Bode & margins formulas

Frequency response: Bode and polar plots, gain and phase margins. 23 formulas for GATE EC, EE, IN Control Systems, each with when to use it. Each is explained in the topic notes.

All 23 Bode & margins formulas

SINUSOIDAL STEADY-STATE RESPONSE

u (t) = A sin ω t \Rightarrow y_{ss} (t) = A ∣ G (j ω) ∣ sin (ω t + ∠ G (j ω))

Valid only for a stable $G (s)$ , after the transients decay. The frequency does not change, only the amplitude and phase.

TIME-CONSTANT FORM: MAGNITUDES ADD IN DB, PHASES ADD

G (s) ∣ G (j ω) ∣_{dB} ∠ G (j ω) = \frac{K \prod _{i} ( 1 + s T _{z i} )}{s ^{N} \prod _{k} ( 1 + s T _{p k} )} = 20 lo g_{10} K + i \sum 20 lo g_{10} 1 + ω^{2} T_{z i}^{2} - 20 N lo g_{10} ω - k \sum 20 lo g_{10} 1 + ω^{2} T_{p k}^{2} = i \sum tan^{- 1} (ω T_{z i}) - 9 0^{\circ} N - k \sum tan^{- 1} (ω T_{p k})

For $K > 0$ and real poles and zeros in the left half plane ( $T_{z i}, T_{p k} > 0$ ). A quadratic factor adds its own magnitude and phase terms (see the Bode table below). $N$ is the system type, the number of poles at the origin. This additive structure is why Bode plots are built factor by factor.

VALUE AT THE NATURAL FREQUENCY

u = 1 : ∣ T (j ω_{n}) ∣ = \frac{1}{2 ζ}, ∠ T (j ω_{n}) = - 9 0^{\circ}

Holds for every $ζ > 0$ . In dB the value is $- 20 lo g_{10} (2 ζ)$ , which is 0 dB exactly when $ζ = 0.5$ .

RESONANT FREQUENCY

ω_{r} = ω_{n} 1 - 2 ζ^{2}

Only for $0 < ζ < 1/ 2 \approx 0.707$ . In that range $ω_{r} < ω_{d} < ω_{n}$ , where $ω_{d} = ω_{n} 1 - ζ^{2}$ .

RESONANT PEAK

M_{r} = \frac{1}{2 ζ 1 - ζ ^{2}}

Only for $0 < ζ < 0.707$ . For $ζ \geq 0.707$ there is no peak, $∣ T ∣$ falls monotonically, and $M_{r} = 1$ at $ω = 0$ . $M_{r}$ depends on $ζ$ alone.

BANDWIDTH OF THE STANDARD SECOND-ORDER SYSTEM

ω_{B} = ω_{n} (1 - 2 ζ^{2}) + 4 ζ^{4} - 4 ζ^{2} + 2

The frequency where $∣ T ∣ = 1/ 2$ (−3 dB relative to $∣ T (0) ∣ = 1$ ). Valid for every $ζ > 0$ . $ω_{B} = ω_{n}$ exactly at $ζ = 1/ 2$ and $ω_{B} = 0.644 ω_{n}$ at $ζ = 1$ .

BANDWIDTH OF A FIRST-ORDER SYSTEM

T (s) = \frac{1}{1 + s τ} \Rightarrow ω_{B} = \frac{1}{τ}

The bandwidth equals the corner frequency. A smaller time constant gives a wider bandwidth and a faster response.

PHASE MARGIN OF THE STANDARD SECOND-ORDER LOOP

ω_{g c} = ω_{n} 1 + 4 ζ^{4} - 2 ζ^{2}, PM = tan^{- 1} \frac{2 ζ}{1 + 4 ζ ^{4} - 2 ζ ^{2}}

For open loop $G (s) = ω_{n}^{2} / [s (s + 2 ζ ω_{n})]$ with unity negative feedback. GM is infinite because the phase never reaches $- 18 0^{\circ}$ . Rough approximation: PM in degrees $\approx 100 ζ$ for $0 < ζ \leq 0.6$ , off by at most about 3.5°. Use the exact formula when the options are close.

DECIBELS, DECADES AND SLOPES

M_{dB} = 20 lo g_{10} ∣ G (j ω) ∣, M (ω_{2}) = M (ω_{1}) + S lo g_{10} \frac{ω _{2}}{ω _{1}}

$S$ is the slope in dB/decade, and the relation holds along one straight segment. A decade is a factor of 10 and an octave a factor of 2, so 20 dB/decade $\approx$ 6.02 dB/octave.

LOW-FREQUENCY ASYMPTOTE AND ITS 0 DB CROSSING

∣ G ∣_{dB, LF} = 20 lo g_{10} K - 20 N lo g_{10} ω, ω_{0 dB} = K^{1/ N} (N \geq 1)

$K$ must be the time-constant-form gain. The line (extended if necessary) reads $20 lo g_{10} K$ at $ω = 1$ . Type 0: the plot starts flat at $20 lo g_{10} K$ .

STATIC ERROR CONSTANTS FROM THE LOW-FREQUENCY ASYMPTOTE

Type 0: K_{p} = K = 1 0^{M_{LF} /20}, Type 1: K_{v} = K = ω_{0 dB}, Type 2: K_{a} = K = ω_{0 dB}^{2}

For unity feedback, where $K_{p} = lim_{s \to 0} G (s)$ , $K_{v} = lim_{s \to 0} s G (s)$ and $K_{a} = lim_{s \to 0} s^{2} G (s)$ . $K$ is the time-constant gain, $M_{LF}$ is the flat low-frequency level in dB (type 0), and $ω_{0 dB}$ is where the low-frequency line, extended if needed, crosses 0 dB.

