# Transient oscillations in transformer?

1. Apr 20, 2013

### guillefix

Hello,

I calculated the transient response of a weakly coupled transformer (M ≠ L1L2), for a heaviside step function input. I get a decaying oscillatory response, even though I didn't include any capacitance, can this be? If interested here is the angular frequency I get:

ω=$\sqrt{\frac{R_{1}R_{2}-\frac{(L_{1}R_{2}+L_{2}R_{1})^{2}}{4(M^{2}-L_{1}L_{2})}}{M^{2}-L_{1}L_{2}}}$

I think the units come out right, which is always a good check

2. Apr 20, 2013

### The Electrician

It would be much easier to help you if you would post a schematic of the transformer model, and the equations you wrote to solve the system. Show how you arrived at the solution from the equations.

Also, the system is perfectly coupled when M = SQRT(L1 L2), not M = L1 L2.

3. Apr 21, 2013

### guillefix

Yes that's what I meant :P

True, I should have posted more detail. The model I've used is quite simple, with just coupled inductors and resistors. The schematic is attached.

The equations I used can also be found in hyperphysics here (with another schematic acutally): http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/imgmag/tracir.gif

What I did was make Vp a heaviside step funciton, and then take the Laplace transform of both sides getting:

M*s*F1-L2*s*F2-R2*F2=M*I1(0)+L2*I2(0)=0
M*s*F2-L1*s*F1-R1*F1=-Fp+M*I2(0)+L1*I1(0)=-V0/s

Where M is the mutual inductance, F1 is the Laplace transform of I1, F2 is the laplace transform of I2, and Fp of Vp and V0 is the size of the heaviside; and the rest can be read in the picture.

I solved the linear system and got for the weakly coupled case:

$F_{1}=\frac{V_{0}(L_{2}s+R_{2})}{s((M^{2}-L_{1}L_{2})s^{2}+(L_{1}R_{2}+L_{2}R_{1})s+R_{1}R_{2})}$

$F_{2}=\frac{V_{0}M}{s((M^{2}-L_{1}L_{2})s^{2}+(L_{1}R_{2}+L_{2}R_{1})s+R_{1}R_{2})}$

And doing the inverse laplace:

$I_{1}=\frac{V_{0}e^{-λt}}{ω(M^{2}-L_{1}L_{2})}(L_{2}sin(ωt)+R_{2}(\frac{1}{2i}(\frac{e^{(iω-λ)t}}{iω-λ}+\frac{^{-(iω+λ)t}}{iω+λ})+\frac{ω}{ω^{2}+λ^{2}}))$

Where $λ=\frac{L_{1}R_{2}+L_{2}R_{1}}{2(M^{2}-L_{1}L_{2})}$ and ω is what I have on my previous post. I2 is obviously similar to second part of I1.

I just realized now that my ω would be imaginary due to $M^{2}-L_{1}L_{2}$ being negative, which means that my sin would become an i*sinh. But I also realized the decaying exponential is not decaying anymore..I think I need to rearange this and see..

Last edited: Apr 21, 2013