Training Control Systems Frequency Response & Root Locus
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Frequency Response & Root Locus

30 min Control Systems

Frequency Response & Root Locus

When specifications are given in the frequency domain — bandwidth, gain margin, phase margin — the Bode plot is the most efficient analysis tool. Plotting $|G(j\omega)|$ and $\angle G(j\omega)$ against $\log\omega$ reveals the loop's ability to reject disturbances and track references at each frequency.

The root locus complements Bode analysis by showing how closed-loop poles move as a single parameter (usually loop gain $K$) is varied. Seeing where the poles travel lets you pick $K$ so the dominant poles sit at the desired $\zeta$ and $\omega_n$.

In this lesson we read key numbers from first- and second-order Bode plots, and sketch the locus for a third-order loop.

Bode Plot Basics

For $G(s) = K/(\tau s + 1)$: corner at $\omega_c = 1/\tau$, slope $-20\,\text{dB/dec}$ beyond, phase going from $0°$ to $-90°$.

For $G(s) = \omega_n^2/(s^2+2\zeta\omega_n s + \omega_n^2)$: resonance peak of height $1/(2\zeta\sqrt{1-\zeta^2})$ near $\omega_n$ for $\zeta < 0.707$; slope $-40\,\text{dB/dec}$ beyond.

Stability Margins

Gain margin: how much loop gain can increase before instability. Phase margin: how much phase lag can be added at crossover before instability. Rules of thumb: $\text{GM} \ge 6\,\text{dB}$, $\text{PM} \ge 45°$.

Example 1 — Corner frequency

$G(s) = 10/(0.05 s + 1)$. Find DC gain and corner frequency.

DC gain $= 10 = 20\,\text{dB}$. Corner $\omega_c = 1/0.05 = 20\,\text{rad/s} \approx 3.18\,\text{Hz}$.

Example 2 — Resonant peak

Second-order $\omega_n = 10, \zeta = 0.1$. Find peak magnitude.

$M_r = 1/(2\zeta\sqrt{1-\zeta^2}) = 1/(0.2\sqrt{0.99}) \approx 5.025$, i.e. about $+14\,\text{dB}$ above the low-frequency asymptote.

Example 3 — Phase margin from approximate crossover

$G(s) = 100/[s(s+10)]$. Estimate crossover frequency and phase margin.

At crossover $|G(j\omega_c)| = 1$. Approximately $100/[\omega_c \sqrt{\omega_c^2 + 100}] = 1$; for $\omega_c$ large compared with $\sqrt{100}=10$, $100/\omega_c^2 \approx 1 \Rightarrow \omega_c \approx 10$.

Phase at $\omega_c = 10$: $-90° - \arctan(10/10) = -90° - 45° = -135°$. Phase margin $= 180° - 135° = 45°$ — acceptable.

Interactive Demo: Bode-Margin Quick Calculator
DC gain (dB) =20.0dB
corner ω_c =20.0rad/s
corner f_c =3.18Hz
phase @ ω_c =-45deg

Practice Problems

1. Define gain margin in one sentence.
2. For $G = K/[s(s+a)]$, what kind of pole shape does the root locus have?
3. What value of phase margin is commonly targeted?
4. What happens to the Bode plot as $\zeta \to 0$?
5. What is the DC gain of $G = 20/(s^2 + 6s + 10)$?
6. State one advantage of frequency-domain design over time-domain design.
Show Answer Key

1. The factor by which loop gain can increase at the phase-crossover frequency before the closed loop becomes unstable.

2. Two branches on the real axis leaving poles at $0$ and $-a$, meeting at $s=-a/2$ and breaking into a vertical line.

3. 45° (sometimes 60° for well-damped designs).

4. The resonant peak grows without bound; at $\zeta=0$ the loop would have an infinite peak at $\omega_n$.

5. $G(0) = 20/10 = 2$.

6. It handles unmodelled high-frequency dynamics and robustness margins more naturally.

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