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Chapter 11 Quiz — Frequency Response and Bode Plots

Test your understanding of transfer functions, Bode magnitude and phase plots, cutoff frequencies, and filter characteristics. Each question includes a detailed explanation.

Question 1

A circuit has transfer function \(H(j\omega) = V_{out}/V_{in} = 5\angle -30°\). If the input is \(v_{in}(t) = 2\cos(\omega t)\) V, what is the output?

  1. \(v_{out}(t) = 10\cos(\omega t - 30°)\) V
  2. \(v_{out}(t) = 5\cos(\omega t - 30°)\) V
  3. \(v_{out}(t) = 2\cos(\omega t - 30°)\) V
  4. \(v_{out}(t) = 10\cos(\omega t + 30°)\) V
Show Answer

Correct Answer: A

The transfer function multiplies the input phasor. The input phasor is \(\mathbf{V}_{in} = 2\angle 0°\). Applying the transfer function: \[\mathbf{V}_{out} = H \times \mathbf{V}_{in} = 5\angle -30° \times 2\angle 0° = 10\angle -30° \text{ V}\] Converting back to time domain: \(v_{out}(t) = 10\cos(\omega t - 30°)\) V. The magnitude scales by 5 and the phase shifts by −30°.

Concept Tested: Transfer Function \(H(j\omega)\)

Question 2

A signal has a magnitude of 0.01 relative to a reference. What is this expressed in decibels?

  1. −20 dB
  2. −10 dB
  3. −40 dB
  4. −30 dB
Show Answer

Correct Answer: C

Using the voltage gain decibel formula: \(|H|_{dB} = 20\log_{10}|H|\). For \(|H| = 0.01\): \[|H|_{dB} = 20\log_{10}(0.01) = 20 \times (-2) = -40 \text{ dB}\] Each factor of 10 in voltage magnitude corresponds to 20 dB. A factor of 1/100 (= 0.01) is −40 dB, representing 100× attenuation.

Concept Tested: Decibel Gain

Question 3

An RC low-pass filter has R = 15 kΩ and C = 10 nF. What is its cutoff frequency?

  1. \(f_c = 1061\) Hz
  2. \(f_c = 1592\) Hz
  3. \(f_c = 6667\) Hz
  4. \(f_c = 150\) Hz
Show Answer

Correct Answer: A

The cutoff frequency of an RC filter is \(f_c = 1/(2\pi RC)\): \[f_c = \frac{1}{2\pi \times 15000 \times 10 \times 10^{-9}} = \frac{1}{2\pi \times 150 \times 10^{-6}} = \frac{1}{9.42 \times 10^{-4}} \approx 1061 \text{ Hz}\] At this frequency, the output is 70.7% of the input (−3 dB), and the phase shift is −45°.

Concept Tested: Cutoff Frequency \(f_c = 1/(2\pi RC)\)

Question 4

A Bode magnitude plot shows the gain as 0 dB for frequencies below 500 Hz, then rolling off at −20 dB/decade above 500 Hz. What type of filter is this and what is the gain at 5000 Hz?

  1. High-pass filter; gain = −20 dB at 5000 Hz
  2. Low-pass filter; gain = −20 dB at 5000 Hz
  3. Low-pass filter; gain = −3 dB at 5000 Hz
  4. Band-pass filter; gain = −20 dB at 5000 Hz
Show Answer

Correct Answer: B

A filter that passes low frequencies (0 dB passband) and attenuates high frequencies is a low-pass filter. At 5000 Hz, which is exactly one decade (10×) above the cutoff of 500 Hz, the asymptotic approximation gives a gain of 0 − 20 dB = −20 dB. The −20 dB/decade slope means each factor-of-10 increase in frequency adds −20 dB of attenuation.

Concept Tested: Bode Magnitude Plot and Roll-Off Rate

Question 5

At the cutoff (−3 dB) frequency of a first-order low-pass filter, what is the phase shift of the output relative to the input?

  1. 0°
  2. −90°
  3. −45°
  4. −180°
Show Answer

Correct Answer: C

For a first-order RC low-pass filter \(H(j\omega) = 1/(1 + j\omega/\omega_c)\), the phase is \(\phi = -\tan^{-1}(\omega/\omega_c)\). At \(\omega = \omega_c\) (the cutoff frequency): \[\phi = -\tan^{-1}(1) = -45°\] The phase ranges from 0° (far below cutoff) to −90° (far above cutoff), passing through exactly −45° at the cutoff frequency.

Concept Tested: Bode Phase Plot at Cutoff Frequency

Question 6

A filter has two poles. What roll-off rate does it produce above its cutoff frequency?

  1. −20 dB/decade
  2. −40 dB/decade
  3. −60 dB/decade
  4. −10 dB/decade
Show Answer

Correct Answer: B

Each pole contributes −20 dB/decade of roll-off. A second-order filter (two poles) therefore produces: \[2 \times (-20 \text{ dB/decade}) = -40 \text{ dB/decade}\] This is also expressed as −12 dB/octave. Second-order filters attenuate stopband frequencies much more sharply than first-order filters. In general, an nth-order filter rolls off at −20n dB/decade.

Concept Tested: Poles and Roll-Off Rate

Question 7

Which filter type is best suited for removing 60 Hz power line hum from an audio signal while preserving all other frequencies?

  1. Low-pass filter with \(f_c = 60\) Hz
  2. High-pass filter with \(f_c = 60\) Hz
  3. Band-pass filter centered at 60 Hz
  4. Band-reject (notch) filter centered at 60 Hz
Show Answer

Correct Answer: D

A band-reject filter (notch filter) attenuates a narrow band of frequencies centered at the target frequency while passing all others with minimal effect. This is exactly what is needed: block 60 Hz hum while preserving the audio above and below it. A low-pass or high-pass filter would remove large portions of the desired audio signal, and a band-pass filter would discard everything except a narrow band around 60 Hz.

Concept Tested: Band-Reject (Notch) Filter Applications

Question 8

A frequency changes from 200 Hz to 2000 Hz. How many decades and how many octaves does this represent?

  1. 1 decade and approximately 3.32 octaves
  2. 2 decades and 6 octaves
  3. 1 decade and 2 octaves
  4. 0.5 decades and 1.66 octaves
Show Answer

Correct Answer: A

A decade is a factor of 10, and an octave is a factor of 2. From 200 Hz to 2000 Hz is exactly a factor of 10, so it is 1 decade. In octaves: \(\log_2(10) = 3.32\) octaves. (Each decade equals approximately 3.32 octaves.) These conversions are fundamental to reading and constructing Bode plots.

Concept Tested: Decades and Octaves

Answers Summary

Question Answer Concept Tested
1 A Transfer Function
2 C Decibel Gain
3 A Cutoff Frequency
4 B Bode Magnitude Plot and Roll-Off Rate
5 C Bode Phase Plot at Cutoff Frequency
6 B Poles and Roll-Off Rate
7 D Band-Reject Filter Applications
8 A Decades and Octaves