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
Laplace transform
asked 2021-01-27
Derive the transform of \(f(t)=\sin kt\) by using the identity \(\sin kt =\frac{1}{2}u (e^{i kt}-e^{-i kt})\)

Answers (1)

2021-01-28
Step 1
Consider the given identity,
\(\sin kt=\frac{1}{(2i)(e^{i kt}-e^{-i kt}}\)
Apply the Laplace transform on both sides,
\(L\left\{\sin kt\right\}=L\left\{\frac{1}{(2i)(e^{i kt}-e^{-i kt})}\right\}\)
Step 2
Apply the linearity of Laplace transform,
\(L\left\{\sin kt\right\}=\frac{1}{(2i)\left[L\left\{e^{ikt}\right\}-L\left\{e^{-ikt}\right\}\right]}\)
\(=\frac{1}{(2i)[\frac{1}{s-ik}-\frac{1}{s-(-ik)}]} \dots \left(L\left\{e^{at}\right\}=\frac{1}{(s-a)}\right)\)
\(=\frac{1}{(2i)\left[\frac{1}{(s-ik)}-\frac{1}{(s+ik)}\right]}\)
\(=\frac{1}{(2i)\left[\frac{(s+ik)-(s-ik)}{(s-ik)(s+ik)}\right]}\)
\(=\frac{1}{(2i)\left[\frac{(2ik)}{(s^2-i^2k^2)}\right]}\)
\(=\frac{k}{(s^2+k^2)}\)
Hence, \(L\left\{\sin kt\right\}=\frac{k}{(s^2+k^2)}\)
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