0
$\begingroup$

If you look at this image, you can see that you have 256 possible binary values by modulating the amplitude (the radius of the trigonometric circle) and the phase (the degrees) of the carrier. enter image description here However, multiple diagrams like this one show that each value is actually two cycles long.

enter image description here

My question is why is there two cycles per value instead of one cycle? The constellation only goes to 360 degrees, so why is two cycles used? Wouldn't you be able to transfer data 2x faster if you used one cycle per value?

$\endgroup$
  • $\begingroup$ It is common to see two (or more) samples per symbol due to pulse-shaping and managing frequency offsets between transmitter and receiver. With pulse shaping considerations alone, you will have some excess bandwidth, so one sample per symbol will not be sufficient. Common carrier tracking and timing recovery techniques (such as the Gardner Timing Error Detector) require 2 samples per symbol. $\endgroup$ – Dan Boschen Jul 2 '16 at 23:05
2
$\begingroup$

It doesn't matter how long you stay on one constellation point. The constellation diagram has NO ties to time whatsoever. The two base vectors $e_x, e_y$ (x-axis and y-axis) represent your real and imaginary part or "basic functions" your signal is made of. For example, if your signal consists of two functions (in this case, QAM, it consists of a real and an imaginary part including the amplitude)

$$ e_x = A\cos(2\pi f_{carrier}t) \\ e_y = A\sin(2\pi f_{carrier}t) \\ $$

If you stay on one constellation point for an hour, that's fine. The constellation diagram only shows you which "constellation" of amplitude and phase you should choose to transmit a specific "constellation" of bits (001 010 110, etc ..).

The question you ask is about the symbol rate. If the symbol rate is half your carrier frequency, then you will always have two cycles before the next symbol will be sent. If your symbol frequency is a quarter of your carrier frequency, you will have four cycles to the next symbol.

I think there are always at least one and a half cycle shown to make it easier on the eye and for the brain to understand because we can follow the sine easily. This wouldn't be the case if you'd change the phase and amplitude every cycle. That being said, there's also the problem of having enough time for the demodulator to detect the phase shifts. This is why the symbol frequency is often a lot lower than the carrier frequency.

$\endgroup$
  • $\begingroup$ The choice of carrier frequency is completely arbitrary really and only for convenience of the characteristics of the channel involved (whether physical or legal). If I was off of DC (baseband) by 0.001 Hz or 10 GHz I can still demodulate the information with the same fidelity (assuming a complex baseband). $\endgroup$ – Dan Boschen Jul 3 '16 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.