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Given an LTI system of the form:
$$ \frac{d}{dt}{x}(t) = {A} {x}(t) \ + \ {B} {u}(t) $$
where ${x}(t)$ is the state variable, ${u}(t)$ an input and ${A}$ and ${B}$ constants, how does one show, by superposition, that the system is linear? I've searched and searched and tried to do it myself and I am unable to do so. Seems more difficult that it seems. Most books seem to gloss over the fact.
I'm going to assume the system has the initial condition $x(0) = 0$, as otherwise the system isn't linear.
Suppose the input $u_1(t)$ yields the output $x_1(t)$ and the input $u_2(t)$ yields the output $x_2(t)$.
We need to prove that the input $u(t) = \alpha u_1(t)+\beta u_2(t)$ yields the output $x(t) = \alpha x_1(t) + \beta x_2(t)$ for any scalars $\alpha$ and $\beta$.
The given input and output pairs must satisfy the differential equation and the initial condition. Hence, $x_1'(t) = Ax_1(t)+Bu_1(t)$ and $x_2'(t) = Ax_2(t)+Bu_2(t)$ for all times $t$ and $x_1(0) = x_2(0) = 0$.
Now, add $\alpha$ times the first equation to $\beta$ times the second equation to get: $\alpha x_1'(t) + \beta x_2'(t) = \alpha(Ax_1(t)+Bu_1(t)) + \beta(Ax_2(t)+Bu_2(t))$
Then, rearrange stuff to get:
$\dfrac{d}{dt}[\alpha x_1(t) + \beta x_2(t)] = A(\alpha x_1(t)+\beta x_2(t)) + B(\alpha u_1(t)+\beta u_2(t))$
$x'(t) = Ax(t)+Bu(t)$
Also, we have $x(0) = \alpha x_1(0)+\beta x_2(0) = \alpha \cdot 0 + \beta \cdot 0 = 0$.
Thus, the signals $u(t) = \alpha u_1(t)+\beta u_2(t)$ and $x(t) = \alpha x_1(t) + \beta x_2(t)$ satisfy the differential equation and the initial condition. Hence, the input $u(t) = \alpha u_1(t)+\beta u_2(t)$ yields the output $x(t) = \alpha x_1(t) + \beta x_2(t)$ for any inputs $u_1(t)$ and $u_2(t)$ and any scalars $\alpha$ and $\beta$.
Therefore, the system is linear, as desired.
here is why, for a simple single-input, single-output (SISO) system, with no hidden states (i.e. completely controllable and completely observable), if superposition is known to be a valid property (called "additivity" by the math guys, you can also show that the scaling property (called "homogeneous of degree 1" by the math guys) of linearity is also the case for real and rational scalers.
you need to do a little hand-waving regarding continuity to extend this to the irrational scalers, because you can always get arbitrarily close to any irrational number with a rational one.
so let's say that $T\{ \cdot \}$ is something that transforms an input $x(t)$ into an output $y(t)$
$$ y(t) = T\{ x(t) \} $$
and this is true for all possible $x_1(t)$ and $x_2(t)$:
$$ T\{ x_1(t) + x_2(t) \} = T\{ x_1(t) \} + T\{ x_2(t) \} $$
now suppose $x_2(t) = x_1(t) = x(t)$
$$ T\{ x(t) + x(t) \} = T\{ x(t) \} + T\{ x(t) \} $$
or
$$ T\{ 2 \ x(t) \} = 2 \ T\{ x(t) \} $$
so you've just proved the linear scaling property for the real scaler $2$. how would you go about proving it for $3$? and then $4$? and then for any positive integer $N$? (it's called "proof by induction" and it's easy.)
so we're able to easily prove that superposition implies linear scaling for positive integers (and easily extended to all real integers):
$$ T\{ N \ x(t) \} = N \ T\{ x(t) \} \quad \quad N \in \mathbb{Z} $$
now how about this:
$$ T\left\{ \frac{1}{M} \ x(t) \right\} = ?? \quad \quad M \in \mathbb{Z} $$
define
$$ x_M(t) \triangleq \frac{1}{M} \ x(t) $$
so we know that
$$ T\left\{ M \ x_M(t) \right\} = M \ T\left\{ x_M(t) \right\} \quad \quad M \in \mathbb{Z} $$
but we also know that
$$ T\left\{ x_M(t) \right\} = T\left\{\frac{1}{M} \ x(t) \right\} $$
and
$$ T\left\{ M \ x_M(t) \right\} = T\left\{M \ \frac{1}{M} \ x(t) \right\} = T\{ x(t) \} $$
then we know that
$$ T\left\{ M \ x_M(t) \right\} = T\{ x(t) \} = M \ T\left\{ \frac{1}{M} \ x(t) \right\} \quad \quad M \in \mathbb{Z} $$
and
$$ T\left\{ \frac{1}{M} \ x(t) \right\} = \frac{1}{M} \ T\{ x(t) \} \quad \quad M \in \mathbb{Z} $$
so now we proved linear scaling for any reciprocal of an integer. sohow do you think we might do this form the general rational scaler $\frac{N}{M}$ where $N,M \in \mathbb{Z}$? i don't think it takes Einstein to see that
$$ T\left\{ \frac{N}{M} \ x(t) \right\} = \frac{N}{M} \ T\{ x(t) \} \quad \quad N,M \in \mathbb{Z} $$
so, just given superposition
$$ T\{ x_1(t) + x_2(t) \} = T\{ x_1(t) \} + T\{ x_2(t) \} $$
we know that scaling is also true
$$ T\{ \alpha x \} = \alpha \ T\{ x(t) \} \quad \quad \alpha = \frac{N}{M} \quad N,M \in \mathbb{Z}$$
and we know that
$$ T\left\{ \sum\limits_{i} \alpha_i x_i(t) \right\} = \sum\limits_{i} \alpha_i T \{ x_i(t) \} $$
where all $\alpha_i$ are rational. for the irrational $\alpha$, one must make an additional (and reasonable) assumption about the transform mapping $T\{\cdot\}$, that it is continuous, so that as rational $\alpha$ gets closer and closer to a irrational target (and changes very little), the behavior of $T\{\cdot\}$ will also change very little. i'm not gonna deal with that now.
but, from just a little thinking, it's clear that, at least for real and rational scalers, superposition implies linear scaling and then the whole of linearity in general. (says nothing about time-invariance, which is a whole 'nother property.)
