Phase shift problem

[-MFra-]

Hello everyone!
I can't determine the origin of the 180-degree offset, between the two results, one in the first worksheet and the other in the second one. As for the creation of the graphs, with the interval chosen, it takes a long time (many minutes). A tip to speed up the drawing? (see photo).
I enclose two worksheets dedicated to the same problem but with different methods.
Thanks to whoever will answer me.

[Werner_E]

Differences in the phase plot might be because of different roundoff errors when you calculate it with different methods.

For an angle of 180.01 deg arg would return -179.99 deg e.g.

Two ways to speed up plotting:

1) get rid of the range for omega and use a "free", undefined variable and let Mathcads quickplot  chose the number of points calculated. You have to set lower and upper limit of the frequency axis manually:

2) Unfortunately we cannot define a log-spaced range, but we can define omega as a log-spaced vector using either logspace or logpts and then use vectorization in the plot

You may now use

instead of arg.. but it shows the very same effect.

phasecor is a function which tries to avoid the jumps at +-pi. I doubt that it will be useful for a non-continuos phase plot where the phase just seems to jump between the values -pi,0 and pi. Anyway, here is what it does in case of your first file

[-MFra-]

I don't think the offset between the two results depends on roundoff errors, but on an incorrect sign somewhere.

[Werner_E]

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@-MFra- wrote:

I don't think the offset between the two results depends on roundoff errors, but on an incorrect sign somewhere.

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Yes, that very likely as the signals seem to be just mirrored.

The imaginary parts of your values seem to have different signs.

[-MFra-]

That's right!

[-MFra-]

Remark: In the collapsed section of the Laplace transforms, read "Mesh" and not "Mash".

[-MFra-]

From a more detailed examination of the network, I find that the fourth equation (V4) the mutual inductances are negative given that the inductor of inductance L4 is wound in the opposite direction to the other three, so the mutual inductance has a negative sign, and in the matrix NL are positive (M41,M14,M42,M24,M34,M43).

[ttokoro]

When omega=0, L is shorted and C is opened. Then the circuit is parallel of G1 and G2. V1=V5 and their phasor angles are same.

When omega=infinite, C is shorted and L is opened. Then the circuit is capacitors only and their phasor angles are also same. V1:V2 is 

[-MFra-]

For me, the limit cases are the following:

[ttokoro]

Yes, that is right. It means the phasors of V1 and V5 are same for omega= 0 and infinite. So the both first phasor plots are not correct if arg(W(w)) is a phase shift of VG1 to VG2.  

[-MFra-]

I created this post hoping that an expert would tell me where I went wrong and what corrections to make. If not, I will publish my solution which I will consider the correct one, and I will accept it as a solution.

[ttokoro]

Now, I can't use Mathcad 15. So, converted it to Prime 9. 
In your sheet I and II, they started from as follows.

V1-(voltage drops by impedanse between node 1 and 2 by ohms law) should be "V2" instead of "0" in your equations.

So I can't understand the meaning of your first equations.

[-MFra-]

Here we are dealing with four inductors traversed by different currents but flowing in the same direction. The voltage on the first inductor is given by the generalized Ohm's law V1(s)=s.L1.I1(s). The inductor L1 generates a vectorial and non-conservative magnetic field in the surrounding space, and so do the other inductors. Thus the first inductor is also immersed in the variable magnetic fields of the other inductors. If the inductors of the case are all wound in the same direction as the first inductor, their magnetic fields induce a voltage in it which depends on the mutual inductances between the first inductor and the others. So we have V1(s)=s.L1.I1(s)+s.M12.I2(s)+s.M13.I3(s)+s.M14.I4(s). In the case under examination M14 is negative because the inductor is wound in the opposite direction of the others and this is indicated by the black spot placed to the right of the inductor as can be seen from the drawing.

[-MFra- · Accepted solution]

Ultimately, I find what results from the attached worksheet whose results are completely heuristic and which I consider as the correct answer to the problem.

[ttokoro]

Thanks, I got it by Prime 9.

[-MFra-]

Those who are interested in the complete analytical study of the active filter in the figure can download the pdf file from the FB group: "Ingegneri Elettronici Italia", in the "File" section.