Note: I'm only concerned with the bandwidth of the signals.

a) If I have a rect function from -1 to 1 (BW = 1), then if I convolve that rect function with itself I get a triangle function from -2 to 2 (BW = 2).

b) What happens to the BW of the triangle function if I convolve the triangle function with itself?

c) Furthermore, what happens to the BW of the convolution of part b) if I convolve that result from part b) with the same triangle function from part a) ?


1 Answer 1


The result of convolving two $BW=2$ triangles is the same as convolving four $BW=1$ rectangles, and has $BW=4$. It is a piece-wise cubic, whose shape approximates a bell shaped curve. If you convolve a $BW=a$ curve with a $BW=b$ curve, the convolution will have $BW=a+b$.

Here I understand "bandwidth" $BW$ to mean half the length of the shortest interval on which the signal is non-zero. If $f$ is supported on interval $[-A,A]$ and $g$ is supported on $[-B,B]$, then $h=f*g$ is supported on $[-(A+B),(A+B)]$. For if $h(x)=\int_{\mathbb{R}} f(y) g(x-y) dy$, the only non-zero contributions to the integral for $h(x)$ come from $y$ values for which $|y|\le A$ and for which $|x-y|\le B.$ By the triangle inequality, for such $y$ we have $|x|\le |y|+|x-y|\le A + B.$ If $|x|>A+B,$ there is no $y$ for which both $f(y)\ne0$ and $g(x-y)\ne0$, so $h(x)=0$ for such $x$.

  • $\begingroup$ Why is the BW of the convolution of two signals just the addition of their BWs? $\endgroup$
    – ItM
    Dec 3, 2017 at 2:57

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