Discussion
What is the convolution of the given two continuous time functions, for all values of t?
\(X(t) = e^{-t^2}\)
\(H(t) = 3t^2\)
Answer: B
The convolution is \(y(t) = \int_{-\infty}^{\infty} x(\tau)h(t - \tau)\,d\tau\) with \(h(t) = 3t^2\).
\[ \begin{aligned} y(t) = 3\int e^{-\tau^2}(t - \tau)^2\,d\tau \\ &= 3\int e^{-\tau^2}\left(t^2 - 2t\tau + \tau^2\right)d\tau \end{aligned} \]
Use \(\int e^{-\tau^2}d\tau = \sqrt{\pi}\), \(\int \tau e^{-\tau^2}d\tau = 0\) (odd function) and \(\int \tau^2 e^{-\tau^2}d\tau = \sqrt{\pi}/2\).
\[ \begin{aligned} y(t) \\ &= 3\left(\sqrt{\pi}\,t^2 + \frac{\sqrt{\pi}}{2}\right) \\ &= 5.317\,t^2 + 2.659 \end{aligned} \]
The term in \(t\) vanishes, which removes the options with a power of \(t\) other than \(t^2\) attached to the wrong constants. The answer is \(5.31t^2 + 2.659\), option B.
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