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Chapter 12 Quiz — Filters and Resonance

Test your understanding of filter design, cutoff frequencies, quality factors, and active versus passive filter tradeoffs. Each question includes a detailed explanation.

Question 1

An RC low-pass filter is required with cutoff frequency 2 kHz. If a 22 nF capacitor is available, what resistor value should be used?

  1. \(R \approx 1.44 \text{ kΩ}\)
  2. \(R \approx 3.62 \text{ kΩ}\)
  3. \(R \approx 7.23 \text{ kΩ}\)
  4. \(R \approx 723 \text{ Ω}\)
Show Answer

Correct Answer: B

Solving the cutoff frequency formula for R: \[R = \frac{1}{2\pi f_c C} = \frac{1}{2\pi \times 2000 \times 22 \times 10^{-9}} = \frac{1}{2.765 \times 10^{-4}} \approx 3617 \text{ Ω} \approx 3.62 \text{ kΩ}\] The nearest standard E24 value is 3.6 kΩ, giving an actual cutoff of approximately 2.01 kHz — acceptably close.

Concept Tested: RC Filter Design, \(f_c = 1/(2\pi RC)\)

Question 2

A series RLC band-pass filter has L = 10 mH and C = 2.53 μF. What is its resonant center frequency?

  1. \(f_0 = 100\) Hz
  2. \(f_0 = 1000\) Hz
  3. \(f_0 = 10\) kHz
  4. \(f_0 = 159\) Hz
Show Answer

Correct Answer: B

The resonant frequency is \(f_0 = 1/(2\pi\sqrt{LC})\): \[f_0 = \frac{1}{2\pi\sqrt{10 \times 10^{-3} \times 2.53 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{25.3 \times 10^{-9}}} = \frac{1}{2\pi \times 1.59 \times 10^{-4}} \approx 1000 \text{ Hz}\] At this frequency the inductive and capacitive reactances cancel, producing maximum current through the load resistor.

Concept Tested: Resonant Frequency of RLC Band-Pass Filter

Question 3

A band-pass filter centered at 5 kHz has a quality factor Q = 10. What is its 3 dB bandwidth?

  1. 50 kHz
  2. 5 kHz
  3. 500 Hz
  4. 50 Hz
Show Answer

Correct Answer: C

Bandwidth is related to Q and center frequency by: \[BW = \frac{f_0}{Q} = \frac{5000}{10} = 500 \text{ Hz}\] The filter passes frequencies from approximately 4750 Hz to 5250 Hz. A higher Q value would produce a narrower, more selective band; a lower Q would give a wider, less selective response.

Concept Tested: Quality Factor and Bandwidth, \(BW = f_0/Q\)

Question 4

Which of the following is a key advantage of active filters over passive LC filters for audio frequency applications?

  1. Active filters can handle higher power and voltage than passive filters
  2. Active filters work at higher frequencies than passive filters
  3. Active filters can provide gain and avoid large inductors
  4. Active filters are simpler and require fewer components
Show Answer

Correct Answer: C

Active filters using op-amps with RC networks can provide gain (amplification while filtering) and completely eliminate inductors. At audio frequencies (20 Hz–20 kHz), the required inductors would be extremely large and expensive (often several henries). Active filters also allow easy cascading without loading effects, since the op-amp provides high input and low output impedance. However, they require a power supply, which passive filters do not.

Concept Tested: Active vs. Passive Filters

Question 5

An RC high-pass filter uses the same component values as a low-pass filter (same R and C), but the output is taken across the resistor instead of the capacitor. How does the cutoff frequency of the high-pass filter compare to that of the low-pass filter?

  1. The high-pass cutoff frequency is half that of the low-pass
  2. The high-pass cutoff frequency is double that of the low-pass
  3. They have the same cutoff frequency
  4. The high-pass cutoff frequency depends on the signal amplitude
Show Answer

Correct Answer: C

Both the RC low-pass (output across C) and the RC high-pass (output across R) have the same cutoff frequency formula: \(f_c = 1/(2\pi RC)\). The difference is in what happens at this frequency: the low-pass has −3 dB with output rolling off above \(f_c\), while the high-pass has −3 dB with output rolling off below \(f_c\). The same RC time constant defines the cutoff in both cases.

Concept Tested: RC Low-Pass vs. High-Pass Filter Cutoff

Question 6

Professional line level in audio equipment is specified as +4 dBu. What RMS voltage does this correspond to? (Recall: 0 dBu = 0.775 V RMS)

  1. 0.775 V RMS
  2. 1.0 V RMS
  3. 1.23 V RMS
  4. 3.1 V RMS
Show Answer

Correct Answer: C

Using the dBu formula \(V = 0.775 \times 10^{dBu/20}\): \[V = 0.775 \times 10^{4/20} = 0.775 \times 10^{0.2} = 0.775 \times 1.585 = 1.229 \text{ V RMS}\] The +4 dBu standard (approximately 1.23 V RMS) is the standard operating level for professional audio equipment, providing adequate headroom above the noise floor.

Concept Tested: dBu Reference Level in Audio

Question 7

A second-order low-pass filter has quality factor Q = 0.707. What does this Q value signify about the filter's passband response?

  1. The filter has a pronounced resonant peak in the passband
  2. The filter has a maximally flat (Butterworth) passband with no peak
  3. The filter is overdamped with a gentle, gradual rolloff
  4. The filter oscillates indefinitely at the cutoff frequency
Show Answer

Correct Answer: B

A quality factor of Q = 0.707 (corresponding to damping ratio ζ = 1/√2 ≈ 0.707) gives a Butterworth response — the flattest possible passband with no peaking before the cutoff frequency. Higher Q values (Q > 0.707) produce a resonant peak before rolloff; lower Q values (Q < 0.5) are overdamped and roll off earlier and more gently. Butterworth is the most common general-purpose design choice.

Concept Tested: Quality Factor and Second-Order Filter Response

Question 8

In a typical audio system, a microphone outputs −50 dBV. How much gain (in dB) is needed to reach the consumer line level of −10 dBV?

  1. 10 dB
  2. 50 dB
  3. 40 dB
  4. 60 dB
Show Answer

Correct Answer: C

Gain in dB is simply the difference between output and input levels (since dBV is already on a logarithmic scale): \[\text{Gain (dB)} = -10 \text{ dBV} - (-50 \text{ dBV}) = -10 + 50 = 40 \text{ dB}\] This corresponds to a voltage gain of \(10^{40/20} = 100\). The microphone preamplifier must amplify the signal 100 times to reach consumer line level.

Concept Tested: dBV Reference Level and Gain Calculations

Answers Summary

Question Answer Concept Tested
1 B RC Filter Design
2 B Resonant Frequency of RLC Filter
3 C Quality Factor and Bandwidth
4 C Active vs. Passive Filters
5 C RC High-Pass vs. Low-Pass Cutoff
6 C dBu Reference Level
7 B Q Factor and Second-Order Filter Response
8 C dBV and Gain Calculations