Inverse Z-Transform of Logarithm Function

We need to find the inverse Z-transform of $X(z) = \log(1 + az^{-1})$ with the Region of Convergence (ROC) given as $|z| > |a|$. This ROC implies a causal signal.

Using Series Expansion

We use the Taylor series expansion for $\log(1+x)$ around $x=0$: $ \log(1+x) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k} $ Substitute $x = az^{-1}$: $ X(z) = \log(1 + az^{-1}) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{(az^{-1})^k}{k} $ $ X(z) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{a^k}{k} z^{-k} $

Identifying the Sequence

The Z-transform of a sequence $x(n)$ is defined as $X(z) = \sum_{n=-\infty}^{\infty} x(n) z^{-n}$. By comparing the derived series for $X(z)$ with the definition:

• The summation starts from $k=1$.
• The term corresponding to $z^{-k}$ is $(-1)^{k+1} \frac{a^k}{k}$.

This implies that the sequence $x(n)$ is non-zero only for $n \ge 1$. By direct comparison, the coefficient for $z^{-n}$ (where $n=k$) is:

Considering Causality (ROC)

The ROC $|z| > |a|$ confirms that the signal is causal. Therefore, $x(n) = 0$ for $n < 0$. Combining this with the result from the series expansion:

This result matches Option 4.