Transfer function of AC Transformer

Click For Summary
SUMMARY

The discussion focuses on deriving the transfer function of an AC transformer using dynamic equations and Laplace transforms. The user starts with the equation P(t) = mx'' + bx' + kx and transforms it into the Laplace domain, resulting in X(s)/P(s) = 1/ms^2 + bs + k. The output voltage Vo(s) is determined to be proportional to the displacement x, leading to the conclusion that Vo(s)/P(s) = m / (Ms^2 + bs + k), where m represents the slope of the graph and M is the mass of the transformer nucleus.

PREREQUISITES
  • Understanding of dynamic equations in mechanical systems
  • Familiarity with Laplace transforms and their applications
  • Knowledge of transfer functions in control systems
  • Basic principles of AC voltage and displacement relationships
NEXT STEPS
  • Study Laplace transform techniques for solving differential equations
  • Learn about transfer functions in electrical engineering
  • Explore the relationship between mechanical displacement and electrical output in transformers
  • Investigate the effects of damping (b) and stiffness (k) on system response
USEFUL FOR

Electrical engineers, control system designers, and students studying transformer dynamics and AC circuit analysis will benefit from this discussion.

Lord Doppler
Messages
9
Reaction score
2
Homework Statement
For the following system you will find its transfer function Vo (s) / P (s). East
system produces a voltage signal as a function of the applied gauge pressure
at the core of a Linear Variable Differential Transformer (LVDT), which
supplies an A.C. voltage of magnitude proportional to the displacement of the nucleus.
(Only consider positive displacements.)
Relevant Equations
P= F/A
Friction force = bx'
I'm solving this exercise, first I did a force diagram for the transformer nucleus and I got this:

∑Fx = ma
P(t) - Fk - Fb = ma
P(t) = mx''+ bx' + kx

So I got that dynamic equation, my question is, after transform that dynamic equation to Laplace Domain how can I relate it with the Output Voltage, I'm not sure, I obtain this:

X(s)/P(s) = 1/ms^2 + bs + k

And I need Vo(s)/P(s)

Thank you for your support

1616609607617.png
 
Physics news on Phys.org
The output voltage (AC) is just proportional to the displacement. Since they said only consider positive displacement, you can ignore the sign change at zero; you're only operating on one side of that graph. So, the output voltage is just a constant times the displacement x, as shown in the graph. You will have to do the inverse transform to solve for x, assuming you want the output voltage in the time domain. But they only asked for the s domain, so you're answer lies in the graph that shows the relationship between x and v.
 
  • Like
Likes   Reactions: Lord Doppler
DaveE said:
The output voltage (AC) is just proportional to the displacement. Since they said only consider positive displacement, you can ignore the sign change at zero; you're only operating on one side of that graph. So, the output voltage is just a constant times the displacement x, as shown in the graph. You will have to do the inverse transform to solve for x, assuming you want the output voltage in the time domain. But they only asked for the s domain, so you're answer lies in the graph that shows the relationship between x and v.
So, the relation I found, Vo = mX like you said, then X=Vo / m. I only replace X in the transfer function I got first?
I obtain that Vo(s)/P(s) = m / Ms^2 + bs + k ; where m is the slope of the graph and M the mass of the transformer nucleus, is correct?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
23
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K