Let the state-space representation of an LTI system be ẋ(t) = A x(t) + B u(t), y(t) = C x(t) + d u(t) where A, B, C are matrices, d is a scalar, u(t) is the input to the system, and y(t) is its output. Let B = [0 0 1]T and d = 0. Which one of the following options for A and C will ensure that the transfer function of this LTI system is $H\left( s \right) = \frac{1}{{{s^3} + 3{s^2} + 2s + 1}} = ?$

Question

Let the state-space representation of an LTI system be ẋ(t) = A x(t) + B u(t), y(t) = C x(t) + d u(t) where A, B, C are matrices, d is a scalar, u(t) is the input to the system, and y(t) is its output. Let B = [0 0 1]T and d = 0. Which one of the following options for A and C will ensure that the transfer function of this LTI system is $H\left( s \right) = \frac{1}{{{s^3} + 3{s^2} + 2s + 1}} = ?$

TuteeHUB · Accepted Answer

$A = \left[ {\begin{array}{*{20}{c}} 0&1&0\ 0&0&1\ { - 3}&{ - 2}&{ - 1} \end{array}} \right]and\;C = \;\left[ {\begin{array}{*{20}{c}} 1&0&0 \end{array}} \right]$

TuteeHUB · Answer

$A = \left[ {\begin{array}{*{20}{c}} 0&1&0\ 0&0&1\ { - 1}&{ - 2}&{ - 3} \end{array}} \right]and\;C = \;\left[ {\begin{array}{*{20}{c}} 1&0&0 \end{array}} \right]$

TuteeHUB · Answer

$A = \left[ {\begin{array}{*{20}{c}} 0&1&0\ 0&0&1\ { - 1}&{ - 2}&{ - 3} \end{array}} \right]and\;C = \;\left[ {\begin{array}{*{20}{c}} 0&0&1 \end{array}} \right]$

TuteeHUB · Answer

$A = \left[ {\begin{array}{*{20}{c}} 0&1&0\ 0&0&1\ { - 3}&{ - 2}&{ - 1} \end{array}} \right]and\;C = \;\left[ {\begin{array}{*{20}{c}} 0&0&1 \end{array}} \right]$

1.	Let the state-space representation of an LTI system be ẋ(t) = A x(t) + B u(t), y(t) = C x(t) + d u(t) where A, B, C are matrices, d is a scalar, u(t) is the input to the system, and y(t) is its output. Let B = [0 0 1]T and d = 0. Which one of the following options for A and C will ensure that the transfer function of this LTI system is \(H\left( s \right) = \frac{1}{{{s^3} + 3{s^2} + 2s + 1}} = ?\)
A.	\(A = \left[ {\begin{array}{{20}{c}} 0&1&0\\ 0&0&1\\ { - 1}&{ - 2}&{ - 3} \end{array}} \right]and\;C = \;\left[ {\begin{array}{{20}{c}} 1&0&0 \end{array}} \right]\)
B.	\(A = \left[ {\begin{array}{{20}{c}} 0&1&0\\ 0&0&1\\ { - 3}&{ - 2}&{ - 1} \end{array}} \right]and\;C = \;\left[ {\begin{array}{{20}{c}} 1&0&0 \end{array}} \right]\)
C.	\(A = \left[ {\begin{array}{{20}{c}} 0&1&0\\ 0&0&1\\ { - 1}&{ - 2}&{ - 3} \end{array}} \right]and\;C = \;\left[ {\begin{array}{{20}{c}} 0&0&1 \end{array}} \right]\)
D.	\(A = \left[ {\begin{array}{{20}{c}} 0&1&0\\ 0&0&1\\ { - 3}&{ - 2}&{ - 1} \end{array}} \right]and\;C = \;\left[ {\begin{array}{{20}{c}} 0&0&1 \end{array}} \right]\)
Answer» B. \(A = \left[ {\begin{array}{{20}{c}} 0&1&0\\ 0&0&1\\ { - 3}&{ - 2}&{ - 1} \end{array}} \right]and\;C = \;\left[ {\begin{array}{{20}{c}} 1&0&0 \end{array}} \right]\)

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