Capacity in AWGN channel for M-PSK modulation

[ipmax]

Hi Community,
I am trying to calculate the capacity of AWGN channel (y = hx + n). I was using the shannon formula C = 0.5log2(1+SNR) where SNR = sigma^2(Tx Signal)/sigma^2(Noise). However, I am wondering how can I calculate the channel capacity for M-PSK constellation. For example QPSK and 16QPSK.

Help me.
Thankyou.

 

[Mityan]

Channel capacity is related to symbols. So for given modulations you should multiply C with bits per symbol, I think.

 

[David83]

Channel capacity is a term used when the input signal is drawn from a Gaussian distribution. For discrete input the term used is achievable rate. The maximum achievable rate of an M-ary constellation is obviously \[log_2M\], because you are transmitting \[log_2M\] bits per channel use, which is achieved without channel coding at high SNR.

 

[ipmax]

I agree with David...however, in reality (fading channels, coding included and different modulation), the number of bits M on the receiver side is not the same as the transmited ones. So, to calculate the bits M, on the receiver side we can use
M = log2(1 + SNR), where
SNR is the Signal to Noise ratio at the receiver side which depends on attenuation, coding and Noise. So ofcourse the SNR decreases due to these factors, as a result the bits M decreases and the overall bit rate as well decreases. Now my question is
How can I calculate SNR at receiver so that it includes the effect of modulation, coding as well as TX power?
Is there some sort of formula? I found this one but its kinda vague

**broken link removed**

Help me
Max

Channel capacity is a term used when the input signal is drawn from a Gaussian distribution. For discrete input the term used is achievable rate. The maximum achievable rate of an M-ary constellation is obviously \[log_2M\], because you are transmitting \[log_2M\] bits per channel use, which is achieved without channel coding at high SNR.

 

[David83]

We ca not use the Shannon formula for a given modulation. This formula is a theoretical limit, which can be achieved only when the input signal is drawn from Gaussian distribution, which obviously not practical, since Gaussian distribution is a continuous function.

The formula in the link says that the number of bits must be less than or equal the capacity, and not equal as you indicated. There is a difference. The capacity by definition is the maximum bit rate that can be supported with negligible bit error rate. So, it is natural to chose m (the constellation size) less than the maximum bit rate, since otherwise, the error will not be negligible.

In my opinion, you can not include the modulation in the formula. I do not think you can include the coding effect neither, but you can calculate the capacity at certain SNR, which may correspond to the coded bit SNR. However, for sure you can include the Tx power, by taking the effect of path loss between the transmitter and the receiver.