• Home
  • Textbooks
  • Discrete-Time Control Systems (Pie)
  • Polynomial Equations Approach to Control Systems Design

Discrete-Time Control Systems (Pie)

Katsuhiko Ogata

Chapter 7

Polynomial Equations Approach to Control Systems Design - all with Video Answers

Educators


Chapter Questions

08:44

Problem 1

Consider polynomials $A(z)$ and $B(z)$ defined by
$$
\begin{aligned}
& A(z)=z^2+a_1 z+a_2=\left(z-\lambda_1\right)\left(z-\lambda_2\right)=z^2-\left(\lambda_1+\lambda_2\right) z+\lambda_1 \lambda_2 \\
& B(z)=b_1 z+b_2=b\left(z-\lambda_3\right)=b z-b \lambda_3
\end{aligned}
$$
where $b=b_1$. Define the Sylvester matrix $\mathbf{E}$ by
$$
\mathbf{E}=\left[\begin{array}{llll}
a_2 & 0 & b_2 & 0 \\
a_1 & a_2 & b_1 & b_2 \\
1 & a_1 & 0 & b_1 \\
0 & 1 & 0 & 0
\end{array}\right]
$$

Show that the determinant of $\mathbf{E}$ can be given by
$$
|\mathbf{E}|=b^2\left(\lambda_1-\lambda_3\right)\left(\lambda_2-\lambda_3\right)
$$

Eduard Sanchez
Eduard Sanchez
Numerade Educator
00:33

Problem 2

Consider the following Diophantine equation:
$$
\alpha(z) A(z)+\beta(z) B(z)=1
$$
where
$$
\begin{aligned}
& A(z)=z^2-0.7 z+0.1 \\
& B(z)=z^2+0.2 z-0.24 \\
& \alpha(z)=\alpha_0 z+\alpha_1 \\
& \beta(z)=\beta_0 z+\beta_1
\end{aligned}
$$

Solve this Diophantine equation for $\alpha(z)$ and $\beta(z)$ and determine coefficients $a_0, a_1$. $\beta_0$, and $\beta_1$.

Lily An
Lily An
Numerade Educator

Problem 3

Consider the plant $Y(z) V U(z)$, where
$$
\frac{Y(z)}{U(z)}=\frac{B(z)}{A(z)}
$$

Assume that $A(z)$ is an $n$ th-degree monic polynomial in $z$ and $B(z)$ is an $m$ th-degree polynomial in 2. Assume also that there are no common factors between $A(z)$ and $B(z)$; that is, the plant is completely state controllable and completely observable.
Consider the following Diophantine equation:
$$
\gamma(z) A(z)+\beta(z) B(z)=F(z)[H(z)-A(z)]
$$
where $H(z)$ is the desired characteristic polynomial for the pole placement part and $F(z)$ is the desired characteristic polynomial for the minimum-order observer. $[H(z)$ is an $n$ th-degree monic polynomial and $F(z)$ is an $(n-1)$ th-degree polynomial.]

Show that if this Diophantine equation is solved for $\beta(z)$ and $\gamma(z)$, then the use of
$$
U(z)=-\frac{\gamma(z)}{F(z)} U(z)-\frac{\beta(z)}{F(z)} Y(z)
$$
will accomplish the desired observed-state feedback regulator system.

Check back soon!

Problem 4

In Example 7-3 a control system was designed such that the desired characteristic equation for the pole placemeat part was
$$
H(z)=(z-0.6-j 0.4)(z-0.6+j 0.4)=z^2-1.2 z+0.52
$$
and the desired characteristic polynomial for the minimum-order observer was
$$
F(z)=z
$$

The Diophantine equation given by Equation (7-13) was solved. $\alpha(z)$ and $\beta(z)$ were determined as follows:
$$
\begin{aligned}
& \alpha(z)=z+0.32 \\
& \beta(z)=24 z-16
\end{aligned}
$$

The constant $K_0$ was determined as 8 . Figure 7-6(a) shows the designed system
Show that the conttol signal $u(k)$ can be given by
$$
\begin{aligned}
& u(k)=-0.32 u(k-1)-24 y(k)+16 y(k-1)+8 r(k), \quad k=1,2,3, \ldots \\
& u(0)=-24 y(0)+8 r(0)
\end{aligned}
$$

Plot $u(k)$ versus $k$ when the input $r(k)$ is a unit-step sequence.

