Question

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.

   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.

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Discrete-Time Control Systems (Pie)
Discrete-Time Control Systems (Pie)
Katsuhiko Ogata 2nd Edition
Chapter 7, Problem 4 ↓

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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) $$  Show more…

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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.
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