Question: "What is the Laplace transform of x(t)=e−tcos⁡(π)t?"

Name: What is the Laplace transform of x(t)=e−tcos⁡(π)t?
Uploaded: 2022-02-23T17:22:57+00:00
Description: What is the Laplace transform of x(t)=e−tcos⁡(π)t?

UkusakazaL

Answered question

2020-10-19

What is the Laplace transform of

x (t) = e^{- t} \cos (π) t

Answer & Explanation

tafzijdeq

Skilled2020-10-20Added 92 answers

Data analysis
To find the Laplace transform of ,

x (t) = e^{- t} \cos π t

Here the proof for standard form has been shown,
Let standard form be,

x (t) = e^{a t} \cos b t

Derivation and Solution

L {e^{- a t} \cos b t} = \int_{0}^{\infty} e^{a t} \cos b t e^{- s t} d t =

= \int_{0}^{\infty} e^{- s (s + a) t} \cos b t d t

Let

u = \cos b t d v = e^{- s (s + a) t}

d u = - b \sin b t d t v = - \frac{1}{s + a} e^{- (s + a) t}

(using integration by parts)

\Rightarrow y = {| - \frac{\cos b t}{s + a} e^{- (s + a) t} |}_{0}^{\infty} - \frac{b}{s + a} \int_{0}^{\infty} e^{- (s + a) t} \sin b t d t

Formula:

\int u d v = u v - \int v^{'} (x) d u

\Rightarrow y = (0 - (- \frac{1}{s + a})) - \frac{b}{s + a} \int_{0}^{\infty} e^{- (s + a) t} \sin b t d t

let

u = \sin b t d v = e^{- (s + a) t} d t

d u = b \cos b t d t v = - \frac{1}{s + a} e^{- (s + a) t}

Again integration by parts

\Rightarrow y = \frac{1}{s + a} - \frac{b}{s + a} [\frac{- e^{- (s + a) t} \sin b t}{s + a} |_{0}^{\infty} + \frac{b}{s + a} \int_{0}^{\infty} e^{- (s + a) t} \cos b t]

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