RHP ZERO

(1 - j ω T) : ∣ \cdot ∣ = 1 + ω^{2} T^{2}, ∠ = - tan^{- 1} (ω T)

Same magnitude as an LHP zero, but the phase is a lag. $T > 0$ .

FIRST-ORDER ALL-PASS

\frac{1 - s T}{1 + s T} : ∣ \cdot ∣ = 1, ∠ = - 2 tan^{- 1} (ω T)

$T > 0$ . Magnitude is 0 dB everywhere and the phase goes from $0^{\circ}$ to $- 18 0^{\circ}$ , even though $n - m = 0$ .

TIME DELAY

e^{- j ω T} : ∣ \cdot ∣ = 1, ∠ = - ω T rad = - \frac{18 0 ^{\circ}}{π} ω T

The magnitude plot is unchanged; the phase lag grows linearly with $ω$ , which curves downward on a log-frequency axis. Remember 1 rad = 57.3°.

INTEGRATOR WITH DELAY

G H = \frac{K e ^{- s T}}{s} : ω_{p c} = \frac{π}{2 T}, GM = \frac{π}{2 K T}, stable ⟺ 0 < K < \frac{π}{2 T}

Negative feedback, $K > 0$ , $T > 0$ . The phase crossover comes from $- 9 0^{\circ} - ω T \cdot \frac{18 0 ^{\circ}}{π} = - 18 0^{\circ}$ . Here $∣ G H ∣ = K / ω$ crosses 1 only once, at $ω_{g c} = K$ , and the phase falls monotonically, so the margins decide stability. Equivalently PM $= 9 0^{\circ} - K T \cdot \frac{18 0 ^{\circ}}{π}$ must be positive.

START AND END POINTS

ω \to 0 : ω \to \infty : G \to K ∠ 0^{\circ} (N = 0), G \to \infty∠ (- 9 0^{\circ} N) (N \geq 1) G \to 0∠ (- 9 0^{\circ} (n - m)) (n > m)

For a minimum-phase function with $K > 0$ in time-constant form. The plot reaches the origin tangent to the axis at angle $- 9 0^{\circ} (n - m)$ . If $n = m$ , it ends at the finite point equal to the ratio of the leading coefficients.

AXIS CROSSINGS

Im G (j ω) = 0 \Rightarrow real-axis crossing; Re G (j ω) = 0 \Rightarrow imaginary-axis crossing

A crossing of the negative real axis happens at the phase crossover frequency $ω_{p c}$ . If the plot crosses at $- a$ , then GM $= 1/ a$ .

LOW-FREQUENCY ASYMPTOTE OF A TYPE-1 POLAR PLOT

G (s) = \frac{K \prod ( 1 + s T _{z} )}{s \prod ( 1 + s T _{p} )} : ω \to 0 lim Re G (j ω) = - K (\sum T_{p} - \sum T_{z})

The plot starts at infinity along −90°, running beside this vertical line. For $K / [s (1 + s T)]$ the asymptote is $Re = - K T$ .

CROSSOVER FREQUENCIES

∣ G (j ω_{g c}) H (j ω_{g c}) ∣ = 1 (0 dB), ∠ G (j ω_{p c}) H (j ω_{p c}) = - 18 0^{\circ}

Both are open-loop quantities, found on the Bode plot or on the polar plot (unit circle and negative real axis).

GAIN MARGIN

GM = \frac{1}{∣ G ( j ω _{p c} ) H ( j ω _{p c} ) ∣}, GM_{dB} = - 20 lo g_{10} ∣ G (j ω_{p c}) H (j ω_{p c}) ∣

If the polar plot crosses the negative real axis at $- a$ , then GM $= 1/ a$ . A positive dB value means stable, in the standard case described below.

PHASE MARGIN

PM = 18 0^{\circ} + ∠ G (j ω_{g c}) H (j ω_{g c})

Use the phase as a negative angle: $- 15 0^{\circ}$ gives PM $= 3 0^{\circ}$ . On the polar plot, PM is the angle from the negative real axis to the point where the plot meets the unit circle.

LARGEST GAIN AND LARGEST DELAY

K_{marginal} = K \cdot GM, T_{m a x} = \frac{PM \times π /180}{ω _{g c}} (PM in degrees)

$K$ is the gain at which GM was computed. $T_{m a x}$ is the delay that uses up the whole phase margin; it assumes a single gain crossover, which a delay does not move.

TYPE-1, THIRD-ORDER LOOP

G H = \frac{K}{s ( 1 + s T _{1} ) ( 1 + s T _{2} )} : ω_{p c} = \frac{1}{T _{1} T _{2}}, ∣ G H (j ω_{p c}) ∣ = \frac{K T _{1} T _{2}}{T _{1} + T _{2}}, K_{marginal} = \frac{T _{1} + T _{2}}{T _{1} T _{2}}

$T_{1}, T_{2} > 0$ . Comes from $tan^{- 1} a + tan^{- 1} b = 9 0^{\circ} ⟺ ab = 1$ for $a, b > 0$ . It agrees with the Routh result for $T_{1} T_{2} s^{3} + (T_{1} + T_{2}) s^{2} + s + K = 0$ . In pole-zero form $G H = K / [s (s + a) (s + b)]$ with $a, b > 0$ : $ω_{p c} = ab$ and $K_{marginal} = ab (a + b)$ .

Where this was asked: 2018 (5), 2020 (4), 2022 (5), 2023 (6), 2024 (4), 2025 (1), 2026 (2). See the 27 previous-year questions on Bode & margins. This shows what was asked, not what will be asked.

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Bode & margins: the notes and 27 solved previous-year questions