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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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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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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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