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Bode plot of (4s+40)/[(s²+4s+25)(s²+50s)]

Short answer

Gain crossover at 0.032 rad/s yields PM = 89.9°. Phase crossover at 5.97 rad/s yields GM = 44.5 dB. Both margins are positive, so the closed-loop system is stable.

Draw Bode plot for the system with transfer function G(s) = (4s + 40)/[(s² + 4s + 25)(s² + 50s)]. Determine gain margin, phase margin and comment on stability of the system according to your plot.

Time-constant form

G(s)=4(s+10)s(s+50)(s2+4s+25)=0.032(1+s10)s(1+s50)(1+0.16s+0.04s2)G(s) = \frac{4(s+10)}{s(s+50)(s^2 + 4s + 25)} = \frac{0.032\left(1 + \dfrac{s}{10}\right)}{s\left(1 + \dfrac{s}{50}\right)\left(1 + 0.16s + 0.04s^2\right)}
  • K=4×1050×25=0.032K = \dfrac{4 \times 10}{50 \times 25} = 0.032, so 20log⁡K=−29.920\log K = -29.9 dB. Type 1.
  • Quadratic: ωn=5\omega_n = 5 rad/s, ζ=4/(2×5)=0.4\zeta = 4/(2 \times 5) = 0.4. Peak correction −20log⁡(0.8)=+1.9-20\log(0.8) = +1.9 dB.

Factors

FactorCorner (rad/s)Slope changePhase
0.032/s0.032/s—−20 dB/dec−90°
quadratic, ζ = 0.45−40 dB/dec0 to −180°
(1+s/10)(1 + s/10)10+20 dB/dec0 to +90°
1/(1+s/50)1/(1 + s/50)50−20 dB/dec0 to −90°

Magnitude asymptotes

RangeSlopeValues
ω < 5−20 dB/dec30.1 dB at 0.001; 10.1 dB at 0.01; 0 dB at 0.032; −29.9 dB at 1; −43.9 dB at 5
5 – 10−60 dB/dec−43.9 → −61.9 dB
10 – 50−40 dB/dec−61.9 → −89.9 dB
ω > 50−60 dB/dec−108.0 dB at 100

Phase

ϕ=−90∘+tan⁡−1ω10−tan⁡−1ω50−tan⁡−10.16ω1−0.04ω2\phi = -90^\circ + \tan^{-1}\frac{\omega}{10} - \tan^{-1}\frac{\omega}{50} - \tan^{-1}\frac{0.16\omega}{1 - 0.04\omega^2}
ω0.010.0320.1155.971050100
φ (deg)−90.0−90.1−90.5−94.9−159.2−180.0−208.2−231.7−246.9

Bode plot

   45 |
      |****
      |   ********
    0 |----------********----------------------------
      |                 ********
      |                        *******
  -45 |                              ******
      |                                   ***
      |                                     *****
  -90 |                                         ***
      |                                            **
      |
 -135 |
      +----------------------------------------------
       0.001    0.01     0.1      1        10       100
Magnitude asymptotes (dB) vs w (rad/s)

  -90 |*****************************
      |                            ****
      |                               **
 -135 |                                **
      |                                 *
      |                                 **
 -180 |----------------------------------*-----------
      |                                  **
      |                                   *****
 -225 |                                       ****
      |                                           ***
      |
 -270 |
      +----------------------------------------------
       0.001    0.01     0.1      1        10       100
Phase (deg); crosses -180 at 5.97 rad/s

Gain and phase margin

  • Gain crossover: on the −20 dB/dec segment, ∣G∣=1|G| = 1 at ωgc=K=0.032\omega_{gc} = K = 0.032 rad/s. There φ = −90.1°, so
PM=180∘−90.1∘=89.9∘PM = 180^\circ - 90.1^\circ = 89.9^\circ
  • Phase crossover: φ = −180° at ωpc=5.97\omega_{pc} = 5.97 rad/s, where the magnitude is −44.5 dB (asymptote ≈ −48.5 dB, plus the resonance lift). So
GM=+44.5 dBGM = +44.5\ \text{dB}

Stability

GM ≈ +44.5 dB and PM ≈ +89.9°, both positive, with ωgc≪ωpc\omega_{gc} \ll \omega_{pc}. The closed-loop system is stable. The very low gain crossover (0.032 rad/s) means the response is very slow; the gain could be raised by about 44 dB (×168) before instability.

Answer: GM ≈ 44.5 dB, PM ≈ 89.9°. The system is stable.

✓ Answer checked by Claude Opus 5.5 (high effort)

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