Exercise 3

Problem 1

Consider the systems $LaTeX: P_1(s) = 87.5 \frac{s + 1/5}{(s-5)(s + 10)}$ $P_1(s) = 87.5 \frac{s + 1/5}{(s-5)(s + 10)}$ and $LaTeX: P_2(s) = 50 \frac{(1/5 - s)(s + 5)}{ (1/5 + s)(5 - s) (s + 10)}$ $P_2(s) = 50 \frac{(1/5 - s)(s + 5)}{ (1/5 + s)(5 - s) (s + 10)}$

Factorize both systems into minimum phase and normalized nonminimum phase components.
Compute step responses for both systems. As the systems are unstable, simulate for $LaTeX: t\in\left\lbrack0,1\right\rbrack$ $t\in\left\lbrack0,1\right\rbrack$ . Are the systems very different?
Plot the nyquists, nichols and bode diagrams for the loop transfer function $L = PC$ where the controller is given by $LaTeX: C=1+\frac{1}{s}$ $C=1+\frac{1}{s}$ . Which (if any) of the closed loops are stable? Why/Why not?
Where applicable, design a low-pass filtered PI controller that does not inject too much noise, gives closed-loop bandwidth of about 15 rad/s, and Ms < 1.4 and Mt < 2.

Problem 2

Show that $LaTeX: \left| \frac{z \bar p - 1}{z - p} \right| = 1$ $\left| \frac{z \bar p - 1}{z - p} \right| = 1$ for all LaTeX: z $z$ on the unit circle.

Problem 3

The purpose of this exercise is to translate the nonminimum phase elements of lecture 4 to discrete time. Factor the following systems into minimum phase and all-pass nonminimum phase elements, $LaTeX: P = P_\text{mp} P_\text{nmp}$ $P = P_\text{mp} P_\text{nmp}$ such that $LaTeX: P_\text{mp}(1) > 0$ $P_\text{mp}(1) > 0$ and $LaTeX: |P_\text{nmp}(z)| = 1, \quad \forall |z| = 1$ $|P_\text{nmp}(z)| = 1, \quad \forall |z| = 1$ . Furthermore, for sampling time LaTeX: h = 0.01 $h = 0.01$ plot the bode diagrams of $LaTeX: P_\text{nmp}$ $P_\text{nmp}$ .

Unstable zero in $z_0 = 1.02$ , $P(z) = z_0 - z$
Unstable pole $p_0 = 1.01$ , $LaTeX: P(z) = \frac{1}{z - p_0}$ $P(z) = \frac{1}{z - p_0}$
unstable pole zero pair with $p_0 = 1.01$ , $z_0 = 1.02$ , $LaTeX: P(z) = \frac{z_0 - z}{z - p}$ $P(z) = \frac{z_0 - z}{z - p}$
Unstable pole zero pair with $p_0 = 1.02$ , $z_0 = 1.01$ , $LaTeX: P(z) = \frac{z_0 - z}{z - p}$ $P(z) = \frac{z_0 - z}{z - p}$

Problem 4

In continuous time, a rhp zero in LaTeX: z_0 $z_0$ limits the bandwidth of the sensitivity function: let $LaTeX: S_\text{req}(s) = \frac{s M_s}{s + aM_s}$ $S_\text{req}(s) = \frac{s M_s}{s + aM_s}$ and require $LaTeX: |S(i\omega)| < |S_\text{req}(i\omega)|$ $|S(i\omega)| < |S_\text{req}(i\omega)|$ for $LaTeX: \omega \in [0, \infty)$ $\omega \in [0, \infty)$ , then $LaTeX: a < z \frac{M_s - 1}{M_s}$ $a < z \frac{M_s - 1}{M_s}$ . Derive a discrete-time version.

Problem 5

Sketch the level curves in the complex plane of $LaTeX: \textrm{Re}(1/z) = c$ $\textrm{Re}(1/z) = c$

Problem 6

Use QFT design on the uncertain system

$LaTeX: P(s) = \frac{k}{(s + a)(s + b)}, \quad k \in [1, 10], \quad a \in [1, 5], \quad b \in [20, 30]$ $P(s) = \frac{k}{(s + a)(s + b)}, \quad k \in [1, 10], \quad a \in [1, 5], \quad b \in [20, 30]$ .

