Question

Driving point impedance of the function \( F(s) = \frac{(s+k_1)(s+k_2)(s+k_3)}{(s+1)(s+2)(s+3)} \) is

• A. 48
• B. 24
• C. 10
• D. 4

Hint:

Driving point impedance can be evaluated at \(s=0\) for DC resistance or as \(s \to \infty\) for high-frequency behavior.
The form \(Z(s) = K \frac{\prod (s-z_i)}{\prod (s-p_j)}\) has zeros at \(z_i\) and poles at \(p_j\). Here, zeros are at \(-k_1, -k_2, -k_3\) and poles at \(-1, -2, -3\).

Answer 1

The Correct Option is D

Let the driving point impedance be \(Z_{dp}(s) = F(s) = \frac{(s+k_1)(s+k_2)(s+k_3)}{(s+1)(s+2)(s+3)}\). 

Since the options are numerical constants, the question likely implies evaluating \(Z_{dp}(s)\) under a specific condition (e.g., at \(s=0\) for DC impedance) or for particular values of \(k_1, k_2, k_3\).

If we consider the DC impedance (at \(s=0\)): 

\(Z_{dp}(0) = \frac{k_1 k_2 k_3}{1 \cdot 2 \cdot 3} = \frac{k_1 k_2 k_3}{6}\).

If this DC impedance is equal to option (d), which is 4:

\(\frac{k_1 k_2 k_3}{6} = 4 \Rightarrow k_1 k_2 k_3 = 24\).

This interpretation requires the product of the zero locations' magnitudes (or specific \(k_i\) values if they represent \(s+k_i\)) to be 24. For example, if \(k_1=2, k_2=3, k_3=4\).

Another common evaluation is at \(s \to \infty\):

As \(s \to \infty\), \(Z_{dp}(s) \approx \frac{s^3}{s^3} = 1\). This is not among the options.

Without further information or clarification on \(k_1, k_2, k_3\) or the specific evaluation point, the question is ambiguous. However, assuming the question intends for the DC impedance to match one of the options and that option (d) is the correct answer "4", this would imply \(k_1 k_2 k_3 = 24\).

Final Answer: \[ \boxed{4 \text{ (assuming DC impedance, with } k_1 k_2 k_3 = 24)} \]