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Solved exercises on Bode plots, Nyquist and stability margins

Nine exercises with solutions and plots: asymptotic Bode, resonant peak, margins, gain for a phase margin, delay margin, lead network, Nyquist of an unstable process, counting encirclements, non-minimum-phase zero.

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Exercises on Bode plots, the Nyquist criterion and stability margins of the kind set in Control Systems courses, in three levels: basic, with drawing and reading plots by hand; intermediate, with margins, delays and gain design; exam, with open-loop unstable processes, counting encirclements, compensating networks and non-minimum-phase systems. The result is given straight away; the full solution, with the plot, opens with a click. The proofs and the MATLAB and Python code are in the in-depth study linked at the top.

Basic level

Exercise 1. Bode by hand

Draw the asymptotic Bode plots of L(s) = \dfrac{100(s + 1)}{s(s + 10)} and estimate the crossover frequency and the phase margin.

Result: Bode form \frac{10(1 + s)}{s(1 + s/10)}; crossover about 100 rad/s; phase margin about 95°.

Solution

Bode form (factors with constant term 1): L = \frac{100\cdot(1 + s)}{s\cdot 10(1 + s/10)} = \frac{10(1 + s)}{s(1 + s/10)}. Gain 10, one integrator, a zero at 1, a pole at 10.

Asymptotic magnitude. For \omega \lt 1: \frac{10}{\omega}, a line at −20 dB per decade that is 20 dB at \omega = 1. Between 1 and 10 the zero cancels the integrator: flat at 20 dB. After 10: again −20 dB per decade, |L| \approx \frac{100}{\omega}, which is 0 dB at \omega = 100.

Phase. It starts from -90° (integrator), rises towards -45° because of the zero, returns to -90° after the pole. At \omega = 100: -90° + \arctan 100 - \arctan 10 = -90° + 89.4° - 84.3° = -84.9°, so \varphi_m \approx 95°.

Exact values (with the code of the in-depth study): \omega_c = 99.5 rad/s, \varphi_m = 95.2°; at \omega = 1 the actual magnitude is 23 dB (the asymptote gives 20: the +3 dB error of the zero).

Actual and asymptotic Bode
The actual magnitude, in blue, follows the dashed asymptotes, departing by about 3 dB at the corner frequencies 1 and 10; the phase, in red, rises towards minus 45 degrees between them and returns to minus 90.

Exercise 2. The resonant peak

A second-order system has \omega_n = 10 rad/s and \zeta = 0.2. Compute the resonant peak, the frequency at which it occurs and the phase at \omega_n.

Result: M_r = 2.55 (8.1 dB) at \omega_r = 9.59 rad/s; phase -90° at \omega_n.

Solution

From the in-depth study: M_r = \frac{1}{2\zeta\sqrt{1 - \zeta^2}} = \frac{1}{0.4\cdot 0.9798} = 2.55, i.e. 20\log_{10}2.55 = 8.1 dB.

\omega_r = \omega_n\sqrt{1 - 2\zeta^2} = 10\sqrt{0.92} = 9.59 rad/s. At \omega = \omega_n the denominator 1 - u^2 + 2i\zeta u equals 2i\zeta: phase -90°, magnitude \frac{1}{2\zeta} = 2.5.

The peak in frequency and the step overshoot (52.7 % for \zeta = 0.2) are two faces of the same lack of damping: an instrument that measures the frequency response of a system allows \zeta to be estimated from the peak.

Resonant peaks for various dampings
The orange curve corresponds to zeta 0.2, with a peak of about 8 dB just before the natural frequency.

Exercise 3. Reading the margins

For L(s) = \dfrac{4}{s(s + 1)(s + 2)} compute the gain margin by hand and check on the Nyquist plot that the closed loop is stable.

Result: \omega_\pi = \sqrt 2, G_m = 1.5 (3.5 dB); \varphi_m = 11.4° at 1.14 rad/s; stable, with small margins.

