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Design a state-variable feedback controller to yield a 20.8% overshoot and settling time of 4 seconds for a plant

$$G(s) = \frac{(s+4)}{(s+1)(s+2)(s+5)}$$

I am studying Design Project via State Space (Chapter 12 - Norman Nise - Control System Engineering) and I am very doubts about the specificity development that transfer function.

Flux-flow signal:

enter image description here

and the first state space equation:

enter image description here

when I look the E.E plant represented in cascade, I understand all above development, except for this expression:

$$y = \textbf{C}_z\textbf{z} = [-1\quad 1\quad 0]\textbf{z}$$

My question: Why the output variable y doesn't have the following expression, consider the derivative (s+4) like this:

$$y = z_1(s+4) = \dot{z}_1 + 4z_1 = 4z_1 + z_2$$

OBS.: \$ z_1 \$ and \$ z_2\$ are a variable of state.

$$y = \textbf{C}_z\textbf{z} = [4\quad 1\quad 0]\textbf{z}$$

I really really confused how can they find the -1 that vector row in the first place.

thanks for your help!

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  • \$\begingroup\$ Without more context it is impossible to answer, unless one of us has that specific book at hand. What is y? And z1 and z2? What's the block diagram of the system? Etc. \$\endgroup\$
    – LorenzoDonati4Ukraine-OnStrike
    Jun 24, 2015 at 0:10
  • \$\begingroup\$ BTW this site is about building a repository of Q&A useful for anyone browsing the site. The question should be comprehensible on its own, without needing someone to buy/have a specific book. Please add all the relevant informations to achieve this goal. \$\endgroup\$
    – LorenzoDonati4Ukraine-OnStrike
    Jun 24, 2015 at 0:12
  • \$\begingroup\$ the few details was describle on the question. This is a example from the book. \$\endgroup\$
    – miguel747
    Jun 24, 2015 at 0:16
  • \$\begingroup\$ If this is really all the example says, it will surely refer to other parts of the book. Edit your answer to add more of that info. We cannot guess what those symbols mean for your book! \$\endgroup\$
    – LorenzoDonati4Ukraine-OnStrike
    Jun 24, 2015 at 0:21
  • \$\begingroup\$ ok. I already done. \$\endgroup\$
    – miguel747
    Jun 24, 2015 at 0:21

1 Answer 1

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There is an unnamed node in the flow diagram between \$z_2\$ and \$z_1\$. Call it \$k\$. The equation for \$k\$ can be written as:

\[ k = z_2 - 5 z_1 \]

Then the equation for y becomes:

\[ y = k + 4 z_1 = (z_2 - 5 z_1) + 4 z_1 = -z_1 + z_2 = [-1\quad 1\quad 0]\;\textbf{z} \]

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