thanks! yes, same length. just curious on why should i divide the total time by number of bits. since my example was 000111000111000.... should'nt I be getting the -000111- period? instead of the entire set? regards.Papabravo said:Are all of the bits the same length, or does the encoding scheme use more time for a zero or vice versa. If all the bits are the same length then you measure the time occupied by a group of bits and divide by the number of bits.
There will be at least two types of errors present. A drift error which is additive, due to error in the clock based on the crystal. The other error will be random and depend on things like temperature, power supply voltage and other factors.
i'm kinda new to data communications. when it is mentioned as 10ns bit-rate is it referring to 10ns per bit or 10ns for the entire bits (one-shot)?beenthere said:Hi,
The bit rate may be considered as the wavelength of the signal. You take the reciprocal of it to get the frequency. For 10 ns, it will be 100 MHz.
Yes, each cycle consists of 6 bits, so the frequency is 16.67MHzPolgi-Wan said:thanks! yes, same length. just curious on why should i divide the total time by number of bits. since my example was 000111000111000.... should'nt I be getting the -000111- period? instead of the entire set? regards.
There is a certain level of ambitguity in what has gone previous, let me clarify a few things. In data communications the bit rate refers to the number of bits transmitted in a specific period of time, for example x-bits per second would refer to the bit rate. When it says 10ns bit rate, it should actually specify explicitly whether it is 10ns per-bit or 10 ns per bit-sequence for it to have any meaning.Polgi-Wan said:i'm kinda new to data communications. when it is mentioned as 10ns bit-rate is it referring to 10ns per bit or 10ns for the entire bits (one-shot)?
thanks!