Discussion
The driving point impedance Z(s) of a network has the pole-zero locations as shown in the following figure. If Z(o) = 3, then Z(s) is :
Answer: B
The impedance is \(Z(s) = \dfrac{2(s+3)}{s^{2}+2s+2}\).
The figure shows poles (crosses) at \(s = -1 \pm j1\), and the mark at \(-3\) on the real axis is the zero. So
\[ \begin{aligned} Z(s) &= \frac{K(s + 3)}{(s + 1 - j)(s + 1 + j)} \\ &= \frac{K(s + 3)}{s^{2} + 2s + 2} \end{aligned} \]
Setting \(s = 0\): \(Z(0) = 3K/2 = 3\), so \(K = 2\).
The denominator \(s^{2} + 2s + 3\) would put the poles at \(-1 \pm j\sqrt{2}\), not \(\pm j1\). The options with \(s - 3\) and \(s^{2} - 2s - \ldots\) place the zero and poles in the right half-plane, which is not what the plot shows.
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