Why we do that in AM demodulation?

[idmond dantes]

Hey all,

"In AM synchronous demodulation, Why we don't divide m(t)coswt by cos(wt)
instead of multiplying by cos(wt), since this can be easily implemented by a simple divider circuit?"
If it's all about thinking mathematically, then it seems like it's more intuitive and a whole lot easier if we just divide by coswt, why we go through all the trouble and multiply then we have to know the trig identity of (cos(a)cos(b)) and then put a LPF after the output...

I asked this question to two professors and I got different answers:

#Professor 1 Reply:
"For your scheme to work you must know exactly what the frequency w is that the transmitter is using, which tends to drift. So try your scheme with dividing by cos (w+delta)t and see whether you can recover the signal." #Professor 2 reply was:
"we don't use the division scheme for two reasons:

a- In the real world, noise is added to the received signal and it is going to look like this, m(t)coswt+n(t). If you then divide by coswt, you will get, m(t)+n(t)/coswt, and since the cosine function ranges from -1 to 1, thus for values of cosine less than 1, the noise term will increase much more, so you will get poor SNR.

b- When the cosine function goes to zero, you will divide by zero and this will cause amplitude spikes which leads to circuit saturation."
I am now confused more than ever. which one is true?

 

[jsgruszynski]

The multiplication is taking a modulated signal down to baseband (centered at 0 Hz) in order to demodulate. Essentially you are taking advantage of the trig identity:

http://en.wikipedia.org/wiki/List_o...#Product-to-sum_and_sum-to-product_identities

The modulation frequency is cleanly removed by multiplying leaving on the side bands which are exactly the audio modulated by the AM.

Multipliers are far easier to implement. At RF frequencies a strict magnitude divider isn't really practical. You may see the term "divider" used in the context of phase locked loops and synthesizers in RF but this isn't the same thing as a divider as you are thinking of.

 

[idmond dantes]

Multipliers are far easier to implement. At RF frequencies a strict magnitude divider isn't really practical. You may see the term "divider" used in the context of phase locked loops and synthesizers in RF but this isn't the same thing as a divider as you are thinking of.

yes, jsgruszynski I know these dividers(used in PLLs and RF synthesizers) are called arithmetic dividers which divide by a number and my demodulation scheme need a divider that can divide by a function like the sinusoid (cosine function). Is such a "divide by function" divider possible to implement it in a circuit or not?

 

[jsgruszynski]

yes, jsgruszynski I know these dividers(used in PLLs and RF synthesizers) are called arithmetic dividers which divide by a number and my demodulation scheme need a divider that can divide by a function like the sinusoid (cosine function). Is such a "divide by function" divider possible to implement it in a circuit or not?

Not at high frequency nor accurately. Not in analog. Maybe digitally but at that point the advantages of just "downconverting" by multiplying pretty much solve the problem already.

 

[idmond dantes]

Not at high frequency nor accurately. Not in analog. Maybe digitally but at that point the advantages of just "downconverting" by multiplying pretty much solve the problem already.

yes, you're right about that because I always get this answer that you cannot do an analog circuit that can divide by a function may be a digital one would be feasible but it would be a more complicated solution than the demodulation by multiplying scheme.

but why with digital, you're saying it won't work at high frequencies and if it did, it won't be accurate, since division can easily be implemented in Digital signal processing even when you divide by a function, you will just have to deal with samples. can you elaborate more on this.
I'm looking forward to hear from you, jsgruszynski.

 

[NascentOxygen]

b- When the cosine function goes to zero, you will divide by zero and this will cause amplitude spikes which leads to circuit saturation."

I like the sound of that. Very difficult to argue around it.

 

[idmond dantes]

I like the sound of that. Very difficult to argue around it.

well, we can still go around the "divide by zero" issue if we did all the demodulation process with digital signal processing and add some piece of code that checks for the case when the cosine function is zero then output some finite nonzero value.

 

[NascentOxygen]

well, we can still go around the "divide by zero" issue if we did all the demodulation process with digital signal processing and add some piece of code that checks for the case when the cosine function is zero then output some finite nonzero value.

Possibly. But how about when the cosine is small. As your professor #ⵒ pointed out, dividing a noise blip by a small number will lead to a large noise spike. That's never desirable.

 

[idmond dantes]

Possibly. But how about when the cosine is small. As your professor #ⵒ pointed out, dividing a noise blip by a small number will lead to a large noise spike. That's never desirable.

