To simplify the algebraic work, always check the limiting behavior as $s \rightarrow 0$ or $s \rightarrow \infty$.
As $s \rightarrow \infty$, the capacitor behaves as a short circuit and inductor as open circuit, leaving only the first parallel resistor, so $Z(s) \rightarrow 1$.
Checking our answer: $\lim_{s\to\infty} \frac{s^2+1.5s+1}{s^2+s} = 1$, which confirms the calculation.
Step 1: Understanding the Question:
The goal of this question is to find the input driving point impedance $Z(s)$ of the given network in the s-domain.
The network consists of a series capacitor $C = 1\text{ F}$, a shunt resistor $R_1 = 1\ \Omega$, a series inductor $L = 2\text{ H}$, and a shunt resistor $R_2 = 1\ \Omega$. Step 2: Key Formula or Approach:
The s-domain impedances of individual components are:
- Resistor: $Z_R(s) = R$
- Inductor: $Z_L(s) = sL$
- Capacitor: $Z_C(s) = \frac{1}{sC}$
We will reduce the network from right to left using parallel and series combinations. Step 3: Detailed Explanation:
• The rightmost branch consists of a shunt resistor $R_2 = 1\ \Omega$.
• This is in series with the inductor $L = 2\text{ H}$ (impedance $2s$). The combined impedance of this branch is:
\[ Z_1(s) = 2s + 1 \]
• This combination is in parallel with the first shunt resistor $R_1 = 1\ \Omega$. The parallel equivalent impedance $Z_p(s)$ is:
\[ Z_p(s) = R_1 \parallel Z_1(s) = \frac{1 \cdot (2s + 1)}{1 + (2s + 1)} = \frac{2s + 1}{2s + 2} = \frac{2s + 1}{2(s + 1)} \]
• Finally, this parallel combination is in series with the input capacitor $C = 1\text{ F}$ (impedance $\frac{1}{s}$). The total input impedance $Z(s)$ is:
\[ Z(s) = \frac{1}{s} + Z_p(s) = \frac{1}{s} + \frac{2s + 1}{2(s + 1)} \]
• Dividing both the numerator and the denominator by 2:
\[ Z(s) = \frac{s^2 + 1.5s + 1}{s(s+1)} \] Step 4: Final Answer:
The s-domain input impedance is $\frac{s^2 + 1.5s + 1}{s(s+1)}$, which corresponds to Option (A).
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