>On Apr 26, 3:59=A0am, Vladimir Vassilevsky <nos...@nowhere.com> wrote:
>> Jerry Avins wrote:
>> > If synchronous demodulation loses the signal
>> > even when I/Q demodulation tracks it, in what way is synchronous
>> > demodulation superior? Have you had a chance to compare the outputs
of
>> > the two detectors when both are working?
>>
>> BTW, in the modern ICs, they demodulate AM in a quasi-synchronous way:
>> the bandpass signal is digitized by comparator and the output of the
>> comparator drives the analog switch which does synchronous
>> rectification. The advantage is much better linearity compared to diode
>> detector. Obviously this could be improved by adding a narrowband
filter
>> in front of the comparator; however it is not worth extra cost for
>> consumer applications.
>>
>> Vladimir Vassilevsky
>> DSP and Mixed Signal Design Consultanthttp://www.abvolt.com
>
>Could all be done with an op-amp! What nonsense. How to make radio
>1000 times more complicated than it need be.
Vlad described a solution consisting of a comparator (i.e. an op-amp with
op-amp crossed out, and comparator written across it in crayon) and an
analogue switch (i.e. one of the simplest structures in mixed signal CMOS).
Your solution requires an op-amp and a diode. Can you explain the 1000
times complexity ratio?
Steve
Reply by HardySpicer●April 27, 20102010-04-27
On Apr 26, 3:59�am, Vladimir Vassilevsky <nos...@nowhere.com> wrote:
> Jerry Avins wrote:
> > If synchronous demodulation loses the signal
> > even when I/Q demodulation tracks it, in what way is synchronous
> > demodulation superior? Have you had a chance to compare the outputs of
> > the two detectors when both are working?
>
> BTW, in the modern ICs, they demodulate AM in a quasi-synchronous way:
> the bandpass signal is digitized by comparator and the output of the
> comparator drives the analog switch which does synchronous
> rectification. The advantage is much better linearity compared to diode
> detector. Obviously this could be improved by adding a narrowband filter
> in front of the comparator; however it is not worth extra cost for
> consumer applications.
>
> Vladimir Vassilevsky
> DSP and Mixed Signal Design Consultanthttp://www.abvolt.com
Could all be done with an op-amp! What nonsense. How to make radio
1000 times more complicated than it need be.
Hardy
Reply by Vladimir Vassilevsky●April 25, 20102010-04-25
HardySpicer wrote:
> On Apr 26, 3:59 am, Vladimir Vassilevsky <nos...@nowhere.com> wrote:
>
>>Jerry Avins wrote:
>>
>>>If synchronous demodulation loses the signal
>>>even when I/Q demodulation tracks it, in what way is synchronous
>>>demodulation superior? Have you had a chance to compare the outputs of
>>>the two detectors when both are working?
>>
>>BTW, in the modern ICs, they demodulate AM in a quasi-synchronous way:
>>the bandpass signal is digitized by comparator and the output of the
>>comparator drives the analog switch which does synchronous
>>rectification. The advantage is much better linearity compared to diode
>>detector. Obviously this could be improved by adding a narrowband filter
>>in front of the comparator; however it is not worth extra cost for
>>consumer applications.
>
> Can you hear the difference though!
You could see the difference. It used to be done in that way in analog TV.
Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com
Reply by HardySpicer●April 25, 20102010-04-25
On Apr 26, 3:59�am, Vladimir Vassilevsky <nos...@nowhere.com> wrote:
> Jerry Avins wrote:
> > If synchronous demodulation loses the signal
> > even when I/Q demodulation tracks it, in what way is synchronous
> > demodulation superior? Have you had a chance to compare the outputs of
> > the two detectors when both are working?
>
> BTW, in the modern ICs, they demodulate AM in a quasi-synchronous way:
> the bandpass signal is digitized by comparator and the output of the
> comparator drives the analog switch which does synchronous
> rectification. The advantage is much better linearity compared to diode
> detector. Obviously this could be improved by adding a narrowband filter
> in front of the comparator; however it is not worth extra cost for
> consumer applications.
>
> Vladimir Vassilevsky
> DSP and Mixed Signal Design Consultanthttp://www.abvolt.com
Can you hear the difference though!
Reply by HardySpicer●April 25, 20102010-04-25
On Apr 25, 5:03�pm, Jerry Avins <j...@ieee.org> wrote:
> On 4/24/2010 8:04 PM, HardySpicer wrote:
>
>
>
> > On Apr 25, 12:39 am, Jerry Avins<j...@ieee.org> �wrote:
> >> On 4/23/2010 5:56 PM, Vladimir Vassilevsky wrote:
>
> >>> Jerry Avins wrote:
>
> >>>> On 4/23/2010 5:33 PM, HardySpicer wrote:
>
> >>>>> On Apr 24, 8:24 am, Vladimir Vassilevsky<nos...@nowhere.com> �wrote:
>
> >>>>>> HardySpicer wrote:
>
> >>>>>>> Synchronous demodulation using a PLL will give you 3dB improvement
> >>>>>>> over ordinary envelope detection.
