Chapter 12 Quiz — Filters and Resonance
Multiple Choice Questions
1. An RC low-pass filter has \(R = 10\ \text{k}\Omega\) and \(C = 10\ \text{nF}\). What is the cutoff frequency?
- A) 159 Hz
- B) 1{,}592 Hz
- C) 15{,}920 Hz
- D) 100 kHz
Answer
B) 1,592 Hz
This is the frequency at which the output voltage falls to \(1/\sqrt{2} \approx 0.707\) of the input, corresponding to -3 dB of attenuation.
2. In an RC high-pass filter, compared with an RC low-pass filter using the same R and C values, the cutoff frequency is:
- A) Twice as high
- B) Twice as low
- C) The same
- D) Dependent on the signal amplitude
Answer
C) The same
Both the RC low-pass and RC high-pass filters have the same cutoff frequency \(f_c = 1/(2\pi RC)\). The only difference is which element — C or R — the output is taken across, which reverses the pass and stop bands but does not change the frequency at which \(|H| = 0.707\).
3. A series RLC circuit has \(L = 50\ \text{mH}\) and \(C = 50\ \text{nF}\). What is the resonant frequency?
- A) 318 Hz
- B) 1{,}004 Hz
- C) 3{,}183 Hz
- D) 10{,}066 Hz
Answer
C) 3,183 Hz
4. An RLC band-pass filter has centre frequency \(f_0 = 5\ \text{kHz}\) and \(Q = 25\). What is the bandwidth?
- A) 20 Hz
- B) 200 Hz
- C) 2{,}000 Hz
- D) 125 kHz
Answer
B) 200 Hz
The filter passes frequencies from approximately 4,900 Hz to 5,100 Hz. A higher Q gives a narrower, more selective band.
5. Which of the following is an advantage of active filters over passive filters at audio frequencies?
- A) They require no power supply
- B) They can handle higher voltages
- C) They eliminate the need for large inductors
- D) They produce less noise than passive circuits
Answer
C) They eliminate the need for large inductors
At audio frequencies (20 Hz–20 kHz), achieving a low cutoff with an RL filter would require inductors in the range of henries — physically large, expensive, and susceptible to magnetic interference. Active filters use only resistors, capacitors, and an op-amp, achieving the same frequency-shaping response with compact, inexpensive components. (Active filters do require a power supply and typically add more noise than a purely passive circuit.)
6. A microphone preamplifier must increase an input of −52 dBV to an output of +2 dBV. What is the required voltage gain in dB, and approximately what linear voltage gain does this represent?
- A) 50 dB, \(\approx 316\times\)
- B) 54 dB, \(\approx 501\times\)
- C) 50 dB, \(\approx 100{,}000\times\)
- D) 26 dB, \(\approx 20\times\)
Answer
A) 50 dB, ≈ 316×
Wait — recalculating carefully: \(+2 - (-52) = 54\) dB, and \(A_V = 10^{54/20} = 10^{2.7} \approx 501\).
Correction — the correct answer is B) 54 dB, ≈ 501×.
The two-stage inverting design with each stage providing gain of ≈22 (≈27 dB) achieves this.
7. The dBu audio level reference is based on:
- A) 1 V RMS into any impedance
- B) 1 mW into any impedance
- C) 0.775 V RMS (the voltage delivering 1 mW into 600 Ω)
- D) The thermal noise voltage at room temperature
Answer
C) 0.775 V RMS
0 dBu is defined as 0.775 V RMS because that is the voltage that produces exactly 1 milliwatt of power in a 600 Ω load, which was the original telephone-industry standard termination impedance. The formula is \(\text{dBu} = 20\log_{10}(V_{rms}/0.775)\).
8. A professional audio amplifier has a maximum output of +24 dBu and a nominal operating level of +4 dBu. What is its headroom?
- A) 4 dB
- B) 20 dB
- C) 24 dB
- D) 28 dB
Answer
B) 20 dB
This 20 dB of headroom (a factor of 10× in voltage above nominal) accommodates loud transients — drum strikes, vocal peaks — without clipping.
9. A bass shelving filter in an audio tone control is best described as:
- A) A band-pass filter centred at 100 Hz
- B) A high-pass filter with -20 dB/decade roll-off
- C) A filter that boosts or cuts frequencies below a corner frequency while leaving higher frequencies unaffected
- D) A notch filter that removes 60 Hz hum
Answer
C) A filter that boosts or cuts frequencies below a corner frequency while leaving higher frequencies unaffected
A shelving filter levels off (shelves) to a constant gain at both low and high frequencies, unlike a simple LP filter whose gain continues to decrease above the cutoff. The bass shelf adjusts only the low-frequency region; midrange and treble are unaltered. The treble shelf does the same at high frequencies.
