I have a problem deriving a formula I will be using in my bachelor thesis. It's about eliminating interfering voltages during measurement.

To understand how the interfering voltage is going to be eliminated, I should explain the method of measurement first. So in this measurement we measured the voltage over a resistor induced by a current, where a transformer was used to push the current. Then we switched the current off and measured the voltage again. After that, we switched the poles on the transformer and thus shifted the phase 180°. Now the current is switched on again and we measure the voltage a third time.

We now have the voltage before switching poles (U_1), the voltage during the current-free pause (U_S) and the current after switching poles (U_2).

The voltage without the interfering voltage (U_M) is now given to

$$U_M = \sqrt{\frac{U_1^2 + U_2^2}{2}-U_S^2}$$

It's given in the german DIN EN 50522:2011-11 without further explanation and my brother and I are not able to derivate it from this wikipedia image. The article (german) is the only thing i found online and in all my literature, but even there is no explanation.

Thanks in advance.

  • \$\begingroup\$ I think you should rephrase the question. At first glance, a current-free pause looks like 0A through the resistor which yield 0V. But I think, you mean (like the wiki page) the voltage across the resistor, which is solely caused by noise/external interferance, when there is no measurement current injected. \$\endgroup\$ – Huisman Mar 18 at 19:36
  • \$\begingroup\$ This formula computes Z from cancelling interferer voltage, and you are measuring current. Z(f)=U(f)/I(f) you are trying to detect Um. Is this valid? \$\endgroup\$ – Sunnyskyguy EE75 Mar 18 at 19:46
  • \$\begingroup\$ There is an explanation on de.wikipedia.org/wiki/Umpolungsverfahren and there are free translators on the internet ... or is the translation obscure? \$\endgroup\$ – Huisman Mar 18 at 19:47
  • \$\begingroup\$ @SunnyskyguyEE75 The next equation on the wiki site calculates Z from \$U_M\$ \$\endgroup\$ – Huisman Mar 18 at 19:49
  • \$\begingroup\$ So if you measure \$I_1=U_1/R\$ can you compute\$ U_1=I_1*R\$ no because that is just the resistor not including Zm. I don't understand why you don't' measure U1,U2,Us \$\endgroup\$ – Sunnyskyguy EE75 Mar 18 at 19:50

For your convenience a translation:

The "polarity reversal method" is a method from measuring engineering for interference suppression in complex impedance measurements with a fixed measuring frequency.

In this method, three measurements are taken to determine the impedance of a DUT at measurement frequency:

  1. Without source (without current injected in OP's case)
  2. With source at measurement frequency (with current injected in OP's case)
  3. With source measurement frequency 180° phase shifted (with negative current injected in OP's case)

Figure 1 shows the complex phasor diagram of the voltage measurement with injected current in a DUT using polarity reversal. Here is:

  • \$U_{S}\$ the voltage caused due to disturbance/interference/noise, measured when the source is switched off
  • \$U_1\$ the voltage measured with the source having 0° phase shift
  • \$U_{2}\$ the voltage measured with the source having 0° phase shift

The voltage \$U_ {M}\$ which is needed to determine the impedance (of the DUT) is calculated from the values ​​of the measured voltages according to the following equation:

$$U_{M}={\sqrt {{\frac {U_{1}^{2}+U_{2}^{2}}{2}}-U_{S}^{2}}} $$

Now, you can calculate the impedance \$ \underline {Z} \$ using the equation:

$$ \underline{Z} = \frac {{\underline {U}}_{M}}{\underline {I}}$$

Important in applying the polarity reversal method is fulfilling the following criteria:

  • The disturbance/inteferance/noise may only occur at one frequency.
  • The disturbance/inteferance/noise must be stable
  • \$\begingroup\$ Reversing the leads will upset the electric fields, perhaps causing non-cancellation. \$\endgroup\$ – analogsystemsrf Mar 19 at 2:39
  • \$\begingroup\$ Thanks for your input, but this is just the wikipedia article in englisch. Im german so i dont have any trouble understanding words or anything like this, I just am not able to get to the formula for $$U_M$$ myself \$\endgroup\$ – prelow Mar 19 at 10:02
  • \$\begingroup\$ Your nationality was unknown, so I gave a "For your convenience a translation" as support for you or others to answer your question. So, not an answer, the remark box is too small and gives less options for layout \$\endgroup\$ – Huisman Mar 19 at 10:25

The area of the parallelogram is the area of the blue region =B*H, which is the interior of the parallelogram enter image description here

Ref Parallelogram enter image description here

Maybe combine two Parallelograms (?)

enter image description here 1. \$~~~U_M =\sqrt{\dfrac{U_1^2+U_2^2}{2}-U_S^2}\$
2. \$~~~2U_M^2+2U_S^2 =U_1^2+U_2^2\$

  • \$\begingroup\$ I'm sorry, but this does not help... Mainly because its not that easy, you cant just say U_S is orthogonal to U_M and therefore pythagoras is not allowed. But if it were, it would'nt help either because I already tried that and was not able to get to the given formula... but thanks again for your time, input and patience, i really appreciate it! \$\endgroup\$ – prelow Mar 19 at 23:42
  • \$\begingroup\$ Do you understand the Wiki REF link? \$\endgroup\$ – Sunnyskyguy EE75 Mar 19 at 23:43
  • \$\begingroup\$ AFAIK it only refs to the wikipedia page? I understand everything on this page, I know what the measurement is for, I know how to carry it out etc. Only thing I don't know is where they get this formula from... \$\endgroup\$ – prelow Mar 19 at 23:46
  • \$\begingroup\$ Sorry I meant en.wikipedia.org/wiki/Parallelogram but since Univ is learning How to learn, I wont solve this for you, just hints \$\endgroup\$ – Sunnyskyguy EE75 Mar 19 at 23:48
  • \$\begingroup\$ Yes, I saw that post and I do fully understand what a parallelogram is and how I can calculate it... BUT: I've tried this w/ my brother, who I think is a very intelligent person, I've tried this with two friends of mine and I even asked my old physics teacher. I've tried to get this formula for hours, i made paintings in MS Paint, on paper, wasted at least 30 sheets of paper just fo calculations but neither me, my brother, one of my friends or my old physics teacher is able to get to the bottom of this... \$\endgroup\$ – prelow Mar 19 at 23:53

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