# What is the maximum data rate a wireless link can support using BPSK for given BER and SNR?

What is maximum data rate a wireless link can support using BPSK for 10^-6 BER, Bandwidth of channel is 200 kHz and SNR 21.335dB ? As per my understanding , Eb/No can be calculated for BPSK for given BER using curves and value found is 11.29dB, and after that we can use relation

Eb/No = S/N * W/R


R comes out to be around 2 Mbps! Is it correct? Need your input!

## 2 Answers

In BPSK, BER is equal to $Q(\sqrt{2E_b/N_0})$, so you need $E_b/N_0>11.3$ to have a BER no larger than $10^{-6}$. Since the bandwidth is 200 kHz, the maximum rate is 400 kb/s. So, since $SNR\approx136$,

$$\frac{Eb}{N_0}=\frac{S}{N}\frac{W}{R}=136\frac{200\cdot10^3}{400\cdot10^3}=\frac{136}{2}=68$$

Since $68>11.3$, you can be sure that you can transmit at 400 kb/s and meet your BER requirement.

Note that I'm assuming baseband transmission (so that the bit rate is twice the bandwidth). Note also that the question, as posed, does not make a lot of sense, since the bandwidth will essentially determine your bit rate.

I finally got the answer.The design here is over specified. The point is we cannot specify all of these things: modulation, BER, SNR, and bandwidth. Or else, it will cause a conflict. For example, if we don't specify BPSK, the resulting computation for finding the maximum data rate will be straightforward, and will also require higher-order signaling. If we leave-out any one of the four items specified as fixed, we wll find what we desire without any conflict.