0
$\begingroup$

Suppose we have a system S and the relationship between $x(t)$ and $y(t)$ is $y(t) = x(t)+1$

However despite the fact the system is made of linear operators the system is not linear.How is this possible?

Can we have a system made of nonlinear operation but the system becomes linear?

$\endgroup$
4
  • 1
    $\begingroup$ Adding a constant is not a linear map. It's affine. Your specific system is easily seen to be non-linear as it maps 0 to something other than 0. $\endgroup$
    – Jazzmaniac
    Commented Sep 8, 2022 at 22:15
  • $\begingroup$ What is a linear map , please explain. $\endgroup$
    – Miss Mulan
    Commented Sep 8, 2022 at 22:17
  • $\begingroup$ en.wikipedia.org/wiki/Linear_map $\endgroup$
    – Jazzmaniac
    Commented Sep 9, 2022 at 7:15
  • 2
    $\begingroup$ Sorry, but I'll be repeating my previous comment: you are not going to learn this stuff by asking one basic question at a time. Maybe something like allsignalprocessing.com/lessons/… ? $\endgroup$
    – Hilmar
    Commented Sep 9, 2022 at 12:11

3 Answers 3

3
$\begingroup$

This system may cause a hallucination because of its similarity to the algebraic equation $y = x + 1 $ which is a linear one indeed as stated in the wikipedia link to System_of_linear_equations

However, signal processing defines linearity according to system theory which defines linearity as an input-output relationship (a.k.a. mapping) possesed by a system $T\{\}$ so that

$$ T\{ a ~x_1(t) + b ~x_2(t) \} = a ~T\{ x_1(t) \} + b ~T\{ x_2(t) \} \tag{1}$$

is satisfied for every input $x_1,x_2$ and every complex constants $a,b$.

One can easily verify that Eq.1 is not satisfied by a system whose input output relation is: $y(t) = x(t) + 1$, simply because of the fixed-constant $1$ present at the output for any input, even for $x(t) = 0$ for all $t$.

Note that this system is also called incrementally linear, which can be very easily linearized, so that another system defined as $y_2(t) = y(t) -1 = x(t)$ will be linear.

$\endgroup$
4
  • $\begingroup$ "... also called incrementally linear ..." or "affine" -- I'm pretty sure they mean the same thing. $\endgroup$
    – TimWescott
    Commented Sep 9, 2022 at 0:26
  • $\begingroup$ @TimWescott affine if you prefer a mathematical jargon, incrementally linear if signal processing... I prefer the latter, as the affine may refer to different mathematical constructs than system input-output relations. $\endgroup$
    – Fat32
    Commented Sep 9, 2022 at 0:43
  • $\begingroup$ It doesn't matter if you call it an algebraic equation or not, $y=x+1$ is not linear. Its graph in a cartesian coordinate system is a line, but that's not the same thing as the equation being linear. $\endgroup$
    – Jazzmaniac
    Commented Sep 9, 2022 at 7:13
  • 1
    $\begingroup$ @Jazzmaniac $$y - x = 1 \\ y + x = 2 $$ is a set of linear equations. $\endgroup$
    – Fat32
    Commented Sep 9, 2022 at 15:03
3
$\begingroup$

A linear system means that the system's output should scale with the input, and the system's output should combine given two inputs. Look at the scaling property, you scale the input and this results in an output:

$$ y_1(t)=\alpha x(t)+1 $$

and now we need to test is this equal to just scaling the output by $\alpha$? No, it is not because if you simply scaled the output the result is:

$$ \alpha y(t)=\alpha (x(t)+1)=\alpha x(t)+\alpha $$

Since $y_1(t) \neq \alpha y(t)$, this is enough to say the system is non-linear and you can do another similar experiment to show the summing of two inputs does not give you the sum of the two outputs. Here is the result of combining two inputs:

$$ y'(t)=(x_1(t)+x_2(t))+1 $$

and here is the result of combining two outputs:

$$ y_1(t)+y_2(t)=x_1(t)+1+x_2(t)+1=x_1(t)+x_2(t)+2 $$

Since $y_1(t)+y_2(t) \neq y'(t)$, the system does not have the combining property either.

$\endgroup$
1
$\begingroup$

Paraphrasing Fat32’s answer: in signal processing, and other linear systems contexts, a linear system is one that satisfies superposition.

One component of superposition is homogeneity: for the system

$$y = f(x)$$

we must have that

$$f(0) = 0$$

Screenshot from linked page:

Homogeneity explanation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.