Consider the following open-loop transfer functions, where G(s) = Y(s) / U(s) . Y(s) is the Laplace transform of the output, and U(s) is the Laplace transform of the input control signal: i. \G(s) = \(\frac{(s+3)}{(s+4)^2}\) ii. G(s) = \(\frac{s}{(s+5)(s+7)}\) iii. G(s) = \(\frac{20 s(s+7)}{(s+3)(s+7)(s+9)}\) iv. G(s) = \(\frac{30(s+2)(s+3)}{(s+4)(s+5)(s+6)}\) v. G(s) = \(\frac{s^2+8 s+15}{\left(s^2+4 s+10\right)\left(s^2+3 s+12\right)}\) For each of these transfer functions, do the following: [Section: 12.2] a. Draw the signal-flow graph in phase-variable form. b. Add state-variable feedback to the signal-flow graph. c. For each closed-loop signal-flow graph, write the state equations. d. Write, by inspection, the closed-loop transfer function, T(s), for your closed-loop signal-flow graphs. e. Verify your answers for T(s) by finding the closed loop transfer functions from the state equations and Eq. (3.73).
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Textbook Solutions for Control Systems Engineering
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
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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