Exercise 3
Problem 1
Consider the systems P1(s)=87.5s+1/5(s−5)(s+10) and
P2(s)=50(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
t∈[0,1]. 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
C=1+1s. 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 |zˉp−1z−p|=1 for all
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, P=PmpPnmpsuch that
Pmp(1)>0 and
|Pnmp(z)|=1,∀|z|=1. Furthermore, for sampling time
h=0.01 plot the bode diagrams of
Pnmp.
- Unstable zero in
z0=1.02 ,
P(z)=z0−z
- Unstable pole
p0=1.01,
P(z)=1z−p0
- unstable pole zero pair with
p0=1.01,
z0=1.02,
P(z)=z0−zz−p
- Unstable pole zero pair with
p0=1.02 ,
z0=1.01,
P(z)=z0−zz−p
Problem 4
In continuous time, a rhp zero in z0 limits the bandwidth of the sensitivity function: let
Sreq(s)=sMss+aMs and require
|S(iω)|<|Sreq(iω)| for
ω∈[0,∞), then
a<zMs−1Ms. Derive a discrete-time version.
Problem 5
Sketch the level curves in the complex plane of Re(1/z)=c
Problem 6
Use QFT design on the uncertain system
P(s)=k(s+a)(s+b),k∈[1,10],a∈[1,5],b∈[20,30].
The specifications are:
maxω|T(iω|<1.2
|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
\omega \in [0, 10] rad/s
|PS(i\omega)| \leq 0.01 for
\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
\textrm{Re}\ s \leq - \alpha for some
\alpha > 0, i.e.
H(s) is analytic on
\textrm{Re } s > -\alpha. Let
h(t) be the corresponding time domain function, i.e.
H = \mathcal L h, where
\mathcal L denotes the Laplace transform. Show that for any
\textrm{Re } s_0 > -\alpha we have
\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
- If
\lim_{s\to0}\textcolor{red}{s}P(s)C(s)=c_1, for some
0 < |c_1| < \infty, then
\lim_{t \to \infty} e(t) = 0 and
\int_0^\infty e(t)dt = \frac{1}{c_1}
- If
\lim_{s \to 0}s^2 P(s)C(s) = c_2, for some
0 < |c_2| < \infty, then
\lim_{t \to \infty} e(t) = 0 and
\int_0^\infty e(t) dt = 0
Hint: ConsiderP(s)C(s) = \tilde L(s) / s where
\lim_{s \to 0}\tilde L(s) = c_1 and apply the final value theorem.
- If
- 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
\int_0^\infty e^{-pt}e(t) dt = 0
\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
\int_0^\infty e^{-qt}e(t)dt = \frac{1}{q}
\int_0^\infty e^{-qt}y(t) = 0
- Discuss implications of B, C and D on overshoot and undershoot.