0
$\begingroup$

I'm given a discrete time signal which is as follows:

$$ x[n] = \alpha^n u[n],\\$$ $$ h[n] = u[n],\\$$

Why is unit function neglected in the following step:

$$ y[n] = \sum_{k=0}^{n} \alpha^{k}\\$$

$\endgroup$
1
$\begingroup$

It isn't. The convolution sum is

$$y[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]=\sum_{k=-\infty}^{\infty}\alpha^ku[k]u[n-k]\tag{1}$$

The step $u[k]$ in $(1)$ can be removed if the index $k$ starts at zero, and the step $u[n-k]$ can be removed by replacing the upper summation limit by $n$. For clarity, one should multiply the result by $u[n]$ because if $n<0$ the result of the convolution is zero:

$$y[n]=u[n]\sum_{k=0}^{n}\alpha^k\tag{2}$$

$\endgroup$

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.