Questions: Bode Plot Construction

Question 1

The transfer function G(s) = 10/((s+1)(s+10)) has two real poles. On the asymptotic Bode magnitude plot, what happens at ω = 10 rad/s?

• A. The slope increases by +20 dB/decade because a zero appears at ω = 10
• B. The slope decreases by an additional −20 dB/decade, giving a total slope of −40 dB/decade beyond this point
• C. The total slope changes to −40 dB/decade starting at ω = 1 rad/s, immediately after the first break
• D. A slope break occurs at ω = 100 rad/s, one decade after the first pole

Answer: (B)

Each real pole at s = −a contributes a −20 dB/decade slope break at its corner frequency ω = a. The pole at s = −1 breaks the slope at ω = 1 rad/s (from 0 to −20 dB/decade); the pole at s = −10 breaks it again at ω = 10 rad/s (adding another −20 dB/decade, for a total of −40 dB/decade). Corner frequencies are at ω = 1/τ = the magnitude of the pole location, not related to each other by a decade. Option C is wrong because the second break occurs at the second pole's corner frequency, not at the first.

Question 2

Why is it essential to write a transfer function in time-constant form G(s) = K·∏(τᵢs+1)/∏(τⱼs+1) before constructing the Bode plot?

• A. Time-constant form makes numerical polynomial factoring easier
• B. Isolating the gain K is necessary because it sets the vertical position of the magnitude plot; different factorizations change both K and the corner frequencies
• C. Time-constant form is required only for the phase plot, not for the magnitude plot
• D. Corner frequencies 1/τᵢ can only be identified after expanding the polynomial into time-constant form

Answer: (B)

The DC gain K must be isolated first because it shifts the entire magnitude curve vertically by 20 log₁₀|K| dB. If you factor differently — for example, pulling out the high-frequency gain instead — you get a different K and different corner frequencies, and your plot will be wrong at all frequencies. For G(s) = 10/((s+1)(s+10)), time-constant form is G(s) = (10/10)·1/((s+1)(s/10+1)) = 1/((s+1)(s/10+1)), with K=1, corners at ω=1 and ω=10. Without this step, DC gain errors propagate through the entire plot.

Question 3

The asymptotic Bode magnitude approximation exactly equals the true magnitude of the transfer function at a real-pole corner frequency.

• T. True
• F. False

Answer: False

At a real-pole corner frequency, the true magnitude is exactly −3 dB below the asymptotic approximation, not equal to it. At ω = 1/τ for a first-order pole, |1/(jω τ + 1)| = 1/√2, which is 20 log₁₀(1/√2) = −3.01 dB. The asymptote assumes 0 dB up to the corner and −20 dB/decade slope after, but in reality the transition is smooth. The maximum error between the asymptote and the true curve is 3 dB, occurring exactly at the corner frequency. For underdamped complex poles, the error can be far larger — the asymptote completely misses the resonant peak.

Question 4

In a Bode plot, the total magnitude (in dB) of a cascaded system G(s) = G₁(s)·G₂(s) equals the sum of the individual magnitude plots of G₁ and G₂ measured in dB.

• T. True
• F. False

Answer: True

This is the key property that makes Bode plot construction tractable. Because dB uses a logarithmic scale: 20 log₁₀|G₁(jω)·G₂(jω)| = 20 log₁₀|G₁(jω)| + 20 log₁₀|G₂(jω)|. Multiplication of transfer functions in the frequency domain becomes addition of their dB magnitudes — and graphical addition is much easier than point-by-point multiplication of complex numbers. The same additivity applies to phase in degrees: ∠(G₁·G₂) = ∠G₁ + ∠G₂. Without logarithms (using linear magnitude), Bode plots would require complex multiplication at every frequency.

Question 5

Explain the key insight that makes Bode plot construction work by graphical addition of component contributions. Then describe what a single real pole at s = −a contributes to both the magnitude and phase plots.

Model answer: The key insight is that the decibel scale converts multiplication into addition: 20 log|G₁G₂| = 20 log|G₁| + 20 log|G₂|. This means each factor in the transfer function contributes independently to the total dB magnitude, allowing graphical superposition. A real pole at s = −a (corner frequency ω = a): contributes 0 dB/decade slope for ω ≪ a, then −20 dB/decade for ω ≫ a, with a −3 dB error at ω = a. Phase transitions from 0° to −90°, spread over two decades centered on ω = a.

The phase contribution of a single first-order pole is: 0° at ω = 0.1a, −45° at ω = a, and −90° at ω = 10a. This two-decade spread means closely spaced poles and zeros have overlapping phase contributions that must be added carefully. The asymptotic phase approximation uses a linear approximation: 0° for ω < 0.1a, linear decrease to −90° at ω = 10a. The magnitude asymptote is simpler: flat, then break at ω = a, then −20 dB/decade. The composite plot for any transfer function is the sum of all individual factor contributions — this is why Bode plots are an engineer's primary tool for frequency-domain analysis.