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Analysis of BPSK signals, BER with AWGN

A radio link sends BPSK signals to a receiver with a bit rate of 100 kbps. It is found that the output of a radio receiver when no signal is present is AWGN with a rms voltage of 1.0 V.

a. What is the usual source of additive white Gaussian noise (AWGN) in a radio receiver?
b. For what percentage of the time is the noise output of the receiver greater than + 2.0 V or more negative than -2.0 V?
c. A 100 kbps BPSK signal is sent to the radio receiver. At the sampling instant at the receiver output the voltage when a logical 1 is sent is +3.0 V and the voltage when a logical zero is sent is -3.0 V. What is the average bit error rate? [Use the Q (z) table below to estimate BER.]
d. What is the average time between errors?
e. The transmitter power is increased so that the sampled voltages at the receiver output are +6.0 V and - 6.0 V. What is the bit rate now and the average time between bit errors?

z Q(z) z Q(z)
0 0.5
2.0 2.280 E-2 5.1 1.701 E-7
2.1 1.791 E-2 5.2 9.981 E-8
2.2 1.394 E-2 5.3 5.799 E-8
2.3 1.075 E-2 5.4 3.372 E-8
2.4 8.220 E-3 5.5 1.902 E-8
2.5 6.227 E-3 5.6 1.073 E-8
2.6 4.674 E-3 5.7 6.000 E-9
2.7 3.476 E-3 5.8 3.320 E-9
2.8 2.562 E-3 5.9 1.820 E-9
2.9 1.871 E-3 6.0 9.979 E-10
3.0 1.354 E-3 6.1 5.310 E-10
3.1 9.702 E-4 6.2 2.827 E-10
3.2 6.889 E-4 6.3 1.490 E-10
3.3 4.847 E-4 6.4 7.778 E-11
3.4 3.378 E-4 6.5 4.021 E-11
3.5 2.332 E-4 6.6 2.058 E-11
3.6 1.595 E-4 6.7 1.043 E-11
3.7 1.081 E-4 6.8 5.236 E-12
3.8 7.252 E-5 6.9 2.603 E-12
3.9 4.821 E-5 7.0 1.281 E-12
4.0 3.174 E-5 7.1 6.244 E-13
4.1 2.070 E-5 7.2 3.014 E-13
4.2 1.337 E-5 7.3 1.440 E-13
4.3 8.558 E-6 7.4 6.816 E-14
4.4 5.423 E-6 7.5 3.194 E-14
4.5 3.404 E-6 7.6 1.482 E-14
4.6 2.117 E-6 7.7 6.810 E-15
4.7 1.303 E-6 7.8 3.098 E-15
4.8 7.948 E-7 7.9 2.396 E-15
4.9 4.800 E-7 8.0 6.226 E-16
5.0 2.872 E-7

Solution Preview

(a) The noise performance of a radio receiver is ultimately most influenced by the front end amplification stage and input filter. It can be shown that the overall system noise of such a receiver is most influenced by the first term in the expression from the "Friis" formula for system noise where we denote a Noise Figure (F1) for the front end amplifier, successive noise contributions to the system noise are then reduced by a factor that includes dividing by the gain of the previous stages. The system noise (FSYS) is given by the Friis formula below.

FSYS = F1 + (F1-1)/G1 + (F2 - 1)/G1*G2 + ...... [Ref 1]

So the main contribution to the AWGN of such a radio receiver is determined by the components in the input filter and 1st amplifier stage.

(b) We are told that this is an AWGN channel so the Noise obeys the statistics of the Standard Normal (Gaussian) Distribution with zero mean SIGMA = 0). The general spread of the Noise voltages about the Mean (μ) for different deviations (SIGMA)is shown in the graph below [Ref 2].

We are told that the rms Noise voltage is 1V, for such a Noise voltage this can be equated to the variance of the ...

Solution Summary

An analysis of BPSK with AWGN are provided. Logical voltages for zero sent volts are determined.

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