What is w_pc and the exact gain margin?

w_pc = 3.46 rad/s (sqrt 12), |G| = 0.625, GM = 4.08 dB

w_pc = 2.71 rad/s, GM = 12.2 dB

Why is this system stable?

GM and PM are both positive (w_gc = 2.71 < w_pc = 3.46)

Because the phase is always above -180 deg

Asked 4/27
2079 Bhadra · 8 marks

Bode plot of 60/[s(s+2)(s+6)] with gain margin

Short answer

Phase crossover occurs at ω=3.46 rad/s with gain margin 4.08 dB. Gain crossover at 2.71 rad/s gives PM ≈ 12.2°. Both margins are positive, so the closed-loop system is stable.

Draw Bode plot for system with open loop transfer function G(s)H(s) = 60/[s(s+2)(s+6)]. Also find GM and comment on stability.

Time-constant form

G(s)H(s) = \frac{60}{s(s+2)(s+6)} = \frac{5}{s\left(1 + \dfrac{s}{2}\right)\left(1 + \dfrac{s}{6}\right)}

$K = 5$ ( $20\log5 = 13.98$ dB), type 1. Corner frequencies: 2 and 6 rad/s.

Magnitude asymptotes

Range	Slope	Values
ω < 2	−20 dB/dec	33.98 dB at 0.1; 13.98 dB at 1; 7.96 dB at 2
2 – 6	−40 dB/dec	7.96 → $7.96 - 40\log3 = -11.12$ dB
ω > 6	−60 dB/dec	−24.4 dB at 10; −42.5 dB at 20

Phase

\phi = -90^\circ - \tan^{-1}\frac{\omega}{2} - \tan^{-1}\frac{\omega}{6}

ω	0.1	0.5	1	2	2.71	3.46	4	6	10	20
φ (deg)	−93.8	−108.8	−126.0	−153.4	−167.8	−180.0	−187.1	−206.6	−227.7	−247.6

Bode plot

   40 |
      |********
      |       ********
   10 |              ********
      |---------------------****---------------------
      |                        ****
  -20 |                            ***
      |                              ****
      |                                 ***
  -50 |                                   ****
      |                                      ***
      |                                        ****
  -80 |                                           ***
      +----------------------------------------------
       0.1            1              10             100
Magnitude asymptotes (dB) vs w (rad/s)

  -90 |*****
      |    *********
      |            ****
 -135 |               ****
      |                  ***
      |                    ***
 -180 |----------------------***---------------------
      |                        ***
      |                          ***
 -225 |                            ****
      |                               ****
      |                                   ********
 -270 |                                          ****
      +----------------------------------------------
       0.1            1              10             100
Phase (deg); crosses -180 at 3.46 rad/s

Gain margin

Phase crossover:

\tan^{-1}\frac{\omega}{2} + \tan^{-1}\frac{\omega}{6} = 90^\circ \;\Rightarrow\; \frac{\omega^2}{12} = 1 \;\Rightarrow\; \omega_{pc} = \sqrt{12} = 3.464\ \text{rad/s}

|G(j3.464)| = \frac{60}{3.464 \times \sqrt{12 + 4} \times \sqrt{12 + 36}} = \frac{60}{3.464 \times 4 \times 6.928} = 0.625

GM = \frac{1}{0.625} = 1.6 = 20\log1.6 = 4.08\ \text{dB}

(The asymptotic plot gives about −1.6 dB at 3.46 rad/s, i.e. GM ≈ 1.6 dB; the exact value is 4.08 dB.)

Phase margin

Exact gain crossover: $\omega_{gc} = 2.71$ rad/s, where φ = −167.8°, so $PM = 12.2^\circ$ .

Stability

GM = +4.08 dB > 0 and PM = +12.2° > 0, and $\omega_{gc} < \omega_{pc}$ . The closed-loop system is stable, but with small margins, so the response will be quite oscillatory.

Check: $s^3 + 8s^2 + 12s + 60$ gives $8 \times 12 = 96 > 60$ . ✓ The gain could rise by a factor of 1.6 (to 96) before instability.

Answer: GM = 4.08 dB at $\omega_{pc}$ = 3.46 rad/s. PM ≈ 12.2°. The system is stable.

✓ Answer checked by Claude Opus 5.5 (high effort)

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