| New Reply |
Why we do that in AM demodulation? |
Share Thread | Thread Tools |
| Feb23-13, 08:48 PM | #1 |
|
|
Why we do that in AM demodulation?
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?
|
| Feb24-13, 05:10 PM | #2 |
|
|
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_of...uct_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. |
| Feb24-13, 07:12 PM | #3 |
|
|
|
| Feb24-13, 07:29 PM | #4 |
|
|
Why we do that in AM demodulation? |
| Feb24-13, 08:07 PM | #5 |
|
|
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. |
| Feb25-13, 12:19 AM | #6 |
|
Recognitions:
|
|
| Feb25-13, 08:20 AM | #7 |
|
|
|
| Feb25-13, 08:55 AM | #8 |
|
Recognitions:
|
|
| Feb25-13, 06:07 PM | #9 |
|
|
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. |
| Feb26-13, 04:27 AM | #10 |
|
|
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. |
| Feb26-13, 04:26 PM | #11 |
|
Recognitions:
|
|
| Feb27-13, 03:47 AM | #12 |
|
|
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. |
| Feb27-13, 11:13 AM | #13 |
|
|
|
| Feb27-13, 11:52 AM | #14 |
|
|
"But the other teacher said x and you said y - you must be wrong sir" |
| Feb27-13, 12:10 PM | #15 |
|
|
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.
|
| Feb27-13, 01:05 PM | #16 |
|
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 |
| Feb27-13, 03:25 PM | #17 |
|
|
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. |
| New Reply |
| Thread Tools | |
Similar Threads for: Why we do that in AM demodulation?
|
||||
| Thread | Forum | Replies | ||
| DSB-AM Demodulation MATLAB | Engineering, Comp Sci, & Technology Homework | 0 | ||
| Query on AM Demodulation | Electrical Engineering | 2 | ||
| IQ demodulation | Electrical Engineering | 21 | ||
| need clarification about FM demodulation | Electrical Engineering | 4 | ||
| Composite FM Demodulation! | General Physics | 0 | ||