17.6: Exercises
- Page ID
- 32163
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercise 17.1
Let \(P(s)=e^{-2 s}-1\) be connected in a unity feedback configuration. Is this system well-posed?
Exercise 17.2
Assume that \(P_{\alpha\) and \(K\) in the diagram are given by:
\[P_{\alpha}(s)=\left(\begin{array}{cc}
\frac{s}{s+1} & \frac{-\alpha}{s+1} \\
\frac{1}{(s+1)} & \frac{1}{s+1}
\end{array}\right), \quad \alpha \in \mathbb{R}, \quad K(s)=\left(\begin{array}{cc}
\frac{s+1}{s(s+5)} & 0 \\
-\frac{s+1}{s(s+5)} & \frac{s+1}{s+5}
\end{array}\right)\nonumber\]
- Is the closed loop system stable for all \(\alpha > 0\)?
- Is the closed loop system stable for \(\alpha = 0\)?
Exercise 17.3
Consider the standard servo loop, with
\[P(s)=\frac{1}{10 s+1}, \quad K(s)=k\nonumber\]
but with no measurement noise. Find the least positive gain such that the following are all true:
- The feedback system is internally stable.
- With no disturbance at the plant output \((d(t) \equiv 0)\), and with a unit step on the command signal \(r(t)\), the error \(e(t) = r(t) - y(t)\) settles to \(|e(\infty)| \leq 0.1\).
- Show that the \(\mathcal{L}_{2}\) to \(\mathcal{L}_{\infty}\) induced norm of a SISO system is given by \(\mathcal{H}_{2}\) norm of the system.
- With zero command \((r(t) \equiv 0),\|y\|_{\infty} \leq 0.1\) for all \(d(t)\) such \(\|d\|_{2} \leq 1\). [ADD NEW Problem]
Exercise 17.4 Parametrization of Stabilizing Controllers
Consider the diagram shown below where \(P\) is a given stable plant. We will show a simple way of parametrizing all stabilizing controllers for this plant. The plant as well as the controllers are finite dimensional.
1. Show that the feedback controller
\[K=Q(I-P Q)^{-1}=(I-Q P)^{-1} Q\nonumber\]
for any stable rational \(Q\) is a stabilizing controller for the closed loop system
2. . Show that every stabilizing controller is given by \(K=Q(I-P Q)^{-1}\) for some stable \(Q\). (Hint: Express \(Q\) in terms of \(P\) and \(K\)).
3. Suppose \(P\) is SISO, \(w_{1}\) is a step, and \(w_{2} = 0\). What conditions does \(Q\) have to satisfy for the steady state value of \(u\) to be zero. Is it always possible to satisfy this condition?
Exercise 17.5
Consider the block diagram shown in the figure below.
(a) Suppose \(P(s)=\frac{2}{s-1}\), \(P_{0}(s)=\frac{1}{s-1}\) and \(Q = 2\). Calculate the transfer function from \(r\) to \(y\).
(b) Is the above system internally stable?
(c) Now suppose that \(P (s) = P_{0}(s) = H(s)\) for some \(H(s)\). Under what conditions on \(H(s)\) is the system internally stable for any stable (but otherwise arbitrary) \(Q(s)\)?
Exercise 17.6
Consider the system shown in the figure below.
The plant transfer function is known to be given by:
\[P(s)=\left[\begin{array}{cc}
\frac{s-1}{s+1} & 1 \\
0 & \frac{s+1}{s+2}
\end{array}\right]\nonumber\]
A control engineer designed the controller \(K(s)\) such that the closed-loop transfer function from \(r\) to \(y\) is:
\[H(s)=\left[\begin{array}{cc}
\frac{1}{s+4} & 0 \\
0 & \frac{1}{s+4}
\end{array}\right]\nonumber\]
(a) Compute \(K(s)\).
(b) Compute the poles and zeros (with associated input zero directions) of \(P (s)\) and \(K(s)\).
(c) Are there pole/zero cancellations between \(P (s)\) and \(K(s)\) ?
(d) Is the system internally stable? Verify your answer
Exercise 17.7
An engineer wanted to estimate the peak-to-peak gain of a closed loop system \(h\) (the input-output map). The controller was designed so that the system tracks a step input in the steady state. The designer simulated the step response of the system and computed the amount of overshoot \((e_{1})\) and undershoot \((e_{2})\) of the response. He/She immediately concluded that
\[\|h\|_{1} \geq 1+2 e_{1}+2 e_{2}\nonumber\]
Is this a correct conclusion? Verify