Let $$g\left( t \right) = {e^{ – \pi {t^2}}}$$ and h(t) is a filter matched to g(t). If g(t) is applied as input to h(t), then the Fourier transform of the output is

$${e^{ - pi {t^2}}}$$
$${e^{ - {{pi {t^2}} over 2}}}$$
$${e^{ - pi left| f ight|}}$$
$${e^{ - 2pi {t^2}}}$$

The correct answer is C. $${e^{ – \pi \left| f \right|}}$$

A filter matched to a signal $g(t)$ is a filter whose impulse response is $h(t) = g(-t)$. The Fourier transform of the output of a filter matched to a signal is the convolution of the Fourier transform of the signal with the Fourier transform of the impulse response. In this case, the Fourier transform of the signal is $G(f) = e^{-\pi f^2}$ and the Fourier transform of the impulse response is $H(f) = e^{-\pi f}$. The convolution of $G(f)$ and $H(f)$ is $G(f) * H(f) = e^{-\pi f^2} * e^{-\pi f} = e^{-\pi \left| f \right|}$.

Option A is incorrect because it is the Fourier transform of the signal, not the output of the filter. Option B is incorrect because it is the Fourier transform of the signal convolved with a Gaussian function. Option D is incorrect because it is the Fourier transform of the signal squared.

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