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From: glen herrmannsfeldt on 23 Apr 2010 19:59
Vladimir Vassilevsky <nospam(a)nowhere.com> wrote:
(snip)

> Even with ideal diode and ideal source, the angle of conduction is
> determined by transcendental equation. Now account for impedance,
> nonlinearity and add noise, and it gets really messy.

(snip)
> Jerry, do you know what was the rationale for choosing 455kHz
> vs 465kHz standard IF ?

Interesting question. I don't see anything in wikipedia that
hints as an answer. It seems obvious that it should be a multiple
of 5kHz, and presumably set such that image frequencies don't come
out in places you don't want them.

Otherwise, my guess is that once one was chosen, and the appropriate
parts started to get popular, there would be incentive to keep
using the same frequency. That might be enough reason.

I am interested to see what Jerry has to say.

-- glen
From: brent on 23 Apr 2010 20:23
On Apr 23, 7:59 pm, glen herrmannsfeldt <g...(a)ugcs.caltech.edu> wrote:
> Vladimir Vassilevsky <nos...(a)nowhere.com> wrote:
>
> (snip)
>
> > Even with ideal diode and ideal source, the angle of conduction is
> > determined by transcendental equation. Now account for impedance,
> > nonlinearity and add noise, and it gets really messy.
>
> (snip)
>
> > Jerry, do you know what was the rationale for choosing 455kHz
> > vs 465kHz standard IF ?
>
> Interesting question.  I don't see anything in wikipedia that
> hints as an answer.  It seems obvious that it should be a multiple
> of 5kHz, and presumably set such that image frequencies don't come
> out in places you don't want them.  
>
> Otherwise, my guess is that once one was chosen, and the appropriate
> parts started to get popular, there would be incentive to keep
> using the same frequency.  That might be enough reason.
>
> I am interested to see what Jerry has to say.
>
> -- glen

I am not sure that a multiple of 5KHz was required. the AM band is 10
KHz wide and when AM was defined it was in the context of a continuous
tunable oscillator. Also, If I am not mistaken, Europe has a 9 KHz
channel spacing.
From: glen herrmannsfeldt on 23 Apr 2010 20:27
Jerry Avins <jya(a)ieee.org> wrote:
(snip)

> Rarely is a carrier to be demodulated sampled at more than twice the
> carrier frequency; that would be a waste. The sampling theorem tells us
> that we have to sample more than twice the frequency corresponding to
> the bandwidth of interest. There are some practical restrictions (the
> second edition of Understanding Digital Signal Processing by Rick Lyons
> has an excellent analysis of them) but in general, sampling 20 KHz wide
> signal on a 356 KHz carrier can be accomplished with a 50 KHz sample
> rate. A bandpass filter assures that the AM signal is not contaminated
> by adjacent channels.

OK, but say one wants to minimize the analog circuitry, and fast
digital circuitry is available, including a fast ADC. That would
seem to go against the analog bandpass filter, but a lot of digital
filtering after the ADC could be provided.

> With bandpass sampling, we need to exclude signals
> both above and below the band being sampled. Even if the signal were
> sampled at 1 MHz, a low-pass filter would be needed to substantially
> eliminate all signals above 500 KHz.

But 1MHz is pretty slow for an ADC by now, isn't it?

> Incidentally, sampling at 1 MHz provides a little over 2 samples per
> carrier cycle, with little chance that either of them will be near a
> carrier peak and hence representative of the envelope. With bandpass
> sampling at 50 KHz, there will be only one sample for every 9 or so
> carrier cycles. There is then no hope of peak detection.

Even a small FPGA should allow for a lot of digital filtering
that can run at 100MHz or so.

-- glen

From: Jerry Avins on 24 Apr 2010 08:27
On 4/23/2010 6:09 PM, gretzteam wrote:
>> On 4/23/2010 5:03 PM, gretzteam wrote:
>>>>
>>>>
>>>> How does the average value of samples of the carrier relate to the
>>>> approximate value of the envelope? Would it help if the "carrier" were
>>>> triangular?
>>>>
>>>
>>> Yes you have a point here! All I've proven so far is that when the
> input
>>> signal contains only a carrier, full scale, then the output of the
> lowpass
>>> filter is pretty much exactly 0.63 (2/pi), which is the average value of
> a
>>> full scale sine wave.
>>
>> How many samples per carrier cycle do you have? How many carrier cycles
>> do you average over? How long does that take, and what does that imply
>> about the highest envelope frequency you can demodulate without
> attenuation?
>
> Ok I'm way oversampled. I'm doing this to learn about it so I don't want to
> have the added difficulty of sample rate (just yet). Here is the current
> system - I should have posted this FIRST!
>
> parameters:
> fs = 4MHz
> carrier: 99kHz
>
> Currently, there is no noise, and no 'information' being modulated. Just a
> carrier sine wave:) One gotta start somewhere!
>
> I then bandpass using a 2nd order bandpass filter centered at the carrier.

