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A system ("A System is any physical set of components that takes a signal, and produces a signal") is described by this equation:

$ \frac{dy(t)}{dt} + 3 \times y(t) = x(t) $

Where $x(t)$ is the input and $y(t)$ the output.

How to determine if this system is a linear one?

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up vote 4 down vote accepted

It is linear because the operators on $y(t)$ are linear, namely differentiation and multiplication. An example of a non-linear system would be $$\frac{dy(t)}{dt} + 3 \times y(t)^2 = x(t)$$ because squaring is not a linear operator. It does NOT satisfy $(y_1+y_2)^2=y_1^2+y_2^2$. This is in contrast to $\frac{d}{dt}(y_1(t)+y_2(t))=\frac{d}{dt}y_1(t)+\frac{d}{dt}y_2(t)$ and $3\times(y_1+y_2)=3\times y_1+3\times y_2$.

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Thank you, Teun. I thought I had to use something like S{ a * x_{1}(t) + b* x_{2}(t)} = a* S{x_{1}(t)} + b * S{x_{2}(t)} (In fact this is what you are using, but I wasn't sure which would be the system output).. – Chris Jan 27 '13 at 22:58
Glad to help! What you're saying is almost correct except for the fact that when we have two solutions $y_1$ and $y_2$ and plug $y_3=ay_1+by_2$ into your differential equation we get: $$\frac{dy_3(t)}{dt} + 3 \times y_3(t) = a\left( \frac{dy_1(t)}{dt} + 3 \times y_1(t)\right) +b\left( \frac{dy_2(t)}{dt} + 3 \times y_2(t)\right)=(a+b)x(t)\neq x(t)$$ unless you pick specific values for $a$ and $b$, so it wouldn't work anymore. This is because the DE is non-homogeneous. – Slugger Jan 27 '13 at 23:02
So the system (and what I mean with system is this: "A System, is any physical set of components that takes a signal, and produces a signal") is not linear? – Chris Jan 27 '13 at 23:18
No it's linear because of the reasons I posted in my original answer (differentiation and multiplication are linear). I just wanted to show that even though the system is linear, it is important to realize that doesn't mean that every linear combination of solutions will give another solution to your system :) – Slugger Jan 27 '13 at 23:22

It depends on your definition of linear.

In the study of ODEs, the system is linear if it can be expressed in the form $L y = x$, where $L$ is a linear differential operator of the form $(Ly)(t) = \sum_{k=0}^n a_k(t) y^{(k)}(t)$, and the coefficients $a_k$ satisfy some continuity condition. Since we can write the above as $Ly = x$ with $(Ly)(t) = y^{(1)}(t)+3 y(t)$, it is clear that the system is linear in this sense.

From a control system perspective, the system is linear (I am omitting many details) if the solution $t \mapsto y_{y_0, x}(t)$ is a linear function of $(y_0,x)$, where $y_0$ is the initial state $y_0$ and $x$ is the input.

The above equation can be written as $\dot{y} = f(y,x)$, where $f(y,x) = -3 y +x$. $f$ is globally Lipschitz in $y$, hence a globally unique solution exists passing through a given initial condition. Since the solution is unique, it is straightforward to verify (by differentiating and checking that it satisfies the ODE) that the system is linear by just checking that if $y_{y_0, x}, y_{y_0', x'}$ are solutions (with initial conditions and inputs $(y_0, x), (y_0', x')$ respectively), then $\alpha y_{y_0, x} + \alpha' y_{y_0', x'}$ is a solution with initial condition $\alpha y_0 + \alpha'y_0'$ and input $\alpha x + \alpha'x'$. Hence the system is linear.

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What do you mean with "control system"? I've updated my question with a definition of the system I mean. Is your answer in compliance with the new meaning of the system so that I can read it to answer to you? – Chris Jan 27 '13 at 23:20
The term 'dynamical system' would be have been more appropriate. The dynamical system focus is mainly concerned with modeling the input-output behavior. Your system is linear in both senses. – copper.hat Jan 28 '13 at 2:55

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