# Show that $g$ is a low pass filter

Let $x(t)$ be a signal.

Let $$g(t) = \int_{-\infty}^{+\infty}x(t-\varepsilon)\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon$$

I would like to show that $g(t)$ is a low pass filter for the signal $x$.

• Try taking the Fourier transform of $g(t)$. Note that the right hand side is a convolution. Also the Fourier transform of a Gaussian is just another Gaussian. Oct 9 '16 at 20:48
• I started studying fourier transformation tonight... Just to solve this problem. I think I can do this in the way you suggest to me. But I don't know how to show that the transformation of g(t) is a low-pass filter... which property should have a low pass filter?
– Sam
Oct 9 '16 at 20:54
• A low-pass filter should kill off large frequency modes of the signal $x$. Thus you should find something like ($\tilde{g}$ being the Fourier transform of $g$ and $k$ frequency / wavenumber) $\tilde{g}(k) = \tilde{x}(k) f(k)$ where $f(k) = 1$ for small $k$ but goes quickly to $0$ for large $k$. Oct 9 '16 at 20:58
• Okay... So basically the transformation of a convolution $(g*f)$ is the product of the transformation of $f$ and $g$. so the trasformation of a gaussian is again a gaussian, here it is, for $k \to \infty$ we go to 0.. And the transformed gaussian will be close to 1 for $k \to 0$ right?
– Sam
Oct 9 '16 at 21:07
• I did not check all the details myself, but that sounds about right. Oct 9 '16 at 21:09

As mentioned in the comments, $g(t)$ is the convolution of your signal with some function (the impulse response of your filter) which is a Gaussian waaveform. That is, $g(t)$ is the output of your filter. You just need to verify that a Gaussian waveform in frequency domain decays as $f\to \infty$ to confirm that your filter is indeed lowpass. This is true as the Fourier transform of a Gaussian is a Gaussian waveform.