8.2: State-Transition Matrix and Asymptotic Stability
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The State-Transition Matrix
Consider the homogenous state equation: \(\dot{\bf x}(t)={\bf Ax}(t),\, \, \, {\bf x}(0)={\bf x}_{0}\).
The solution to the homogenous equation is given as: \({\bf x}(t)=e^{{\bf A}t} {\bf x}_{0}\), where the state-transition matrix, \(e^{{\bf A}t}\), describes the evolution of the state vector, \(x\left(t\right)\).
The state-transition matrix of a linear time-invariant (LTI) system can be computed in the multiple ways including the following:
- \(e^{{\bf A}t} ={\rm \mathcal L}^{-1} [(s{\bf I}-{\bf A})^{-1} ]\)
- \(e^{{\bf A}t} =\sum _{0}^{\infty } \frac{{\bf A}^{i} t^{i} }{i\, !}\)
- By using the modal matrix (see below)
- By using the fundamental matrix
- By using the Cayley–Hamilton theorem
Characteristic Polynomial of A
The characteristic polynomial of \({\bf A}\) is an \(n\)th order polynomial obtained as the determinant of \((s{\bf I}-{\bf A})\), i.e., \(\Delta (s)=|s{\bf I}-{\bf A}|\). The roots of the characteristic polynomial are the eigenvalues of \({\bf A}\).
The transfer function, \(G\left(s\right)\), is expressed as:
\[G\left(s\right)={\bf c}^T\frac{adj\left(s{\bf I}-{\bf A}\right)}{\left|s{\bf I}-{\bf A}\right|}{\bf b}=\frac{n\left(s\right)}{d\left(s\right)} \nonumber \]
where \({adj\left(s{\bf I}-{\bf A}\right)}\) represents the adjoint matrix of \(s{\bf I}-{\bf A}\).
Assuming no pole-zero cancelations in \(G\left(s\right)\), the characteristic polynomial matches the denominator polynomial.In the event of pole-zero cancelations, the order of the denominator polynomial, \(d\left(s\right)\), is less than \(n\), i.e., the zeros of \(d\left(s\right)\) form a subset of the eigenvalues of \({\bf A}\).
State-Transition Matrix in MATLAB
The state-transition matrix may be obtained by using the symbolic variable 't' defined by using the 'syms' command from the MATLAB Symbolic Math Toolbox. Assuming that a symbolic variable 't' and an \(nxn\) numeric matrix \({\bf A}\) have been defined, the state-transition matrix can be obtained by issuing the matrix exponential command as: \(\rm expm(t*{\bf A})\).
The Modal Matrix
Consider the homogenous state equation: \(\dot{\bf x}(t)={\bf Ax}(t),\, \, \, {\bf x}(0)={\bf x}_{0}\).
Assume that \({\bf A}\) has a full set of eigenvectors, that obey \(\left({\lambda }_i{\bf I}-{\bf A}\right){\bf v}_i=0,\ \ i=1,\dots ,n\), where \({\lambda }_i\) denotes an eigenvalue of \({\bf A}\).
Let \(\bf M=\left[v_1,\ v_2,\cdots ,v_n\right]\) denote the modal matrix containing eigenvectors of \({\bf A}\), \(e^{\mathit{\bf \Lambda}t}=diag\left(\left[e^{{\lambda }_1t},e^{{\lambda }_2t},\cdots ,e^{{\lambda }_nt}\right]\right)\) define a diagonal matrix of the natural response modes of \({\bf A}\), and \({\bf c}\) denote a constant vector; then, the general solution to the homogenous state equation is given as:
\[{\bf x}_{h} (t)={\bf M}e^{{\bf \Lambda} t} {\bf c} \nonumber \]
Assuming a set of initial conditions: \({\bf x} _{h} (0)= {\bf x} _{0} = {\bf Mc}\); we have, \( {\bf c} = {\bf M} ^{-1} {\bf x} _{0}\). Then, a particular solution to the homogenous system is given as:
\[{\bf x}_{h} (t)={\bf M}e^{{\bf \Lambda} t} {\bf M}^{-1} {\bf x}_{0} \nonumber \]
The state-transition matrix can be computed from the modal matrix as:
\[e^{{\bf A}t} ={\bf M}e^{{\bf \Lambda} t} {\bf M}^{-1} \nonumber \]
Modal Matrix in MATLAB
The modal matrix is obtained by using the ‘eig’ command in MATLAB. The eigenvectors of \({\bf A}\) obtained from MATLAB are normalized to unity. The 'eig' command also provides a diagonal matrix of eigenvalues of \({\bf A}\). Given the modal matrix \(\bf M\) of eigenvectors and the diagonal matrix \(\bf D\) of eigenvalues, the state-transition matrix is obtained as \(\rm M*expm(t*D)/M\).
