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# Comparing BPSK, QPSK, 4PAM, 16QAM, 16PSK, 64QAM and 32PSK

by on July 8, 2008

I have written another article in DSPDesginLine.com. This article can be treated as the third post in the series aimed at understanding Shannon’s capacity equation.

For the first two posts in the series are:

The article summarizes the symbol error rate derivations in AWGN for modulation schemes like BPSK, QPSK, 4PAM, 16QAM, 16PSK, 64QAM and 32PSK.

The article in DSPDesignline.com details the following:

• Based on the knowledge of bandwidth requirements for each type of modulation scheme, the capacity in bits/seconds/Hz is listed.
• Using the knowledge that the symbol to noise ratio $\frac{E_s}{N_0}$ is $k=\log_2(M)$ times the bit to noise ratio $\frac{E_b}{N_0}$, the symbol error rate vs $\frac{E_b}{N_0}$ curves are plotted.
• Using symbol error rate versus $\frac{E_b}{N_0}$ plots, the $\frac{E_b}{N_0}$ required for achieving symbol error rate of $10^{-5}$is identified.
• Upon having the capacity and $\frac{E_b}{N_0}$ requirement, the requirements for BPSK, QPSK, 4PAM, 16QAM, 16PSK, 64QAM and 32PSK are mapped on to the Shannon’s capacity vs Eb/No curve.
• Assuming Gray coded modulation mapping, each symbol error causes one bit out of $k=\log_2(M)$ bits to be in error. So, the relation between symbol error and bit error is,$P_b \approx \frac{Ps}{k}$.
• Using this assumption, the Bit Error Rate (BER) for BPSK, QPSK, 4PAM, 16QAM, 16PSK, 64QAM and 32PSK are listed and the BER vs Eb/No curve plotted.