User:Egm6321.f12.team7.parikh.a/HW15.3

=HW - Linear Time Variant System =

Given
$$\displaystyle \mathbf {x}(t) = exp [{\mathbf {A}(t-t_0)}]\mathbf{x}(t_0) + \int^t_{t0} exp[{ \mathbf{A}(t-\tau)}]\mathbf{B}\mathbf{u}(\tau)d\tau $$3.8.1

Find
Generalize $$\displaystyle \mathbf {x}(t) = exp [{\mathbf {A}(t-t_0)}]\mathbf{x}(t_0) + \int^t_{t0} exp[{ \mathbf{A}(t-\tau)}]\mathbf{B}\mathbf{u}(\tau)d\tau $$ to the case of linear time varient system.

Verify that your expression is indded the solution for

$$\displaystyle \dot\mathbf{X}(t)= \mathbf{A}(t)\mathbf{x}(t)+\mathbf{B}(t)\mathbf{u}(t) $$

Solution
$$\displaystyle \mathbf{X}(t)= \mathbf{A}(t)\mathbf{x}(t)+\mathbf{B}(t)\mathbf{u}(t) $$ Let A(t)= A and B(t) = B Where $$\mathbf{x} $$ is a n x 1 vector, $$\mathbf{u}$$ is 'm x 1' vector, $$\mathbf{A}$$ is a 'n x n' matrix and $$\mathbf{B}$$ is a 'n x m' matrix. The equation is following that $$\displaystyle \dot x(t) = Ax(t) + Bu(t)$$ Rearrange the above equation. Thus, $$\displaystyle \underbrace{1}_{N}\cdot \dot x(t) - \underbrace{Ax(t)}_{M} = Bt(u) $$ $$\displaystyle h(t) = e^{-\int \frac{1}{N}(N_t - M_x)dt}$$3.8.2

Where : $$\displaystyle N =1, N_t = 0, M_x = - A$$ From the above equation, intergrating factor is that $$\displaystyle h(t) = e^{\int^t-Ads}=e^{-At}$$3.8.3

$$\displaystyle e^{-At}(\dot x(t)-Ax(t)) = \mathbf{B}\mathbf{u}(t)e^{-At}$$ [LHS] : $$\displaystyle e^{-At}(\dot x(t)-Ax(t)) = [e^{-At}x(t)]'$$ Rearraing the eq $$\displaystyle [e^{-At}x(t)]' = \mathbf{B}\mathbf{u}(t)e^{-At}$$3.8.4

Interval is $$\displaystyle t_0< \tau < t$$ The equation is following that $$\displaystyle \int^t_{t_0} [e^{-A\tau}x(\tau)]'d\tau = \int^t_{t_0}Bu(\tau)e^{-A\tau}d\tau $$ $$\displaystyle e^{-At}x(t)-e^{-At_0}x(t_0) = \int^t_{t_0}e^{-A\tau}Bu(\tau)d\tau$$ Dividing by $$\displaystyle e^{-At}$$ And rearranging for x(t) $$\begin{align} \displaystyle x(t)&=\frac {\int^t_{t_0}e^{-A\tau}\mathbf{B}\mathbf{u}(\tau)d\tau}{e^{-At}} + e^{A(t-t_0)}x(t_0) \\&={\int^t_{t_0}e^{A(t-\tau)}\mathbf{B}\mathbf{u}(\tau)d\tau} + e^{A(t-t_0)}x(t_0)\end{align} $$

SC-L1-ODE-VC can be based on the SC-L1-ODE-CC.