How can we evaluate the matched filter's impulse response of a Gaussian function $x(t) = \exp(-\frac{t^2}{2})$. As far as we know, for a signal of finite duration $T$, the impulse response of it's matched filter is $h(t) = x(T-t)$. But as Gaussian function is of infinite duration, how can we find the matched filter ?

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    $\begingroup$ gonna have to truncate it somewhere. because $e^{-t^2}$ dies off quickly, that shouldn't be hard. $\endgroup$ Commented Apr 30, 2018 at 7:01
  • $\begingroup$ No signal that extends out to $\infty$ can have a causal matched filter. With a causal filter, at any finite time $T$, there is still signal that has not been processed by the filter as yet, and so by waiting a little bit longer, you can always improve performance. That being said, one can define the matched filter for any $x(t)$ as the filter with impulse response $x(-t)$ (see, e.g. this answer). For your Gaussian function, try something like $$h(t) = \begin{cases} \exp\left(-\frac{(t-3)^2}{2}\right), &t \geq 0,\\0, &t<0\end{cases}$$ $\endgroup$ Commented May 1, 2018 at 19:04

1 Answer 1


If you look at:


the derivation does not depend on the filter being either causal or have a finite duration. The derivation is fully satisfactory for convolution for time index $k$ over $-\infty,\infty$. It's all about what filter will maximize the Schwarz inequality.

In this case $$ x(t)=\exp(-\frac{(t-0)^2}{2}) = x(0-t) $$

Functions that are symmetric are their own matched filter.

The $T$ parameter enters the discussion in terms of when does one sample the filter for the purpose of detection. If $x(t)$ was delayed by some unknown time(nonsynchronous detection), I would use the matched filter to estimate it's delay.


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