The specifications are:

$LaTeX: \max_\omega |T(i\omega| < 1.2$ $\max_\omega |T(i\omega| < 1.2$
$LaTeX: |S(i\omega)| \leq \left| \frac{0.02s^3 + 1.28s^2 + 14.96s + 48}{s^2 + 14.4\textcolor{red}{s} + 169}\right|$ $|S(i\omega)| \leq \left| \frac{0.02s^3 + 1.28s^2 + 14.96s + 48}{s^2 + 14.4\textcolor{red}{s} + 169}\right|$ for $LaTeX: \omega \in [0, 10]$ $\omega \in [0, 10]$ rad/s
$LaTeX: |PS(i\omega)| \leq 0.01$ $|PS(i\omega)| \leq 0.01$ for $LaTeX: \omega \in [0, 50]$ $\omega \in [0, 50]$ rad/s

HINT: This is Example 1 in the manual for the QFTIT tool.

Problem 7

Let $H(s)$ be a strictly proper transfer function that has all its poles in the half plane $LaTeX: \textrm{Re}\ s \leq - \alpha$ $\textrm{Re}\ s \leq - \alpha$ for some $LaTeX: \alpha > 0$ $\alpha > 0$ , i.e. $H(s)$ is analytic on $LaTeX: \textrm{Re } s > -\alpha$ $\textrm{Re } s > -\alpha$ . Let $h(t)$ be the corresponding time domain function, i.e. $LaTeX: H = \mathcal L h$ $H = \mathcal L h$ , where $LaTeX: \mathcal L$ $\mathcal L$ denotes the Laplace transform. Show that for any $LaTeX: \textrm{Re } s_0 > -\alpha$ $\textrm{Re } s_0 > -\alpha$ we have $LaTeX: \int_0^\infty e^{-s_0 t}h(t) dt = \lim_{s\to s_0} H(s)$ $\int_0^\infty e^{-s_0 t}h(t) dt = \lim_{s\to s_0} H(s)$ . Hint: Look at the definition of the (one-sided) Laplace transform.
For a process $P(s)$ and controller $C(s)$ . Consider our standard error feedback configuration, i.e. the 2-DOF configuration with $F = I$ . Suppose that the closed loop is stable. Show that for a reference step input
1. If $LaTeX: \lim_{s\to0}\textcolor{red}{s}P(s)C(s)=c_1$ $\lim_{s\to0}\textcolor{red}{s}P(s)C(s)=c_1$ , for some $LaTeX: 0 < |c_1| < \infty$ $0 < |c_1| < \infty$ , then $LaTeX: \lim_{t \to \infty} e(t) = 0$ $\lim_{t \to \infty} e(t) = 0$ and $LaTeX: \int_0^\infty e(t)dt = \frac{1}{c_1}$ $\int_0^\infty e(t)dt = \frac{1}{c_1}$
2. If $LaTeX: \lim_{s \to 0}s^2 P(s)C(s) = c_2$ $\lim_{s \to 0}s^2 P(s)C(s) = c_2$ , for some $LaTeX: 0 < |c_2| < \infty$ $0 < |c_2| < \infty$ , then $LaTeX: \lim_{t \to \infty} e(t) = 0$ $\lim_{t \to \infty} e(t) = 0$ and $LaTeX: \int_0^\infty e(t) dt = 0$ $\int_0^\infty e(t) dt = 0$
  Hint: Consider $LaTeX: P(s)C(s) = \tilde L(s) / s$ $P(s)C(s) = \tilde L(s) / s$ where $LaTeX: \lim_{s \to 0}\tilde L(s) = c_1$ $\lim_{s \to 0}\tilde L(s) = c_1$ and apply the final value theorem.
Suppose that a process $P(s)$ has an unstable pole at $s = p$ such that $\textrm{Re}\ p > 0$ . Show that if the closed loop is stable, then for a reference step input
1. $LaTeX: \int_0^\infty e^{-pt}e(t) dt = 0$ $\int_0^\infty e^{-pt}e(t) dt = 0$
2. $LaTeX: \int_0^\infty e^{-pt}y(t)dt = \frac{1}{p}$ $\int_0^\infty e^{-pt}y(t)dt = \frac{1}{p}$
  Hint: Interpolation constraints, use the result in A
Suppose instead that $P(s)$ has an unstable zero at $s = z$ such that $\textrm{Re}\ p > 0$ . Show that if the closed loop is stable, then for a reference step input
1. $LaTeX: \int_0^\infty e^{-qt}e(t)dt = \frac{1}{q}$ $\int_0^\infty e^{-qt}e(t)dt = \frac{1}{q}$
2. $LaTeX: \int_0^\infty e^{-qt}y(t) = 0$ $\int_0^\infty e^{-qt}y(t) = 0$
Discuss implications of B, C and D on overshoot and undershoot.