We know that for two signals $x[n]$ and $h[n]$ such that : $$ y_{1}[n]=(x*h)[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]=\sum_{k=-\infty}^{\infty}h[k]x[n-k] $$ We can deduce that :$$y_{2}[n]=x[n+2]*h[n]=y_{1}[n+2]$$ and $$y_{3}[n]=x[n]*h[n+2]=y_{1}[n+2]$$ but can we know what : $$ y_{4}[n]=x[n+2]*h[n+2] $$ gives in terms of $y_{1}[n]$? I am unaware if I might be asking a really stupid question or am I not seeing it? Is it $y_{1}[n+4]$?


1 Answer 1


Whenever you have doubts regarding the properties of the convolution operator, you should resort to its definition.

Let $x_s[n] = x[n+2]$ and $h_s[n] = h[n+2]$. Then:

$$y_4[n] = x_s[n] * h_s[n] = \sum_{k=-\infty}^\infty x_s[k]h_s[n-k]$$

If we replace with the definitions of our discrete functions:

$$y_4[n] = \sum_{k=-\infty}^\infty x[k+2]h[n-k+2]$$

See what happens if we make the change of variables $m = k+2$:

$$y_4[n] = \sum_{m=-\infty}^\infty x[m]h[n-(m-2)+2]=\sum_{m=-\infty}^\infty x[m]h[n+4-m]$$

Does that final expression sound familiar? You were right indeed in your OP, as that is exactly what you thought:

$$y_4[n] = \sum_{m=-\infty}^\infty x[m]h[n+4-m] = y_1[n+4]$$


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