Asymptotic Stability
The asymptotic stability refers to the long-term behavior of the natural response modes of the system. These modes are also reflected in the state-transition matrix, \(e^{{\bf A}t}\).
Consider the homogenous state equation: \(\dot{\bf x}(t)={\bf Ax}(t),\, \, \, {\bf x}(0)={\bf x}_{0}\).
The homogenous state equation is said to be asymptotically stable if \({\mathop{\lim }\limits_{t\to \infty }} {\bf x}(t)=0\).
Since \({\bf x}(t)=e^{{\bf A}t} {\bf x}_0\), the homogenous state equation is asymptotically stable if \(\displaystyle\lim_{t\to \infty} e^{{\bf A}t}=0\).
Further, using modal decomposition, \(e^{{\bf A}t} ={\bf M}e^{ {\bf \Lambda} t} {\bf M}^{-1}\), the homogenous system is asymptotically stable if \(\displaystyle\lim_{t\to \infty} e^{{\bf \Lambda} t} =0\).
Since \(e^{\mathit{\bf \Lambda}t}=diag\left(\left[e^{{\lambda }_1t},e^{{\lambda }_2t},\cdots ,\;e^{{\lambda }_nt}\right]\right)\), the above condition implies that \(Re\left[{\lambda }_i\right]<0,\ i=1,\dots ,n\), where \({\lambda }_i\) represents a root of the characteristic polynomial: \(\Delta (s)=|s{\bf I-A}|\).
The computation of modal and state-transition matrices is illustrated separately when the characteristic polynomial, \(\mathit{\Delta}(s)\), has real or complex roots.
For the mass-spring-damper model, let \(m=1,\; k=2,\; {\rm and}\, b=3\); then, the characteristic polynomial is: \(\Delta (s)=s^{2} +3s+2\), which has real roots: \(s_{1} ,s_{2} =-1,-2\).
The natural response modes are: \(\{ e^{-t} ,e^{-2t} \}\).
The modal matrix of eigenvectors is obtained as: \({\bf M}=\left[\begin{array}{cc} {-1} & {-1} \\ {1} & {2} \end{array}\right]\). The diagonal matrix of eigenvalues is: \(\mathit{\bf \Lambda}=\left[ \begin{array}{cc} -1 & 0 \\ 0 & -2 \end{array} \right]\).
Since \({\bf A}={\bf M}\mathit{\bf \Lambda}{\bf M}^{-1}\), we have: \(e^{{\bf A}t}={\bf M}e^{\mathit{\bf \Lambda}t}{\bf M}^{-1}\), which computes as: \(e^{{\bf A}t} =\left[\begin{array}{cc} {2e^{-t} -e^{-2t} } & {e^{-t} -e^{-2t} } \\ {2e^{-2t} -2e^{-t} } & {2e^{-2t} -e^{-t} } \end{array}\right]\).
Further, since \({\mathop{\lim }\limits_{t\to \infty }} e^{{\bf A}t} =0\), the homogenous state equation is asymptotically stable.
For the mass-spring-damper model, , let \(m=1,\; k=2,\; {\rm and}\, b=2\); then, the characteristic polynomial is: \(\Delta (s)=s^{2} +2s+2\), which has complex roots: \(s_{1} ,s_{2} =-1\pm j1\). The natural response modes are: \(\{ e^{-t} \sin t,e^{-t} \cos t\}\).
The complex eigenvectors are given as: \(\left[\begin{array}{c} {1} \\ {1\pm j1} \end{array}\right]\). Let \({\bf V}=\left[ \begin{array}{cc} 1 & 1 \\ -1+j & -1-j \end{array} \right]\); then, \({\bf V}^{-1}{\bf AV}={\rm diag}(s_1,s_2)\).
In this case, to avoid the complex algebra, we construct a modal matrix from the real and imaginary parts of the eigenvectors. Accordingly, let \({\bf M}=\left[ \begin{array}{cc} 1 & 0 \\ -1 & 1 \end{array} \right]\).
Define \({\bf M}^{-1}{\bf AM}=\left[ \begin{array}{cc} -1 & 1 \\ -1 & -1 \end{array} \right]=\mathit{\bf \Gamma}\); then, \({\bf A}={\bf M}\mathit{\bf \Gamma}{\bf M}^{-1}\) and \(e^{{\bf A}t}={\bf M}e^{\mathit{\bf \Gamma}t}{\bf M}^{-1}\), which computes as: \(e^{{\bf A}t} =\left[\begin{array}{cc} {e^{-t} \left(\cos t+\sin t\right)} & {e^{-t} \sin t} \\ {-2e^{-t} \sin t} & {e^{-t} \left(\cos t-\sin t\right)} \end{array}\right]\).
Further, \({\mathop{\lim }\limits_{t\to \infty }} e^{{\bf A}t} =0\); hence, the homogenous state equation is asymptotically stable.