Chapter 11 Practice Problems
Practice Problems
Problem 1 — Transfer Function of an RC Filter
An RC low-pass filter has R = 10 kΩ and C = 15.9 nF.
(a) Derive the transfer function \(H(j\omega) = V_{out}/V_{in}\) (output taken across C).
(b) Calculate the cutoff frequency \(f_c\) in Hz.
(c) Calculate \(|H|\) and \(\angle H\) at f = \(f_c\), at \(f = f_c/10\), and at \(f = 10f_c\).
(d) Express the magnitude at \(f = 10f_c\) in dB.
Solution
(a) Transfer function (voltage divider with \(Z_C = 1/(j\omega C)\)):
(b) Cutoff frequency:
(c) At \(f = f_c\): \(\omega RC = 1\), so \(H = 1/(1+j) = 0.707\angle -45°\)
At \(f = f_c/10\): \(\omega RC = 0.1\), \(H = 1/(1+j0.1)\), \(|H| = 1/\sqrt{1.01} = 0.995\), \(\angle H = -\arctan(0.1) = -5.7°\)
At \(f = 10f_c\): \(\omega RC = 10\), \(H = 1/(1+j10)\), \(|H| = 1/\sqrt{101} = 0.0995\), \(\angle H = -\arctan(10) = -84.3°\)
(d) At \(f = 10f_c\): \(|H|_{dB} = 20\log_{10}(0.0995) = 20 \times (-1.002) = -20.0\text{ dB}\). Asymptotic approximation predicts −20 dB/decade × 1 decade = −20 dB. ✓
Problem 2 — Bode Plot Sketching
A transfer function is \(H(j\omega) = \dfrac{10}{(1 + j\omega/100)(1 + j\omega/10{,}000)}\).
(a) Identify the DC gain (in dB) and the two pole frequencies.
(b) Describe the asymptotic Bode magnitude plot: slopes in each region.
(c) At f = 100 Hz, f = 1,000 Hz, and f = 10,000 Hz, what does the asymptotic approximation predict for \(|H|_{dB}\)?
(d) What is the phase at \(\omega = \omega_{p2}/10\) (one decade below the second pole)?
Solution
(a) DC gain: at \(\omega = 0\), \(H(0) = 10\), so \(|H|_{dB} = 20\log_{10}(10) = 20\text{ dB}\). Pole frequencies: \(\omega_{p1} = 100\text{ rad/s}\) (\(f_{p1} = 15.9\text{ Hz}\)) and \(\omega_{p2} = 10{,}000\text{ rad/s}\) (\(f_{p2} = 1{,}592\text{ Hz}\)).
(b) Asymptotic regions: - Below 15.9 Hz: flat at 20 dB - 15.9 Hz to 1,592 Hz: −20 dB/decade (one active pole) - Above 1,592 Hz: −40 dB/decade (both poles active)
(c) Asymptotic approximation: - At f = 100 Hz (1 decade above \(f_{p1}\)): \(20 - 20 = 0\text{ dB}\) - At f = 1,000 Hz: \(0 - 20\log(1000/15.9) \approx 0 - 20\times1.8 = -36\text{ dB}\)... More precisely, at \(f_{p2} = 1592\text{ Hz}\): \(20 - 20\log(1592/15.9) = 20 - 40 = -20\text{ dB}\) - At f = 10,000 Hz (1 decade above \(f_{p2}\)): \(-20 - 20 = -40\text{ dB}\)
(d) Phase contribution at \(\omega = \omega_{p2}/10\): each pole contributes approximately −5.7° at one decade below its corner frequency. Total phase ≈ −90° (from first pole, fully past its 45° midpoint by now) − 5.7° (from second pole) ≈ −95.7°.
Problem 3 — High-Pass Filter Design
Design an RC high-pass filter with cutoff frequency \(f_c = 500\) Hz.
(a) Write the transfer function for an RC high-pass filter.
(b) Select C = 100 nF and calculate the required R.
(c) Calculate \(|H|\) and phase at f = 50 Hz, 500 Hz, and 5 kHz.
