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Potential divider involving reflected impedances

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ramuna

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RefImp_1.png **broken link removed** **broken link removed** **broken link removed** **broken link removed** **broken link removed** **broken link removed** Hello,

I have a potential divider problem involving reflected impedances. I have scanned my worksheets re this problem and uploaded them for your perusal.

Many thanks in advance for your help!

Regards,RefImp_1.pngRefImp_2.pngRefImp_3.pngRefImp_4.png
 
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Hello,

I have a potential divider problem involving reflected impedances. I have scanned my worksheets re this problem and uploaded them for your perusal.

Many thanks in advance for your help!

Regards,

The symbolic expression for the voltage divider is very complicated, and there is no hope of comparing my result with yours. A good way to get a reasonable confidence that my result is the same as yours is to substitute numerical values for the components, evaluate the expression for the voltage divider ratio, then compare my value to yours.

I used these values for the components, with the impedance of inductors set to j w L, and the impedance of capacitors set to 1/(j w C), and w=1 so that jw = j:

w= 1; s =j w; L1 = 3; L2 = 5; L3 = 7; L4 = 11;
C1 = 13; C2 = 17; C3 = 19; C4 = 23;
M1 = .6 Sqrt[L1 L2]; M2 = .7 Sqrt[L1 L3]; M3 = .8 Sqrt[L1 L4];
M4 = .5 Sqrt[L2 L3]; M5 = .3 Sqrt[L2 L4]; M6 = .9 Sqrt[L3 L4];

I get .00549726 for the divider ratio with those values.

If all the mutual inductances are made zero, then I get .429032
 
NP1.png NP2.png NP3.png NP4.png NP1.png NP2.png NP3.png NP4.png
The symbolic expression for the voltage divider is very complicated, and there is no hope of comparing my result with yours. A good way to get a reasonable confidence that my result is the same as yours is to substitute numerical values for the components, evaluate the expression for the voltage divider ratio, then compare my value to yours.

I used these values for the components, with the impedance of inductors set to j w L, and the impedance of capacitors set to 1/(j w C), and w=1 so that jw = j:

w= 1; s =j w; L1 = 3; L2 = 5; L3 = 7; L4 = 11;
C1 = 13; C2 = 17; C3 = 19; C4 = 23;
M1 = .6 Sqrt[L1 L2]; M2 = .7 Sqrt[L1 L3]; M3 = .8 Sqrt[L1 L4];
M4 = .5 Sqrt[L2 L3]; M5 = .3 Sqrt[L2 L4]; M6 = .9 Sqrt[L3 L4];

I get .00549726 for the divider ratio with those values.

If all the mutual inductances are made zero, then I get .429032

I discovered an error in my workings. I have uploaded the scanned new worksheets with the corrections. I have input the new formulae and your test values into Matlab and unfortunately get different results compared to your own. I have also uploaded a text file with the Matlab entries, so that you can cut&paste from it.
 

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I discovered an error in my workings. I have uploaded the scanned new worksheets with the corrections. I have input the new formulae and your test values into Matlab and unfortunately get different results compared to your own. I have also uploaded a text file with the Matlab entries, so that you can cut&paste from it.

Your method isn't taking all the reflected impedances into account. For example, when you reflect the X2 subcircuit into the upper half, you haven't taken into account the fact that the X4 subcircuit needs to be reflected into the X2 subcircuit BEFORE you reflect the X2 subcircuit. And, in fact, BOTH the X3 and X4 subcircuits reflect their impedances into the X2 subcircuit BEFORE you reflect it into the upper half. Also, the X3 subcircuit must be reflected into X4 BEFORE you reflect the X4 subcircuit into the X2 subcircuit, which is then reflected into the top half.

And, this is just the beginning. You have to account for all these various interactions, which greatly complicate things.

I think your only hope is to solve the full circuit at once using mesh analysis, which is what I did.
 
Your method isn't taking all the reflected impedances into account. For example, when you reflect the X2 subcircuit into the upper half, you haven't taken into account the fact that the X4 subcircuit needs to be reflected into the X2 subcircuit BEFORE you reflect the X2 subcircuit. And, in fact, BOTH the X3 and X4 subcircuits reflect their impedances into the X2 subcircuit BEFORE you reflect it into the upper half. Also, the X3 subcircuit must be reflected into X4 BEFORE you reflect the X4 subcircuit into the X2 subcircuit, which is then reflected into the top half.

And, this is just the beginning. You have to account for all these various interactions, which greatly complicate things.

I think your only hope is to solve the full circuit at once using mesh analysis, which is what I did.
Thank you for pointing me in the right direction ! Now, where did I put that drawing board :-(
 
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