Question

The continuous signal f(t) = cos(πt/2) sampled at 1 second intervals starting from t = 0.

(a) Find the Laplace transform of the sampled signal f*(t)

Answer 1

$f^*(s)=\int\limits_0^1\cos\frac{\pi t}{2}e^{-ts}dt=-\frac{1}{s}\int\limits_0^1\cos\frac{\pi t}{2}de^{-ts}=$

$[-\frac{1}{s}\cos\frac{\pi t}{2}e^{-ts}]_0^1+\frac{1}{s}\int\limits_0^1e^{-ts}d\cos\frac{\pi t}{2}=$ $\frac{1}{s}-\frac{\pi}{2s}\int\limits_0^1\sin\frac{\pi t}{2}e^{-ts}dt$

$\int\limits_0^1\sin\frac{\pi t}{2}e^{-ts}dt=-\frac{1}{s}\int\limits_0^1\sin\frac{\pi t}{2}de^{-ts}=$$[-\frac{1}{s}\sin\frac{\pi t}{2}e^{-ts}]_0^1+\frac{1}{s}\int\limits_0^1e^{-ts}d\sin\frac{\pi t}{2}=$

$=-\frac{1}{s}e^{-s}+\frac{\pi}{2s}\int\limits_0^1\cos\frac{\pi t}{2}e^{-ts}dt=-\frac{1}{s}e^{-s}+\frac{\pi}{2s}f^*(s)$

Hence, $f^*(s)=\frac{1}{s}-\frac{\pi}{2s}\left(-\frac{1}{s}e^{-s}+\frac{\pi}{2s}f^*(s)\right)$

$(1+\frac{\pi^2}{4s^2})f^*(s)=\frac{1}{s}+\frac{\pi}{2s^2}$

Therefore, $f^*(s)=\frac{\frac{1}{s}+\frac{\pi}{2s^2}}{1+\frac{\pi^2}{4s^2}}=\frac{2\pi+4s}{\pi^2+4s^2}$