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I'd like to get some input on whether the assumption I'm making is correct. For background: we want to prove the continuous linear time varying system $$\dot x(t)=A(t)x(t)+ Bu(t), x(t_0) = x_0 $$ $${y = C(t)x(t) + D(t)u(t)}$$ is causal and linear. We know the unique solution to the first equation is $$x(t) = \Phi(t,t_0)x_0 +\int_{t_0}^t{\Phi(t,\tau)B(\tau)u(\tau)d\tau}.$$ Where $\Phi(t,t_0)$ is the system's state transition matrix https://en.wikipedia.org/wiki/State-transition_matrix

I've already proven the system is causal. Now I want to prove it is linear. I want to show that an input $u = \alpha u_1+\beta u_2$ corresponds to an output $y = \alpha y_1+\beta y_2$. I've asserted that $u_1$ coupled with initial condition $x_1(0)$ produces a unique output $x_1(t)$, and $u_2$ coupled with initial condition $x_2(0)$ produces a unique output $x_2(t)$ Is it correct that if $\hat x(t)=\alpha x_1(t)+\beta x_2(t)$, then $\hat x(0) = \alpha x_1(0)+\beta x_2(0)$? Or does this fact rely on the fact that the system is linear (which I haven't proven yet)?

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This is not a complete answer, but it might nudge you in the right direction.

For a transformation, $\mathcal{A}: V \to B$, linearity means $$ \mathcal{A}\left( {x + \alpha y} \right) = \mathcal{A}\left( x \right) + \alpha \mathcal{A}\left( y \right) $$ where $ x,y \in V $ and $ \mathcal{A}\left( x \right),\alpha \mathcal{A}\left( y \right) \in B $.

Then the LTV system described above is by linear by definition. All matrices $A(t)$ act linearly on the states $x(t)$. And from the convolution formula for $x(t)$ you have shown, the evolution of states can also be described in a linear fashion. Then observing the output equation, we see the time evolution of the states $x(t)$ is also operated on, in a linear fashion, by the matrix $C(t)$, and so the output is linear too. This should be enough, but I think you want to show linearity of the input output relation.

Now is there a direct linear correspondence between the inputs and the outputs? There should be, but it is in general hard to show because the dynamics encoded by $A(t)$, though linear, might be so coupled that it is impossible to get it into the form shown above. However, in the special case that the system is diagonalizable, there exists a transformation $P$ such that $ D = {P^{ - 1}}AP $ where $D$ is a diagonal matrix. Then propagating the change of variable $\bar x = P^{-1}x$ throughout the state space description might put it a more amenable form.

At least for a LTI system the change of variable gives, $$ \begin{gathered} \dot{ \bar x} = {P^{ - 1}}AP\bar x + {P^{ - 1}}Bu \hfill \\ \bar y = CV\bar x + Du \hfill \\ \end{gathered} $$

and so the solution becomes, $$ {\bar x}\left( t \right) = {{\bar x}_0}{P^{ - 1}}{e^{Dt}}P + \int\limits_0^t {{P^{ - 1}}{e^{D(t - s)}}P {\bar B}} u\left( s \right) ds $$ Then the $e^{Dt}$ matrices are diagonal matrices, and the states are completely decoupled so it becomes easier to show linearity of the input to state relation though showing input to output linearity may still not be trivial.

As for $\hat x\left( t \right) = \alpha {x_1}\left( t \right) + \beta {x_2}\left( t \right) \Rightarrow \hat x\left( 0 \right) = \alpha {x_1}\left( 0 \right) + \beta {x_2}\left( 0 \right) $, I think it is probably true and does rely on linearity. Let $x_1(t)$ and $x_2(t)$ be two systems with two different dynamics. Then at least in the case where you define $\dot {\hat x}$ as $$ \dot {\hat x} = \left[ {\begin{array}{*{20}{c}} {{A_1}}&0 \\ 0&{{A_2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}} \\ {{{\dot x}_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {{B_1}}&0 \\ 0&{{B_2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{u_1}} \\ {{u_2}} \end{array}} \right] $$ we have, $$ \hat x\left( t \right) = \alpha \left( {{\phi _1}\left( {t,{t_o}} \right){x_1}\left( 0 \right) + \int\limits_{{t_0}}^t {{\phi _1}\left( {t,s} \right){B_1}\left( s \right){u_{_1}}\left( s \right)ds} } \right) $$ $$ + \beta \left( {{\phi _2}\left( {t,{t_o}} \right){x_2}\left( 0 \right) + \int\limits_{{t_0}}^t {{\phi _2}\left( {t,s} \right){B_2}\left( s \right){u_2}\left( s \right)ds} } \right)$$ and so, $$ \hat x\left( 0 \right) = \alpha {x_1}\left( 0 \right) + \beta {x_2}\left( 0 \right)$$

For a more general definition of $\dot {\hat x}$ things will get more complicated. Maybe it is best to show that each (loosely speaking) component behaves linearly, and the components are all coupled linearly, so the entire thing behaves linearly.

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  • $\begingroup$ I suppose if I characterize the solution this way $\hat x\left( t \right) = \alpha \left( {{\phi _1}\left( {t,{t_o}} \right){x_1}\left( 0 \right) + \int\limits_{{t_0}}^t {{\phi _1}\left( {t,s} \right){B_1}\left( s \right){u_{_1}}\left( s \right)ds} } \right) + \beta \left( {{\phi _2}\left( {t,{t_o}} \right){x_2}\left( 0 \right) + \int\limits_{{t_0}}^t {{\phi _2}\left( {t,s} \right){B_2}\left( s \right){u_2}\left( s \right)ds} } \right)$ then the assumption does hold, as the integral disappears at time $t = t_0$. Thanks. $\endgroup$ – Timothy Hensel Sep 14 '16 at 14:38

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