AWGN bit error probability

[johnkaz77]

Hello,I'm writing a project for a class but i'm having some difficulties in the theoretical part and i would like some help...In a transmition through AWGN channel with FSK modulation given that the bit error probability without encoding is
Pa= (1/2) *[e^(-Eb/(2*No))]
how can we prove that bit error probability using repetition code (n,1) is
Pb= (1/2) *[e^(-Eb/(2*n*No))] ???

 

[kalyanasv]

If you consider an orthogonal FSK signal as s(t)=√A cos(wt+φ)
I presume your using non coherent detection. for whcih your equation holds.

Then the Pa= 1/2*e^(-A/4(sigma^2)) where sigma^2 is your noise.

here the Noise is typically passed through the bandpass filter at the receiver.
this implies let sigma^2 = N*H Where N is noise and H is filter bandwidth.

Pa= 1/2*e^(-A/4(N*W))

This corresponds to your first equation Eb is bit energy

Pa= 1/2*e^(-Eb/2(N))

once you code your equation is Ec is coded symbol energy

Pb= 1/2*e^(-Ec/2(N))

If you add redundancy of n essentially what you have done is:

Ec/N = Eb/N *(Rate=1/n)

Using which you get Pb.

To understand the equation better you need to
you need to pass the above signal through a envelope detector and get a bessel function and then approximate it to the expression you have at hand..which can be found in any good communication book.