Cosine for synchronous demodulation

In summary, for synchronous demodulation, the signal x(t) is multiplied by the carrier frequency xc(t) to recover the modulating signal xm(t). The modulation index, m, and the carrier frequency, fc, and the modulation frequency, fm, are also given in the problem.
  • #1
ƒ(x)
328
0

Homework Statement



Synchronous demodulate x(t).

Homework Equations



xc(t) = cos(2pi*fc*t), fc is the carrier frequency
xm(t) = cos(2pi*fm*t), fm is the modulation frequency

x(t) = xc(t)*(1+m*xm(t)), m is the modulation index
m = .8
fc = 2000 hz
fm = 200 hz

The Attempt at a Solution



I really don't even know where to start. I know synchronous demodulation means I multiply x(t) by a function, but how do I come up with that function? I realize that I could divide x(t) by (1+m*xm(t)), but that doesn't seem to be what the problem is asking.
 
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  • #2
ƒ(x) said:
xc(t) = cos(2pi*fc*t), fc is the carrier frequency
xm(t) = cos(2pi*fm*t), fm is the modulation frequency

x(t) = xc(t)*(1+m*xm(t)), m is the modulation index
m = .8
fc = 2000 hz
fm = 200 hz

The Attempt at a Solution



I really don't even know where to start. I know synchronous demodulation means I multiply x(t) by a function, but how do I come up with that function? I realize that I could divide x(t) by (1+m*xm(t)), but that doesn't seem to be what the problem is asking.
No, that would involve the receiver being able to predict the modulation perfectly, so there would be little point in sending it in the first place.

I don't know where to start, either. But isn't synchronous something about multiplying by the carrier? So why not try multiplying x(t) by xc(t) and see what you can get from that? The whole idea is to recover xm(t), presumably?
 
  • #3
NascentOxygen said:
No, that would involve the receiver being able to predict the modulation perfectly, so there would be little point in sending it in the first place.

I don't know where to start, either. But isn't synchronous something about multiplying by the carrier? So why not try multiplying x(t) by xc(t) and see what you can get from that? The whole idea is to recover xm(t), presumably?

I think I'm supposed to do x(t) * xc(t) and then the next step is to use a step function of sorts. The point is to recover xc(t) since that's the original signal.

Edit: My mistake, I was misreading the problem. You're correct. I'm supposed to recover xm(t).
 

1. What is cosine for synchronous demodulation?

Cosine for synchronous demodulation is a mathematical technique used in signal processing to extract the original signal from a modulated carrier wave. It involves multiplying the modulated signal with a cosine wave of the same frequency as the carrier, effectively removing the carrier and leaving behind the original signal.

2. How does cosine for synchronous demodulation work?

Cosine for synchronous demodulation works by taking the product of the modulated signal and a cosine wave with the same frequency as the carrier. This results in the carrier wave being canceled out, leaving only the original signal. This technique works because the cosine wave is in phase with the carrier and thus can effectively remove it.

3. When is cosine for synchronous demodulation used?

Cosine for synchronous demodulation is commonly used in radio and telecommunications to recover the original signal from a modulated carrier wave. It is also used in digital communication systems to extract data from a modulated signal.

4. What are the advantages of using cosine for synchronous demodulation?

One of the main advantages of using cosine for synchronous demodulation is that it is a simple and efficient technique for extracting the original signal from a modulated carrier wave. It also has a low error rate and can be easily implemented in hardware or software.

5. Are there any limitations to using cosine for synchronous demodulation?

While cosine for synchronous demodulation is a useful technique, it does have some limitations. It is most effective when the carrier and the cosine wave are perfectly in phase, which may not always be the case in real-world scenarios. Additionally, it may not work well with signals that have a low signal-to-noise ratio.

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