# Control Systems/Time Variant System Solutions

## General Time Variant Solution

The state-space equations can be solved for time-variant systems, but
the solution is significantly more complicated than the time-invariant
case. Our time-variant state equation is given as follows:

$$x'(t) = A(t)x(t) + B(t)u(t)$$

We can say that the general solution to time-variant state-equation is
defined as:

$$x(t) = \phi(t, t_0)x(t_0) + \int_{t_0}^{t} \phi(t,\tau)B(\tau)u(\tau)d\tau$$

The function $\phi$ is called the **state-transition matrix**, because
it (like the matrix exponential from the time-invariant case) controls
the change for states in the state equation. However, unlike the
time-invariant case, we cannot define this as a simple exponential. In
fact, $\phi$ can\'t be defined in general, because it will actually be a
different function for every system. However, the state-transition
matrix does follow some basic properties that we can use to determine
the state-transition matrix.

In a time-variant system, the general solution is obtained when the
state-transition matrix is determined. For that reason, the first thing
(and the most important thing) that we need to do here is find that
matrix. We will discuss the solution to that matrix below.

### State Transition Matrix

The state transition matrix $\phi$ is not completely unknown, it must
always satisfy the following relationships:

$$\frac{\partial \phi(t, t_0)}{\partial t} = A(t)\phi(t, t_0)$$

$$\phi(\tau, \tau) = I$$

And $\phi$ also must have the following properties:

:   {\| class=\"wikitable\"

\|- \|1.\|\|$\phi(t_2, t_1)\phi(t_1, t_0) = \phi(t_2, t_0)$ \|-
\|2.\|\|$\phi^{-1}(t, \tau) = \phi(\tau, t)$ \|-
\|3.\|\|$\phi^{-1}(t, \tau)\phi(t, \tau) = I$ \|-
\|4.\|\|$\frac{d\phi(t_0, t_0)}{dt} = A(t)$ \|}

If the system is time-invariant, we can define $\phi$ as:

$$\phi(t, t_0) = e^{A(t - t_0)}$$

The reader can verify that this solution for a time-invariant system
satisfies all the properties listed above. However, in the time-variant
case, there are many different functions that may satisfy these
requirements, and the solution is dependent on the structure of the
system. The state-transition matrix must be determined before analysis
on the time-varying solution can continue. We will discuss some of the
methods for determining this matrix below.

## Time-Variant, Zero Input

As the most basic case, we will consider the case of a system with zero
input. If the system has no input, then the state equation is given as:

$$x'(t) = A(t)x(t)$$

And we are interested in the response of this system in the time
interval T = (a, b). The first thing we want to do in this case is find
a **fundamental matrix** of the above equation. The fundamental matrix
is related

### Fundamental Matrix

Given the equation:

$$x'(t) = A(t)x(t)$$

The solutions to this equation form an *n*-dimensional vector space in
the interval T = (a, b). Any set of *n* linearly-independent solutions
{x~1~, x~2~, \..., x~n~} to the equation above is called a **fundamental
set** of solutions.

A **fundamental matrix FM** is formed by creating a matrix out of the
*n* fundamental vectors. We will denote the fundamental matrix with a
script capital X:

$$\mathcal{X} = \begin{bmatrix}x_1 & x_2 & \cdots & x_n\end{bmatrix}$$

The fundamental matrix will satisfy the state equation:

$$\mathcal{X}'(t) = A(t)\mathcal{X}(t)$$

Also, *any matrix that solves this equation can be a fundamental matrix*
if and only if the determinant of the matrix is non-zero for all time
*t* in the interval T. The determinant must be non-zero, because we are
going to use the inverse of the fundamental matrix to solve for the
state-transition matrix.

### State Transition Matrix

Once we have the fundamental matrix of a system, we can use it to find
the state transition matrix of the system:

$$\phi(t, t_0) = \mathcal{X}(t)\mathcal{X}^{-1}(t_0)$$

The inverse of the fundamental matrix exists, because we specify in the
definition above that it must have a non-zero determinant, and therefore
must be non-singular. The reader should note that this is only one
possible method for determining the state transition matrix, and we will
discuss other methods below.

