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$T(x[n]) = ax[n] + b$

$ T(\alpha_{1}x_{1}[n] + \alpha_{2}x_{2}[n]) = \alpha_{1}ax_{1}[n]+ \alpha_{2}ax_{2} [n] +b $

$ \alpha_{1}T(x_{1}[n] ) + \alpha_{2}T(x_{2}[n]) = \alpha_{1}(ax[n]+b) + \alpha_{2}(ax_{2} [n] +b) $

therefore they are not equal so the system is non-linear?

or would it be something like this:

$ T(\alpha_{1}x_{1}[n] + \alpha_{2}x_{2}[n]) = \alpha_{1}ax_{1}[n]+\alpha_{1}b + \alpha_{2}ax_{2} [n] + \alpha_{2}b $

$ \alpha_{1}T(x_{1}[n] ) + \alpha_{2}T(x_{2}[n]) = \alpha_{1}(ax[n]+b) + \alpha_{2}(ax_{2} [n] +b) $

and therefore they are equal so the system is linear?

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    $\begingroup$ This is an example of incrementally linear system. $\endgroup$ – Fat32 Nov 26 '18 at 20:04
  • $\begingroup$ Also called \emph{affine} systems, as affine objects are often "linear+constant". $\endgroup$ – Laurent Duval Sep 8 at 8:35
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In the following, I suggest that, before using the generic $T(\alpha_1 x_1+\alpha_2 x_2)$ versus $\alpha_1 T( x_1)+\alpha_2T( x_2)$, it can be more informative to try with simpler partial tests, or try counterexamples, based on your intuition.

I don't really understand the motivation behind the second group of equations, or why $b$ gets multiplied by $\alpha_i$.

Here, you are using the generic version of the linearity test, which is good, but can raise some doubts, as apparent from your question. You can, in complement, use other simpler tests, that can show that the system is not linear, as a double check. Those are counter-examples. For instance:

  • is the output of the zero signal zero? For a linear system , $T(\vec{0})=0$.
  • is a single output with a zero coefficient zero? This tests if $T(0.\vec{x})=0.T(\vec{x}) = 0$
  • is a single output linear? This tests if $T(\alpha.\vec{x})=\alpha.T(\vec{x})$
  • is a simple addition linear? This tests if $T(\vec{x}+\vec{y})=T(\vec{x})+T(\vec{y})$

Those, if not passed, prove that the system is non-linear. And instead of using the generic version, they can show to your (clever) professor that you have some intuition about what is going on, and the risk of errors is reduced. I personally appreciate a lot when students use minimal arguments: they go straight to the point, spend less time on such questions, to focus on more involved ones.

Of course, if the system is linear, more is required.

Here, your system is non-linear... unless $b=0$.

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Though Laurent gave the standard answer of the linear system. We shall also note that this is an incrementally linear system and it could be made linear by the following argumentation:

Let the first system (your system) be given by the following I/O relationship:

$$ y[n] = \mathcal{T}\{ x[n] \} = a~x[n] + b $$ where $a$ and $b$ are constants.

This system, $\mathcal{T}$, is clearly not linear. However if you define the following second system, $\mathcal{S}$, as:

$$ y[n] = \mathcal{S}\{ x[n] \} = \mathcal{T}\{ x[n] - b/a \}. $$

Then $\mathcal{S}$ will be a linear system as easily demonstrated :

$$ y[n] = \mathcal{S}\{ x[n] \} = \mathcal{T}\{ x[n] - b/a \} = ( x[n] - b/a) \cdot a + b = a~x[n]. $$ Hence effectively the second system is

$$ y[n] = \mathcal{S}\{ x[n] \} = a ~x[n] $$ a linear one.

So by this method, you can convert any incrementally linear system to an equivalent linear one. Most typically $b$ can be seen as a bias term that prevented the former system to be linear, and when properly removed the linearity is reclaimed...

Of course, strictly speaking, $S$ is not the same system with $T$, it's another system.

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  • $\begingroup$ Aren't you trying to prove that a linear system is an affine one without an offet, somehow? :) How would you do with polynomial versions? Horner's rule? $\endgroup$ – Laurent Duval Nov 26 '18 at 22:21
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    $\begingroup$ Or probably I'm trying to avoid the term affine ;-) @LaurentDuval $\endgroup$ – Fat32 Nov 26 '18 at 22:49
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    $\begingroup$ That's ah, fine $\endgroup$ – Laurent Duval Nov 26 '18 at 22:53

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