(d) What is the roll-off rate in the stopband (below \(f_c\))?
Solution
(a) Transfer function (output across R, capacitor in series):
(b) Required R:
Use nearest standard value: 3.3 kΩ (gives \(f_c = 1/(2\pi \times 3300 \times 100\times10^{-9}) = 482\text{ Hz}\)).
(c) At each frequency (using \(f_c = 500\text{ Hz}\) for simplicity):
- f = 50 Hz (0.1\(f_c\)): \(\omega RC = 0.1\), \(|H| = 0.1/\sqrt{1.01} = 0.0995\), \(\angle H = 90° - \arctan(0.1) = 84.3°\)
- f = 500 Hz (\(f_c\)): \(|H| = 1/\sqrt{2} = 0.707\), \(\angle H = +45°\)
- f = 5 kHz (10\(f_c\)): \(\omega RC = 10\), \(|H| = 10/\sqrt{101} = 0.995\), \(\angle H = 5.7°\)
(d) In the stopband below \(f_c\), the high-pass filter rolls off at +20 dB/decade going toward higher frequencies, or equivalently −20 dB/decade going toward lower frequencies. First-order filter, one pole.
Problem 4 — Identifying Filter Type from Transfer Function
A circuit has the transfer function:
(a) What type of filter is this? (Examine behavior at \(\omega \to 0\), \(\omega = \omega_0\), and \(\omega \to \infty\).)
(b) At \(\omega = \omega_0\), what determines the peak magnitude?
(c) How does this compare to a first-order band-pass filter?
Solution
(a) Evaluate at extremes:
- \(\omega \to 0\): numerator \(\to 0\), denominator \(\to 1\): \(H \to 0\) (attenuated)
- \(\omega = \omega_0\): numerator = \(j\), denominator = \(1 + 2j - 1 = 2j\): \(H = j/(2j) = 1/2\) (peak)
- \(\omega \to \infty\): numerator grows as \(\omega\), denominator grows as \(\omega^2\): \(H \to 0\) (attenuated)
This is a second-order band-pass filter centered at \(\omega_0\).
(b) At \(\omega = \omega_0\) the denominator reduces to \(2j\zeta\) (where \(\zeta\) is the damping ratio). Peak magnitude = \(1/(2\zeta) = Q\). Higher Q gives higher peak gain and narrower bandwidth.
(c) A first-order band-pass filter cannot achieve the sharp roll-off of this second-order design. The second-order filter rolls off at −20 dB/decade on each side of the passband versus −20 dB/decade for the first-order RC band-pass.
Problem 5 — Cascaded Filters
Two RC filters are cascaded: Stage 1 is a high-pass filter with \(f_{c1} = 200\) Hz, Stage 2 is a low-pass filter with \(f_{c2} = 8\) kHz.
(a) What is the overall filter type?
(b) What is the passband (approximate frequency range where the response is within −3 dB)?
(c) In dB, what is the combined roll-off rate for frequencies above 80 kHz?
(d) If both stages have unity passband gain, what is the combined gain (in dB) at f = 20 Hz?
Solution
(a) A high-pass filter cascaded with a low-pass filter creates a band-pass filter that passes frequencies between 200 Hz and 8 kHz.
(b) Approximate passband: 200 Hz to 8 kHz. At exactly these frequencies, each stage contributes −3 dB, so the overall response is −6 dB at the edges. The −3 dB passband is slightly narrower than 200 Hz to 8 kHz.
(c) Above 80 kHz (one decade above \(f_{c2} = 8\text{ kHz}\)): Stage 2 (LPF) rolls off at −20 dB/decade. Stage 1 (HPF) is fully in its passband and contributes 0 dB/decade. Combined: −20 dB/decade.
(d) At 20 Hz (one decade below \(f_{c1} = 200\text{ Hz}\)): Stage 1 (HPF) attenuates by −20 dB/decade × 1 decade = −20 dB. Stage 2 (LPF) is fully in its passband. Combined: −20 dB.