DSB Demodulation

Thread Starter

mentaaal

Joined Oct 17, 2005
451
Hey guys, in studying DSB demodulation, it is said that the demodulator has a to generate a synchronis carrier signal to multiply with the DSB (using a multiplier) to extract the modulating signal and when the generated carrrier signal is not synchronis, the extracted signal is distorted.

I tried seeing this for myself mathetatically and the output doesnt look distorted, phase shifted yes. So are my calculations wrong somewhere?

 

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Ron H

Joined Apr 14, 2005
7,063
You apparently demodulated with a carrier that is phase shifted but still synchronous. Nonsynchronous means the demodulating frequency is different from the original carrier frequency.
 

mik3

Joined Feb 4, 2008
4,843
Yes, RonH is right. You have to multiply the DSB signal by a signal like this:

V*cos[(ω+Δω)*t+Δφ]

where

Δω is the difference of the DSB carrier frequency with the demodulating signal frequency

Δφ is the phase difference of the DSB carrier with the demodulating signal
 

Thread Starter

mentaaal

Joined Oct 17, 2005
451
Sorry i described my question incorrectly. Would it be fair to say that the phase difference between the generated carrier and the original carrier only effects the amplitude of the extracted signal and not the actual frequency of it? Why is there a requirement then for the generated carrier to be phase locked to the original carrier? The necessity for the generated carrier to have the same frequency is clear to me but the phase requirement is not clear to me at all!
 

KL7AJ

Joined Nov 4, 2008
2,229
Sorry i described my question incorrectly. Would it be fair to say that the phase difference between the generated carrier and the original carrier only effects the amplitude of the extracted signal and not the actual frequency of it? Why is there a requirement then for the generated carrier to be phase locked to the original carrier? The necessity for the generated carrier to have the same frequency is clear to me but the phase requirement is not clear to me at all!
Yes, your interpretation is absolutely correct. From a practical standpoint, a conventional AM transmitter/receiver system never has issues in this regard, because the very carrier you're using to create the DSB signal is the one that's used as a "self-demodulator" with a simple diode detector. All the phase relationships have no choice but to "come out right." In practice, you can greatly reduce the transmitted carrier, as long as there's enough left to phase lock a local oscillator to in the receiver. (Partially suppressed carrier technology is used in a lot of shortwave broadcast transmitters)

For communications purposes, the phasing is not an issue anyway. However for instrumentation purposes (such as measuring doppler shift as in an ionosonde) the absolute phase can be important.

Early single sideband transmitters used a phasing method of removing one or the other sidebands. The technique fell into disuse in the late 1950s, but has experienced a great revival with DSP techniques.

Eric
 

Ron H

Joined Apr 14, 2005
7,063
Eric, I have no experience with DSB-SC, so I may be completely off base here. From what I read, phasing of the demod carrier is important. Also, I ran a sim in LTspice where I balance-modulated a 1MHz carrier with 10kHz. I then demodulated with a 1MHz carrier while playing with the phase relationship between the two carriers. The demodulated output amplitude was a function of this phase relationship, and in fact was zero when the demod carrier was leading the mod carrier by 90°.
I did a little research and (re)discovered the Costas loop, which uses a phaselocked loop to regenerate the proper frequency and phase for demodulation. The output of the loop is the demodulated signal.
 

Thread Starter

mentaaal

Joined Oct 17, 2005
451
I have the costas loop in my notes as well actually. So for small phase lag, the output is fiine if you neglect the amplitude attenuation effects. But perhaps just with particular angles like as you say 90°, there could be problems.

I did however do the same maths, but with a frequency shift as well and the effects could clearly be seen.
 

KL7AJ

Joined Nov 4, 2008
2,229
I have the costas loop in my notes as well actually. So for small phase lag, the output is fiine if you neglect the amplitude attenuation effects. But perhaps just with particular angles like as you say 90°, there could be problems.

I did however do the same maths, but with a frequency shift as well and the effects could clearly be seen.
Another way to eliminate (or at least deal with) the carrier phase issue is to use a quadrature demodulator. This method is becoming a de-facto standard in a lot of H.F. receivers. The quadrature detector allows you to extract the amplitude informationl, regardless of phase...which for most communications purposes is all you need. :)

eric
 

Thread Starter

mentaaal

Joined Oct 17, 2005
451
Cool thanks for that, one last question if you dont mind:

I was looking over the Phase locked loop schematic and noticed that the derivations (at least in my notes) are all given with the output of the Voltage controlled oscillator being Cos(Wct + ψ). Is using this form adequate (As opposed to using cos((ωc+λω)t+ψ) because if the frequency shifted by a bit as well as the phase, it would still appear as a phase shift and be compensated accordingly?

Thanks for the help
 

Ron H

Joined Apr 14, 2005
7,063
Cool thanks for that, one last question if you dont mind:

I was looking over the Phase locked loop schematic and noticed that the derivations (at least in my notes) are all given with the output of the Voltage controlled oscillator being Cos(Wct + ψ). Is using this form adequate (As opposed to using cos((ωc+λω)t+ψ) because if the frequency shifted by a bit as well as the phase, it would still appear as a phase shift and be compensated accordingly?

Thanks for the help
If the loop is locked, there should be no Δω.
Or maybe I misunderstood your question.:eek:
 

Thread Starter

mentaaal

Joined Oct 17, 2005
451
Well... the loop has to get locked in the first place right? If there is a slight offset in the VCO's frequency it gets trimmed away back to the correct frequency. So what i am wondering about is the reason why the Δω is not required is because if there was a slight frequency offset, it would appear to the vco as a phase shift?
 

KL7AJ

Joined Nov 4, 2008
2,229
I might want to mention that, other than the mathematical exercise, suppressed carrier DSB is almost never used. (I did have an old Central Electronics 100V.....a classic ham radio "boat anchor" if there ever was one, that would produce DSB). However that mode would pretty much leave you with the sound of one hand clapping in the dark, if you tried it these days.

In fact, synchronous modulation of full-carrier A.M. is only available in the most expensive general coverage receivers, and that, only recently. It is primarily used to reduce selective fading, where you ordinarily might have undesirable phase shift (most frequently due to different ionospheric propagation times of the upper sideband and lower sideband frequencies. It's also useful for demodulating the newer partially-suppressed (exhalted carrier) carrier shortwave stations, with less distortion.

eric
 

Thread Starter

mentaaal

Joined Oct 17, 2005
451
Well the extent i care about it is such that i can understand the maths of it ;-)

Am i right in what i said before, that the system will see any change in frequency as a change in phase shift as well and adjust accordingly? It sounds plausable
 

Ron H

Joined Apr 14, 2005
7,063
Well the extent i care about it is such that i can understand the maths of it ;-)

Am i right in what i said before, that the system will see any change in frequency as a change in phase shift as well and adjust accordingly? It sounds plausable
A step in frequency is a ramp in phase. PLLs correct for frequency and phase error. It's their job.:D
 
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