# Discrete Time Convolution Evaluation

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 $$m<0$$?

• Isn't this a duplicate of your previous question dsp.stackexchange.com/questions/71458/… ? – Gilles Nov 16 '20 at 10:04
• No...that was a typo error...this is the same question but i'm facing an issue in the evaluation for n<-1 case – Orpheus Nov 16 '20 at 10:05
• 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. – Gilles Nov 16 '20 at 10:10

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 $$k=n+1$$.
• 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 – Orpheus Nov 16 '20 at 10:57
• @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 $\sum_{n+1}^{\infty}\ldots$. – Matt L. Nov 16 '20 at 11:20
• 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 – Orpheus Nov 16 '20 at 12:06
• 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 – Orpheus Nov 16 '20 at 12:47
• @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. – Matt L. Nov 16 '20 at 12:58