Solution to 2001 Problem 39

We can use phasor analysis to determine which of the circuits is a high-pass filter. A high-pass filter is defined as a circuit where the magnitude of the output signal and the magnitude of the input signal are identical when the input frequency goes to \infty, and the magnitude of the output signal is 0 when input frequency goes to 0.

The voltage gain (which means the phasor input voltage divided by the phasor output voltage) for the four circuits are

\begin{align*}G_I &= \frac{R}{j \omega L + R} \\G_{II} &= \frac{j \omega L}{j \omega L + R} \\G_{III} &= \frac{R}...
For a high-pass filter, the magnitude of \omega should go to 1 as \omega \to \infty and to 0 as \omega \to 0. This is only true for G_{II} and G_{III}. The other two circuits are low-pass filters. Therefore, answer (D) is correct.

return to the 2001 problem list

return to homepage

Please send questions or comments to where X = physgre.