>
> >>>>>> This is wrong.
>
> >>>>> It's in the textbooks...read it!
>
> >>>> What is ordinary envelope detection? Peak detection?
>
> >>> Doesn't matter; It is very simple. Think of |I| vs sqrt(I^2 + Q^2)
>
> >> Still, when someone claims "3dB improvement", I want to know what is
> >> improved upon.
>
> >> Jerry
> >> --
> >> "I view the progress of science as ... the slow erosion of the tendency
> >> � �to dichotomize." --Barbara Smuts, U. Mich.
> >> �����������������������������������������������������������������������
>
> > If I can remember that far back I believe it is in received SNR
> > (baseband). Look up synchronous demodulation vs
> > � envelope detection. Of course you digital guys have I and Q and make
> > things even more complicated. Do you remember when we just used a cats
> > whisker!
> > The I and Q method vs a PLL - I have no idea. How do you get I and Q -
> > do you need a PLL to get I and Q? If so then I expect it is the same
> > result.
>
> > It was Taub and Shilling or some such that had the details but most
> > older coms books will have the info. People are so locked into digital
> > nowadays that they forget the basics
> > which all comes from analogue. You cannot understand digital without
> > understanding analogue first. I don't mean you Jerry of course but
> > many of the new breed of engineer that gets taught exclusively
> > digital.
> > (does happen!!) For example, what is an exclusive OR in a PLL? To me
> > it is a multiplier!!
>
> I/Q demodulation is pure envelope detection, even for carriers not much
> higher than the highest modulating frequency. What you seem to refer to
> as envelope detection is peak detection, which is merely a damn good
> approximation under typical conditions of operation.
>
> There are a number of ways to make an analytic (I-Q) signal. A Hilbert
> transformer is one, Clay Turner's Tips & Tricks article is better. Often
> the best produces both I and Q in the sampling process. Synchronous
> demodulation does improve on peak detection, but not on sqrt(I^2+Q^2).
> So does exalted carrier, for the same fundamental reason.
>
> Jerry
> --
> "I view the progress of science as ... the slow erosion of the tendency
> � to dichotomize." --Barbara Smuts, U. Mich.
> �����������������������������������������������������������������������
Ok so we are comparing apples with oranges.
Hardy
Reply by Jerry Avins●April 25, 20102010-04-25
On 4/25/2010 11:31 AM, gretzteam wrote:
>>> Now, I'm not building an AM radio, and I'm mostly interested in very
> low
>>> frequency (up to 20Hz), but I need quite good performance. My sampling
> rate
>>> and carrier are fixed and won't change. This is not under my control
> and
>>> were decided because of other stuff that this system is doing.
>>>
>>> So another way to ask the original question:
>>> When taking the absolute value of a digital signal, what really
> happens?
>>> I'm trying to see this from a frequency domain perspective.
>>
>> I'd have to do the math -- a Laplace transform -- to give a definitive
>> answer even for a continuous signal, and the discrete-time case is more
>> involved. There seems to be a conflict among authorities. In the one
>> hand, the waveform is what one gets from a push-push doubler, and should
>> contain no odd harmonics, not even the fundamental. On the other hand,
>> the ITT Reference Data for Radio Engineers (4th edition; 1949) gives a
>> formula that I don't believe involving finite values of coefficients for
>> all harmonics, including the fundamental.
>>
>>> My original experiment - reinventing the wall-wart I guess - showed
> that
>>> the carrier is shifted to DC pretty well.
>>
>> I wouldn't say that. There's a strong second harmonic, so you could also
>> say that the carrier is shifted up. What we know is that the average
>> value is extracted and that harmonics are produced. Each of those
>> harmonics will exhibit sidebands.
>>
>>> I'm trying to get a feel for
> what
>>> happens to the sidebands, how much distortion is introduced when I move
> out
>>> of DC? I understand abs() is nonlinear, but there might be some
> analysis
>>> possible?
>>>
>>> Now, since multiplying by sin/cos is a perfect shift, I know the
> sidebands
>>> won't get affected, and my problem becomes designing a good lowpass
> filter.
>>
>> What do you mean by "not affected"? they get shifted along with the
> carrier.
>>
>> Jerry
>
> Yes but the sidebands don't get distorted at all during the sin/cos
> multiplication, not matter what other stuff is in the spectrum from 0 to
> fs/2, right? Everything just gets shifted, and performance will depend on
> the lowpass filter.
> However, can we say that for the abs() case? Can we say that the sidebands
> are shifted along with the carrier back to DC without distortion? I don't
> care if everything else becomes all distorted, cause it will get lowpass
> filtered.
