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Bode plot of (20s+200)/[(s²+?s+25)(s²+40s)]

Short answer

Gain crossover at 0.2 rad/s yields PM ≈ 89°. Phase crosses -180° at 5.88 rad/s with magnitude -28.3 dB, giving GM ≈ 28.3 dB. Both margins positive, so closed-loop stable.

Draw Bode plot for the system with transfer function G(s) = (20s + 200)/[(s² + [?]s + 25)(s² + 40s)]. Determine gain margin, phase margin and comment on stability of the system according to your plot.

One coefficient in the quadratic is unreadable in the paper ("s² + [?]s + 25"). It is taken here as 4, giving s2+4s+25s^2 + 4s + 25 (as in the similar 2072 question). Then ωn=5\omega_n = 5 rad/s and ζ=4/(2×5)=0.4\zeta = 4/(2 \times 5) = 0.4.

Time-constant form

G(s)=20(s+10)s(s+40)(s2+4s+25)=0.2(1+s10)s(1+s40)(1+0.16s+0.04s2)G(s) = \frac{20(s+10)}{s(s+40)(s^2 + 4s + 25)} = \frac{0.2\left(1 + \dfrac{s}{10}\right)}{s\left(1 + \dfrac{s}{40}\right)\left(1 + 0.16s + 0.04s^2\right)}

K=0.2K = 0.2 (20log⁡0.2=−13.9820\log0.2 = -13.98 dB), type 1.

Corner frequencies and slopes

RangeFactor addedSlope
ω < 50.2/s0.2/s−20 dB/dec
5 – 10quadratic (ω_n = 5)−60 dB/dec
10 – 40zero at 10−40 dB/dec
ω > 40pole at 40−60 dB/dec

Magnitude asymptotes

ω0.010.10.2151040100
dB26.06.00−14.0−28.0−46.0−70.1−94.0

The quadratic peak correction at ωn\omega_n is −20log⁡(2ζ)=+1.9-20\log(2\zeta) = +1.9 dB.

Phase

ϕ=−90∘+tan⁡−1ω10−tan⁡−1ω40−tan⁡−10.16ω1−0.04ω2\phi = -90^\circ + \tan^{-1}\frac{\omega}{10} - \tan^{-1}\frac{\omega}{40} - \tan^{-1}\frac{0.16\omega}{1 - 0.04\omega^2}
ω0.10.21255.88102040100
φ (deg)−90.5−91.0−95.2−102.4−160.6−180.0−211.0−221.1−233.2−251.6

Bode plot

   30 |*
      | ******
      |      *******
    0 |------------******----------------------------
      |                  ******
      |                       *******
  -30 |                             ****
      |                                ***
      |                                  ***
  -60 |                                    ****
      |                                       ***
      |                                          **
  -90 |                                           ***
      +----------------------------------------------
       0.01       0.1        1           10         100
Magnitude asymptotes (dB) vs w (rad/s)

  -90 |*************************
      |                        ****
      |                            **
 -135 |                             **
      |                              *
      |                              **
 -180 |-------------------------------*--------------
      |                                **
      |                                 ****
 -225 |                                    *****
      |                                        *****
      |                                            **
 -270 |
      +----------------------------------------------
       0.01       0.1        1           10         100
Phase (deg); crosses -180 at 5.88 rad/s

Margins

  • Gain crossover: the −20 dB/dec line cuts 0 dB at ω = K = 0.2 rad/s (exact 0.200). There φ = −91.0°, so PM = 89.0°.
  • Phase crossover: φ = −180° at ωpc=5.88\omega_{pc} = 5.88 rad/s, where the exact magnitude is −28.3 dB. So GM = +28.3 dB.

Stability

GM > 0 and PM > 0, and ωgc\omega_{gc} (0.2) < ωpc\omega_{pc} (5.88). The closed-loop system is stable with large margins, though it will be slow because the bandwidth is small.

Answer (with s2+4s+25s^2 + 4s + 25): GM ≈ 28.3 dB, PM ≈ 89°. The system is stable.

✓ Answer checked by Claude Opus 5.5 (high effort)

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