In this case, we could increase our signal power to compensate for the amplified noise caused by dividing by a small number.

I believe that this whole thing "demodulation by division" actually does more harm than good. After all, what we gained from this :

1- we complicated our AM demodulator design by shifting the whole demodulation process to the digital domain to avoid the divide by zero problem.


2- we had to increase our signal power to compensate for the amplified noise that our scheme was the reason that it is amplified, in the first place.

 

[sophiecentaur]

Hey all,

"In AM synchronous demodulation, Why we don't divide m(t)coswt by cos(wt)
instead of multiplying by cos(wt), since this can be easily implemented by a simple divider circuit?"
If it's all about thinking mathematically, then it seems like it's more intuitive and a whole lot easier if we just divide by coswt, why we go through all the trouble and multiply then we have to know the trig identity of (cos(a)cos(b)) and then put a LPF after the output...

I asked this question to two professors and I got different answers:

#Professor 1 Reply:
"For your scheme to work you must know exactly what the frequency w is that the transmitter is using, which tends to drift. So try your scheme with dividing by cos (w+delta)t and see whether you can recover the signal." #Professor 2 reply was:
"we don't use the division scheme for two reasons:

a- In the real world, noise is added to the received signal and it is going to look like this, m(t)coswt+n(t). If you then divide by coswt, you will get, m(t)+n(t)/coswt, and since the cosine function ranges from -1 to 1, thus for values of cosine less than 1, the noise term will increase much more, so you will get poor SNR.

b- When the cosine function goes to zero, you will divide by zero and this will cause amplitude spikes which leads to circuit saturation."
I am now confused more than ever. which one is true?

Interesting . . . .This question could only, possibly, have been raised by the 'new' generation.
Whatever the arguments of your teachers, the historical answer is just 'Implementation'. Synchronous detection was implemented with analogue circuitry and a multiplication function is easily achievable with a simple mixer circuit - even just a diode will do what you want.
Those answers you got are not contradictory, I think - they just looked at two aspects of your suggestion.
If you want to do the demodulation 'numerically', sample by sample, then you do need to avoid the 'divide by zero' situation but that is very easy to achieve by making sure that the Cos function you are generating never contain samples at the zero crossings - that is just a matter of what phase you choose for your 'local oscillator' synthesis. Once you have locked your LO to the received carrier, the effect of the noise bursts around the zero crossings will, I think, generate noise which is related to the sampling frequency, rather than the base band frequencies so, once filtered, it wouldn't appear in the demodulated signal. This will be true, even for any samples from the division process which are 'full amplitude' (limiting value). The higher the sampling frequency, the less the effect of this noise energy - and this is the same situation as when you use low bit digital sampling and very much over-sample; the quantisation noise power is spread over a bigger and bigger bandwidth, away from the base bandwidth.

 

[NascentOxygen]

In this case, we could increase our signal power to compensate for the amplified noise caused by dividing by a small number.

You'd request the distant radio station to increase their transmitter's power so that your reception was better? How practical does this sound?

I believe that this whole thing "demodulation by division" actually does more harm than good. After all, what we gained from this :

The myriad advantages, if any, have yet to be revealed in this thread.

 

[sophiecentaur]

You'd request the distant radio station to increase their transmitter's power so that your reception was better? How practical does this sound?The myriad advantages, if any, have yet to be revealed in this thread.

It would be safe to say that the 'big disadvantage' concerning the noise at the zero crossings, is probably non existent - for the reasons I gave above. Which leaves us with 'possible advantages'.

Not being very familiar with DSP techniques, I would be grateful if someone could justify the view that Division is easier than Multiplication, before we go any further. That certainly doesn't apply in Analogue.

 

[rbj]

I asked this question to two professors and I got different answers:

#Professor 1 Reply:
"For your scheme to work you must know exactly what the frequency w is that the transmitter is using, which tends to drift. So try your scheme with dividing by cos (w+delta)t and see whether you can recover the signal."


#Professor 2 reply was:
"we don't use the division scheme for two reasons:

a- In the real world, noise is added to the received signal and it is going to look like this, m(t)coswt+n(t). If you then divide by coswt, you will get, m(t)+n(t)/coswt, and since the cosine function ranges from -1 to 1, thus for values of cosine less than 1, the noise term will increase much more, so you will get poor SNR.

b- When the cosine function goes to zero, you will divide by zero and this will cause amplitude spikes which leads to circuit saturation."



