# Manchester Encoding Probability of Error

1. Feb 2, 2013

### Meteora

Greetings,

I am studying Analog & Digital Communications in one of my lectures and I'm stuck with a subject. I would be much pleased if someone is willing to help me on the subject below;

Probability of error of a digital signal sent by using non return to zero encoding is easy
to find such that we can apply -A+N for output if 0 is sent and A+N if 1 is sent for a single signal transmitted in the interval 0<t<T. Since this interval will only consist of an rectangle of height A or either height -A of cases 1 and 0 being sent respectively, we can use the mean and variance of Y (which is the output Y=-A+N or Y=A+N depending on the data being transmitted), normalize it and use the Q() function to find the error rate. Howeve Manchester encoding is different in sense that to send signal 1 0<t<T/2 is A and T/2<t<T is -A and viceaversa for 0.

How am I supposed to do the math of this? When I was working with NRZ I built my calculations on the fact that x(t)=-A+w(t) in the case of 0 being transmitted (w(t) being white gaussian noise. How can I build my new x(t) fuction which the rest I'm sure I can handle.

2. Feb 2, 2013

### rbj

do you have your synchronization taken care of? are you pretty certain when $t=0$ is and when $t=T/2$ and $t=T$ are?

the additive noise error function is unchanged if the code is different. so you have to multiply by a $[-1,1]$ pulse or its negative, integrate (over one period T) and see if the result is positive or negative (the threshold will be at 0). then assume how big the variance is of the added noise and compute what the p.d.f. is of the integrated result. a tail of that p.d.f. will cross over zero and it's the error of the tail of the p.d.f. that crosses over that is your probability of error.

3. Feb 2, 2013

### Meteora

Sync is out of content. So okay first of all thanks for responding to my thread. Now what I understood so far is as follows;

If i am sending a "0" with manchester code I expect it to be -A in half period and +A in the other half period. What is the possibility that it would be decided as a "1" in the receiver? In my mind there is no such possibility since the treshold is λ=0. Because the -A+N (where N is noise) part will be greater than λ however A+N part will be much greater than λ. So what I have in my receiver will be an useless signal looking like a manchester signal just lifted up above the zero treshold.

So instead of misinterpreting the signal it shouldn't interpret any signal at all (unless we change the treshold) if I am thinking correct. Another possiblity is that it's probability of error could be this not interpreting. Which would only be equal to P(y(t)>λ) which is just a single NRZ probability.

Why I thought that a 1 could never be received as a 0 is because of the following reason;

Since my additive noise doesnt change in time, if -A+N is above zero A+N could never be below zero which is the condition for receiver to decide on the signal as a "1" which was transmitted as a "0".

Again looking forward for responses.

4. Feb 3, 2013

### Meteora

I still need help with this problem. I have an exam tomorrow with this to be asked for sure.