Relating Transformer Parameter I2 to V1 (Non-ideal) Updated

[The Electrician]

You should avoid the use of the laplace transform altogether. That also precludes convolution. Instead, either use phasors as I described earlier, or, since you're a math major not an engineering one, solve the equations classically.

The powerful laplace transform enables transient as well as steady-state solutions but this is of no interest here.

Did you not read the posts following #19?

 

[ZenSerpent]

Sir rude man, yup, failure to perform convolution is a serious mistake. But if I were to perform convolution, I'll get an expression as hairy as the one posted by The Electrician in the post above. So, I think phasor analysis is the way to go. And I now gained some good insight just now. And I noticed that the intuition is much similar to the concepts discussed in complex number analysis. Now everything appears to be clear to me.

 

[ZenSerpent]

The Electrician, so the expression I mentioned in #29 is correct right? One thing I noticed is that all the units cancel out. Why is this?

 

[The Electrician]

The Electrician, so the expression I mentioned in #29 is correct right? One thing I noticed is that all the units cancel out. Why is this?

sin(ω t) has units of volts. That's the units of your final answer.

 

[rude man]

I explained this in post #23

Yes, but don't say "I(s), V(s) etc. Avoid "s" completely. Say I(jw), V(jw) etc. which are phasors.

 

[The Electrician]

Yes, but don't say "I(s), V(s) etc. Avoid "s" completely. Say I(jw), V(jw) etc. which are phasors.

On this page: https://en.wikipedia.org/wiki/Electrical_impedance, a little over halfway down under the heading "Generalized s-plane impedance" is the justification for using s to stand for jω. It's common usage for "circuits that are driven with a steady-state AC signal".

You even do it yourself, for example here: https://www.physicsforums.com/threads/transfer-function-for-rc-and-rl-circuits.738708/#post-4663732

and here: https://www.physicsforums.com/threa...cy-response-of-a-circuit.726774/#post-4593798

 

[ZenSerpent]

sin(ω t) has units of volts. That's the units of your final answer.

Well this sounds counter-intuitive to me as, v1max is in volts and sin wt is dimensionless.

So the amplitude of the input sine wave, which is v1 max doesn't really matter at all in the i2(t)?

 

[The Electrician]

Well this sounds counter-intuitive to me as, v1max is in volts and sin wt is dimensionless.

So the amplitude of the input sine wave, which is v1 max doesn't really matter at all in the i2(t)?

The expression V1max*sin(ω t) derives its units from the amplitude variable V1max. Plain old sin(ϖ t) is really 1*sin(ϖ t), and derives its units from the amplitude 1--that is, 1 volt. Does that make sense to you?

 

[ZenSerpent]

So instead of

i2(t) = |z| sin(wt - arg(z))

It needs to be

i2(t) = V1max*|z| sin(wt - arg(z))

Or I'll just have to consider
i2(t) = |z| sin(wt - arg(z)) with sin(wt -arg(z)) having a unit of volts?

 

[ZenSerpent]

What if the initial amplitude is v1 max =/= 1. Will I not be able to see its value in the expression in i2(t). I apologize as this still confuses me up to now.

 

[The Electrician]

So instead of

i2(t) = |z| sin(wt - arg(z))

It needs to be

i2(t) = V1max*|z| sin(wt - arg(z))

Or I'll just have to consider
i2(t) = |z| sin(wt - arg(z)) with sin(wt -arg(z)) having a unit of volts?

If V1max equals 1 (and therefore isn't explicitly shown), it still has units of volts, so your final result will have units of volts. In other words, your final result has units of volts whatever the value of V1max.

 

[ZenSerpent]

So the bottom line of all these is that we cannot see V1max in the function of i2(t) = |z| sin(wt - arg(z)) but the unit of sin(wt - arg(z)) is in volts since there will be a hidden 1 volt in there regardless of what the value of V1max is, right? (Even if V1max is not equal to 1.)

 

[The Electrician]

You should be able to see V1max in i2(t) = |z| sin(wt - arg(z)) because to show exactly where it is, we should write i2(t) = |z|*V1max*sin(wt - arg(z)). The only reason I didn't show it is because I assumed the input excitation voltage was 1 volt, just for simplicity. Whenever a sine wave voltage function is used, if there isn't an explicit multiplier for the peak voltage, assume it is unity.

You know that the standard mathematical sine function has a peak value of 1, even though the multiplier of 1 is not explicitly shown. If you want to show a sine function with a peak value of 3, you will have to write out the 3, as 3*sin(ω t), but its a convention to not show the multiplier if it's 1.

We don't need to assume that there's a hidden 1 if V1max is not 1 because then the numerical value of V1max will be explicitly present and the V1max expression is what carries the units. It's only when V1max is 1 that it won't (by convention) be shown as a multiplier of 1*.

I suppose I should have shown it in post #23, but I assumed it would be clear that I was using a multiplier of 1 for simplicity in my work, and that it carried units of volts.

 

[ZenSerpent]

That's what I'm trying to say several posts ago. Woah! Thank you so much The Electrician and rude man.

 

[ZenSerpent]

Sir Electrician. There is still a problem. I applied what you did in Mathematica using the expressions for the transformer. Attached are photos for the phasor one and the other one for the Laplace. I scaled both 10000x. It turns out that they don't look identical.

 

[The Electrician]

One big problem is that your expression for M, namely M = 3*SQRT(L1 L2) is not possible. The 3 you have there is the coupling coefficient, and the maximum physically possible value is 1. I wouldn't even choose it to be very close to 1 or you might have numerical problems; I wouldn't set to greater than .95 I set it to .5 in what follows.

I can't tell if you got the right Inverse Laplace tranform, but you shouldn't keep all those exponentials when you plot the result. They are likely to cause numerical problems. Here's what I got; I only show the first two parts with the steady state part first in red, and the first exponential. Only keep the steady state part. There's a minus sign in front of the whole thing due to the particular choice of dots on the transformer windings; if one dot is moved to the other side, it will change the sign of the result:

Using the expressions for magnitude and phase derived from the phasor solution here's what I get for 1/magnitude and phase using your numerical values. I calculated the reciprocal of the magnitude because that's what you need to multiply the transform expression by to make the red trace fit the plot:

Here's a plot of the input sine wave including a phase angle of -1.09383 radians. I left its magnitude as 1 in this plot, and multiplied the Laplace expression by 1/magnitude. You can see that both blue and red plots are on top of each other. So the result from the phasor solution is identical to the result from the Inverse Laplace transform:

 

[ZenSerpent]

Got it. But I now have this question. Notice that arg(z) is already negative. That means the general form should be i2(t) = V1max*|z|*sin(wt + arg(z))? Not i2(t) = V1max*|z|*sin(wt - arg(z)) for those circuits that involve inductance?

 

[ZenSerpent]

Alright, I get it. When we expanded the complex numbers into the real and imaginary part, we disregarded the negative sign in computing for its arg( ) but we promised to put the negative sign in sin(wt - arg(z)). So generally speaking, it's just |z| sin(wt + arg(z)) if we included the negative sign in the computation of arg(z).