Frequency response: Bode plots

Why it matters

Frequency response tells you how a process or loop treats disturbances of different speeds: slow ones pass through, fast ones are filtered, and every element adds phase lag. Bode plots make this visible, handle dead time exactly (which Routh cannot), and are the basis of the Bode and Nyquist stability criteria and of frequency-based controller tuning.

Key ideas

Frequency response. If a stable linear system receives u = A·sin(ωt), then after the transient dies out the output is a sinusoid of the same frequency, with

• amplitude ratio AR = |G(jω)| (output amplitude/input amplitude; often normalised by the steady-state gain K), and
• phase angle φ = ∠G(jω) (negative means the output lags). Both are obtained simply by substituting s = jω in the transfer function. This is the "substitution rule".

Bode diagram. Two plots against log ω:

• log AR (or 20·log₁₀AR in decibels) on a log scale, and
• φ in degrees on a linear scale. For elements in series, ARs multiply (so log ARs, or dB, add) and phase angles add. A complex loop is therefore drawn by adding the plots of simple elements.

Building blocks.

• Gain K: AR = K at all ω; φ = 0 (or −180° if K < 0).
• First-order lag 1/(τs + 1): AR = 1/√(1 + ω²τ²); φ = −tan⁻¹(ωτ). Asymptotes: AR = 1 for ω < 1/τ and slope −1 (−20 dB/decade) above the corner frequency ω_c = 1/τ. At the corner, AR = 0.707 (−3 dB) and φ = −45°; φ goes from 0 to −90°.
• First-order lead (τs + 1): the mirror image, slope +1 and φ from 0 to +90°.
• Integrator 1/s: AR = 1/ω (slope −1 everywhere), φ = −90°.
• Second-order 1/(τ²s² + 2ζτs + 1): high-frequency slope −2 (−40 dB/decade), φ from 0 to −180° (−90° at ωτ = 1). For ζ < 0.707 the AR curve shows a resonant peak above 1.
• Dead time e^(−θs): AR = 1 at all ω; φ = −ωθ (rad), unbounded. It does not appear on the AR plot at all, but it dominates the phase plot at high frequency.
• PI controller K_c(1 + 1/(τ_I s)): AR = K_c·√(1 + 1/(ωτ_I)²), φ = −tan⁻¹(1/(ωτ_I)), from −90° at low ω to 0 at high ω.
• PD controller K_c(1 + τ_D s): AR rises with slope +1 above 1/τ_D; φ adds lead up to +90°.

General patterns. A transfer function with n poles and m zeros (no dead time) has a high-frequency AR slope of −(n − m) on log–log axes and a limiting phase of −90°·(n − m). Only systems whose phase can reach −180° (third order and higher, or any dead time) can be made unstable by proportional gain alone, which links directly to the next topic.

Formulas

AR = |G(j·ω)|, φ = ∠G(j·ω)

• ω angular frequency (rad/s); ω = 2π/P for a period P.

AR_dB = 20·log₁₀(AR)

AR = K/√(1 + ω²·τ²), φ = −tan⁻¹(ω·τ)

• First-order lag; corner frequency ω_c = 1/τ.

AR = 1/√((1 − ω²τ²)² + (2ζωτ)²), φ = −tan⁻¹(2ζωτ/(1 − ω²τ²))

• Second-order lag (φ taken between 0 and −180°).

AR = 1, φ = −ω·θ (rad) = −57.3·ω·θ (degrees)

• Dead time θ (s).

AR_total = AR₁·AR₂·…, φ_total = φ₁ + φ₂ + …

• Elements in series.

Worked examples

Example 1 (standard): two lags. G(s) = 2/((s + 1)(0.5s + 1)), time in s. Find AR (also in dB) and φ at ω = 1 rad/s.

1. Lag 1 (τ = 1 s): AR₁ = 1/√(1 + 1) = 0.7071; φ₁ = −tan⁻¹(1) = −45.0°.
2. Lag 2 (τ = 0.5 s): AR₂ = 1/√(1 + 0.25) = 0.8944; φ₂ = −tan⁻¹(0.5) = −26.57°.
3. AR = 2·0.7071·0.8944 = 1.265; in dB: 20·log₁₀(1.265) = 2.04 dB.
4. φ = −45.0 − 26.57 = −71.57°.

AR ≈ 1.265 (2.04 dB), φ ≈ −71.6° at ω = 1 rad/s

Example 2 (GATE level): first-order lag plus dead time. G(s) = 5·e^(−0.5s)/(2s + 1), time in min. Find the frequency at which the phase is −180°, and the AR there.

1. φ(ω) = −0.5ω (rad) − tan⁻¹(2ω). Set φ = −π: 0.5ω + tan⁻¹(2ω) = π.
2. Solve by trial: at ω = 3.4, 1.700 + 1.4248 = 3.125 (too low); at ω = 3.45, 1.725 + 1.4269 = 3.152 (too high). Interpolating, ω ≈ 3.431 rad/min.
3. Check: 0.5·3.431 = 1.7155 rad (98.3°) and tan⁻¹(6.862) = 1.4261 rad (81.7°); sum 180.0°.
4. AR = 5/√(1 + (2·3.431)²) = 5/√(1 + 47.09) = 5/√48.09 = 5/6.934 = 0.721.

ω ≈ 3.43 rad/min, AR ≈ 0.721. Without dead time, the phase of a single lag never reaches −180°; the dead time supplies the extra 98°.

Example 3 (short): PI controller. For K_c = 2, τ_I = 1 min at ω = 1 rad/min: AR = 2·√(1 + 1) = 2.83, φ = −tan⁻¹(1) = −45°.

Common mistakes

• Using frequency in Hz (cycles per time) where ω in rad per time is needed; ω = 2πf.
• Mixing radians and degrees when adding the dead-time phase −ωθ to the lag phase in degrees.
• Expecting dead time to change the AR plot; it changes only the phase.
• Adding ARs instead of multiplying them (or adding dB values but multiplying phases).
• Using the asymptote value instead of the true value at the corner: the true AR there is 0.707, not 1.

For GATE CH

Expect: AR and phase of a given transfer function at a given frequency, the corner frequency, slopes of asymptotes, and the frequency at which the phase reaches −180° for systems with dead time (by trial). Questions also ask for the amplitude of a sensor reading for a sinusoidal input. Practise trial-and-error solution of phase equations with consistent units.

Quick check

1. What is the corner frequency of 3/(4s + 1) (time in min)?
2. What is the high-frequency slope of 1/((s + 1)(2s + 1)(3s + 1)) on log–log axes?
3. What AR and phase does e^(−2s) give at ω = 0.5 rad/s?
4. Convert an AR of 0.1 to dB.

Answers: 1. 0.25 rad/min. 2. −3 (−60 dB/decade). 3. AR = 1, φ = −1 rad = −57.3°. 4. −20 dB.