Show that $\dot x = x + u, y = x$ is time-invariant.

Question

I am given a system,

$\dot x = x + u, y = x$

where $u$ is the input, $x$ is the state, $y$ is the output.

I want to show that this system is time-invariant.

To do this, I know that $y(t) = H(u(t)) = \int_0^t e^{t - \tau} u(\tau) d\tau$, where $H$ denotes the input-output map.

I need to show $y(t-T) = H(u(t - T))$ for arbitrary time $T$.

$y(t-T) = \int_0^{t- T} e^{(t - T) - \tau} u(\tau) d\tau$

But $H(u(t - T)) = \int_0^t e^{t - (\tau - T)} u(\tau - T) d\tau$

I am not sure how to continue to show $y(t-T) = H(u(t - T))$.

Answer 1

You are almost there, as you mentioned above, we can solve for $x$ in the first equation using the integrating factor $e^{-t}$, in other words, $\dot x e^{-t} - xe^{-t} = ue^{-t} \Rightarrow \frac{d}{dt}\left(xe^{-t}\right) = ue^{-t} \Rightarrow x(t) - x(0) = e^{t}\int_{0}^{t}e^{-\tau}u(\tau)d\tau$. Now assuming the system started from the rest (i.e. $x(0) = 0$) we get $x = \int_{0}^{t}e^{t-\tau}u(\tau)d\tau$. Now we have to show a time shift in input $u$ results the same time shift in the output $H(u)$. We have

$$y(t-T) = \int_{0}^{t-T}e^{t-T-\tau}u(\tau)d\tau \\ H(u(t-T)) = \int_{0}^{t}e^{t-\tau}u(\tau-T)d\tau $$

the second equation comes from solving the first system with the input $u(t-T)$, in other words by applying $H$ on the input $u(t-T)$. Now by the change of variable $s=\tau-T$ in the second integral we get

$$H(u(t-T)) = \int_{-T}^{t-T}e^{t-T-s}u(s)ds = \int_{-T}^{0}e^{t-T-s}u(s)ds + \int_{0}^{t-T}e^{t-T-s}u(s)ds$$

we assumed the system was at rest at $t=0$, that implies $u(t)=0, t\le 0$, therefore the first integral term equates to zero and we have $H(u(t-T)) = \int_{0}^{t-T}e^{t-T-s}u(s)ds = y(t-T)$ since $\tau$ and $s$ are dummy variables.

Answer 2

There seems to be a bit of confusion with the notation, which I will try to fix here.

Let us define the operator $H:U\mapsto Y$, where $U,Y$ are the input and output function spaces, as

$$y(t)=H(u)(t):=\int_0^te^{t-s}u(s)ds.$$

Note that the operator $H$ takes a function as an "input", here $u$, and "output" a function, here $y$, that is $y=H(u)$. Additionally, $H(u)(t)$ is that "output function" $y$ evaluated at $t$, that is, $y(t)$.

Now, define the delayed function $u_T$ as $u_T(t)=u(t-T)$ if $t\ge T$ and 0 otherwise.

Then, we have that

$$y_T(t):=H(u_T)(t)=\int_0^te^{t-\tau}u_T(\tau)d\tau=\int_T^te^{t-\tau}u(\tau-T)d\tau$$

and, on the other hand, we have that

$$y(t-T)=H(u)(t-T)=\int_0^{t-T}e^{t-T-s}u(s)ds.$$

Now what happens to the first integral when we consider the change of variables $s=\tau-T$?