Z-Transform in System Analysis Questions and Answers

Answer: A

The Z-transform of a sequence is \(X(z) = \sum x(n) z^{-n}\). A right hand finite duration sequence written as \(\{1, 2, 4, 0, 7\}\) starts at \(n = 0\) with the first value, so the samples are at \(n = 0, 1, 2, 3, 4\).

\[ \begin{aligned} X(z) \\ &= 1 + 2z^{-1} + 4z^{-2} + 0 \cdot z^{-3} + 7z^{-4} \\ &= 1 + \frac{2}{z} + \frac{4}{z^2} + \frac{7}{z^4} \end{aligned} \]

The zero sample at \(n = 3\) drops out, so the last term has \(z^4\), not \(z^3\), which excludes the third and fourth options. The options with minus signs change the signs of the samples, which are all positive. The answer is option A.

Answer: D

The first term is a right sided sequence and the second is a left sided one, and each has a standard transform.

\[ a^n u(n) \leftrightarrow \frac{z}{z - a},\ |z| \gt |a| \]

\[ b^n u(-n-1) \leftrightarrow -\frac{z}{z - b},\ |z| \lt |b| \]

Adding them gives

\[ X(z) = \frac{z}{z - a} - \frac{z}{z - b} \]

The minus sign comes from the left sided term, so the options with a plus are wrong, and the options without \(z\) in the numerator are the transforms of delayed sequences. The answer is option D.

Answer: A

The Z transform of the unit ramp \(n\,u(n)\) is found from the geometric series by differentiating. Since \(\sum_{n=0}^{\infty} z^{-n} = \dfrac{z}{z - 1}\) for \(|z| \gt 1\), multiplying by \(n\) in the time domain corresponds to \(-z\,\dfrac{d}{dz}\) in the z domain.

\[ \begin{aligned} Z\{n\,u(n)\} \\ &= -z\frac{d}{dz}\left(\frac{z}{z-1}\right) \\ &= \frac{z}{(z-1)^2},\quad |z| \gt 1 \end{aligned} \]

The region of convergence is outside the unit circle because the signal is right sided. The options with \(|z| \lt 1\) are for left sided signals, and \(z/(z-1)\) is the transform of the unit step. The answer is option A.