Question: "a) Find the Laplace transformation of the following..."

avissidep

Answered question

2020-10-27

a) Find the Laplace transformation of the following function by using definition of Laplace transformation:

f (t) = \cos (p t) where t \geq 0

Answer & Explanation

Nathanael Webber

Skilled2020-10-28Added 117 answers

Step 1
Given

f (t) = \cos (p t)

Using the definition of Laplace Transform we have,

L [f (t)] = F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t

\Rightarrow L [\cos p t] = \int_{0}^{\infty} e^{- s t} \cos p t d t

Now we will use Integration by parts to solve it

Let I = \int_{0}^{\infty} e^{- s t} \cos p t d t

I = \cos p t \int e^{- s t} d t - \int (- p \sin p t) \frac{e^{- s t}}{- s} d t

I = \cos p t [\frac{e^{- s t}}{- s}] - \frac{p}{s} \int \sin p t e^{- s t} d t

I = - \frac{1}{s} \cos p t e^{- s t} - \frac{p}{s} [\sin p t (\frac{e^{- s t}}{- s}) - \int p \cos p t (\frac{e^{- s t}}{- s})]

I = {[- \frac{1}{s} \cos p t e^{- s t} - \frac{p}{(s^{2})} \sin p t e^{- s t} - \frac{(p^{2})}{(s^{2})} I]}_{0}^{\infty}

\Rightarrow I [1 + \frac{(p^{2})}{(s^{2})}] = - \frac{1}{s} {[\cos p t e^{- s t} + \frac{p}{s} \sin p t e^{- s t}]}_{0}^{\infty}

\Rightarrow I [\frac{(s^{2} + p^{2})}{s^{2}}] = - \frac{1}{s} [(0 + 0) - (1 + 0)]

\Rightarrow I [\frac{(s^{2} + p^{2})}{s^{2}}] = \frac{1}{s}

\Rightarrow I = \frac{s}{(s^{2} + p^{2})}

Step 2
Hence,

I = \int_{0}^{\infty} e^{- s t} \cos p t d t = L [\cos p t] = \frac{s}{(s^{2} + p^{2})}

Rightarrow By, definition of Laplace Transform of (f (t) = \cos p t) is \frac{s}{(s^{2} + p^{2})}

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