# 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:

1. $$e^{{\bf A}t} ={\rm \mathcal L}^{-1} [(s{\bf I}-{\bf A})^{-1} ]$$
2. $$e^{{\bf A}t} =\sum _{0}^{\infty } \frac{{\bf A}^{i} t^{i} }{i\, !}$$
3. By using the modal matrix (see below)
4. By using the fundamental matrix
5. 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: $$\dot{\bf x}(t)={\bf Ax}(t),\, \, \, {\bf x}(0)={\bf x}_{0}$$.

##### Asymptotic Stability

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.

##### Example $$\PageIndex{1}$$: Polynomial with Real 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.

##### Example $$\PageIndex{2}$$: Polynomial with Complex Roots

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.

This page titled 8.2: State-Transition Matrix and Asymptotic Stability is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Kamran Iqbal.