Radiated Emissions from 15 UUT's
Dave
electronics equipment.
I want to avoid testing the system as a whole due to the difficult and
time-consuming setup.
The whole system needs to pass CISPR11 Class-A emissions (40 dBuv/m @
10 meters).
By my calculations each unit needs to be at least 23.53 dB below the
limit.
This assumes the worse case where the E-fields from all 15 units add
together.
V1 = E-Field from 1 unit
V2 = E-Field from 15 units
dB = 20 Log(V2/V1)
23.53 = 20 Log(15/1)
Is this the correct way to calculate the total emissions?
This doesn't feel quite right...
Wimpie
Hello Dave,
Are you talking about in installation (larger then 10m) or a small
setup that contains all the devices? For now I assume a small setup.
The E-fields do only sum when they are from in-phase-synchronized
sources and do couple with the measuring antenna. In this case your
23.5 dB is valid for both Average and QP readings. If some of the
emissions couple in other phase, of course the increase will be less
then 23.5 dB. The phase issue will be present for sure when you have a
relative large setup with cables running everywhere and at UHF
frequencies (due to multiple (housing) reflection, etc). So 23.5 dB
will be the absolute worst case situation.
When the sources are non-synchronized, but continuous wave (for
example all devices have a 10 MHz clock [own crystals]), it is likely
that all sources will be within the receiver bandwidth (9 kHz for
0.15 to 30 MHz). In your case the long-term average power will be 15
times above the power of one signal. This would result in just 11.8 dB
increase (for the average detector), however there is QP-detector used
also. As you have a relative low number of signals (15), the
occurrence that all signals are temporary in phase is too high, so the
QP detector will follow the peak envelope. Tthis results in a somewhat
less then 23.5 dB higher reading then for one unwanted emission
only.
When the non-synchronized sinusoidal sources are very close together
(say <1Hz), the envelope varies very slowly and in that case the
Average detector is able to follow the envelope, hence the output may
increase with more then 11.8 dB (with a strong statistical component
as the measurement has to start when all emissions are (almost) in
phase).
When your problem area is already noise-like and non-synchronized,
both Average and QP detector will show 11.8 dB increase (of course
assuming that all individual emission would give same readings).
When the problem is in pulse-like emission with certain PRF (so QP
will be higher then Average reading), the increase depends on both
PRF, duty cycle of the signal and the synchronicity between the pulse-
like unwanted emissions. The increase can be anything between just
some dB to 20 dB for the QP detector and about 11.8 dB for the average
detector.
So the actual increase of unwanted emission depends on the nature of
signals and their relation.
Best regards,
Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me very likely
Dave
Hi Wim thanks for the response.
The actual installation size will vary, say 4m x4m for a starting
point. The turntable in the EMC test chamber is 12 feet diameter so
everything has to fit on it during testing. The antenna is 10m
distance.
I noticed you used some power 10log() calculations in your reply. I
was not sure if I should use power or voltage calculations. Since the
final result is in V/M using 20log() seemed appropriate. Please
explain why you and when you would use power versus voltage math.
The bottom line for my testing is the QP reading. The nature of the
emission will be non-synchronized digital coming from multiple CPU
boards and stepper motors and a CAN comm bus. So there is a variety
of noise sources. I think the probability of all emissions being in
phase and adding is next to zero. Therefore I agree that your long-
term average emission is most likely. I can't grasp the concept why
that would be power 10log() and not voltage 20log(). Please
elaborate.
Thanks,
Dave
glen herrmannsfeldt
(snip)
> I noticed you used some power 10log() calculations in your reply. I
> was not sure if I should use power or voltage calculations. Since the
> final result is in V/M using 20log() seemed appropriate. Please
> explain why you and when you would use power versus voltage math.
The rule in optics, is:
Add the amplitude (voltage or current) for coherent sources,
Add the intensity (power) for incoherent sources.
That is the difference he was showing.
If you have N sources, all in phase, the voltage increases as N,
and the power as N**2.
If you have N source with no definite phase relationship, and
sufficiently different frequency, then on the average, even
over a fairly short time, the power will increase as N.
-- glen