Why we do that in AM demodulation?

jim hardy

That's an excellent computational answer.

I have to think demodulation infers real time processing ergo not a lot of time for fancy arithmetic -

but i am absolutely inexperienced in DSP. I can only marvel at its capabilities and daydream about understanding the appnotes.

Usually fundamental principles apply to any operation whether it be done numerically or analog. We implemented a minimum denominator in one analog application..



Thanks sophie - that is food for thought.

old jim

sophiecentaur

You would be amazed what they can do with numbers these days Jim! If you can do it in Excel or Basic, you can do it real-time at GHz speeds. Some of these young lads have never wound a coil in their lives!!!

sophiecentaur

Actually, I just re-read the OP and I really don't know what a "simple divider circuit" consists of - surely he doesn't mean a flip-flop???? That's the simplest 'divider circuit' I could think of. DSP circuits are in no way "simple".
Could this all be a huge misunderstanding?

0xDEADBEEF

Some food for thought.

1) In the high frequency ranges you cannot sample your signal fast enough for your method to work and you cannot amplify at these frequencies, so you use a reference and a non-linear mixer to produce a signal in a range that can be handled more easily. Of course this is not really a problem in the ranges where AM is still used.

2) The signal from your method will look like **** on the short time scales. You substitute the multiplication with cos(x) by a multiplication with 1/cos(x). This signal has terrible spikes. The signal to noise in real space is the worst at the zero crossings, and this is the moment were you weight your signal the strongest. Any noise and non linear distortion will produce strong noise at the carrier frequency and its harmonics. Therefore you will definitely need a low pass filter.

3) Mathematically the multiplication with a sine wave and low pass filtering is a scalar product in Hilbert space, which is very similar to a division. It is very well understood in Fourier space. The stronger the signal is low pass filtered after the mixer, the more the noise bandwidth is reduced, I believe that this property cannot be retained using the division.

4) I think it is interesting that a multiplication and a division can result in the same signal I have to think about that some more.

sophiecentaur

A mathematical identity is no guarantee of a working circuit, surely. The analytical world can do cancelling top and bottom and can assume ω and ω are the same. You can also 'prove' 1=0 with a bit of sleight of hand and a hidden divide by zero.

But the two functions do not produce the same signal aamof. You need to low-pass filter the produces of a multiplier to eliminate the other (RF) components, in practice.

jim hardy

I took it to mean something akin to AD634, an impressive transconductance analog multiplier IC which can be wired to do division.
http://www.analog.com/static/importe...eets/AD534.pdf
From its datasheet a caution about small denominators:

in other words, dont try gain>100.

My experience with time division approach for steam flow was similar.
While time division proved more precise than transconductance approach with small denominators, it just isn't feasible to resolve fluid flow signals(ΔP across an orifice) below about a 100::1 turndown.
I would imagine the same applies to a radio signal plucked from the air.

But i am respectful of DSP's capabilities.

old jim

btw = thanks...

idmond dantes

by "them", you mean the members or the professors?

idmond dantes

I didn't mean the divider to be simple, literally. It just occurred to mean that since the multiplication scheme requires multiplication and then filtering while the division scheme requires only one operation, i.e division, it would be much simpler in Implementation.
Clearly I was wrong about that.

idmond dantes

why there will be noise at only the carrier frequency and it's harmonics, assuming that the noise added by the channel is a white noise?

idmond dantes

Why the noise will be related the sampling frequency, rather than the base band frequencies? why you think this would happen?

idmond dantes

do you think that there are out there some analog dividers that can divide by functions like (cosine) as there exist analog dividers that can divide by arithmetic numbers because one answer I got to my question above was that 'there no analog dividers that can divide by a function like cosine, neither there are circuits that can generate a sec function(1/cosine) for a multiplier (to be used as a divider) and the only dividers that exists right now are only dividers that can divide by arithmetic numbers' . is that true, Jim?

sophiecentaur

PF members have had ages for a response. The Profs were probably grabbed by the elbow (as you do) and the question posed as they were dashing off for their coffee. Remember, a response was optional for PF members. The profs were on the spot and had to respond somehow.

jim hardy

I woke up in the middle of the night thinking about this thread.

First an explanation about my closed-mindedness on the issue -
i come from the world of process control where anything above a couple hertz is noise, so my divider experience is basically with DC circuits. Ergo i am unaccustomed to mathematical analysis of radio signals. The Fourier expansion of AM modulation was for me an eye opener into another world.

My time-division divider worked very well for process signals because at small denominators it naturally becomes quite slow which attenuates the natural noise in a small flow signal. For square root extraction we tested it alongside a transconductance divider which we beat hands down..

That AD534 has response out to tens of khz (and please forgive my typo where i called it AD634). Faster devices exist.

So if you fed an AD534 a denominator that's got a base DC value plus an AC coupled cosine function in it, you'd be dividing by (A + Bcos(ωt)) which would not reach zero so long as B<A.
That eliminates the divide by zero bugaboo. But it's not so simple anymore

Surely someplace in the resulting polynomial is the term you seek. Analog division followed by AC coupling and filtering might well work for you..
It'd sure be an enlightening experiment.

take a look at AD834..similar device but with but 250 mhz bandwidth.
http://www.analog.com/static/importe...eets/AD835.pdf
Its datasheet does not describe feedback to perform division - but the 534's does.




old jim

sophiecentaur

OMG Jim. you're taking your work home with you again!!
Aren't you supposed to be retired?

jim hardy

Yes, old firehorse syndrome i guess ...

thouht maybe i'd learn something new !

old jim

0xDEADBEEF

Imagine the 1/cos function. Those spikes are terrible! And you multiply your function with it. If there is any signal at all there due to white or whatever noise, it will produce a giant signal. So you will see the same spikes again after dividing your signal. These spikes have twice the period of your carrier so they will be full of harmonics of the carrier frequency.

And finally we have the answer why this stuff isn't done. Three things will kill the scheme:

1) Noise. If your signal is f(t)cos(kt)+e(t) where e is noise, you will have divergences in the spikes of the 1/cos function, because e(t) is surely not zero at the crossings. As I said: you amplify your signal infinitely in the region with the worst signal to noise ratio.

2) Phase and frequency. If your local oscillator is not perfectly phase and frequency locked against the sending oscillator, the zero crossings will be off, the spikes will not get cancelled and will dominate whatever comes out

3) Offset voltages. It is pretty much impossible to have components without any offset voltage. So even if your frequency and phase would match perfectly, your zero crossings will be off again producing spikes.

Infinite spikes are simply a bad idea. This type of problem happens also in the Wiener deconvolution, where you also divide the blurring of the signal away, but due to the same signal to noise problems you need to dampen the method at the frequencies were the noise is large compared to the signal.

NascentOxygen

That is a very clear-eyed way of expressing it.

Multiplying by cos x looks much smoother than multiplying by this:

Attached Thumbnails