Z-transform: Convolution <=> Multiplication giving strange results

Question

I try two slightly different routes to an answer, and get two different answers. (This is from a past exam paper.)

Find h[n] if $ H(z) = \frac{1}{(1-az^{-1})^2} $

First try:

$ H(z) = H_1(z)\cdot H_2(z) $,

$ H_1(z) = a^{-1}z $, $ \qquad H_2(z) = \frac{az^{-1}}{(1-az^{-1})^2}$

According to Wikipedia,

$ \mathcal{Z^{-1}}(a^{-1}z) = a^{-1} \delta[n+1]$

$ \mathcal{Z^{-1}}(\frac{az^{-1}}{(1-az^{-1})^2}) = na^n $, $ \qquad n > 0$

Therefore, since multiplication in the Z domain is convolution in the time domain,

$ h[n] = a^{-1} \delta[n+1] \ast na^n = a^{-1}(n+1)a^{n+1} = (n+1)a^n $.

$ (n+1)a^n $ is the answer given in the solution. However...

Second try:

$ H(z) = H_3(z)^2, \qquad H_3(z) = \frac{1}{(1-az^{-1})} $

$ \therefore h[n] = h_3[n] \ast h_3[n] = a^n \ast a^n = \sum_{k=-\infty}^{\infty} a^k \cdot a^{n-k} $

But:

$ \sum_{k=-\infty}^{\infty} a^k \cdot a^{n-k} \neq (n+1)a^n $

Answer 1

You are doing fine. You only have a problem with the convolution. We define the convolution (Cauchy product) of two sequences as

$$ a_n*b_n = \sum_{k=0}^{n} a_i b_{n-i}. $$

So, in your case, we have

$$ a^n*a^n = \sum_{k=0}^{n} a^i a^{n-i}=a^n\sum_{i=0}^{n}1 = (n+1) a^n.$$