Solution

Phase -90° - \arctan\omega - \arctan\frac{\omega}{2} = -180° requires \arctan\omega + \arctan\frac{\omega}{2} = 90°, i.e. \omega\cdot\frac{\omega}{2} = 1 (two complementary angles have reciprocal tangents): \omega_\pi = \sqrt 2.

|L(i\sqrt 2)| = \frac{4}{\sqrt 2\cdot\sqrt 3\cdot\sqrt 6} = \frac{4}{6} = \frac{2}{3}, so G_m = 1.5: the critical gain is 4\cdot 1.5 = 6 (the same that Routh gives on s^3 + 3s^2 + 2s + K: 3\cdot 2 = K).

On the Nyquist plot, L has no poles in the right half-plane (P = 0) and the curve crosses the negative half-axis at -\frac{2}{3}, without encircling -1: N = 0, Z = 0, stable.

Nyquist of 4 over s times s plus 1 times s plus 2
The curve crosses the negative real half-axis at minus two thirds, close to the critical point but to its right.
Margins on the Bode plots
The phase margin, in green, is about 11 degrees; the gain margin, in purple, about 3.5 dB.

Intermediate level

Exercise 4. The gain for a phase margin

For L(s) = \dfrac{K}{s(s + 1)(s + 2)} find the gain that gives a phase margin of 45°. What is the gain margin then?

Result: K = 1.34, with \omega_c = 0.562 rad/s; G_m = 4.48 (13 dB).

Solution

The gain does not change the phase. First find the frequency where the phase is -135°: \arctan\omega + \arctan\frac{\omega}{2} = 45°. With the tangent addition formula: \frac{\omega + \omega/2}{1 - \omega^2/2} = 1, i.e. \omega^2 + 3\omega - 2 = 0, \omega = \frac{-3 + \sqrt{17}}{2} = 0.562.

Then choose K so that the magnitude there is 1: K = \omega\sqrt{1 + \omega^2}\sqrt{4 + \omega^2} = 0.562\cdot 1.147\cdot 2.077 = 1.34.

Gain margin: the critical gain is 6 (exercise 3), so G_m = 6/1.34 = 4.48. The closed-loop overshoot turns out to be 24 %, consistent with the rule \zeta \approx \varphi_m/100 = 0.45.

Bode plots of the loop function
Lowering the gain shifts the magnitude plot rigidly down, the crossover frequency moves left where the phase is further from minus 180, and the phase margin increases.

Exercise 5. How much delay can be tolerated

For L(s) = \dfrac{2}{s + 1} compute the phase margin and the maximum delay that can be inserted into the loop before instability.

Result: \omega_c = \sqrt 3, \varphi_m = 120°; maximum delay 1.21 s.

Solution

|L(i\omega)| = \frac{2}{\sqrt{1 + \omega^2}} = 1 for \omega_c = \sqrt 3. Phase -\arctan\sqrt 3 = -60°, margin 180° - 60° = 120°.

A delay \tau adds -\omega\tau to the phase without changing the magnitude: the crossover stays at \sqrt 3, and instability arrives when \omega_c\tau = 120° = 2.094 rad, i.e. \tau = \frac{2.094}{1.732} = 1.21 s.

Without delay the first-order system is stable for any gain; with a delay, it is not. This is why systems controlled over a network (vehicles, teleoperated robots) must take latency into account.

The delay on the Bode plots
With a delay of 0.6 seconds the phase, in red, falls well below that of the system without delay, in grey, and the phase margin is roughly halved.

Exercise 6. A lead network

The network C(s) = \dfrac{1 + 10s}{1 + s} is a lead network with \alpha = 10. Find the maximum phase lead, the frequency at which it occurs and the gain at that point.

Result: 54.9° at \omega = 0.316 rad/s; gain \sqrt{10}, i.e. 10 dB.

Solution

With C = \frac{1 + \alpha Ts}{1 + Ts} (here T = 1), the phase is \arctan(\alpha T\omega) - \arctan(T\omega). Differentiating and setting to zero gives the maximum at the geometric mean of the two corner frequencies: \omega_m = \frac{1}{T\sqrt\alpha} = \frac{1}{\sqrt{10}} = 0.316.

There \sin\varphi_{max} = \frac{\alpha - 1}{\alpha + 1} = \frac{9}{11}, \varphi_{max} = 54.9°, and |C| = \sqrt\alpha = 3.16 (10 dB, half of the total 20 dB).

