Exact meaning of Transfer Function

S

Thread Starter

student

I have a device whose input is x and output is y. y and x are related by the equation
y = 1/(1-e^(-x)).

How I got this equation is described as follows: I Injected input y beginning from minimum and kept recording x for each y. I incremented y bit by bit and plotted x. I found it to be a S-shaped curve. I searched the internet and found a general S-shaped curve's equation.

Most textbooks relate equations with respect to time. Mass Spring systems, Heat transfer equations, etc. because all of those equations are described by time derivatives of changing parameters. In my equation there is no 't'. Can this system still have its transfer function? Should time 't' be always a part of the transfer function?
 
More than likely, this equation is weighted by time in the controller. This time weighting is usually taken into account with the controllers scan times. I am sure if you look into it, the controller operates this equation at a specific interval. Therefore there is really no need to include a variable for time (t).
 
J

Joao Paulo Massignan

This equation define the STATIC relation between your input and output. You just measured the static gain of your system.

If you desire determine the DYNAMIC of your system, you must measure the relation between x(t) and y(t), that's the Transfer Function (to linear system). You'll have some problems, because your system look like nonlinear. But for more help, what is your system?
 
This is the equation of a valve positioner with its valve side tubing removed and connected to a digital pressure readout, instead.

Regards,
 
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