Short notes

Bode Plot Short Notes

Hand over a transfer function G(jω) = 10/[jω(1+jω/5)(1+jω/20)] to a control engineer and within three minutes they'll have the asymptotic Bode plot on graph paper: a −20 dB/decade line below ω=5, an additional −20 dB/decade slope (total −40) after ω=5, and another −20 (total −60) after ω=20, with a low-frequency magnitude of 20 log(10) = 20 dB normalised to ω=1. That straight-line approximation technique is the entire Bode plot construction procedure.

EEE, ECE

How it works

Magnitude plot: each factor contributes a slope change at its corner frequency. A gain K contributes 20 log K dB (horizontal line). A pole at ω_c contributes a −20 dB/decade slope break at ω = ω_c; a zero contributes +20 dB/decade. An integrator 1/s contributes −20 dB/decade across all frequencies. Phase plot: each first-order pole contributes −45° at its corner frequency, ranging from 0° (one decade below) to −90° (one decade above). The actual magnitude error at the corner frequency is ±3 dB for a first-order factor.

Key points to remember

Gain margin (GM) = −20 log|G(jω_pc)| in dB, where ω_pc is the phase crossover frequency (where phase = −180°). Phase margin (PM) = 180° + ∠G(jω_gc), where ω_gc is the gain crossover frequency (where |G| = 0 dB). For a stable system, GM > 0 dB and PM > 0°. Typically PM of 30°–60° is desired; below 30° the transient response is highly oscillatory. A second-order system with ζ = 0.5 gives PM ≈ 52°. The Bode gain formula for Type 1 system: the −20 dB/decade line extended to ω = 1 gives 20 log K_v.

Exam tip

Every Anna University paper asks you to draw the Bode magnitude and phase plots for a given G(jω) and find gain and phase margins — tabulate the corner frequencies and slope contributions before drawing, because a missed corner frequency ruins the entire plot.

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