Calculating Baud Rate.

[Sir_Pogo]

Question:

It is determined that 100cycles of frequency 433MHz are required to recognize a symbol with a certain wireless receiver. Two symbols form a bit, fi + fm and fi - fm. What is the baud rate?

I was thinking that since T=1/f then you would have:
100cycles=1/433MHz => 100*433 = 43300hz
so then the baud rate would be 43.3khz??

Is that correct or am I wrong here?

[zgozvrm]

First of all, Hz = cycles per second, so when you divide cycles per second into 1, you end up with seconds per cycle (the period).

If you have a frequency of 433MHz, then the period is 1/f = 1/(433MHz) = 2.3nS per cycle. In other words, it takes 2.3nS to complete 1 cycle.

Therefore 100 cycles of a 433MHz signal would take 100 * 2.3nS = 0.23 micro-seconds.

Two symbols form a bit, fi + fm and fi - fm.

I'm not sure what you mean by this statement. Since a bit is a 1 or 0, it is the smallest amount of data that you can deal with. Therefore, how can 2 symbols form a bit? Perhaps, you meant that 2 bits form a symbol? If that is the case, then since it took 100 cycles to be able to "see" a symbol, that would mean that it took 0.23 micro-seconds to see 2 bits, your bit rate would then be 2 bits/0.23 micro-seconds = approximately 8700 bps.

Still, this is not the baud rate. Baud rate is defined as the maximum rate at which the signal can change states (symbols per second). The values are usually pretty close, but in some instances the bit rate can exceed the baud rate by a factor of 4.

Based on my assumption of what your definition of a symbol is, I would guess that the baud rate is 9600.

[Jeff Reid]

Wiki article:

http://en.wikipedia.org/wiki/Baud

[mugaliens]

Question:

It is determined that 100cycles of frequency 433MHz are required to recognize a symbol with a certain wireless receiver. Two symbols form a bit, fi + fm and fi - fm. What is the baud rate?

I was thinking that since T=1/f then you would have:
100cycles=1/433MHz => 100*433 = 43300hz
so then the baud rate would be 43.3khz??

Is that correct or am I wrong here?

2.165 MHz would be my guess, but that doesn't see to jibe with the base-2-log approach...