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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

The Question

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  • $\begingroup$ Second framed expression equals to y[n] $\endgroup$
    – Krlngc
    Nov 15, 2014 at 16:33

1 Answer 1

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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:

$$y[n]=\sum_{k=-\infty}^{\infty}a^ku[k]b^{n-k}u[n-k]=u[n]\sum_{k=0}^{n}a^kb^{n-k}=u[n]b^n\sum_{k=0}^{n}\left(\frac{a}{b}\right)^k=\\=u[n]b^n\frac{1-\left(\frac{a}{b}\right)^{n+1}}{1-\frac{a}{b}}=u[n]b^{n+1}\frac{1-\left(\frac{a}{b}\right)^{n+1}}{b-a}=u[n]\frac{b^{n+1}-a^{n+1}}{b-a}$$

which equals (1), as it should.

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  • $\begingroup$ n+1 is not supposed be in the equation. You're right. Thank you for your help. $\endgroup$
    – Krlngc
    Nov 16, 2014 at 11:24

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