> >> Apply a good widow -- I like Nuttall for this --
> >
> > I hope you have good old Mrs. Nuttall's consent!
> >
> > SCNR, and thanks for the actual content ;)
>
> Oops! Window; window. What's "SCNR"?
>
> Hi everybody
> thanks to the support I received on a previous post I have been able to get
> working my implementation of a 90 degrees phase shifter,
> all I need now is a good guy which can calculate and post for me the
> coefficients for a 67 order Hilbert with Hamming windowing, because my tool
> just reach 35, or some reference to a free tool on the net for this job.
> ( I'm not planning to buy a 1K$ one)
>
> Thank you in advance.
> Giuseppe
I assume, from what you have written, that you are monitoring mains
power. A snag with what you are trying to do is you need to make the
Hilbert transform very precise in its 90 degree shift. If, say, you want
0.5% (consumer grade metering) accuracy at 0.5PF you need the shift to
be within about 0.02 degrees of 90 degrees. For industrial metering it
needs to be rather better than that.
One common way to deal with this problem is to phase lock to the voltage
waveform. It is fairly straightforward to get the loop to lock rather
precisely. It is then trivial to obtain a signal shifted 90 degrees from
the voltage waveform. However, mirroring any distortion in the original
signal is less easy. The voltage waveform seldom has much distortion,
but it can have a little.
Another method is to measure the mains frequency accurately. It doesn't
change very fast, so there are several trivially complex ways to do
this. Then you can calculate the exact number of samples delay which
represent a 90 degree shift at the current frequency, and delay by that
much. The delay will consist of a whole number of samples, which a
simple pipeline will give you, and a fraction, which a single step FIR
will give you. The frequency response of the single step FIR is not
flat, but this is seldom of consequence for power measurements. The
voltage waveform seldom has significant high harmonic content. You can
simply store a table of FIR coefficients for the various fractional
delays you will need for varying mains frequencies. The compute
requirement for this is far lower than for your Hilbert approach, and
the results are generally superior. If you are using an MCU, rather than
a DSP, (the normal case for power measurement) you will have no problems
with the compute load of this method.
Regards,
Steve
Reply by Steve Underwood●July 4, 20052005-07-04
Jerry Avins wrote:
> Martin Eisenberg wrote:
>
>> Jerry Avins wrote:
>>
>>
>>> Number the taps from -33 to +33, with n = 0 in the middle. Set
>>> all even-numbered taps equal to zero and the others to 1/n,
>>> including sign. Apply a good widow -- I like Nuttall for this --
>>> and a gain factor that you can determine experimentally.
>>
>>
>>
>> I hope you have good old Mrs. Nuttall's consent!
>>
>> SCNR, and thanks for the actual content ;)
>
>
> Oops! Window; window. What's "SCNR"?
>
> Jerry
I guess widowing must have something to do with military DSP. :-)
Steve
Reply by Jerry Avins●July 4, 20052005-07-04
Jerry Avins wrote:
> Giuseppe Sbarra wrote:
>
>> ... I have no idea on how calculate windowing.
>> Please, a little more help.
>> Thank
>> Giuseppe
>
>
> For hand calculation, a von Hann window is a good compromise. Counting
> from the beginning this time,
>
> w[k+1] = 0.5 * ( 1 - cos( 2*pi*k/(n-1) ), k = 0, 1, .. n-1
>
> Compute the window for n = 69 and discard the zero-value end points.
> Note that the values of w[k] are symmetric about the center and that
> half the filter's coefficients are zero. You therefore need to compute
> only 17 -- did I get that right? -- window coefficients.
See http://www.lohninger.com/helpcsuite/weightingwindow.htm for other
windows. Those further down the list (more terms) will show less ripple
in the passband.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Jerry Avins●July 4, 20052005-07-04
Martin Eisenberg wrote:
> Jerry Avins wrote:
>
>
>>Number the taps from -33 to +33, with n = 0 in the middle. Set
>>all even-numbered taps equal to zero and the others to 1/n,
>>including sign. Apply a good widow -- I like Nuttall for this --
>>and a gain factor that you can determine experimentally.
>
>
> I hope you have good old Mrs. Nuttall's consent!
>
> SCNR, and thanks for the actual content ;)
Oops! Window; window. What's "SCNR"?
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Jerry Avins●July 4, 20052005-07-04
Giuseppe Sbarra wrote:
> ... I have no idea on how calculate windowing.
> Please, a little more help.
> Thank
> Giuseppe
For hand calculation, a von Hann window is a good compromise. Counting
from the beginning this time,
w[k+1] = 0.5 * ( 1 - cos( 2*pi*k/(n-1) ), k = 0, 1, .. n-1
Compute the window for n = 69 and discard the zero-value end points.
Note that the values of w[k] are symmetric about the center and that
half the filter's coefficients are zero. You therefore need to compute
only 17 -- did I get that right? -- window coefficients.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Martin Eisenberg●July 4, 20052005-07-04
Jerry Avins wrote:
> Number the taps from -33 to +33, with n = 0 in the middle. Set
> all even-numbered taps equal to zero and the others to 1/n,
> including sign. Apply a good widow -- I like Nuttall for this --
> and a gain factor that you can determine experimentally.
I hope you have good old Mrs. Nuttall's consent!
SCNR, and thanks for the actual content ;)
--
Quidquid latine dictum sit, altum viditur.
Reply by Giuseppe Sbarra●July 3, 20052005-07-03
"Jerry Avins" <jya@ieee.org> ha scritto nel messaggio
news:Os2dnWiA2_aKpVXfRVn-gA@rcn.net...
> Giuseppe Sbarra wrote:
> > Hi everybody
> > thanks to the support I received on a previous post I have been able to
get
> > working my implementation of a 90 degrees phase shifter,
> > all I need now is a good guy which can calculate and post for me the
> > coefficients for a 67 order Hilbert with Hamming windowing, because my
tool
> > just reach 35, or some reference to a free tool on the net for this job.
> > ( I'm not planning to buy a 1K$ one)
> >
> > Thank you in advance.
> > Giuseppe
>
> Number the taps from -33 to +33, with n = 0 in the middle. Set all
> even-numbered taps equal to zero and the others to 1/n, including sign.
> Apply a good widow -- I like Nuttall for this -- and a gain factor that
> you can determine experimentally.
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
>
Clear and useful, but I have no idea on how calculate windowing.
Please, a little more help.
Thank
Giuseppe
Reply by Jerry Avins●July 3, 20052005-07-03
Giuseppe Sbarra wrote:
> Hi everybody
> thanks to the support I received on a previous post I have been able to get
> working my implementation of a 90 degrees phase shifter,
> all I need now is a good guy which can calculate and post for me the
> coefficients for a 67 order Hilbert with Hamming windowing, because my tool
> just reach 35, or some reference to a free tool on the net for this job.
> ( I'm not planning to buy a 1K$ one)
>
> Thank you in advance.
> Giuseppe
Number the taps from -33 to +33, with n = 0 in the middle. Set all
even-numbered taps equal to zero and the others to 1/n, including sign.
Apply a good widow -- I like Nuttall for this -- and a gain factor that
you can determine experimentally.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������