# To and from Z - Calculating Rs and Cs from Z(real) & Z(Imag)?

• posted

Hi

Can anyone show by example how to calculate the R and C values if you have both real and imaginary parts of the impedance Z? I do not have the R and C values but I have the Z(Real) and Z(Imag). Also is there say a BASIC program that can say produce the correct R & C values from a Bode plot?

Cheers

Wayne

• posted

I read in sci.electronics.design that Wayne wrote (in ) about 'To and from Z

- Calculating Rs and Cs from Z(real) & Z(Imag)?', on Fri, 24 Sep 2004:

For R and C in series, R = Z(real) and C = 1/(2[pi]fZ(Imag)). f is the frequency in Hz; you need to know that. And the results assume sine-wave signals.

Not unless you know something else about the impedances. A Bode plot can only tell you the product RC:

RC = 1/(2[pi]f3),

where f3 is the frequency at which the response is 3 dB down or up. And this works only if the Bode plot shows a 20 dB/decade ultimate slope, implying a first-order filter network. You hardly need a program to do that.

These three equations are just rearrangements of bog-standard elementary a.c. theory equations. Maybe you need a bit of math brush-up.

To analyse more complex Bode plots, you might need a program, but there are fairly simple graphical methods that actually give you an *insight* into how the network is behaving, which is very valuable. I can't give you any references, but a *good* textbook on Bode plots should deal with 'straight-line approximation', which is the key phrase.

```--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.```
• posted

R is equal to Re(Z). The imaginary part correspond to an inductor(if it is positive) or a capacitor(if negative) according to the folowwing equation:

C = 1/2*pi*Freq*Z(imag) L = Z(imag)/2*pi*Freq

• posted

Thanks All.

If I have the Phase then say 40deg and the magnitude say 10 then how can I convert that into capasitance. I have the following circuit:

___SIGGEN_____ ¦ ¦ ¦--/\/\/\/\-----¦ ¦---¦ R_shunt Cx ¦ ¦

----V?----

I am measuring the voltage drop accross R_shunt and comparing that with the siggen so I have V_R_shunt - SIGGEN = Z and on a scope I can measure the phase. In other words I have the equiverlant info as you would find when creating a Lissajour plot. How can I find the capacitance of Cx with the info I have here?

Cheers

Wayne

• posted

the

have

and

--------------- Total Zreal =10 (magnitude) Zimag=20 (amgnitude)

R=Zreal If you know the actual series R (it is not shunt as you drew it). Assuming the Zimag is purely capacitive then Zimag=-1/(2*pi*frequency*C) and C can be found. If purely inductive, it will be Zimag=+2*pi*freqency*L

If,for example the series R (as you have drawn it) was 5 ohms then the Zreal would be 10-5=5 ohms and Z (as drawn) would be root(5^2+20^2) =20.6 ohms at a phase angle of -76 degrees (- for capacitance)

Also your expression above is wrong. Siggen =VR(series)+ VCx where the sum is a phasor sum- the relative phase angles must be taken into account. With the siggen and VR values only , you must assume a pure capacitance (which may be in error). Measure the signal voltage, the voltage across the resistor and the voltage across the capacitor and then you can graphically or numerically find the real and imaginary parts of the capacitance impedance.

```--
Don Kelly
dhky@peeshaw.ca```
• posted

If one has the Real and the Imaginary terms, then the Real term is the resistive component, and the Imaginary term (depending on the sign) is capacitive or inductive term. If one has phase and magnitude, one can easily calculate the Real and imaginary terms. Most bode plots have only the magnitude term; certain slopes (say

6dB/octave) are related to certain phase angles (say 90 degrees) but only in the limit (ie not accurate for all parts of the curve with that slope).
• posted

Alternatively, throw away your formulae and just plot the point(s) on a Smith Chart.

```--

"What is now proved was once only imagin'd." - William Blake, 1793.```
• posted

Given a magnitude of 10 and a phase angle of 40 degress, the real part will be equal to 10 * cos 40 and the imaginary part is 10 * sin 40. This works out to be 7.66 + j6.43. In this case, the reactive component is actually an inductor and not a capacitor.

• posted

I read in sci.electronics.design that Renante Solar wrote (in ) about 'To and from Z - Calculating Rs and Cs from Z(real) & Z(Imag)?', on Mon, 27 Sep 2004:

The OP specified 45 degrees and didn't say whether it was +45 degrees or

-45 degrees, but specified a capacitor in his circuit.

```--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.```
• posted

can I

with

measure

find

you

R
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and

Zreal

ohms at

into

voltage

the

------- The real part, I assume, includes the series resistor. The sign of the phase angle is not given so the imaginary part would be +/- 6.4 either inductive or capacitive. He said it was a capacitor so angle must be -40.

```--
Don Kelly
dhky@peeshaw.ca```

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