For G(s)H(s) = 50(s+10)/[(s+1)(s+100)], what are the type and the low-frequency magnitude?

Type 0, K = 5, so 13.98 dB flat at low frequency

Type 1, magnitude falls at -20 dB/dec

What is the gain margin and why?

Infinite, because phase dips to only about -55 deg and tends to -90 deg, never crossing -180 deg

Zero, because the phase reaches -180 deg at 5.64 rad/s

At the gain crossover 5.64 rad/s the phase is -53.7 deg. What is the PM and conclusion?

PM = 53.7 deg, marginally stable

Asked 3/27
2080 Baisakh · 10 marks

Bode plot of 50(s+10)/[(s+1)(s+100)]

Short answer

Phase never reaches –180°, so GM = ∞. At gain crossover ω=5.64 rad/s phase is –53.7°, giving PM ≈ 126°. The closed-loop system is stable.

Draw Bode plot for given open loop transfer function G(s)H(s) = 50(s+10)/[(s+1)(s+100)]. Also comment on stability from the plot.

Time-constant form

G(s)H(s) = \frac{50(s+10)}{(s+1)(s+100)} = \frac{5\left(1 + \dfrac{s}{10}\right)}{(1+s)\left(1 + \dfrac{s}{100}\right)}

It is a type-0 system with $K = 5$ , so $20\log5 = 13.98$ dB.

Corner frequencies

Factor	Corner (rad/s)	Slope change
$1/(1+s)$	1	−20 dB/dec
$(1 + s/10)$	10	+20 dB/dec
$1/(1 + s/100)$	100	−20 dB/dec

Magnitude asymptotes

Range	Slope	Value at ends
ω < 1	0	13.98 dB
1 – 10	−20 dB/dec	13.98 → −6.02 dB
10 – 100	0	−6.02 dB
ω > 100	−20 dB/dec	−26.02 dB at 1000

The asymptotic gain crossover is where $13.98 - 20\log\omega = 0$ , i.e. ω = 5 rad/s. The exact value is 5.64 rad/s.

Phase

\phi = \tan^{-1}\frac{\omega}{10} - \tan^{-1}\omega - \tan^{-1}\frac{\omega}{100}

ω	0.1	1	2	5	5.64	10	20	50	100	1000
φ (deg)	−5.2	−39.9	−53.3	−55.0	−53.7	−45.0	−35.0	−36.7	−50.1	−84.8

The phase dips to a minimum of about −56.8° near ω ≈ 3.4 rad/s and tends to −90° at high frequency. It never reaches −180°.

Bode plot

   15 |*************
      |            ****
      |               ****
    0 |------------------***-------------------------
      |                     ***************
      |                                   ***
  -15 |                                     ****
      |                                        ****
      |                                           ***
  -30 |
      |
      |
  -45 |
      +----------------------------------------------
       0.1        1          10          100        1000
Magnitude asymptotes (dB) vs w (rad/s)

    0 |***
      |  ******
      |        ****             *******
  -45 |           ****     ******     ****
      |              *******             ****
      |                                      ******
  -90 |                                           ***
      |
      |
 -135 |
      |
      |
 -180 |----------------------------------------------
      +----------------------------------------------
       0.1        1          10          100        1000
Phase (deg); never reaches -180

Stability from the plot

Phase margin: at $\omega_{gc} = 5.64$ rad/s, φ = −53.7°, so $PM = 180^\circ - 53.7^\circ = 126.3^\circ$ .
Gain margin: the phase never crosses −180°, so $\omega_{pc} = \infty$ and GM = ∞.

Both margins are positive and large, so the closed-loop system is stable (and stays stable for any increase in gain). Check: the closed loop $s^2 + 151s + 600 = 0$ has both roots in the LHP. ✓

Answer: PM ≈ 126°, GM = ∞. The system is stable.

✓ Answer checked by Claude Opus 5.5 (high effort)

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