Analog Intuition

Part 5 · Control Loops

Phase margin at \(f_c\)

Key point: phase margin is the phase headroom at the 0 dB crossing of \(T\). Bandwidth \(f_c\) is that crossing frequency. You read both on the same Bode of loop gain.

Loop-gain Bode · read at 0 dB

Type II \(T(s)\) · plant one pole · \(K\) and \(f_z\) solved to hit PM at \(f_c\)

PM → —
∠T at fc → —

PM = 180° + ∠T. Read both numbers at the purple 0 dB crossing.

Block diagram: Vref into summing junction, through Compensator C(s) and Plant P(s) to Vout, with Feedback F(s) returning to the summer
Loop gain \(T=P\,C\,F\). Read PM on \(T(j\omega)\) at 0 dB, not on \(T_{\mathrm{cl}}\).

Bode of T · 10 Hz–1 MHz

|T| (dB) Phase 0 dB fc −180°

2nd-order peek · not the plant

Step Target = 1 Load-step → next page
Dig deeper — why \(180° + \angle T\)

Negative feedback already spends 180° going around the loop (the minus at Σ). If the rest of \(T\) spends another 180° at the frequency where \(|T|=1\), the returned signal reinforces the error. Phase margin is the leftover headroom at that one frequency:

\[ \mathrm{PM} = 180° + \angle T(j\omega_c) \qquad |T(j\omega_c)| = 1 \]

The lab is a one-pole plant plus a Type II compensator \(C(s)=(1+s/\omega_z)/[s(1+s/\omega_c)]\). A zero below \(f_c\) buys phase; a pole above takes some back. \(K\) is solved so the 0 dB crossing sits at the bandwidth you set. A 120° preset asks for more phase than this model can give (~93°) — that ceiling is the lesson.

The thumbnail is a 2nd-order stand-in from \(\zeta\approx\mathrm{PM}/100\). The next page puts the same PM and \(f_c\) on a load step. Closed-loop \(T_{\mathrm{cl}}=L/(1+LF)\) was Part 4 — we still design by shaping \(T\).