Reply by Fred Marshall March 22, 20072007-03-22
"Jerry Avins" <jya@ieee.org> wrote in message 
news:WLGdncIg6oFcb5zbnZ2dnUVZ_uvinZ2d@rcn.net...
> Fred Marshall wrote: >> "Jerry Avins" <jya@ieee.org> wrote in message >> news:du-dnZHzAag4rpzbnZ2dnUVZ_rXinZ2d@rcn.net... >>> Fred Marshall wrote: >>>> <redpathdu@yahoo.co.uk> wrote in message >>>> news:1174403295.380400.60650@l75g2000hse.googlegroups.com... >>>>> Hi Fred, >>>>> >>>>> Finally got it fixed. Output waveforms look good, except for one >>>>> thing...the output, though spectrally correct is very low compared to >>>>> the input...there is a great deal of attenuation. Can this attenuation >>>>> be quantified? >>>>> >>>>> Duncan >>>> You didn't say but I think it's probably a factor of "N" where N is the >>>> decimation factor. After all, (N-1)/N of the energy has been removed, >>>> leaving 1/N, by ignoring all but every Nth sample. >>> Why doesn't the zero-order hold put the energy back? Suppose a 4 KHz >>> bandlimited signal sampled at both at 8 KHz and at 44.1. Is there a >>> difference in loudness when they're played back? >>> >>> Jerry >> >> Jerry, >> >> It's not clear that the "waveform" he mentions is analog/continuous.... > > Nevertheless, decimation lowers the sample rate and therefore the > effective "duration" of each sample. > > Does a 100 HZ full-scale sinewave show more energy when sampled at 20 KHz > than it does when samples at 500 Hz? > > Jerry
Yes. When one is dealing with finite-length sequences. When the finite length is measured in time. When the number of samples is reduced, the integral of the samples squared is reduced over the same time period. Or, if you like, the integral of the samples squared per unit time is reduced. I think that's likely what's being referred to nere. At least it's a good possibility. Fred
Reply by Jerry Avins March 22, 20072007-03-22
Fred Marshall wrote:
> "Jerry Avins" <jya@ieee.org> wrote in message > news:du-dnZHzAag4rpzbnZ2dnUVZ_rXinZ2d@rcn.net... >> Fred Marshall wrote: >>> <redpathdu@yahoo.co.uk> wrote in message >>> news:1174403295.380400.60650@l75g2000hse.googlegroups.com... >>>> Hi Fred, >>>> >>>> Finally got it fixed. Output waveforms look good, except for one >>>> thing...the output, though spectrally correct is very low compared to >>>> the input...there is a great deal of attenuation. Can this attenuation >>>> be quantified? >>>> >>>> Duncan >>> You didn't say but I think it's probably a factor of "N" where N is the >>> decimation factor. After all, (N-1)/N of the energy has been removed, >>> leaving 1/N, by ignoring all but every Nth sample. >> Why doesn't the zero-order hold put the energy back? Suppose a 4 KHz >> bandlimited signal sampled at both at 8 KHz and at 44.1. Is there a >> difference in loudness when they're played back? >> >> Jerry > > Jerry, > > It's not clear that the "waveform" he mentions is analog/continuous....
Nevertheless, decimation lowers the sample rate and therefore the effective "duration" of each sample. Does a 100 HZ full-scale sinewave show more energy when sampled at 20 KHz than it does when samples at 500 Hz? Jerry -- Engineering is the art of making what you want from things you can get. &macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
Reply by Fred Marshall March 21, 20072007-03-21
"Jerry Avins" <jya@ieee.org> wrote in message 
news:du-dnZHzAag4rpzbnZ2dnUVZ_rXinZ2d@rcn.net...
> Fred Marshall wrote: >> <redpathdu@yahoo.co.uk> wrote in message >> news:1174403295.380400.60650@l75g2000hse.googlegroups.com... >>> Hi Fred, >>> >>> Finally got it fixed. Output waveforms look good, except for one >>> thing...the output, though spectrally correct is very low compared to >>> the input...there is a great deal of attenuation. Can this attenuation >>> be quantified? >>> >>> Duncan >> >> You didn't say but I think it's probably a factor of "N" where N is the >> decimation factor. After all, (N-1)/N of the energy has been removed, >> leaving 1/N, by ignoring all but every Nth sample. > > Why doesn't the zero-order hold put the energy back? Suppose a 4 KHz > bandlimited signal sampled at both at 8 KHz and at 44.1. Is there a > difference in loudness when they're played back? > > Jerry
Jerry, It's not clear that the "waveform" he mentions is analog/continuous.... Fred
Reply by Jerry Avins March 21, 20072007-03-21
Fred Marshall wrote:
> <redpathdu@yahoo.co.uk> wrote in message > news:1174403295.380400.60650@l75g2000hse.googlegroups.com... >> Hi Fred, >> >> Finally got it fixed. Output waveforms look good, except for one >> thing...the output, though spectrally correct is very low compared to >> the input...there is a great deal of attenuation. Can this attenuation >> be quantified? >> >> Duncan > > You didn't say but I think it's probably a factor of "N" where N is the > decimation factor. After all, (N-1)/N of the energy has been removed, > leaving 1/N, by ignoring all but every Nth sample.
Why doesn't the zero-order hold put the energy back? Suppose a 4 KHz bandlimited signal sampled at both at 8 KHz and at 44.1. Is there a difference in loudness when they're played back? Jerry -- Engineering is the art of making what you want from things you can get. &macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
Reply by Fred Marshall March 21, 20072007-03-21
<redpathdu@yahoo.co.uk> wrote in message 
news:1174403295.380400.60650@l75g2000hse.googlegroups.com...
> Hi Fred, > > Finally got it fixed. Output waveforms look good, except for one > thing...the output, though spectrally correct is very low compared to > the input...there is a great deal of attenuation. Can this attenuation > be quantified? > > Duncan
You didn't say but I think it's probably a factor of "N" where N is the decimation factor. After all, (N-1)/N of the energy has been removed, leaving 1/N, by ignoring all but every Nth sample. Fred
Reply by March 20, 20072007-03-20
Hi Fred,

