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

W

#### Wayne

Jan 1, 1970
0
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

J

#### John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Wayne <[email protected]>
wrote (in said:
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).

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.
Also is there say
a BASIC program that can say produce the correct R & C values from a
Bode plot?

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.

R

#### Renante Solar

Jan 1, 1970
0
Wayne said:
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

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

W

#### Wayne

Jan 1, 1970
0
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

D

#### Don Kelly

Jan 1, 1970
0
Wayne said:
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
---------------
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.

R

#### Robert Baer

Jan 1, 1970
0
Don said:
---------------
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.

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).

P

#### Paul Burridge

Jan 1, 1970
0
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).

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

R

#### Renante Solar

Jan 1, 1970
0
Don Kelly said:
---------------
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.

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.

J

#### John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Renante Solar
ogle.com>) about 'To and from Z - Calculating Rs and Cs from Z(real) &
Z(Imag)?', on Mon, 27 Sep 2004:
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.

The OP specified 45 degrees and didn't say whether it was +45 degrees or
-45 degrees, but specified a capacitor in his circuit.

D

#### Don Kelly

Jan 1, 1970
0
Renante Solar said:
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.

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