Given,

Y[z] = log(1 + az^–1)

$\begin{aligned} &\frac{d Y(z)}{d z}=\frac{d}{d z}\left(\log \left(1+a z^{-1}\right)\right) \\ &=\frac{-a z^{-2}}{1+a z^{-1}} \\ &\frac{-z d Y(z)}{d z}=\frac{a z^{-1}}{1+a z^{-1}} \end{aligned}$

Taking the inverse z- transform of above equation:

$\begin{aligned} &z^{-1}\left[-z \frac{d}{d z} Y(z)\right]=z^{-1}\left[\frac{a z^{-1}}{1+a z^{-1}}\right] \\ &n y(n)=z^{-1}\left[\frac{a z^{-1}}{1+a z^{-1}}\right] \end{aligned}$

We know that,

$a^{n} u(n) \longleftrightarrow \frac{1}{1-a z^{-1}}|z|>|a|$

Hence, $(-a)^{n} u(n) \longleftrightarrow \frac{1}{1+a z^{-1}}$

$a(-a)^{n} u(n) \longleftrightarrow \frac{a}{1+a z^{-1}}$

By shifting property,

$a(-a)^{n-1} u(n-1) \longleftrightarrow \frac{a z^{-1}}{1+a z^{-1}}$

$-(-a)^{n} u(n-1) \longleftrightarrow \frac{a z^{-1}}{1+a z^{-1}}$ $true$

Therefore,

$n y(n)=-(-a)^{n} u(n-1)$

$\begin{aligned} &y(n)=\frac{-(-a)^{n}}{n} u(n-1) \\ &y(n)=\frac{1}{n}(-1)^{n+1} a^{n} u(n-1) \end{aligned}$

$y(n)=\frac{(-1)^{n+1}}{n} a^{n} u(n-1)$

Hence, the correct option is (B).

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Classification of Signals

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