Question

Derive the Laplace transform of the following function
\(\sin(h(at))\)

Answer 1

Step 1 We have to derive the Laplace transform of function:
\(\sin(h(at))\)
We know from definition of hyperbolic function,
\(\sin hx=\frac{e^x-e^{-x}}{2}\)
If x=at then,
\(\sin h x =\frac{e^x-e^{-x}}{2}\)
\(\sin h at =\frac{e^{at}-e^{-at}}{2}\)
\(=\frac{1}{2}(e^{at}-e^{-at})\)
Step 2
Now finding Laplace transform,
\(L(\sin h at)=L(\frac{1}{2}(e^{at}-e^{-at})\)
\(=\frac{1}{2}(L{e^{at}}-L(e^{-at}))\)
We know the Laplace transform of exponential function,
\(L(e^{ax})=\frac{1}{(s-a)}\)
if a=-a then \(L(e^{-ax})=\frac{1}{(s+a)}\)
Putting Laplace transform of exponential function, we get
\(\frac{1}{2}(L{e^{at}}-L(e^{-at}))=\frac{1}{2}((\frac{1}{(s-a)})-(\frac{1}{(s+a)})\)
\(=\frac{1}{2}\frac{(s+a-(s-a))}{(s-a)(s+a)}\)
\(=\frac{1}{2}\frac{(s+a-s+a)}{(s^2-a^2)} \text{ since } (\sin ce (a+b)(a-b)=a^2-b^2)\)
\(=\frac{1}{2}\frac{2a}{(s^2-a^2)}\)
\(=\frac{a}{s^2-a^2}\)
Hence, Laplace transform of \(\sin h at\) is \(\frac{a}{(s^2-a^2)}\)