Consider the following system : \(x(t) = e^{-\pi t^{2}}\) is applied to a block \(h(t)\) whose output is \(y(t)\).
Impulse response \(h(t)\) is a filter matched to \(x(t)\). The Fourier transform of the output \(y(t)\) shall be :
MH SET - 2018, SET Exam - 2018,
Answer: B
The Fourier transform of the output is \(e^{-2\pi f^{2}}\).
The Gaussian pulse is its own Fourier transform: \(x(t) = e^{-\pi t^{2}}\) has \(X(f) = e^{-\pi f^{2}}\). A filter matched to \(x(t)\) has \(h(t) = x(-t)\) (taking the delay as zero), so \(H(f) = X^{*}(f) = e^{-\pi f^{2}}\), real and even here. Then
\[ \begin{aligned} Y(f) &= X(f)H(f) = e^{-\pi f^{2}} \cdot e^{-\pi f^{2}} \\ &= e^{-2\pi f^{2}} \end{aligned} \]
\(e^{-\pi f^{2}}\) is the input spectrum alone, as if the filter passed everything. \(e^{-\pi f^{2}/2}\) would be a wider spectrum than the input, which filtering by a Gaussian cannot produce, and \(e^{-\pi|f|}\) is the transform of a Cauchy-shaped pulse, not of anything Gaussian.
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