Exercise 2.4

Exercise 2.4#

An astronaut working in space needs to control their position using thrusters, which generate force by expelling gas. A control system is designed to achieve this, based on the following control law:

\[ u = K_1 (r - y) - K_1 K_2 \frac{dy}{dt}, \]

where:

  • \(u\) is the thrust force generated by the thrusters (control input),

  • \(r\) is the desired position of the astronaut (setpoint),

  • \(y\) is the current position of the astronaut (measured output). \end{itemize}

The astronaut, including equipment, has a total mass of 100 kg. Your task is to determine suitable values for the controller parameters \(K_1 > 0\) and \(K_2 > 0\) such that the following requirements are met:

  1. If the desired position is given as \(r(t) = \textrm{ramp}(t)\), there should be a time \(T_0\) such that the tracking error satisfies \(|r(t) - y(t)| < 1\) for all \(t \geq T_0\).

  2. The damping ratio of the closed-loop system is 0.7. This ensures smooth and stable motion without excessive oscillations.

Solution#

Let’s start by writing the dynamical model of the astronaut. From the second Newton’s law \(F = m \cdot a\). The force \(F\) here is the thrust force (\(u\)), while the accelerations is the second derivative of the the position (\(\ddot{y}\)). Hence \(m \ddot{y} = u\), or equivalently

\[ \ddot{y} = \frac{1}{m} u \]

The transfer function corresponding to this ODE is

\[ Y(s) = \frac{0.01}{s^2} U(s) \]

In Laplace domain, the control law is

\[ U(s) = K_1 \underbrace{(R(s) - Y(s))}_{E(s)} - K_1 K_2 s Y(s) \]

Replacing this \(U(s)\) in \(Y(s)\) we get

\[ Y(s) = \frac{0.01}{s^2} \left[ K_1 R(s) - K_1 Y(s) - K_1 K_2 s Y(s)\right] \]

We then isolate \(Y(s)\),

\[ Y(s) = \frac{\frac{0.01}{s^2} K_1}{1 + K_1 \frac{0.01}{s^2} + K_1 K_2 \frac{0.01}{s}} R(s) = \frac{0.01 K_1}{s^2 + 0.01 K_1 K_2 s + 0.01 K_1} R(s) \]

Question 1#

It can be verified that any \(K_1 > 0\), \(K_2 > 0\) guarantees the stability of the closed-loop (Descartes’ rule of signs).

We can therefore apply the Final Value Theorem, recalling that \(E(s) = R(s) - Y(s)\), where \(R(s) = \frac{1}{s^2}\).

\[ \lim_{t \to \infty} e(t) = \lim_{s \to 0} s \left( 1 - \frac{0.01 K_1}{s^2 + 0.01 K_1 K_2 s + 0.01 K_1} \right) \frac{1}{s^2} = \lim_{s \to 0} \frac{s(s + 0.01 K_1 K_2)}{s^2 + 0.01 K_1 K_2 s + 0.01 K_1} \cdot \frac{1}{s} = \frac{0.01 K_1 K_2}{0.01 K_1} = K_2 \]

Hence, \(K_2 \leq 1\) ensures that the error is asymptotically smaller than \(1\).

Question 2#

We have seen that the denominator of the closed-loop transfer function is \(s^2 + 0.01 K_1 K_2 s + 0.01 K_1\).

For two complex-conjugate poles, the damping \(\xi \geq 0\) and the natural frequency \(\omega_n > 0\) are defined as the values such that

\[ s^2 + 2 \xi \omega_n s + \omega_n^2 \]

In our case

\[\begin{split} \begin{dcases} 2 \xi \omega_n = 0.01 K_1 K_2 \\ \omega_n^2 = 0.01 K_1 \end{dcases} \quad \implies \quad \begin{dcases} \omega_n = 0.1 \sqrt{K_1} \\ \xi = \frac{0.01 K_1 K_2}{0.2 \sqrt{K_1}} = 0.05 \sqrt{K_1} K_2 \end{dcases} \end{split}\]

Forcing the damping ration \(\xi\) to be smaller than \(0.7\) we get

\[ \sqrt{K_1} \leq \frac{14}{K_2} \quad \implies \quad K_1 \leq \frac{196}{K_2^2} \]