Exercise 1

1.Nichols diagrams

Sketch the Nichols diagrams for the following systems

LaTeX: 
\frac{1}{s(s+1)(s+10)}, 
\quad \frac{1}{s-1},
\quad \frac{\exp{(-s)}}{1+s},
\quad \frac{1-s}{s(1+s)},
\quad \frac{1}{s^2+2 \zeta s + 1}, \; (\zeta \textrm{ small})
1s(s+1)(s+10),1s1,exp(s)1+s,1ss(1+s),1s2+2ζs+1,(ζ small)

For what feedback gains is the closed loop system stable?

 

 

2. Analysis of design

Consider the design in Handin 1B:

Output error feedback control of LaTeX: G(s)=(s+1)/s^2G(s)=(s+1)/s2 with closed loop poles given by the design polynomials LaTeX: s^2+2\zeta\omega s+\omega^2s2+2ζωs+ω2 with LaTeX: \omega=10, \zeta=0.71ω=10,ζ=0.71 for both the state feedback and the observer pole placement. 

 

Construct code, in language of your choice, to plot a) Gang of Four, b) Nyquist diagram c) Nichols diagram d) Root locus for the design.

You can use the matlab code ex01hint.m Download ex01hint.m to get started.

 

3 Cancelling slow process poles

Consider the following system with a slow process pole in LaTeX: s = -0.1s=0.1
LaTeX: G(s) =\frac{1}{10s+1}G(s)=110s+1
controlled with the controller
LaTeX: C(s) =\frac{10(10s + 1)}{s}C(s)=10(10s+1)s

which cancels the slow pole and puts the closed loop pole in LaTeX: s = −10s=10. Is fast
control achieved? Plot the Gang of Four and study the time response of an
input load disturbance.

 

4 Taking care of an oscillating load disturbance

Consider a system with transfer function

LaTeX: G(s) = \frac{1}{(s+1)^3}.
G(s)=1(s+1)3.

Assume a disturbance LaTeX: l(t)=\sin \omega tl(t)=sinωt with LaTeX: \omega = 0.05ω=0.05 acts on the process input.

 

a)  Investigate the error obtained with a PI

controller with LaTeX: k=1k=1 and LaTeX: T_i=2Ti=2. Try to trim the PI controller to reduce the impact of the disturbance. What is the trade-off ?

 

b) Construct an LTI controller LaTeX: C\left(s\right)C(s) so that the error due to the sinusoidal disturbance, after initial transients, is less than 0.001,  i.e. LaTeX: |y(t)|<0.001|y(t)|<0.001 for large LaTeX: tt. What is the main trade-off ?

Hint: Should the controller have low or high gain at LaTeX: \omega=0.05ω=0.05?

 

Plot GoF, Bode,  Nichols and Nyquist plots, root loci. Also show time domain  responses to reference step change and impact of the input disturbance LaTeX: l(t)=\sin \omega tl(t)=sinωt.

 

5 A non-fractional transfer function related to heat propagation

The transfer function for one-dimensional heat propagation in a semi-infinite metal rod, without dissipation, is given by (see Example 9.6 in Åström-Murray 2020)

LaTeX: G(s) = e^{-\sqrt{sT}}G(s)=esT.

a) Determine the critical gain under i) proportional control and ii) integral control.

b) The transfer function LaTeX: G(s)G(s) relates temperatures at points within a given distance LaTeX: xx. The time constant LaTeX: T=\frac{C_V}{\lambda} x^2T=CVλx2 is related to material properties of the metal. Measurements of LaTeX: G(i\omega)G(iω) using sinusoidal temperature variations were used by Ångström in 1861 to determine thermal conductivity LaTeX: \lambdaλ. One problem that Ångström had was that the experimental transfer function turned out to be closer to

LaTeX: G_\mu(i\omega) = e^{-\sqrt{(i\omega+\mu)T}}Gμ(iω)=e(iω+μ)T

where the parameter  LaTeX: \mu>0μ>0 is related to heat losses due to radiation and convection during the experiments, which might be hard to determine. Show that if

LaTeX: A(\omega) = \log |G_\mu(i\omega)|  \quad \textrm{and} \quad \varphi(\omega) = \mathrm{arg}\,G_\mu(i\omega)A(ω)=log|Gμ(iω)|andφ(ω)=argGμ(iω)

then the following remarkable relation holds

LaTeX: A(\omega)\cdot \varphi(\omega) = \frac{\omega T}{2}A(ω)φ(ω)=ωT2.

Note that this relation is independent of LaTeX: \muμ, making it easy to estimate LaTeX: TT from experimental data.

c)  Suggest a way to simulate such a system (for example the step response of G(s) under PI control). Hint: Use FFT and IFFT.

----------------------------------------------------------------------------------------------------------

On the exercise we will also discuss the results of Handin 1 and familiarize ourselves with Handin 2: The Flexible Servo Benchmark problem