We know that convolution in time is equivalent to multiplication in frequency (Fourier).
$$x\_1(t) \ast x\_2(t) \leftrightarrow X\_1(\omega)X\_2(\omega) \tag1$$
However, for a sampled signal, this property only holds if the frequency resolution $\Delta f = \frac{1}{T}$ for the sampled signals are the same, i.e. the signals have to have the same duration. This is illustrated below.

[![enter image description here](https://i.stack.imgur.com/1ooEL.png)](https://i.stack.imgur.com/1ooEL.png)

If the sampled signals $x\_1(t)$ and $x\_2(t)$ initially don't have the same duration, then we can zero pad the signals, such that they do. This is what is done in the lower graph pair and the samples now line up in the spectrum. Both $x\_1(t)$ and $x\_2(t)$ have been zero padded, such that they have a length of `length(x1)+length(x2)-1` , which turns out to be the length of the convolution `length(conv(x1,x2))` .

**My question is**, why does the length of the signals have to be the length of the convolution? If $x\_1(t)$ is shorter than $x\_2(t)$ why isn't it sufficient just to extend $x\_1(t)$ to the length of $x\_2(t)$? Then, both signals have the same length and the samples would still line up in the frequency spectrum, guaranteeing the property mentioned in (1), right?