Discussion
The voltage transfer function of the network shown in the figure below is :
Answer: C
The transfer function is \(\frac{1}{1 + 2s^{2}}\).
The 4 Ω resistor is connected directly across the input, so it carries its own current but does not change the voltage applied to the rest of the network. The output is then a voltage divider of the 2 H inductor and the 1 F capacitor:
\[ \begin{aligned} \frac{V_{out}}{V_{in}} &= \frac{1/sC}{sL + 1/sC} = \frac{1}{1 + s^{2}LC} \\ &= \frac{1}{1 + 2s^{2}} \end{aligned} \]
\(\frac{1}{1 + 2s}\) would come from a resistor and capacitor, not an inductor and capacitor. \(1 + 4s\) and \(6 - s\) grow with frequency, which a passive divider of this kind cannot do.
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