You realize that there's nothing for the bandpass filter to remove.

> Then take the absolute value.
> Then lowpass filter using a 2nd order CIC filter all the way down to
> something ridiculous like 10-50Hz.

You learned that the numbers are correctly manipulated, but little more.

> The output matches surprisingly well the 2*A/pi formula depending on the A
> of the carrier.

It doesn't surprise me. The effective sample rate is even higher than it
might seem. Successive carrier cycles have their sampled points in
slightly different places along the curve. You get information about
every point along the curve (well, almost) by overlaying enough cycles.

> Now if I do a frequency sweep, using a full scale sine wave from 0 to 2MHz
> (fs/2), and plot the obtained average value after it settled, I get the
> shape of the bandpass filter! I guess this was to be expected, which is why
> I asked if 'method-1' was only dependent on the performance of the bandpass
> filter.

The sweep probes the bandpass filter. Overall output falls when the
filter attenuates the signal. How could it be otherwise?

> I guess so far my 'information' is only at DC, but it works well.
>
> Now, ff I modulate a 2Hz signal, I also see it at the output, with a DC
> offset.
>
> And this is where I decided to post, since I didn't know how to measure
> performance of the system when there IS information.
>
> Does this make sense?

It makes sense in that it fits together in an understandable way. It
won't get you to a practical demodulator without a lot of tweaking.

Jerry
--
"I view the progress of science as ... the slow erosion of the tendency
to dichotomize." --Barbara Smuts, U. Mich.
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From: Jerry Avins on 24 Apr 2010 08:36
On 4/23/2010 7:23 PM, Tim Wescott wrote:
> Jerry Avins wrote:
>> On 4/23/2010 4:24 PM, Tim Wescott wrote:
>>> Jerry Avins wrote:
>>>> On 4/23/2010 1:52 PM, gretzteam wrote:
>>>>>>> Use it, but understand it. Understand the implication of in-band
>>>>>>> interference. Understand the need to exclude out-of-band signals
>>>>>>> from
>>>>>>> the demodulation process. (The baseband low-pass filter can't remove
>>>>>>> aliases.)
>>>>>>
>>>>>> I am assuming that he is properly prepping the signal prior to the
>>>>>> multiplication by sin/cos and will pick appropriate filters at
>>>>>> baseband.
>>>>>
>>>>>
>>>>> Ok I must admit that I'm more confused than before! Why do you still
>>>>> need a
>>>>> bandpass filter for method 2? Isn't multiplying by sin/cos shifting
>>>>> the
>>>>> carrier frequency to DC?
>>>>
>>>> What Brent said. Keep in mind that you not only shift the carrier to
>>>> baseband, you also shift everything else down by a similar amount.
>>>> Where do the aliases of the out-of-band signals go?
>>>>
>>>>> About method 1 having the problem of peak values not being close to
>>>>> full
>>>>> scale, can we say that this is not a problem when fs>> carrier?
>>>>
>>>> When the carrier is adequately oversampled, method 1 works. I leave it
>>>> to you to determine what "adequate" means. How many samples per
>>>> carrier cycle are needed to ensure that one is at least 95% of either
>>>> peak? Is that a reasonable expenditure of resources?
>>>
>>> Except that by his original description he's not peak-seeking -- he's
>>> averaging the absolute value. That _ought_ to work better, but I don't
>>> know by how much.
>>
>> How does the average value of samples of the carrier relate to the
>> approximate value of the envelope? Would it help if the "carrier" were
>> triangular?
>
> Well, the RF signal (not the carrier) is carrier * (audio signal +
> offset) -- so you can find a scaled value of the audio signal either by
> finding the peaks (as in traditional AM receivers) or by rectifying and
> averaging.
>
> I suspect (but would have to play with it to find out) that the rectify
> and average is not as harshly nonlinear, and therefore would stand a
> lower sampling (or carrier) rate.

Even so, it takes more oomph to adequately upsample for averaging than
to generate an analytic signal. Unless there is a frequency lock between
the carrier and the sample clock, rectifying effectively doubles the
chance of finding a point near the carrier peak and also doubles the
number of points averaged over. That's a wash.

Jerry
--
"I view the progress of science as ... the slow erosion of the tendency
to dichotomize." --Barbara Smuts, U. Mich.
�����������������������������������������������������������������������
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