# Basic Understanding of Convolution Operation

I would really appreciate any help in understanding this.

If $$y[n] = x[n]*r[n]\tag{1}$$ then what is $$y[n-n_0]$$ for any system?

Is it $$y[n-n_0] = x[n-n_0]*r[n-n_0]\tag{2}$$ or $$y[n-n_0] = x[n-n_0]*r[n]\tag{3}$$ or $$y[n-n_0] = x[n]*r[n-n_0]\tag{4}$$

Will that result (answer to the above) change when we now have an LTI system with $$h[n]=r[n]$$ being its impulse response?

Thank you very much

Since $$y[n-n_0] = y[n] \ast \delta [n - n_0]$$ and convolution is both associative and distributive, is it not the case that

$$y[n-n_0] = \left(x[n] \ast r[n] \right) \ast \delta[n-n_0] = x[n] \ast r[n-n_0] = x[n-n_0] \ast r[n]?$$

Why not taking the definition of convolution and see what happens?

$$y[n-n_0] = \sum_k x[k]r[n-n_0-k]$$ which gives $$y[n-n_0] = x[n]*r[n-n_0]$$

Also, since $$x[n]*r[n] = r[n]*x[n]$$, $$y[n-n_0] = \sum_k r[k]x[n-n_0-k]$$ which gives $$y[n-n_0] = r[n] * x[n-n_0]$$

So both (3) and (4) are correct.

I will go with $$(4)$$, since it is possible to build a system with $$h[n-n_0]$$

• Thank you very much @jomegaA, a follow-up question will be why is (3) not true as well since the convolution operation is commutative.
– site
Feb 2 '20 at 23:52
• Practically I may have no influence on the input but I can influence the design of my system. Feb 3 '20 at 4:37