Check back soon!

Problem 5

Consider a plant defined by
$$
\begin{aligned}
\mathbf{x}(k+1) & =\mathrm{Gx}(k)+\mathrm{H} u(k) \\
y(k) & =\mathrm{Cx}(k)
\end{aligned}
$$
where The pulse transfer function for the plant can be written as
$$
\frac{Y(z)}{U(z)}=\frac{B(z)}{A(z)}
$$

Determine polynomials $A(z)$ and $B(z)$.
Using the polynomial equations approach, design a control system for the plant.

It is desired that the block diagram configuration of the designed system is the same as that of Figure 7-4. In solving the Diophantine equation
$$
\alpha(z) A(z)+\beta(z) B(z)=F(z) H(z)
$$
assume that $H(z)$ and $F(z)$ are, respectively, as follows:
$$
H(z)=z^3, \quad F(z)=z^2
$$

Obtain the unit-step response and unit-ramp response of the designed control system. The sampling period $T$ is $1 \mathrm{sec}$.

Check back soon!

Problem 6

Consider the same plant as given in Problem B-7-5. Using the polynomial equations approach, design a control system for the plant. Use the block diagram configuration shown in Figure 7-5. Assume the following $H(z)$ and $F(z)$ :
$$
H(z)=z^3, \quad F(z)=z^2
$$

Obtain the unit-step response and unit-ramp response of the designed control system. Assume the sampling period $T$ to be $1 \mathrm{sec}$.

Check back soon!

Problem 7

Consider the plant defined by
$$
\begin{aligned}
\mathbf{x}(k+1) & =\mathbf{G} \mathbf{x}(k)+\mathbf{H} u(k) \\
y(k) & =\mathbf{C x}(k)
\end{aligned}
$$
where
$$
\mathbf{G}=\left[\begin{array}{ccc}
0 & 0 & -0.25 \\
1 & 0 & 0 \\
0 & 1 & 0.5
\end{array}\right], \quad \mathbf{H}=\left[\begin{array}{l}
1 \\
0 \\
1
\end{array}\right], \quad \mathbf{C}=\left[\begin{array}{lll}
1 & 0 & 0
\end{array}\right]
$$

Design a control system for the plant. For the pole placement part, we want to have three closed-loop poles at the origin, or
$$
H(z)=z^3
$$
and for the characteristic equation for the minimum-order observer, we want to have
$$
F(z)=z^2
$$

Use the polynomial equations approach to the design

Check back soon!

Problem 8

Consider the plant defined by
$$
\frac{Y(z)}{U(z)}=\frac{0.6 z+0.5}{(z-1)^2}
$$
Using the polynomial equations approach, design a control system such that the system will behave like the following model, $G_{\text {model: }}$
$$
G_{\text {medet }}=\frac{2 z-1}{z^2}
$$

Obtain the unit-step response and unit-ramp response of the designed system (which is the same as $G_{\text {model }}$ ) The sampling period $T$ is $1 \mathrm{sec}$.

Check back soon!

Problem 9

Consider the plant defined by
$$
\frac{Y(z)}{U(z)}=\frac{0.01873(z+0.9356)}{(z-1)(z-0.8187)}
$$

Using the polynomial equations approach, design a control system such that the system will behave like the following model, $G_{\text {madal }}$
$$
G_{\text {matal }}=\frac{0.32}{z^2-1.2 z+0.52}
$$

Obtain the unit-step response and unit-ramp response of the control system (which is the same as the model, $\left.G_{\text {modet }}\right)$. The sampling period $T$ is $0.2 \mathrm{sec}$.

Check back soon!