Question
Download Solution PDFThe system matrix of a linear time-invariant continuous-time system is
Given by \(A\left[ {\begin{array}{*{20}{c}} 0&1\\ { - 4}&{ - 5} \end{array}} \right]\). What are the roots of the characteristic equation?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
For the root of the characteristic equation:
[sI - A] = 0
Where
I = Unit matrix
\({\rm{I}} = \left[ {\;\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)
A = Given matrix
Explanation:
Given
\(A=\left[ {\begin{array}{*{20}{c}} 0&1\\ { - 4}&{ - 5} \end{array}} \right]\)
For the root of the characteristic equation:
[sI - A] = 0
\({\left[ {\left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 0&1\\ -4&-5 \end{array}} \right]} \right]^{ }}=0\)
\(\left[ {\begin{array}{*{20}{c}} s&-1\\ { 4}&{ s+ 5} \end{array}} \right]=0\)
s(s + 5) + 4 = 0
s2 + 5s + 4 = 0
(s+4)(s+1) = 0
s = -4, -1
Last updated on Jun 23, 2025
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