# Convolution effects width of the signal?

Let's say there are two signal with different frequency: \begin{align} X_1(\omega) &= 0\quad\text{for}\quad \lvert \omega\rvert > 1000\pi\\ X_2(\omega) &= 0\quad\text{for}\quad\lvert \omega\rvert > 2000 \pi\\ \text{And}\quad y(t) &= x_1(t) \star x_2(t) \end{align}

• I understand that convolution in time domain means multiplication in frequency domain, but it does not affect the length of the signal. But in this case, what is the maximum frequency of $Y(\omega)$?
• If I were to find maximum sampling interval $T_s$ then which maximum frequency do I have to use?

## 2 Answers

If $y(t)$ is the signal resulting from the convolution of $x_1(t)$ with $x_2(t)$ then it will have the same bandwidth as $x_1(t)$. $x_1(t)$ has (presumably one-sided) frequency support in $[0, 500]$ Hz and $x_2(t)$ has frequency support in $[0, 1000]$ Hz. So, when performing pointwise multiplication in the frequency domain, the higher frequency components in $x_2(t)$ between $500$ Hz and $1000$ Hz will be multiplied by zero (the value of $X_1(f)$ at those frequencies). The Nyquist frequency required to capture $y(t)$ without distortion should be any frequency greater than $1000$ Hz (twice the highest frequency in the signal).

Leaving technicalities apart, the bandwidth (understood here as the intervals for the spectrum is non zero) of the convolution of two signals is the product of bandwidth.

In your case, one bandwidth is included into the other, thus the product is that on the smallest bandwidth.

On other terms, here, $x_1$ rules!