# Discrete-time Convolution Convergence Issue

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.

• 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 Nov 15 '20 at 13:41