# Is the system linear if there is an independant gravity term

1. ### silentwf

37
Like the topic,if in my system derivation, there is an independent gravitational force independent of the state variable, is my system linear? I believe it is not but my instructor is telling is it is. I'm thinking that if there is is an extra gravitational term,the output of the system does not obey f(ax ) = af(x).

### Staff: Mentor

Linear in which variable?

3. ### silentwf

37
Um, could you explain further by which variable you mean? I think I have the misconception that if there are any terms in the equation of motion not associated to input or output, the system is nonlinear, since the equation itself, when input or output is multiplied by a constant, does not follow the law of superposition.

### Staff: Mentor

Simple example: ##y=a^2x##

y and x have a linear relationship - if you multiply x by 2, y gets multiplied by 2 as well. The system is linear, if you consider "a" as a constant parameter, and look at x and y. You can add multiple solutions as superpositions.

y and a do not have a linear relationship - if you multiply a by 2, y gets multiplied by 4. The system is not linear, if you consider "x" as a constant parameter, and look at a and y. You cannot add multiple solutions as superposition.

5. ### silentwf

37
Okay, um then I should make it something like this: if my system is equation is something like:
$f - mg - u = 0$ where f is the input and u is a disturbance (both are functions of time).
having a scalar multiple of my force would not increase the system's response in a linear way. does this mean that this system is not linear?
$f(a*t) - mg - u(t) = 0$

### Staff: Mentor

A scalar multiple of the force would not lead to a scalar multiple of the mass (if u does not scale in the same way), indeed.

7. ### silentwf

37
So the system is non-linear? It's kind of strange to think of it in this way, so I'm not really sure.

### Staff: Mentor

I think it is non-linear with the interpretation as in post #5.