# How to work Maths is the time sample domain

Discussion in 'Electronic Design' started by Joseph Goldburg, Jul 18, 2004.

1. ### Joseph GoldburgGuest

Hi All,

I want to make an impedance measuring device from 0 to 20 KHz.

If I generate |V| at zero phase and I measure |I| at some phase theta

Then |Z| at delta phase = |V| / ( |I| theta) = |V /I| at
(- theta)

But both V(t) and I(t) are digitally sampled.

V(nT) = |V| sin (2 x Pi x nT) and I(nT) = |I| sin (2 x Pi x nT)

n = sample number etc

NOTE --- I can digitally sample V as V(nT) and I as I(nT) as a 8bit

Question

How do I get Z = V / I from digital samples and calculate the
magnitude and phase.

I will fix the frequency of V(t) = Sin (2 x Pi X f x t) then sample V

You thoughts would be appreciated.

Regards
Joseph

2. ### Don PearceGuest

So far you have only got the magnitude of Z. The actual value is R +
jX. From your samples you can recover the peak values of both V and I.
You can also count the number of samples between the two waveforms
crossing zero in the same direction. This will give you the phase
relationship between them. Now do a vector calculation on I to
establish the proportion that is in phase, or at ninety degrees to the
volt age waveform and do a V/I calculation for each. These will
give you R and X for the R+jX calculation.

d
Pearce Consulting
http://www.pearce.uk.com

3. ### Tim WescottGuest

Looking for zero crossings only works if you sample really fast.

If you sample V and I several times over a cycle, and demodulate them
with inphase and quadrature sine waves:

Vi = (1/N)sum(V(n) * cos(2*pi/M * n));
Vq = (1/N)sum(V(n) * sin(2*pi/M * n));
(and the same for I)

Then you can treat them like complex numbers, so

Z = ((Vi*Ii + Vq*Iq) + j(Vq*Ii - Vi*Iq))/(Ii^2 + Iq^2).

I'm prone to typos so you'll want to check to make sure my complex
arithmetic is all OK, and to make yourself understand it.

4. ### Don PearceGuest

Alternatively, you average out the v-crossing intervals and i-crossing
intervals to get a mean phase shift.
What is the source of your i and q references?

d
Pearce Consulting
http://www.pearce.uk.com

5. ### Joseph GoldburgGuest

Hi Tim,

I would like to understand the maths behind this - can you name the
algorith you are using.

Regards
Joseph

6. ### John WoodgateGuest

I read in sci.electronics.design that Tim Wescott
com>) about 'How to work Maths is the time sample domain', on Sat, 17
Jul 2004:
.... which would be possible only if you define M, N and n.

7. ### Paul KeinanenGuest

Since the waveforms are sampled only with 8..10 bits, my guess is that
10..20 samples/cycle should be enough. Close to the zero crossing of a
sine wave, the waveform can be approximated with a straight line,
quite accurately within +/-5 degrees, but I guess that even within
the +/-30 degree range the approximation would be acceptable in this
case.

Take the last sample below zero and first sample above zero and draw a
straight line between these two points and calculate were the line
crosses the zero line between the sampling point. This will give the
_fractional_ sampling period after the first sample.

This way the phase shift between the current and voltage can be
measured to within a few degrees even with 10..20 samples/cycle.

Assuming the sampling frequency is not synchronised with the
measurement signal, the sampling points around the zero crossing do
not always hit the same spots, thus averaging several measurements
should improve the resolution.

Calculating the phase shift between the positive going current and
voltage transitions and then separately between negative going
transitions may help to detect any DC bias in the measurement system.

Paul

8. ### Ken SmithGuest

Since he is just going for the phase difference between the voltage and
the current he can just pick a reference at random. Whatever he picks
will add the same value to the two values he needs to subtract.

9. ### Don PearceGuest

While this is true, he still needs a constant phase relationship,
which means frequency locking to the signal - which implies some sort
of decoder to generate the i and q. In hardware, you could have a
Costas loop.

d
Pearce Consulting
http://www.pearce.uk.com

10. ### Ken SmithGuest

I assume he is creating one of these signals so he already has the
frequency. If not then yes he needs to frequency lock. A PLL would do
this for him and the error in the phase detector would drop out.

Based on what was implied elsewhere, I think he has a micro that can do
the locking if needed.

11. ### Tim WescottGuest

You'd have to average a _lot_ of crossings -- my method will get you an
answer in one cycle who's accuracy is only dependent on system noise.

The source of the i and q references is the same timebase he's using to
generate his V.

12. ### Tim WescottGuest

Actually it's only possible if _Joseph_ defines M, N and n. _I'm_ just

M is the number of samples in a cycle (conventional wisdom says 3 or
more, but could be anything that builds you up a complete picture of the
wave).

N is the number of samples in a vector (should end on a cycle, so than
N*M is an integer).

n is the particular sample in the vector.

13. ### John LarkinGuest

The easy way to do this is to generate a sine wave (applied voltage)
at frequency F, and take adc (current) samples at 4F. Call a set of 4
sequential samples a,b,c, and d. Then let

I = a-c is the angle-zero (in-phase, "I") p-p value

Q = b-d is the 90 degree (quadrature, "Q") p-p value.

You can take a lot of samples and average; this washes out both noise
and any DC offset, and extends the effective precision of your ADC.

(Or, equivalently, make the stimulus F the current source and sample
resulting voltage at 4F.)

Then just do a little trig.

John