Convolution of Two Impulse Signals

Question

I have encountered convolution of two different impulse signals.

x[n] = (1/2)^n . u[n-2] * u[n]
x[n] = u[n] * [n]

u[n] = discrete impulse signal
. = product operation
* = convolution operation

For the first one, I found this solution:

x[n] = 1/4 if n = 2
x[n] = 0 if n != 2

For the second one, I found impulse signal itself

Edit: Are my answers are true ? My professor told me that the answer for the first one is wrong, but he did not say the correct answer.

Answer 1

Your answer is right, assuming you posted the question right.

But you better use the standard notation as Dilip Sarwate already indicated; $u[n]$ is the unit-step and $\delta[n]$ is the unit impulse. Then

$$ 0.5^n \delta[n-2] \star \delta[n] = 0.5^2 \delta[n-2] = \begin{cases} { 0.25 ~~~, ~~~n= 2 \\ 0.00 ~~~,~~~n \neq 2 } \end{cases} $$

you can get the answer for the second case, exactly in the same way.

Answer 2

This is basically an exercise to test the student's understanding of the concept that the unit impulse is effectively the unit in convolutions, that is: $$\delta \star x = x$$ for all signals $x$. Perhaps a systems explaination might help. If an LTI system has impulse response $h$, then we know that the output of the system when $x$ is the input is $y = h \star x$. So, $\delta \star x$ can be thought of as the output of an LTI system with impulse response $\delta$ when the input to the LTI system is $x$. What LTI system has output $\delta$ when its input is the unit impulse $\delta$?? It is just the canonical straight wire with (no) gain that audio enthusiasts dream about! And so, $\delta \star x = x$ for all $x$.

With this, it it is easy to verify that the OP's answer to the first question is correct (but maybe his professor wanted to see the answer as $\left(\frac 12\right)^n \delta[n-2]$ or $\left(\frac 12\right)^2 \delta[n-2]$ instead of what the OP wrote) while the second answer $\delta[n]\star [n] = \delta[n]$ is incorrect, it should be $[n]$.