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Quiz: Frequency Response and Bode Plots

Test your understanding of sinusoidal steady-state analysis and Bode plot construction.

1. When a sinusoidal input is applied to a stable LTI system, the steady-state output is:

  1. A constant DC value
  2. A sinusoid at the same frequency with possibly different amplitude and phase
  3. A sinusoid at double the input frequency
  4. An exponentially decaying signal
Show Answer

The correct answer is B. For a stable LTI system, the steady-state response to a sinusoidal input is a sinusoid at the same frequency, but with amplitude scaled by $|G(j\omega)|$ and phase shifted by $\angle G(j\omega)$. This is the basis of frequency response analysis.

Concept Tested: Sinusoidal Steady State, Frequency Response

2. The frequency response $G(j\omega)$ is obtained from the transfer function by:

  1. Setting $s = 0$
  2. Setting $s = j\omega$
  3. Taking the derivative with respect to $s$
  4. Finding the poles and zeros
Show Answer

The correct answer is B. The frequency response is found by substituting $s = j\omega$ into the transfer function: $G(j\omega) = G(s)|_{s=j\omega}$. This gives a complex function of frequency whose magnitude and phase characterize the system's sinusoidal response.

Concept Tested: Substitution s=jw, Frequency Transfer Func

3. In a Bode magnitude plot, the vertical axis is typically scaled in:

  1. Linear amplitude units
  2. Radians
  3. Decibels (dB)
  4. Seconds
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The correct answer is C. Bode magnitude plots use decibels: $|G|{dB} = 20\log|G|$. This logarithmic scale converts multiplication into addition, making it easy to combine gains of cascaded systems and to show wide dynamic ranges.

Concept Tested: Decibel, Bode Magnitude Plot

4. The corner frequency (break frequency) of a first-order system $G(s) = 1/(\tau s + 1)$ is:

  1. $\omega_c = \tau$
  2. $\omega_c = 1/\tau$
  3. $\omega_c = 2\pi\tau$
  4. $\omega_c = \tau^2$
Show Answer

The correct answer is B. The corner frequency is $\omega_c = 1/\tau$ rad/s, where the magnitude is 3 dB below the low-frequency asymptote and the phase is -45°. This is where the asymptotic approximation transitions from flat (0 dB/decade slope) to -20 dB/decade.

Concept Tested: Corner Frequency, First-Order Factor

5. A decade represents a frequency ratio of:

  1. 2:1
  2. 10:1
  3. 100:1
  4. e:1 (approximately 2.718:1)
Show Answer

The correct answer is B. A decade is a 10:1 frequency ratio (e.g., 1 rad/s to 10 rad/s, or 100 Hz to 1000 Hz). On a logarithmic frequency axis, decades appear as equal intervals. An octave is a 2:1 ratio.

Concept Tested: Decade

6. The Bode magnitude plot of a pure integrator ($G(s) = 1/s$) has a slope of:

  1. 0 dB/decade (flat)
  2. -20 dB/decade
  3. +20 dB/decade
  4. -40 dB/decade
Show Answer

The correct answer is B. A pure integrator has magnitude $|1/j\omega| = 1/\omega$, which decreases by a factor of 10 (or -20 dB) for each decade increase in frequency. The phase is constant at -90°.

Concept Tested: Integrator Bode Plot

7. Bandwidth of a system is typically defined as the frequency where the magnitude drops to:

  1. 0 dB
  2. -3 dB (half-power point)
  3. -20 dB
  4. -90 dB
Show Answer

The correct answer is B. Bandwidth is the frequency range where the gain is within 3 dB of its maximum (low-frequency) value. At -3 dB, the power is half of maximum (hence "half-power point"), and the amplitude is $1/\sqrt{2} \approx 0.707$ of maximum.

Concept Tested: Bandwidth, Half-Power Point, Cutoff Frequency

8. The resonant peak in a second-order system's frequency response increases as:

  1. Damping ratio increases toward 1
  2. Damping ratio decreases toward 0
  3. Natural frequency decreases
  4. The DC gain decreases
Show Answer

The correct answer is B. The resonant peak magnitude $M_r = 1/(2\zeta\sqrt{1-\zeta^2})$ increases as damping ratio decreases. For $\zeta < 0.707$, there is a resonant peak above unity gain. At $\zeta = 0$, the peak would be infinite (undamped resonance).

Concept Tested: Resonant Peak, Damping Ratio

9. A low-pass filter allows:

  1. Only high frequencies to pass
  2. Frequencies below the cutoff to pass while attenuating higher frequencies
  3. A narrow band of frequencies to pass
  4. Only frequencies that match the resonant frequency
Show Answer

The correct answer is B. A low-pass filter passes low frequencies with little attenuation while progressively attenuating higher frequencies above the cutoff. A first-order RC circuit is a classic low-pass filter with -20 dB/decade rolloff.

Concept Tested: Low-Pass System

10. In constructing a Bode plot for a transfer function, the asymptotic magnitude contributions from individual factors:

  1. Multiply together
  2. Add together (in dB)
  3. Average out
  4. Cancel each other
Show Answer

The correct answer is B. In decibels, multiplication becomes addition: $|G_1 \cdot G_2|{dB} = |G_1|$. This allows building complex Bode plots by graphically adding the contributions of individual first-order, second-order, integrator, and constant terms.} + |G_2|_{dB

Concept Tested: Bode Plot Construction, Asymptotic Approximation