Chapter 11 Practice Problems — Frequency Response and Bode Plots
These problems are meant to build your problem-solving intuition. A hint is provided for each — try the problem on your own first before reading it.
Problem 1 — First-Order RC Low-Pass Filter
An RC low-pass filter has R = 10 kΩ and C = 15.9 nF.
(a) Derive the transfer function \(H(j\omega) = V_{out}/V_{in}\) using the voltage divider rule with impedances.
(b) Find the cutoff frequency \(f_c\) in Hz.
(c) At \(f = f_c\), what is \(|H|\) in dB?
(d) Sketch the asymptotic Bode magnitude plot, labeling the corner frequency and the slope beyond \(f_c\).
Hint
(a) Treat the capacitor as impedance \(Z_C = 1/(j\omega C)\). Apply the voltage divider: \(H = Z_C / (R + Z_C)\). Simplify to the form \(H = 1/(1 + j\omega/\omega_c)\).
(b) The cutoff frequency is where \(|H| = 1/\sqrt{2}\), which occurs at \(\omega_c = 1/(RC)\), so \(f_c = 1/(2\pi RC)\).
(c) At \(\omega = \omega_c\): \(|H| = 1/\sqrt{2}\). Convert to dB: \(20\log_{10}(1/\sqrt{2}) = -3\) dB.
(d) The asymptotic Bode plot is flat (0 dB) below \(f_c\), then falls at −20 dB/decade above \(f_c\). The exact response is 3 dB below the asymptote exactly at \(f_c\).
Problem 2 — High-Pass Filter Transfer Function
An RC high-pass filter has R = 4.7 kΩ and C = 33 nF. The output is taken across R.
(a) Write the transfer function \(H(j\omega)\).
(b) Find \(f_c\).
(c) At \(f = f_c/10\) (one decade below cutoff), estimate \(|H|\) in dB using the asymptotic approximation.
(d) What is the phase angle of H at \(f = f_c\)?
Hint
(a) Output is across R, not C. The voltage divider gives \(H = R/(R + 1/(j\omega C))\). Multiply numerator and denominator by \(j\omega C\) and simplify to the standard high-pass form \(H = (j\omega/\omega_c)/(1 + j\omega/\omega_c)\).
(b) Same formula: \(f_c = 1/(2\pi RC)\).
(c) The asymptote below \(f_c\) rises at +20 dB/decade. At one decade below cutoff, the asymptote predicts −20 dB. (The exact value differs by ~3 dB only near the corner.)
(d) At \(\omega = \omega_c\), the high-pass transfer function has angle \(\angle H = 90° - \arctan(1) = 45°\). Think about which way the angle shifts for a high-pass vs. low-pass filter.
Problem 3 — Bode Plot from Transfer Function
A circuit has transfer function:
(a) What type of filter is this?
(b) Identify the corner frequency in rad/s and Hz.
(c) Sketch the asymptotic Bode magnitude plot from 10 to 100,000 rad/s. Label slopes and key values.
(d) At \(\omega = 100\) rad/s and \(\omega = 10{,}000\) rad/s, estimate \(|H|\) in dB.
Hint
(a) Examine what happens at \(\omega \to 0\) and \(\omega \to \infty\). A numerator with \(j\omega\) means the gain is low at DC and increases with frequency.
(b) The corner is where the denominator term \(j\omega/\omega_c = 1\), so \(\omega_c = 1000\) rad/s. Convert with \(f_c = \omega_c/(2\pi)\).
(c) Below \(\omega_c\): the \(j\omega/\omega_c\) term dominates and the slope is +20 dB/decade. Above \(\omega_c\): numerator and denominator both grow with \(\omega\), so the slope becomes 0 dB (flat).
(d) Use the asymptotic slopes: at 100 rad/s (one decade below \(\omega_c\)), the asymptote gives −20 dB. At 10,000 rad/s (one decade above), the response is near 0 dB.
Problem 4 — Identifying Filter Type and Order
You measure the frequency response of an unknown filter and observe: - At 100 Hz: \(|H| \approx 0\) dB - At 10 kHz: \(|H| \approx -40\) dB - The phase shifts from 0° to −180°
(a) What type of filter is this (low-pass, high-pass, band-pass, band-reject)?
(b) What is the filter order? Justify from the roll-off rate.
(c) Estimate the cutoff frequency \(f_c\). Show your reasoning from the given data.
Hint
(a) The gain is flat at low frequency and rolls off at high frequency. What type of filter passes low frequencies and attenuates high ones?
(b) The roll-off between 100 Hz and 10 kHz is two decades (100× in frequency). Compute the dB change per decade. Recall that a first-order filter rolls off at −20 dB/decade; each additional order adds another −20 dB/decade.
(c) The cutoff is where the roll-off begins (−3 dB point). With a −40 dB/decade slope, the response is −40 dB at 10 kHz. Work backwards: at −3 dB, how far up from −40 dB are you, and how many decades backward does that place \(f_c\)?
Problem 5 — Poles, Zeros, and Gain
A transfer function has: - A zero at \(\omega = 0\) (i.e., a factor of \(j\omega\) in the numerator) - A pole at \(\omega_1 = 500\) rad/s - A pole at \(\omega_2 = 5000\) rad/s - DC gain factor K = 500
(a) What is the slope of the Bode magnitude plot at very low frequencies (\(\omega \ll 500\))?
(b) What is the slope at very high frequencies (\(\omega \gg 5000\))?
(c) What type of filter does this represent overall?
(d) Estimate \(|H|\) in dB at \(\omega = 500\) rad/s.
Hint
(a) At low \(\omega\): the denominator terms approach 1, and the numerator is \(j\omega\). So \(|H| \approx K\omega\) — this is a +20 dB/decade slope. Each zero contributes +20 dB/decade; each pole contributes −20 dB/decade.
(b) At high \(\omega\): count all poles and zeros. One zero: +20 dB/decade. Two poles: −40 dB/decade. Net slope = +20 − 40 = −20 dB/decade.
(c) Gain is low at DC (due to the zero), peaks in the mid-band, and rolls off at high frequency. This is a band-pass response.
(d) At \(\omega = 500\) (the first pole), evaluate the numerator and denominator magnitudes separately, then combine. The zero contributes \(|j\omega| = 500\), the first pole term gives \(|1 + j1| = \sqrt{2}\), and the second pole term \(|1 + j(500/5000)| \approx |1 + j0.1|\).