I'm finding it very difficult to tackle questions in exams involving calculation of Probability of error and Bit error rate for various digital modulation schemes involving AWGN Channel. Firstly is BER and Pe calculation in exams the same? I clearly understood upto the signal representation of the different Keying Techniques (ASK,FSK,PSK,BPSK) ie S(t). Why did we go for Energy/Energy Difference calculations of these signals? What exactly are constellation diagrams and why do we need them? Even in constellation diagrams we convert the Energy back to amplitude maybe because we need to calculate the magnitude of noise at the boundary of the signals that will cause an error. I understand that the channel here adds White Guassian Noise.Which is the best method to tackle Pe/BER problems, is it by drawing the constellation diagrams and then the boundary or are there other methods or standard formula's to be learned? These are a few questions asked in the exam that I'm preparing. Please help me with them also. I mainly have ASK, FSK, PSK, BPSK in syllabus.

Let Q(√g ) be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral density N0/2. The parameter g is a function of bit energy and noise power spectral density.

A system with two independent and identical AWGN channels with noise power spectral density N0/2 are in parallel. The BPSK demodulator receives the sum of outputs of both the channels. If the BER of this system is Q(b√g), then the value of b is?(Ans given is 1.414)

A BPSK scheme operating over an AWGN channel with noise power spectral density of N/2, uses equiprobable signals S1(t)(t) = √(2E/T)sin(t) and S2(t) = -√(2E/T)sin(t) over the symbol interval (0, T). If the local oscillator in a coherent receiver is ahead in phase by 45 degrees with respect to the received signal, the probability of error in the resulting system is (Answer key given is Q(√(E/N))

Also please suggest where I can find good problems of the same cadre for practice.


1 Answer 1


1.414 is the square root of 2. If the same signal is sent over the 2 independent channels, then your received Noise PSD is the same, but your signal is now twice as powered. You add the signals, but the noise is not additive because it is IID / uncorrelated. Twice as much power is a 3dB Gain. I'm a little hazy on BER curves, but if that Q(sqrt(g)) is the likelyhood function, then instead of g you have 2*g, and Sqrt(2*g) = Sqrt(2)*sqrt(g)

Yeah. The argument under the square root is the received Signal to noise ratio in Eb/No (energy-per-bit/Noise-Power-in-1Hz-bandwidth). One receiver and one channel is R = S + N. In this example R = S1 + S2 + N. If S1 and S2 are equal then you have 2S + N, thus twice the SNR.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.