# 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, 2019 at 15:49
• I cannot make sure that my answers are true or not Apr 7, 2019 at 16:00
• what is your answer ? Apr 7, 2019 at 16:00
• My answer is specified above. Apr 7, 2019 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, 2019 at 22:38

## 2 Answers

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.

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 $$[n]$$.