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Let me confirm how the demodulation process of QAM is.

$$s(t)=I(t)sin(2πft)+Q(t)cos(2πft)$$

Then we have to guess \$I(t)\$ and \$Q(t)\$ in order to know what the said bit formation of the demodulated signal is. And we have already known what \$s(t)\$ is. And we can use the below formula

$$sin^2(2πft)+cos^2(2πft)=1$$

In case of QPSK, we can say

$$I^2(t) + Q^2(t) = constant$$

But when we look at QAM, in case of 16 QAM, there are 2 circles whose center is (0,0) and in case of 64 QAM, there are 4 ones, and in case of 256 QAM, there are 8 circles. So we cannot say what is the fixed amplitude.

In such a case how should we work out solution of \$I(t)\$ and \$Q(t)\$ accurately using what kind of expressions or formulas if we do the solution by the pure software? Could you please teach what the needed equations are?

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You need to deduce the amplitude, carrier phase difference and carrier frequency error with help from the sender.

There are three basic methods:

  1. Fixed preamble

    This is useful if you have intermittent transmissions. Every transmission starts with a fixed symbol pattern, which is used to determine channel gain, frequency and phase offset.

  2. Pilot symbols

    If you have a continuous transmission, the fixed symbols are spaced out in between data symbols.

  3. Pilot tone

    A separate carrier is transmitted adjacent to the QAM signal and is used as a reference.

Please take a look at my answer to QAM's performance and SN to get an overview of the various error terms in the receiver equation.

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I am not sure if you are finding problems understanding the basic idea behind demodulation or if you are having trouble understanding how it is actually implemented.I will go ahead and explain the basic idea in this answer using a practical example.

The diagram below is of 16QAM

16QAM

Say we wanted to transmit the bits \$0010\$, from the 16QAM diagram we can see that this corresponds to \$I(t) = 1\$ and \$Q(t) = 3\$ so we will transmit \$s(t)\$ which in our case will be given by

$$ s(t) = I(t)\sin(2\pi ft) + Q(t) \cos(2\pi ft) \\ s(t) = \sin(2\pi ft) + 3 \cos(2\pi ft) \\ $$

At the transmitter end we receive \$s(t)\$ but we do not know what \$I(t)\$ and \$Q(t)\$.We need \$I(t)\$ and \$Q(t)\$ to figure out what the actual data sent was.

To get \$I(t)\$ we mulitiply \$s(t)\$ by \$\sin(2\pi ft)\$ and then Low pass filter, or mathematically speaking, we compute

$$ s(t)\sin(2\pi ft) = I(t)\sin ^2(2\pi ft) + Q(t) \cos(2\pi ft) \sin(2\pi ft) \\ = \frac{I(t)}{2} - \frac{I(t)}{2}\cos(4 \pi ft) + \frac{Q(t)}{2} \sin( 4 \pi ft) $$

we then apply a low pass filter and multiply by 2 to get I(t), ie

$$ I(t) = 2 \times \text{LPF} ( s(t)\sin(2\pi ft) )\\ = 2 \times \text{LPF }( \frac{I(t)}{2} - \frac{I(t)}{2}\cos(2x) + \frac{Q(t)}{2} \sin( 4 \pi ft) )\\ = 2 \frac{I(t)}{2} $$

It can also be shown that Q(t) can also be obtained by multiplying by cos followed by a Low pass filter.Now we have the values of I(t) and Q(t) that ware sent to us (assuming no noise, perfect synchronisation e.t.c) We now have to use values of I(t), Q(t) and our constellation diagram to get the bit sequence that was sent.

Hope this overview helps

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  • \$\begingroup\$ I have noticed that s(t) (basic QAM signal) is sometimes written as a sum of I and Q component, and sometimes as a difference between the I and Q component. Is is correct to write it either way because we get the sam result? Please note I am not electrical engineer. \$\endgroup\$ – Quirik Sep 30 '17 at 13:53
  • \$\begingroup\$ @Navi It does not matter as long there is consistency between the transmitter and the receiver i.e receiver needs to know the exact form the signal \$s(t)\$ for it to function properly. \$\endgroup\$ – KillaKem Sep 30 '17 at 15:50
  • \$\begingroup\$ Thanks a lot. That's what I tought because mathematically we always get the same waves. Only condition is consistency on form of the signal between Tx and Rx, as you pointed out. \$\endgroup\$ – Quirik Sep 30 '17 at 15:57

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