Finally got it fixed. Output waveforms look good, except for one
thing...the output, though spectrally correct is very low compared to
the input...there is a great deal of attenuation. Can this attenuation
be quantified?

Duncan


Reply by Fred Marshall March 19, 20072007-03-19
<redpathdu@yahoo.co.uk> wrote in message 
news:1174320586.873587.206550@p15g2000hsd.googlegroups.com...
> Hi Fred, > > Looked at the results both on scope and spec analyzer. One thing I > need to ask you about is when you apply the unit sample to the > polyphase decimator. I apply the data at the input via a commutator > arrangement. However, the unit sample only gets applied to filter5. > The rest of the filters are filled with zeros. Therefore the output of > the whole polyphase filter is just based on the convolution of the > unit sample with the coefficients of filter5. The rest of the filters > in the structure have no effect because their outputs are always zero. > Is this correct? > > Duncan
Duncan, I was afraid of that! In fact, why should it be filter5 instead of any other? Only the phase of the commutator seems would determine that - or the absolute temporal placement of the unit sample. This works better (to visualize) *before* the decimation. After all, the timing of the decimation is arbitrary isn't it? So, the unit sample response out of the filter/decimated should be just that.... Let's see now.... it causes one to ask: "what is the unit sample response after decimation?" The answer should be that it remains a sinc-shaped response so that it coincides with being lowpassed. Is that what you see? Yes! because the sinc shape is quite broad for a narrow lowpass and resampling it (decimating it) should cause the shape to remain. Depending on the phase of the sampling/decimation, you may or may not hit the sinc peak is all. Fred
Reply by March 19, 20072007-03-19
Hi Fred,

Looked at the results both on scope and spec analyzer. One thing I
need to ask you about is when you apply the unit sample to the
polyphase decimator. I apply the data at the input via a commutator
arrangement. However, the unit sample only gets applied to filter5.
The rest of the filters are filled with zeros. Therefore the output of
the whole polyphase filter is just based on the convolution of the
unit sample with the coefficients of filter5. The rest of the filters
in the structure have no effect because their outputs are always zero.
Is this correct?

Duncan


Reply by Fred Marshall March 18, 20072007-03-18
<redpathdu@yahoo.co.uk> wrote in message 
news:1174085261.719268.134920@l77g2000hsb.googlegroups.com...
> Hi Fred, > > That shed some light on it. I applied a unit sample to the standard > FIR and looked at it on the spectrum analyzer...fine, polyphase > interpolator...fine, polyphase decimator...not fine at all. Now > looking at the code. Will keep you posted. > > Cheers > Duncan
I might rather have looked at the sample sequence or the output on an oscilloscope (i.e. vs. time). Looking at the output of a spectrum analyzer might obscure things a bit. But, if you can interpret it, then fine. Fred
Reply by March 16, 20072007-03-16
Hi Fred,

That shed some light on it. I applied a unit sample to the standard
FIR and looked at it on the spectrum analyzer...fine, polyphase
interpolator...fine, polyphase decimator...not fine at all. Now
looking at the code. Will keep you posted.

Cheers
Duncan