# Complex measurement problem

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#### pixel

This is one practical measurement problem. I have resistor divider Rs, Rp and capacitance Cp in parallel with Rp, connected to gnd.
Here are circuit equations:
Iin = Vin*(1+s*Cp*Rp)/(Rp+Rs+s*Cp*Rs*Rp)
Iin = Vin*(Rp+Rs+Rs(wCpRp)^2+jw*Cp*Rp*Rp)/((Rp+Rs)^2+(w*Cp*Rs*Rp)^2)
v[Cp]=v[Rp]=(Rp*Vin)/(Rp+Rs+s*Cp*Rs*Rp)
i[Rp] = v[Cp]/Rp
i[Cp]=v[Cp]*s*Cp

I can measure only admitance vector Y (or Iin), i.e. A(w) and B(w) values at frequenciues w1 and w2. How can be Rp, Rs and Cp identiified ?

#### tom_hanks

##### Full Member level 5
Rp and Rs are frequency independent component...
so they are easy to identifed
using current divide formula

while for CP you need use this formela

find out >
VC1 at freq=w1
VC2 at freq=w2

Cp=(Vc1-Vc2) * Y

#### pixel

I can not measure voltage at output, because I dont have acess. Such case is for example in some biomedical measurement, or chemical process. Only I can measure is input current amplitude and phase at some frequency. If you write equations you will see that it is not so easy, but also quite interesting task to find solution.

#### teteamigo

We should know the greatness order of Cp, to apply a frequency that made impedance of Cp equal to ZERO.

Measure Input resistor in this condition.

Then measure current DC

Calculate Rs and Rp

The Cp may be calculate by measure the time constant-ζ (ideal for Cp higher) of transient response. Or then by made the RC series equivalent of parallel Rp//Cp, then you can use the ressonance frequency criterium to determine Cp.

#### tom_hanks

##### Full Member level 5
It s look ery tricky...

i guess u need to use a Frequency and wave generator...

and see the responce of input change....

and find out the resonant frequency...
with resonant frequency u can get the component values..

tom

#### pixel

I have found one method... Here Y is differentaiated, and gives more suitable equations, but problem is how to measure accuratelly dY/dw... Doeas anyone have some simpler solution?

#### tomhive

##### Member level 2
pixel said:
I have found one method... Here Y is differentaiated, and gives more suitable equations, but problem is how to measure accuratelly dY/dw... Doeas anyone have some simpler solution?
Hi,
why u wants to know the individualy RS,RP & C, once u have input Admt. or current mag. if u cant acces to those.

tomhive
bye

#### pixel

Because thoose values are important characteristics of some electrochemical process.

#### Borber

We have 3 unknowns. We need 3 equations to solve the problem. If you measure Y at 3 frequencies you have those 3 equations.

### pixel

points: 2

#### pixel

I have at first also had such idea, but I could not solve equations...
maybe question for math forum... I think that they are not independant...

#### Element_115

First do a DC measurement.
Say @ 1Vdc, find the current ?mA--> then Rs+Rp = 1v/?mA! (C is Open at DC)

Now use a Funtion Genorator and put a sine wave into the circuit.
Increase the frequency until you get a maximum current! At this point
you can "Assume" the Cap is a short. Now if you can get the AC current
you can find out what Rp is.

Rs = Vac/Iac

Rp = (Rs+Rp) - Rs.

Now get a nominal frequency and calculate C.

1Vrms (@ Fo) /Iac = Rs+ [(Rp*2piFoC)/( Rp*2piFoC)]

Vac is known, Iac measured, Rs&Rp calculated---> Solve for C

Cheers

#### zorro

Actually, there are 3 unknowns and 4 equations: A(w1), B(w1), A(w2), B(w2). The problem can be solved as a "best fit". The system can be ill conditioned: for instance, if B<<A for both frequencies, C cannot be practically calculated.

Pixel, can you choose the 2 frequencies?

Regards
Z

### pixel

points: 2

#### pixel

zorro said:
Actually, there are 3 unknowns and 4 equations: A(w1), B(w1), A(w2), B(w2). The problem can be solved as a "best fit". The system can be ill conditioned: for instance, if B<<A for both frequencies, C cannot be practically calculated.

Pixel, can you choose the 2 frequencies?

Regards
Z
I can choose two frequencies between zero and pole in current characteristics, but I can not measure DC, or too high frequencies.

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