The state-variable formulation of a system is ẋ = Ax + Bu; y = [
![The state-variable formulation of a system is ẋ = Ax + Bu; y = [](/img/relate-questions.png)
A. <span class="math-tex">\(\frac{{s + 2}}{{{s^2} + 5s - 6}}\)</span>
B. <span class="math-tex">\(\frac{{2s + 5}}{{{s^2} + 5s + 6}}\)</span>
C. <span class="math-tex">\(\frac{{2s - 5}}{{{s^2} + 5s - 6}}\)</span>
D. <span class="math-tex">\(\frac{{s + 1}}{{{s^2} + 5s + 6}}\)</span>
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Concept:
X’ = Ax + Bu → (1)
Y = Cx + Du → (2)
Apply Laplace transform, to equation (1)
S x(s) = A x(s) B u(s)
⇒ B u(s) = (SI - A) X(s)
⇒ X(s) = (SI - A)-1 B u(s)
Now apply Laplace transform to equation (2)
Y(s) = C x(s) + D u(s)
⇒ Y(s) = C[(SI - A)-1] B u(s) + D u(s)
\(\Rightarrow \frac{{Y\left( s \right)}}{{U\left( s \right)}} = C{\left( {SI - A} \right)^{ - 1}}B + D\)
Calculation:
From the given state-space representation,
\(A = \left[ {\begin{array}{*{20}{c}} { - 3}&1\\ 0&{ - 2} \end{array}} \right],B = \left[ {\begin{array}{*{20}{c}} 2\\ 1 \end{array}} \right],C = \left[ {\begin{array}{*{20}{c}} 1&0 \end{array}} \right],D = 0\)
\(\left( {sI - A} \right) = \left[ {\begin{array}{*{20}{c}} {s + 3}&{ - 1}\\ 0&{s + 2} \end{array}} \right]\)
\({\left( {sI - A} \right)^{ - 1}} = \frac{1}{{\left( {{\rm{s}} + 3} \right)\left( {{\rm{s}} + 2} \right)}}\left[ {\begin{array}{*{20}{c}} {s + 2}&1\\ 0&{s + 3} \end{array}} \right]\)
\( = \frac{1}{{{s^2} + 5s + 6}}\left[ {\begin{array}{*{20}{c}} {s + 2}&1\\ 0&{s + 3} \end{array}} \right]\)
\(TF = \left[ {\begin{array}{*{20}{c}} 1&0 \end{array}} \right]\frac{1}{{{s^2} + 5s + 6}}\left[ {\begin{array}{*{20}{c}} {s + 2}&1\\ 0&{s + 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2\\ 1 \end{array}} \right] + 0\)
\( = \frac{{2s + 5}}{{{s^2} + 5s + 6}}\)