Exercise 3.3

A vehicle is characterized by the following transfer function, which relates the throttle input \(U(s)\) to the velocity \(Y(s)\),

\[ G(s) = \frac{s+10}{(s+1)^2(s+5)} \]

We need to design a controller that regulates the speed of the vehicle.

1. Start by designing a proportional controller \(F(s) = K\). For what values of \(K\) is the closed-loop stable? How does the closed-loop performance change when \(K\) is changed?

2. Consider now \(K=10\). Compute the gain margin \(A_m\) and the phase margin \(\varphi_m\).

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Solution

Question 1

Since the order of the closed-loop transfer function \(G_c(s) = \frac{F(s) G(s)}{1 + F(s) G(s)}\) is three, we can’t easily provide an analytical solution. We therefore inspect the root locus.

% Check if the `asymptotic_bode` function is available, otherwise download it
if ~exist('asymptotic_bode')
    urlwrite('https://raw.githubusercontent.com/bonassifabio/1RT485/refs/heads/main/book/res/matlab/asymptotic_bode.m', 'asymptotic_bode.m');
end

s = tf('s');
G =  1 * (s + 10) / ((s + 1)^2 * (s + 5));

rlocus(G); hold on; % Plot the root locus

% Plot the closed-loop poles for specific values of K
K_points = [1, 24, 100];
for k=K_points
    Go = k * G;
    [num, den] = tfdata(Go, "vector");
    poles = roots(num + den);
    p = plot(real(poles), imag(poles), 's', 'MarkerSize', 10, 'MarkerFaceColor', 'auto', 'DisplayName', sprintf('K=%.2f', k));
    set(p, 'markerfacecolor', get(p, 'color'));
end
legend('Location', 'northwest'); hold off;

From the root locus, we can see that the closed-loop poles are stable for \(K < 24\).

Question 2

Phase margin

Let’s start by computing the phase margin, which is defined as

\[ \varphi_m = 180^o + \angle G_o(i \omega_c) \]

where \(G_o(s) = F(s) G(s)\) is the open-loop transfer function and \(\omega_c\) is the critical frequency, that is, the frequency where the open loop transfer function has unitary amplitude, \(\lvert G_o(i \omega_c) \lvert = 1\). Remember that \(G_o(i \omega)\) is a complex number!

\[ G_o(s) = 10 \frac{s + 10}{(s+1)^2(s+5)} \]

To compute \(\omega_c\) we replace \(s = i \omega\) for different values of \(\omega\), until we find a value where the modulus of \(G_o(i \omega)\) is approximately \(1\).

Go_modulus = @(w) abs(10 * (1i * w + 10) / ((1i * w + 1)^2 * (1i * w + 5)));

Go_modulus(1)
Go_modulus(10)
Go_modulus(4)

ans = 9.8547

ans = 0.1252

ans = 0.9894

Therefore \(\omega_c \approx 4\) rad/s.

Note

You can also find the critical frequency from the Bode diagram, as the interception between the plot of \(\lvert G_o(i \omega) \lvert_{\text{dB}}\) and the 0dB axis!

figure; asymptotic_bode(10 * G)

We can now evaluate the phase margin as

\[ \varphi_m = 180 + \angle G_o(i 4) \]

% We create a function that computes the angle of Go given the frequency omega
% We use the mod function to convert the angle to the range [-360, 0]
Go_angle = @(w) mod(rad2deg(angle(10 * (1i * w + 10) / ((1i * w + 1)^2 * (1i * w + 5)))), 360) - 360;

phase_margin = 180 + Go_angle(4)   % Conver the angle from radians to degrees

phase_margin = 11.2141

While a phase margin of \(11^\circ\) is bigger than \(0^\circ\), so it is enough to guarantee the closed-loop stability, in practical applications it is too low to be accepted.

• If \(\varphi_m > 90^\circ\), the dominant closed-loop pole is real valued and stable.

• If \(0^\circ < \varphi_m < 90^\circ\), the dominant closed-loop poles are complex-conjugate, with damping coefficient \(\xi = \cos(\varphi_m)\).

• If \(\varphi_m \leq 0^\circ\), the system is unstable.

Warning

A phase margin below \(30^\circ\) indicates that the dominant poles have a low damping coefficient. In general, the phase margin should be at least \(45^\circ\) for adequate closed-loop performance.

Gain margin

We can then proceed computing the gain margin, which is defined as

\[ A_m = \frac{1}{\lvert G_o(i \omega_p) \lvert} \]

or, in decibels,

\[ A_m = -\lvert G_o(i \omega_p) \lvert_{\text{dB}} \]

Here \(\omega_p\) is the phase crossover frequency, i.e., the frequency at which the phase plot first crosses the \(-180^\circ\) axis. To find \(\omega_p\) we need to solve

(2)\[ \angle G_o(i \omega) := \varphi(0) + \sum_{i} \textrm{atan} \left( \frac{\omega}{p_i} \right) - \sum_{i} \textrm{atan} \left( \frac{\omega}{z_i} \right) = -180^\circ \]

The initial phase \(\varphi(0)\) is \(0^\circ\) if the static gain is positive, decremented by \(90^\circ\) for each pole in the origin and incremented by \(90^\circ\) for each zero in the origin. In (2), \(p_i\) is the \(i\)-th pole of \(G_o(s)\) and \(z_i\) is the \(i\)-th zero.

To compute \(\omega_p\) we try replacing some candidate value until we find a value where \(\angle(G_o(i \omega)) = -180^\circ\). Note that the phase might cross the \(-180^\circ\) axis multiple times, and we must be sure that we get the smallest possible of such values. The easiest way to find a reasonable value is to sketch the asymptotic Bode diagram (see above).

It is apparent that the phase is crossing is somewhere between \(4\) rad/s and \(10\) rad/s. To find the exact value, we try evaluate \(\angle G_o(i \omega)\) in that range

Go_angle(5)
Go_angle(7)
Go_angle(6)

ans = -175.8151

ans = -183.2101

ans = -180.3060

Therefore, \(\omega_p \simeq 6\) rad/s. We can finally evaluate the gain margin as

Am = 1 / Go_modulus(6)

Am = 2.4780

20 * log10(Am)

ans = 7.8819

The gain margin is \(A_m \simeq 2.5 \simeq 7.8 \text{ dB}\). Since it is bigger than \(1\) (i.e. \(0\) dB), the closed-loop is stable.

We can confirm our results using MATLAB’s margin command.

figure;
margin(10 * G); grid on;

Alternative solution to Question 1

Because \(G(s)\) is stable, we can solve Question 1 also computing the gain margin of \(G(s)\).

[A_m, Phi_m] = margin(G)

A_m = 24.0063

Phi_m = 85.3027

The gain margin is \(A_m = 24\), which means that we can take \(0 \leq K \leq 24\) without compromising the closed-loop stability.