In design, \omega_m is placed at the new crossover frequency, to use all the lead where it is needed; the gain \sqrt\alpha must be compensated.

Bode plots of the lead network
The magnitude rises by 20 dB between the frequencies 0.1 and 1; the phase has a bell shape with a maximum of about 55 degrees at the frequency 0.316.

Exam level

Exercise 7. Nyquist with an unstable process

The process L(s) = \dfrac{K}{s - 1} has an unstable pole. Draw the Nyquist plot for K = 2 and use the criterion to determine for which K the closed loop is stable.

Result: stable for K \gt 1; for K = 2 the plot encircles -1 once anticlockwise.

Solution

L(i\omega) = \frac{K}{i\omega - 1} = \frac{K(-1 - i\omega)}{1 + \omega^2}. For \omega = 0 it is -K, for \omega \to \pm\infty it tends to 0. Real and imaginary parts satisfy \left(x + \frac{K}{2}\right)^2 + y^2 = \frac{K^2}{4}: a circle centred at -\frac{K}{2} with diameter from -K to 0.

The process has P = 1. If K \gt 1 the circle contains -1, traversed (for \omega from -\infty to +\infty) anticlockwise: N = -1, so Z = N + P = 0, stable. If K \lt 1, -1 stays outside: N = 0, Z = 1, unstable.

Direct check: the closed loop has pole 1 - K, negative for K \gt 1. With an unstable process, the plot must go round -1, in the right direction, for the loop to be stable.

Nyquist of an unstable process
For K equal to 2 the curve is a circle from minus 2 to 0 that contains the point minus 1 and goes round it once anticlockwise.

Exercise 8. Counting encirclements

For L(s) = \dfrac{60}{s(s + 1)(s + 5)} use the Nyquist criterion to count the unstable closed-loop poles.

Result: two clockwise turns round -1: Z = 2, closed loop unstable with a pair of poles in the right half-plane.

Solution

From the lesson, for \frac{K}{s(s + 1)(s + 5)} the plot crosses the negative half-axis at -\frac{K}{30} at the frequency \sqrt 5. With K = 60 the crossing is at -2, to the left of -1.

Following the complete curve (for positive and negative \omega, closed at infinity clockwise because of the integrator), the point -1 is encircled twice clockwise: N = 2. The process is stable, P = 0, so Z = 2.

Check with Routh on s^3 + 6s^2 + 5s + 60: 6\cdot 5 = 30 \lt 60, two sign changes in the first column, two unstable poles.

Nyquist with too high a gain
The curve crosses the negative real half-axis at minus 2, to the left of the critical point, which is encircled twice clockwise.

Exercise 9. A non-minimum-phase zero

The system G(s) = \dfrac{1 - s}{(1 + s)^2} has a zero at +1. Compute the step response, its minimum value and the time at which it is reached, and compare the phase with that of \dfrac{1 + s}{(1 + s)^2}.

Result: y(t) = 1 - e^{-t} - 2te^{-t}; minimum -0.213 at t = 0.5 s; the phase falls to -270° instead of stopping at -90°.

Solution

Y = \frac{1 - s}{s(1 + s)^2} = \frac{1}{s} - \frac{1}{s + 1} - \frac{2}{(s + 1)^2} (residues: at 0 it is 1; for the double pole, G_2 = \frac{1 - s}{s}, G_2(-1) = -2, G_2'(-1) = -1). So y = 1 - e^{-t} - 2te^{-t}.

y'(t) = e^{-t}(2t - 1): negative for t \lt 0.5. The response starts downwards, reaches y(0.5) = 1 - 2e^{-0.5} = -0.213 and then rises towards 1.

In frequency: |1 - i\omega| = |1 + i\omega|, same magnitude; but \arg(1 - i\omega) = -\arctan\omega instead of +\arctan\omega. The total phase goes from 0 to -270°. The extra phase lag reduces the margin and limits the achievable bandwidth: with an unstable zero at z, the crossover frequency cannot exceed about z/2.

Step response of a non-minimum-phase system
The response, in blue, falls to minus 0.21 at half a second and then rises towards 1; dashed, the response of the system with the stable zero, which rises at once.

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