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 $$y[n]=a^{n}\sum_{k=m}^{+\infty} a^k$$ where $m=n+1$ could be $<0$ or $>0$ and I tried to evaluate for $m>0$ which is the same as $n>-1$, which evaluated as:

\begin{align} y[n]&=a^{n}\sum_{k=m}^{+\infty} a^m\\ &=a^{n}\left[\left(\sum_{k=0}^{+\infty} a^m\right)-\left(\sum_{k=0}^{m-1} a^m\right)\right]\\ &=a^{n}\left[\left(\frac{1}{1-a^{-1}}\right)-\left(\frac{1-a^{-m}}{1-a^{-1}}\right)\right]\\ &=\frac{a^{n-m}}{1-a^{-1}}\\ &=\frac{a^{-1}}{1-a^{-1}}\tag{since $n-m=-1$} \end{align}

but when I evaluated for $m<0$ which is $n\le -1$ I am facing a problem:

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

How do I evaluate the first summation? I mean am I to consider $k=-m$ since $m<0$?

  • $\begingroup$ Isn't this a duplicate of your previous question dsp.stackexchange.com/questions/71458/… ? $\endgroup$
    – Gilles
    Nov 16, 2020 at 10:04
  • $\begingroup$ No...that was a typo error...this is the same question but i'm facing an issue in the evaluation for n<-1 case $\endgroup$
    – Orpheus
    Nov 16, 2020 at 10:05
  • $\begingroup$ Re-posting same questions is not advised, if you have a typo it's best to edit the original question so the answers are all linked to the intended question. $\endgroup$
    – Gilles
    Nov 16, 2020 at 10:10

1 Answer 1


The multiplication of the two unit step sequences $u[k]\cdot u[-n+k-1]$ is only non-zero if both sequences are non-zero. This means that the condition $k\ge 0$ as well as the condition $k\ge n+1$ must be satisfied. So you have two cases: for $n<=-1$ you have to evaluate the sum with the lower limit $k=0$, and for $n>-1$, you have to evaluate the sum with a lower limit $k=n+1$.

  • $\begingroup$ This is the case for n<= -1:$$y[n]= a^n((\sum_{k=n+1}^{-1} a^{-k})+(\sum_{k=0}^{\infty} a^{-k}))$$ where the second summation evaluates to $$\frac{1}{1-a^{-1}}$$ but somehow the first summation seems to give me a problem $\endgroup$
    – Orpheus
    Nov 16, 2020 at 10:57
  • $\begingroup$ @Orpheus: The first sum doesn't make sense because the lower limit only equals $n+1$ if $n+1\ge 0$. So for $n<-1$ you get $\sum_{k=0}^{\infty}\ldots$, and otherwise you get $\sum_{n+1}^{\infty}\ldots$. $\endgroup$
    – Matt L.
    Nov 16, 2020 at 11:20
  • $\begingroup$ can you explain to me as to how ya found the limits? I mean how is it that the lower limit is 0 for n<=-1....what I understood is that there are two cases: for n<=-1 and n>-1 and both the cases have the same expression: $$y[n]=a^{n}\sum_{k=n+1}^{+\infty} a^k$$ how did ya get the limits becuase when I put the limits the answer seems exact....so which implies that my understanding of the concept is flawed $\endgroup$
    – Orpheus
    Nov 16, 2020 at 12:06
  • $\begingroup$ What you said earlier made perfect sense to the point where ya said $u[k] u[-n+k-1]$ is zero only if $k \ge 0$ as wells as $k \ge n+1$ is valid. That is the reason I set the limits from n+1 to $\infty$ and decided to evaluate for the case $n > -1$ and $n \le -1$ using the same expression $\endgroup$
    – Orpheus
    Nov 16, 2020 at 12:47
  • 1
    $\begingroup$ @Orpheus: You never get negative indices $k$ in the sum. The smallest index is $\max\{0,n+1\}$. So for $n>-1$ you can use $n+1$ as the lower limit, but otherwise you can't because $k$ can never become negative due to $u[k]$ in the original sum. $\endgroup$
    – Matt L.
    Nov 16, 2020 at 12:58

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