# Convolution of Two Impulse Signals

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.

• what is the question ? Apr 7 '19 at 15:49
• I cannot make sure that my answers are true or not Apr 7 '19 at 16:00
• what is your answer ? Apr 7 '19 at 16:00
• My answer is specified above. Apr 7 '19 at 16:01
• $u[n]$ is generally used to denote the unit step function, not the unit impulse function which is usually denoted $\delta[n]$. Please don't introduce new notation unnecessarily. Apr 7 '19 at 22:38

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