0
$\begingroup$

Is my impulse response right? By definition,the impulse response is the output when the input is a impulse signal,so

$y[n]=\sum\limits ^{n}_{k=-\infty}\frac{1}{2^{n-k}}\ x[k]$,the impulse response of $y[n]=\sum\limits ^{n}_{k=-\infty}\frac{1}{2^{n-k}}\ \delta[k]$ ,Is my thinking right?

$\endgroup$
1
$\begingroup$

Of course, if you have a formula describing the system's response to an arbitrary input $x[n]$, you will obtain the response to an impulse if you choose $x[n]=\delta[n]$.

However, I suppose that the idea of the exercise is to obtain a closed-form expression for the impulse response. You should recognize that the given input-output relation is just a convolution sum, and from that you can just write down the impulse response in a very simple form, without using an infinite sum.

You can also find that simple expression from your second formula by realizing that $\delta[k]$ is only non-zero for exactly one value of $k$. Depending on the value of $n$, that value may or may not be reached within the given summation limits.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.