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?


1 Answer 1


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
    Commented Nov 15, 2020 at 13:41

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