# How to show that y[n] = x[n] * h[n] turns into the Y(z) = X(z).H(z)?

I'm trying to show that $y[n]=x[n]*h[n]$ turns into $Y(z) = X(z)H(z)$ in Z-domain by first applying convolution then by taking the inverse Z-transform of the $Y(z)$, stating that it's the same sequence after all. However, in power series expansion an extra term appears which kills the equation. You can find the question below; I added it as an image.

I'd be grateful if you could help. Thank you

• Second framed expression equals to y[n] Nov 15, 2014 at 16:33

The correct sequence $y[n]$ is the one you obtained via the $\mathcal{Z}$-transform:
$$y[n]=\frac{1}{a-b}\left[a^{n+1}-b^{n+1}\right]u[n]\tag{1}$$
I'm not sure where that $n+1$ term is coming from in your time-domain derivation, so I'll write out the convolution sum:
$$y[n]=\sum_{k=-\infty}^{\infty}a^ku[k]b^{n-k}u[n-k]=u[n]\sum_{k=0}^{n}a^kb^{n-k}=u[n]b^n\sum_{k=0}^{n}\left(\frac{a}{b}\right)^k=\\=u[n]b^n\frac{1-\left(\frac{a}{b}\right)^{n+1}}{1-\frac{a}{b}}=u[n]b^{n+1}\frac{1-\left(\frac{a}{b}\right)^{n+1}}{b-a}=u[n]\frac{b^{n+1}-a^{n+1}}{b-a}$$