Transfer function with initial conditions (DE)

[MechEEE]

I have a differential equation of the form y''(t)+y'(t)+y(t)+C = 0. I think this implies that there are non-zero initial conditions. Is it possible to write a transfer function for this system?
This post:
https://www.physicsforums.com/threads/transfer-function-with-non-zero-initial-conditions.852028/
sort of helps me, but I still need more help. How to actually use this method to write a transfer function? I don't see how to continue with the suggested substitution there, either, if that is part of the strategy. Thanks.

 

[berkeman]

Welcome to PF. Is this question for your schoolwork?

 

[MechEEE]

No, not for schoolwork. I'm getting ahead at my workplace and trying some self-learning.

 

[berkeman]

Okay, but even self-study questions go in our schoowork forums. I'll move your thread there now. You will get good help as long as you show good efforts.

I'm curious -- why do you say that the form of that differential equation suggests non-zero initial conditions?

 

[MechEEE]

Okay, but even self-study questions go in our schoowork forums. I'll move your thread there now. You will get good help as long as you show good efforts.

I'm curious -- why do you say that the form of that differential equation suggests non-zero initial conditions?

I think because it can be re-written as y''(t)+y'(t)+y(t)+y(0) = 0, where y(0) evaluates to nonzero.

 

[DaveE]

I have a differential equation of the form y''(t)+y'(t)+y(t)+C = 0. I think this implies that there are non-zero initial conditions.

If y(t) and it's derivatives were all 0 at t=0, you would be left with C=0. So, yes, I think the initial conditions are y''(0)+y'(0)+y(0) = -C.

Is it possible to write a transfer function for this system?

As I understand it transfer functions have an output divided by an input. So you could certainly take the laplace transform but what is the input in your case? This looks more like a transient response problem to me.

 

[MechEEE]

Yes, I see. I think I did intend to put an input in there. So let me re-formulate my equation as:
y''(t)+y'(t)+y(t)+C = u(t)
where u(t) is an input. If I take the laplace of both sides I get
s²Y(s)+sY(s)+Y(s)+(C/s) = U(s).
But now I have a term on the left side that doesn't have Y(s) in it, so I don't know how to solve for Y(s)/U(s) to get the transfer function. In all the examples I've seen, you can factor out a common Y(s) from the entire LHS to solve for Y(s)/U(s). Here I'm not sure what to with the C/s term that results from taking the laplace of that constant.

 

[DaveE]

Yes, I see. I think I did intend to put an input in there. So let me re-formulate my equation as:
y''(t)+y'(t)+y(t)+C = u(t)
where u(t) is an input. If I take the laplace of both sides I get
s²Y(s)+sY(s)+Y(s)+(C/s) = U(s).
But now I have a term on the left side that doesn't have Y(s) in it, so I don't know how to solve for Y(s)/U(s) to get the transfer function. In all the examples I've seen, you can factor out a common Y(s) from the entire LHS to solve for Y(s)/U(s). Here I'm not sure what to with the C/s term that results from taking the laplace of that constant.

Ignore it. We set the ICs to zero for the transfer function. So you would normally think of a linear system as having a two part response; the transient response with ICs but no driving function, plus the driven response, with no ICs. In linear systems they don't affect each other.

 

[MechEEE]

Thank you, DaveE.