In the dc-dc converter of 67, Chapter 4 (Van Dijk, 1995), | StudySoup

Textbook Solutions for Control Systems Engineering

Chapter 12 Problem 34

Question

Problem 22 in Chapter 3 introduced the model for patients treated under a regimen of a single day of Glargine insulin (Tarín, 2005). The model to find the response for a specific patient to medication can be expressed in phase-variable form with

\(\mathbf{A}=\left[\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -501.6 \times 10^{-6} & -128.8 \times 10^{-3} & -854 \times 10^{-3} \end{array}\right]\);

\(\mathbf{B}=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right] ; \quad \mathbf{C}=\left[\begin{array}{lll} 0.78 \times 10^{-4} & 41.4 \times 10^{-4} & 0.01 \end{array}\right]\)

\(\mathbf{D}=0\)

The state variables will take on a different significance in this expression, but the input and the output remain the same. Recall that u = external insulin flow, and y = plasma insulin concentration.

a. Obtain a state-feedback gain matrix so that the closed-loop system will have two of its poles placed at -1/5 and the third pole at -1/2.

Solution

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The first step in solving 12 problem number 34 trying to solve the problem we have to refer to the textbook question: Problem 22 in Chapter 3 introduced the model for patients treated under a regimen of a single day of Glargine insulin (Tarín, 2005). The model to find the response for a specific patient to medication can be expressed in phase-variable form with\(\mathbf{A}=\left[\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -501.6 \times 10^{-6} & -128.8 \times 10^{-3} & -854 \times 10^{-3} \end{array}\right]\);\(\mathbf{B}=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right] ; \quad \mathbf{C}=\left[\begin{array}{lll} 0.78 \times 10^{-4} & 41.4 \times 10^{-4} & 0.01 \end{array}\right]\)\(\mathbf{D}=0\)The state variables will take on a different significance in this expression, but the input and the output remain the same. Recall that u = external insulin flow, and y = plasma insulin concentration.a. Obtain a state-feedback gain matrix so that the closed-loop system will have two of its poles placed at -1/5 and the third pole at -1/2.
From the textbook chapter DESIGN VIA STATE SPACE you will find a few key concepts needed to solve this.

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full solution

Title Control Systems Engineering 7 
Author Norman J. Nise
ISBN 9781118170519

In the dc-dc converter of 67, Chapter 4 (Van Dijk, 1995),

Chapter 12 textbook questions

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