Some of the harmonics caused by taking the magnitude exceed fs/2 and
alias. The higher the sample rate, the less of a problem that is.
Jerry
--
"I view the progress of science as ... the slow erosion of the tendency
to dichotomize." --Barbara Smuts, U. Mich.
�����������������������������������������������������������������������
Reply by Jerry Avins●April 25, 20102010-04-25
On 4/25/2010 11:59 AM, Vladimir Vassilevsky wrote:
>
>
> Jerry Avins wrote:
>
>> If synchronous demodulation loses the signal even when I/Q
>> demodulation tracks it, in what way is synchronous demodulation
>> superior? Have you had a chance to compare the outputs of the two
>> detectors when both are working?
>
> BTW, in the modern ICs, they demodulate AM in a quasi-synchronous way:
> the bandpass signal is digitized by comparator and the output of the
> comparator drives the analog switch which does synchronous
> rectification. The advantage is much better linearity compared to diode
> detector. Obviously this could be improved by adding a narrowband filter
> in front of the comparator; however it is not worth extra cost for
> consumer applications.
That's the digital equivalent of exalted carrier operation that I
mentioned earlier in this thread.
Jerry
--
"I view the progress of science as ... the slow erosion of the tendency
to dichotomize." --Barbara Smuts, U. Mich.
�����������������������������������������������������������������������
Reply by Vladimir Vassilevsky●April 25, 20102010-04-25
Jerry Avins wrote:
> If synchronous demodulation loses the signal
> even when I/Q demodulation tracks it, in what way is synchronous
> demodulation superior? Have you had a chance to compare the outputs of
> the two detectors when both are working?
BTW, in the modern ICs, they demodulate AM in a quasi-synchronous way:
the bandpass signal is digitized by comparator and the output of the
comparator drives the analog switch which does synchronous
rectification. The advantage is much better linearity compared to diode
detector. Obviously this could be improved by adding a narrowband filter
in front of the comparator; however it is not worth extra cost for
consumer applications.
Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com
Reply by gretzteam●April 25, 20102010-04-25
>My point is, I know they did not go to great lengths to build thisFFT/
>PLL detection scheme if I/Q demodulation would give the same result.
>I cannot tell you right now what the math is, but I know synchronous
>detection is better in some respects.
Maybe they try to avoid I/Q demodulation because it involves two
multipliers running 'fast', and twice the lowpass filtering? This is kind
of what I'm trying to avoid, if I can get decent performance with another
scheme.
However, a bandpass filter is not free either, so maybe it all boils down
to the same thing!
Thanks,
Reply by gretzteam●April 25, 20102010-04-25
>> Now, I'm not building an AM radio, and I'm mostly interested in very
low
>> frequency (up to 20Hz), but I need quite good performance. My sampling
rate
>> and carrier are fixed and won't change. This is not under my control
and
>> were decided because of other stuff that this system is doing.
>>
>> So another way to ask the original question:
>> When taking the absolute value of a digital signal, what really
happens?
>> I'm trying to see this from a frequency domain perspective.
>
>I'd have to do the math -- a Laplace transform -- to give a definitive
>answer even for a continuous signal, and the discrete-time case is more
>involved. There seems to be a conflict among authorities. In the one
>hand, the waveform is what one gets from a push-push doubler, and should
>contain no odd harmonics, not even the fundamental. On the other hand,
>the ITT Reference Data for Radio Engineers (4th edition; 1949) gives a
>formula that I don't believe involving finite values of coefficients for
>all harmonics, including the fundamental.
>
>> My original experiment - reinventing the wall-wart I guess - showed
that
>> the carrier is shifted to DC pretty well.
>
>I wouldn't say that. There's a strong second harmonic, so you could also
>say that the carrier is shifted up. What we know is that the average
>value is extracted and that harmonics are produced. Each of those
>harmonics will exhibit sidebands.
>
>> I'm trying to get a feel for
what
>> happens to the sidebands, how much distortion is introduced when I move
out
>> of DC? I understand abs() is nonlinear, but there might be some
analysis
>> possible?
>>
>> Now, since multiplying by sin/cos is a perfect shift, I know the
sidebands
>> won't get affected, and my problem becomes designing a good lowpass
filter.
>
>What do you mean by "not affected"? they get shifted along with the
carrier.
>
>Jerry
Yes but the sidebands don't get distorted at all during the sin/cos
multiplication, not matter what other stuff is in the spectrum from 0 to
fs/2, right? Everything just gets shifted, and performance will depend on
the lowpass filter.
However, can we say that for the abs() case? Can we say that the sidebands
are shifted along with the carrier back to DC without distortion? I don't
care if everything else becomes all distorted, cause it will get lowpass
filtered.
Thanks.