# Bandwidth and BER of FM0 encoding

What is the bandwidth (i.e., the lowest possible bandwidth) and BER for FM0 encoding?

To be precise, this is my example of an FM0 encoded waveform:

There is always a transition between each bit. For a zero there is an additional transition in the middle. For a one there is no transition (that's similar to differential Manchester code).

From the plot above I would assume that the minimum required bandwidth is 80kHz (if the waveform would be interpreted as NRZ waveform the bandwidth would be 80kHz). In my experiments I am surprised that 40kHz seems sufficient. I had expected that FM0 would just be an inefficient encoding.

Regarding the BER, I found the article "Some Interesting Observations for Certain Line Codes With Application to RFID" which derives the BER. However, from this article it is not clear if E/N0 relates to the symbol or to the bit. Actually, for this encoding it is unclear to me what exactly is the bit and what the symbol (for me, the bit is equivalent with the symbol).

I simulated the BER in MATLAB and found a deterministic discrepancy between simulation and the formula reported in the paper. The way I decode it in my simulations: I look at each halfsymbol. If two consecutive samples are the same, the output bit is a one; otherwise zero.

Due to this observation, I wonder if I can only get the proper BER if I employ a more sophisticated decoding scheme.

40kHz would be the two sided bandwidth, in which case sampling requirement is 40khz minimum

• I appreciate your reply but it really doesn't answer the question at all. The principal question is simple: "For X bits per second and FM0 encoding, what is the minimum bandwidth required and what is the BER?". The rest of my question is for clarification. [As to your answer: If we assume this waveform is simple NRZ, then one bit has a width of 12.5us (=1/80kHz). The required bandwidth is certainly 80kHz (see ece.tamu.edu/~spalermo/ecen689/…, slide 16)]
– divB
Commented Apr 12, 2020 at 0:10