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Bode plot of 40(s+2)/[(2+s+25s²)(1+2s)s]

Short answer

Phase crosses -180° at 0.275 rad/s with magnitude 58.8 dB, giving GM = -58.8 dB. Phase margin is -124.5°. Both margins are negative, so the closed-loop system is unstable.

Draw Bode plot for unity feedback system with open loop transfer function G(s) = 40(s+2)/[(2 + s + 25s²)(1+2s)s]. Find GM and comment on stability.

Time-constant form

G(s)=40(s+2)s(1+2s)(25s2+s+2)=40(1+s2)s(1+2s)(1+0.5s+12.5s2)G(s) = \frac{40(s+2)}{s(1+2s)(25s^2 + s + 2)} = \frac{40\left(1 + \dfrac{s}{2}\right)}{s(1+2s)\left(1 + 0.5s + 12.5s^2\right)}

Quadratic factor: ωn2=2/25\omega_n^2 = 2/25, so ωn=0.283\omega_n = 0.283 rad/s; 2ζωn=1/252\zeta\omega_n = 1/25, so ζ=0.0707\zeta = 0.0707 (very lightly damped).

Factors

FactorCorner (rad/s)Slope changePhase
K=40K = 40—32.04 dB0°
1/s1/s—−20 dB/dec−90°
quadratic, ζ = 0.0710.283−40 dB/dec0 to −180° (sharp)
1/(1+2s)1/(1+2s)0.5−20 dB/dec0 to −90°
(1+s/2)(1 + s/2)2+20 dB/dec0 to +90°

Magnitude asymptotes

RangeSlopeValues
ω < 0.283−20 dB/dec72.0 dB at 0.01; 52.0 dB at 0.1; 43.0 dB at 0.283
0.283 – 0.5−60 dB/dec43.0 → 28.2 dB
0.5 – 2−80 dB/dec28.2 → −20.0 dB
ω > 2−60 dB/dec−61.9 dB at 10

Resonant correction at ωn\omega_n: −20log⁡(2ζ)=+17-20\log(2\zeta) = +17 dB, so the true peak is about 59 dB near 0.28 rad/s.

Phase

ϕ=−90∘−tan⁡−12ω+tan⁡−1ω2−tan⁡−10.5ω1−12.5ω2\phi = -90^\circ - \tan^{-1}2\omega + \tan^{-1}\frac{\omega}{2} - \tan^{-1}\frac{0.5\omega}{1 - 12.5\omega^2}
ω0.010.10.20.250.2750.2830.30.51210
φ (deg)−91−102−117−139−180−201−242−294−304−300−278

Bode plot

   90 |
      |****
      |   *************
   45 |               *********
      |                       *****
      |                           ***
    0 |-----------------------------****-------------
      |                                ****
      |                                   *****
  -45 |                                       *****
      |                                           ***
      |
  -90 |
      +----------------------------------------------
       0.01           0.1            1              10
Magnitude asymptotes (dB); true peak ~59 dB at 0.28

  -90 |***************
      |              *******
      |                    **
 -150 |                     *
      |---------------------*------------------------
      |                      *
 -210 |                      *
      |                      *
      |                      *
 -270 |                       *                    **
      |                       *****       **********
      |                           ********
 -330 |
      +----------------------------------------------
       0.01           0.1            1              10
Phase (deg); crosses -180 at 0.275 rad/s

Gain margin

The phase crosses −180° at ωpc=0.275\omega_{pc} = 0.275 rad/s, just below ωn\omega_n. Because of the resonance, the magnitude there is very large: ∣G∣=871.6|G| = 871.6 (58.8 dB).

GM=−58.8 dB(i.e. 1/871.6)GM = -58.8\ \text{dB} \quad (\text{i.e. } 1/871.6)

Phase margin

Gain crossover is at ωgc=1.16\omega_{gc} = 1.16 rad/s (asymptotic estimate 1.12), where φ = −304.5°.

PM=180∘−304.5∘=−124.5∘PM = 180^\circ - 304.5^\circ = -124.5^\circ

Stability

GM is negative (in dB), PM is negative, and ωpc<ωgc\omega_{pc} < \omega_{gc}. The closed-loop system is unstable.

Check: 50s4+27s3+5s2+42s+80=050s^4 + 27s^3 + 5s^2 + 42s + 80 = 0 gives Routh first column 50,27,−72.8,…50, 27, -72.8, \dots. There are two sign changes, so there are 2 RHP poles. ✓

Answer: GM ≈ −58.8 dB, PM ≈ −124.5°. The system is unstable.

✓ Answer checked by Claude Opus 5.5 (high effort)

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