I'm trying to solve a problem on convolution from Alan V.Oppenheim:

Find the convolution output $y[n]$ for the following signals:

$$x[n]= u[n]\quad\text{and}\quad h[n]=a^{-n}u[-n-1], \ a>1 $$

I started the evaluation:

$$y[n]=\sum_{k=-\infty}^{+\infty} u[k]a^{-n+k}u[-n+k-1]$$

considering that $u[k]=1$ for $k\ge0$ and $u[-n+k-1]=1$ for $k\ge n+1$ which I evaluated to $$y[n]=a^{-n}\sum_{k=n+1}^{+\infty} a^k$$ where $n$ could be $<0$ or $>0$ and I tried to evaluate for $n>0$: where I faced an issue:

I subsittuted $k-1$ to $m$ and reframed the equation as:

$$y[n]=a^{-n+1}\sum_{m=n}^{+\infty} a^m=a^{-n+1}\left[\left(\sum_{m=0}^{+\infty} a^m\right)-\left(\sum_{m=0}^{n-1} a^m\right)\right]$$

now for $a>1, \ \displaystyle \sum_{m=0}^{+\infty} a^m$ will not converge. How will I evaluate this?


In the third edition of Oppenheim and Schafer's Discrete-Time Signal Processing, the corresponding exercise is given with $h[n]=a^nu[-n-1]$, $a>1$ (note the positive sign in the power of $a$). In that case everything works out fine. So either there's a typo in your edition or you made a mistake copying the exercise.

  • $\begingroup$ I know this will sound stupid....but there is no mistake other than me copying the problem...anyways thanks a lot...I will let you know how it turns out $\endgroup$ – Orpheus Nov 15 '20 at 13:41

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.