### Example: 2-Dimensional System

{e\^{-2t}}`</math>`{=html}$= \begin{bmatrix} {e}^{t}&-\frac{1}{2}\,{e}^{3t}\\0&{e}^{t}\end{bmatrix}$

The state-transition matrix is given by:

$$\phi(t, t_0) = \mathcal{X}(t)\mathcal{X}^{-1}(t_0) = \begin{bmatrix}e^{-t} & -\frac{1}{2} e^{t} \\ 0 & e^{-t}\end{bmatrix} \begin{bmatrix} {e}^{t_0}&\frac{1}{2}\,{e}^{3t_0}\\0&{e}^{t_0}\end{bmatrix}$$

$$\phi(t, t_0) = \begin{bmatrix} e^{-t + t_0} & \frac{1}{2}(e^{t + t_0} - e^{-t + 3t_0}) \\ 0 & e^{-t+t_0}\end{bmatrix}$$}}

### Other Methods

There are other methods for finding the state transition matrix besides
having to find the fundamental matrix.

Method 1:If A(t) is triangular (upper or lower triangular), the state transition matrix can be determined by sequentially integrating the individual rows of the state equation.

```{=html}
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```

Method 2:If for every τ and t, the state matrix commutes as follows:

:   

    :   $A(t)\left[\int_{\tau}^{t}A(\zeta)d\zeta\right]=\left[\int_{\tau}^{t}A(\zeta)d\zeta\right]A(t)$

:   Then the state-transition matrix can be given as:
    $$\phi(t, \tau) = e^{\int_\tau^tA(\zeta)d\zeta}$$

:   The state transition matrix will commute as described above if any
    of the following conditions are true:
    1.  A is a constant matrix (time-invariant)
    2.  A is a diagonal matrix
    3.  If $A = \bar{A}f(t)$, where $\bar{A}$ is a constant matrix, and
        f(t) is a scalar-valued function (not a matrix).

:   If none of the above conditions are true, then you must use **method
    3**.

Method 3:If A(t) can be decomposed as the following sum:

:   

    :   $A(t) = \sum_{i = 1}^n M_i f_i(t)$

:   Where *M*~i~ is a constant matrix such that M~i~M~j~ = M~j~M~i~, and
    *f*~i~ is a scalar-valued function. If A(t) can be decomposed in
    this way, then the state-transition matrix can be given as:
    $$\phi(t, \tau) = \prod_{i=1}^n e^{M_i \int_\tau^t f_i(\theta)d\theta}$$

It will be left as an exercise for the reader to prove that if A(t) is
time-invariant, that the equation in **method 2** above will reduce to
the state-transition matrix $e^{A(t-\tau)}$.

### Example: Using Method 3

e\^{\\begin{bmatrix}0 & t-\\tau \\\\ -t+\\tau &
0\\end{bmatrix}}`</math>`{=html}

The first term is a diagonal matrix, and the solution to that matrix
function is all the individual elements of the matrix raised as an
exponent of *e*. The second term can be decomposed as:

$$e^{\begin{bmatrix}0 & t-\tau \\ -t+\tau & 0\end{bmatrix}} = e^{\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}(t-\tau)} = \begin{bmatrix}\cos(t-\tau) & \sin(t-\tau)\\ -\sin(t-\tau) & \cos(t-\tau)\end{bmatrix}$$

The final solution is given as:

$$\phi(t, \tau) =$$$\begin{bmatrix}e^{\frac{1}{2}(t^2-\tau^2)} & 0 \\ 0 & e^{\frac{1}{2}(t^2-\tau^2)}\end{bmatrix}\begin{bmatrix}\cos(t-\tau) & \sin(t-\tau)\\ -\sin(t-\tau) & \cos(t-\tau)\end{bmatrix}$$= \begin{bmatrix}e^{\frac{1}{2}(t^2-\tau^2)}\cos(t-\tau) & e^{\frac{1}{2}(t^2-\tau^2)}\sin(t-\tau)\\ -e^{\frac{1}{2}(t^2-\tau^2)}\sin(t-\tau) & e^{\frac{1}{2}(t^2-\tau^2)}\cos(t-\tau)\end{bmatrix}$}}

## Time-Variant, Non-zero Input

If the input to the system is not zero, it turns out that all the
analysis that we performed above still holds. We can still construct the
fundamental matrix, and we can still represent the system solution in
terms of the state transition matrix $\phi$.

We can show that the general solution to the state-space equations is
actually the solution:

$$x(t) = \phi(t, t_0)x(t_0) + \int_{t_0}^{t} \phi(t,\tau)B(\tau)u(\tau)d\tau$$