10. For a 16-bit PCM digital audio system, what is the approximate theoretical dynamic range?
- A) 48 dB
- B) 96 dB
- C) 120 dB
- D) 144 dB
Answer
B) 96 dB
The theoretical dynamic range of an \(n\)-bit PCM system is approximately \(6.02n + 1.76\) dB:
This is often rounded to 96 dB (6 dB per bit). CD audio uses 16-bit encoding, giving approximately 96 dB of dynamic range — enough for quiet whispers to loud orchestral fortissimos.
Practice Problems
Problem 1: RC Low-Pass Filter Design
Design an RC low-pass filter with a cutoff frequency of \(f_c = 3.4\ \text{kHz}\) (the upper edge of the telephone voice band). Use \(C = 10\ \text{nF}\). Find the required R, select the nearest E24 standard value, and calculate the actual cutoff frequency with the standard value.
Solution
Step 1 — Ideal R:
Step 2 — Nearest E24 standard value:
E24 values near 4.68 kΩ: 4.7 kΩ (the E24 series includes 4.7 kΩ as a standard value).
Choose \(R = 4.7\ \text{k}\Omega\).
Step 3 — Actual cutoff frequency:
Result: The actual cutoff is 3.39 kHz — 0.4% below the 3.4 kHz target. Excellent agreement; no adjustment needed.
Problem 2: RLC Band-Pass Filter Design
Design a series RLC band-pass filter with centre frequency \(f_0 = 455\ \text{kHz}\) (a common AM radio intermediate frequency) and bandwidth \(BW = 10\ \text{kHz}\). Calculate Q, the LC product, and suitable component values using \(L = 220\ \mu\text{H}\).
Solution
Step 1 — Quality factor:
Step 2 — LC product:
Step 3 — Capacitor (given \(L = 220\ \mu\text{H}\)):
Nearest E12 value: 560 pF.
Step 4 — Resistor:
Nearest E24 value: 13 Ω or 15 Ω. Use 13 Ω (gives slightly wider bandwidth for better AM reception).
Verification:
Very close to the 455 kHz target. Actual BW with 13 Ω: \(R/(2\pi L) = 13/(2\pi \times 220 \times 10^{-6}) \approx 9.4\ \text{kHz}\).
Problem 3: Decibel Level Calculations
A recording studio chain has the following stages: microphone output = −55 dBu; preamplifier gain = +48 dB; equaliser gain = +3 dB; power amplifier gain = +26 dB. (a) What is the signal level at the preamplifier output? (b) After the equaliser? (c) At the power amplifier output? (d) If the power amplifier clips at +30 dBu, does any stage clip?
Solution
Calculating levels stage by stage (adding gains in dB):
(a) Preamplifier output:
(b) After equaliser:
(c) Power amplifier output:
(d) Clipping check:
The power amplifier clips at +30 dBu. Its output is +22 dBu — 8 dB below clipping. No stage clips.
Headroom at power amplifier output: \(30 - 22 = 8\ \text{dB}\). This is below the professional standard of 20 dB; if a loud transient arrives 8 dB above the nominal level, the amplifier will clip. The system designer might reduce the equaliser gain or preamplifier gain by ~12 dB to restore adequate headroom.
Problem 4: Active Filter Gain and Cutoff Design
Design a first-order inverting active low-pass filter with DC gain = +20 dB (gain magnitude = 10) and cutoff frequency \(f_c = 800\ \text{Hz}\). Choose \(C_f = 22\ \text{nF}\) and determine \(R_f\), then determine \(R_i\). Select E24 standard values for all resistors.
Solution
Step 1 — Recall the transfer function:
DC gain magnitude: \(|A_{DC}| = R_f / R_i = 10\)
Cutoff frequency: \(f_c = 1/(2\pi R_f C_f)\)
Step 2 — Find \(R_f\) from \(f_c\):
Nearest E24 value: \(R_f = 9.1\ \text{k}\Omega\).
Step 3 — Find \(R_i\):
E24 value: \(R_i = 910\ \Omega\) (910 Ω is a standard E24 value). ✓
Step 4 — Verify actual cutoff:
Result: DC gain = 9,100/910 = 10× (+20 dB) ✓. Actual cutoff = 795 Hz (0.6% below 800 Hz target) ✓. Both specifications are met with standard E24 components.