I am now confused more than ever. which one is true?

why can't both be true?

 

[sophiecentaur]

why can't both be true?

Divide and conquer. A well known student trick!
"But the other teacher said x and you said y - you must be wrong sir"

 

[sophiecentaur]

Actually, I could be cynical and say that your question may have taken them totally on the hop and they came out with the first thing that came into their heads. Your idea will almost certainly not come into the syllabus (it's very novel) and they (unlike PF!) probably would rather you hadn't asked it.

 

[jim hardy]

I think the question has been answered several times but perhaps not in direct enough words.Professor 2 is certainly right.

Since cos(anything) is never greater than 1, niether is ((cos(anything))^2. With multiplication, as input gets smaller so does output.

So a multiplier will never be asked to deliver a large result because its gain is always less than 1.
But a divider must be capable of quite high gain, hence it is capable of producing quite large numbers.I have built analog dividers. To keep them stable with small denominators is well nigh impossible.

When designing any machine why would you intentionally make it inherently unstable?



old jim

 

[sophiecentaur]

That is an excellent 'analogue' answer Jim and those ideas have coloured other responses.
However, when this is done numerically, you can do all sorts of arithmetic tricks and I'm not so sure that it is a complete no-no. Stability doesn't need to be a problem as with your ancient steam analogue systems. Though god knows why anyone would actually want to demodulate this way. The OP obviously came from way outfield amongst the theoretical daisies but there should always be a really good answer to the 'why not practical?' question.
You could limit the output amplitude of the division (even using non-linear quantising) and that will only give you short bursts of 'dodgy' (limiting value) samples. Some intelligent AGC could ensure that there are never more than a few of these and this corresponds to a short enough noise burst so that most of the spectral content is actually out of band. I'm assuming that the processing would need to be heavily over-sampled so I can't help thinking that the gut reaction may be mis-placed - as with bit slice coders and decoders which work at hundreds of times over-sampling.
It's such an innocent little question, though and the algebra is right enough. It reminds me of the rules for minimising errors in evaluating formulae with early calculators : multiply then divide then multiply then divide etc. to avoid too many zeros and too few sig figs building up.

 

[jim hardy]

It's such an innocent little question, though and the algebra is right enough. It reminds me of the rules for minimising errors in evaluating formulae with early calculators : multiply then divide then multiply then divide etc. to avoid too many zeros and too few sig figs building up.

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]

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

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]

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.

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]

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?

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/imported-files/data_sheets/AD534.pdf
From its datasheet a caution about small denominators:

The increase in noise level and reduction in bandwidth preclude operation much beyond a ratio of 100 to 1.

in other words, don't 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]

Actually, I could be cynical and say that your question may have taken them totally on the hop and they came out with the first thing that came into their heads. Your idea will almost certainly not come into the syllabus (it's very novel) and they (unlike PF!) probably would rather you hadn't asked it.

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

 

[idmond dantes]

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?

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]

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.

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]

Once you have locked your LO to the received carrier, the effect of the noise bursts around the zero crossings will, I think, generate noise which is related to the sampling frequency, rather than the base band frequencies so, once filtered, it wouldn't appear in the demodulated signal. This will be true, even for any samples from the division process which are 'full amplitude' (limiting value). The higher the sampling frequency, the less the effect of this noise energy - and this is the same situation as when you use low bit digital sampling and very much over-sample; the quantisation noise power is spread over a bigger and bigger bandwidth, away from the base bandwidth.

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

 

[idmond dantes]

I think the question has been answered several times but perhaps not in direct enough words.


Professor 2 is certainly right.

Since cos(anything) is never greater than 1, niether is ((cos(anything))^2. With multiplication, as input gets smaller so does output.

So a multiplier will never be asked to deliver a large result because its gain is always less than 1.
But a divider must be capable of quite high gain, hence it is capable of producing quite large numbers.


I have built analog dividers. To keep them stable with small denominators is well nigh impossible.

When designing any machine why would you intentionally make it inherently unstable?



old jim

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]

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

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]

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?

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/imported-files/data_sheets/AD835.pdf
Its datasheet does not describe feedback to perform division - but the 534's does.

old jim

 

[sophiecentaur]

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

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

 

[jim hardy]

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

Yes, old firehorse syndrome i guess ...

thouht maybe i'd learn something new !

old jim

 

[0xDEADBEEF]

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?

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 canceled 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]

You substitute the multiplication with cos(x) by a multiplication with 1/cos(x).

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

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