Derive the transform of f(t)=sin kt by using the identity sin kt =frac{1}{2}u (e^{i kt}-e^{-i kt})

Question

Derive the transform of

f (t) = \sin k t

by using the identity

\sin k t = \frac{1}{2} u (e^{i k t} - e^{- i k t})

Answer 1

Step 1
Consider the given identity,

\sin k t = \frac{1}{(2 i) (e^{i k t} - e^{- i k t}}

Apply the Laplace transform on both sides,

L {\sin k t} = L {\frac{1}{(2 i) (e^{i k t} - e^{- i k t})}}

Step 2
Apply the linearity of Laplace transform,

L {\sin k t} = \frac{1}{(2 i) [L {e^{i k t}} - L {e^{- i k t}}]}

= \frac{1}{(2 i) [\frac{1}{s - i k} - \frac{1}{s - (- i k)}]} \dots (L {e^{a t}} = \frac{1}{(s - a)})

= \frac{1}{(2 i) [\frac{1}{(s - i k)} - \frac{1}{(s + i k)}]}

= \frac{1}{(2 i) [\frac{(s + i k) - (s - i k)}{(s - i k) (s + i k)}]}

= \frac{1}{(2 i) [\frac{(2 i k)}{(s^{2} - i^{2} k^{2})}]}

= \frac{k}{(s^{2} + k^{2})}

Hence,

L {\sin k t} = \frac{k}{(s^{2} + k^{2})}

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