t]}_{0}^{π} = {[(\frac{- e^{- s t}}{s}) \sin t - \frac{e^{- s t}}{s^{2}} \cos t]}_{0}^{π} - \frac{1}{s^{2}} \int_{0}^{π} e^{- s t} \sin t d t = {[(\frac{- e^{- s t}}{s}) \sin t - \frac{e^{- s t}}{s^{2}} \cos t]}_{0}^{π} - \frac{y}{s^{2}} [∵ y = \int_{0}^{π} (\sin t) e^{- s t} d t] \Rightarrow y + \frac{y}{s^{2}} = {\{- e^{- s t} [\frac{\sin t}{s} + \frac{\cos t}{s^{2}}]\}}_{0}^{π} \Rightarrow y + \frac{y}{s^{2}} = \{- e^{- π s} [\frac{\sin (π)}{s} + \frac{\cos (π)}{s^{2}}] + e^{- 0 s} [\frac{\sin (0)}{s} + \frac{\cos (0)}{s^{2}}]\} \Rightarrow \frac{y (s^{2} + 1)}{s^{2}} = \{- e^{- π s} [\frac{1}{s^{2}}] + \frac{1}{s^{2}}\} \Rightarrow \frac{y (s^{2} + 1)}{s^{2}} = \frac{1}{s^{2}} (1 - e^{- π s}) \Rightarrow y (s^{2} + 1) = (1 - e^{- π s}) \Rightarrow y = \frac{1 - e^{- π s}}{s^{2} + 1} ∴ L {f (t)} = \frac{1}{1 - e^{- s π}} \{\frac{1 - e^{- π s}}{s^{2} + 1}\} = \frac{1}{s^{2} + 1} \Rightarrow L {f (t)} = \frac{1}{s^{2} + 1}$

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