What is laplacetransform cos(t-pi/2)\H(t-pi/2) ?

The answer to laplacetransform cos(t-pi/2)\H(t-pi/2) is (e^{-(pis)/2}s)/(s^2+1)

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laplacetransform cos(t-pi/2)\H(t-pi/2)

laplace transform

cos (t - \frac{π}{2}) H (t - \frac{π}{2})

\frac{e ^{- \frac{π s}{2}} s}{s ^{2} + 1}

Solution steps

L {cos (t - \frac{π}{2}) H (t - \frac{π}{2})}

Use Laplace Transform table:

L {H (t - c) f (t)} = e^{- cs} L {f (t + c)}

For

cos (t - \frac{π}{2}) H (t - \frac{π}{2}) : f (t) = cos (t - \frac{π}{2}), c = \frac{π}{2}

= e^{- \frac{π}{2} s} L {cos ((t + \frac{π}{2}) - \frac{π}{2})}

L {cos ((t + \frac{π}{2}) - \frac{π}{2})} : \frac{s}{s ^{2} + 1}

= e^{- \frac{π}{2} s} \frac{s}{s ^{2} + 1}

= \frac{e ^{- \frac{π s}{2}} s}{s ^{2} + 1}

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What is laplacetransform cos(t-pi/2)\H(t-pi/2) ?
The answer to laplacetransform cos(t-pi/2)\H(t-pi/2) is (e^{-(pis)/2}s)/(s^2+1)