Question

Determine the z-transform corresponding to the given pole-zero plot.

• A. \[ X(z)=G\frac{(z+r\cos\omega_0)} {(z-re^{j\omega_0})(z-re^{-j\omega_0})}, \qquad |z|<r \]
• B. \[ X(z)=G\frac{(z+r\cos\omega_0)} {(z+re^{j\omega_0})(z-re^{-j\omega_0})}, \qquad |z|<r \]
• C. \[ X(z)=G\frac{z(z-r\cos\omega_0)} {(z-re^{j\omega_0})(z-re^{-j\omega_0})}, \qquad |z|>r \]
• D. \[ X(z)=G\frac{z(z+r\cos\omega_0)} {(z+re^{j\omega_0})(z-re^{-j\omega_0})}, \qquad |z|>r \]

Hint:

For pole-zero plots: \[ X(z)=G\frac{\prod(z-\text{zero})} {\prod(z-\text{pole})} \] and the ROC is determined separately from the shaded region. \[ |z|>r \Rightarrow \text{Right-sided (causal) sequence} \] \[ |z|<r \Rightarrow \text{Left-sided (anti-causal) sequence} \]

Answer 1

The Correct Option is C

Concept: A pole-zero plot directly provides the locations of poles, zeros, and the Region of Convergence (ROC). The z-transform can be written as \[ X(z)=G \frac{\prod (z-z_i)} {\prod (z-p_i)} \] where \(z_i\) are zeros and \(p_i\) are poles.

Step 1: Identify the poles from the figure.
From the pole-zero plot, the poles are located at \[ p_1=re^{j\omega_0}, \qquad p_2=re^{-j\omega_0}. \] Therefore the denominator becomes \[ (z-re^{j\omega_0}) (z-re^{-j\omega_0}). \]

Step 2: Identify the zeros from the figure.
The figure shows
• One zero at the origin.
• One zero on the real axis at \[ z=r\cos\omega_0. \] Hence the numerator is \[ z(z-r\cos\omega_0). \]

Step 3: Determine the ROC.
The shaded region lies outside the circle of radius \(r\). Therefore, \[ ROC:\quad |z|>r. \] This corresponds to a right-sided (causal) sequence.

Step 4: Construct the z-transform.
Combining numerator, denominator and ROC, \[ X(z) = G \frac{z(z-r\cos\omega_0)} {(z-re^{j\omega_0})(z-re^{-j\omega_0})}. \] Thus, \[ \boxed{ X(z)= G\frac{z(z-r\cos\omega_0)} {(z-re^{j\omega_0})(z-re^{-j\omega_0})}, \quad |z|>r } \] which matches option (C).