Calculating Baud Rate: Frequency 433MHz, 100 Cycles/Symbol

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Discussion Overview

The discussion revolves around calculating the baud rate for a wireless communication system operating at a frequency of 433MHz, where 100 cycles are required to recognize a symbol. Participants explore the relationship between cycles, symbols, and bits, as well as the definitions of baud rate and bit rate.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the baud rate as 43.3 kHz based on the formula T=1/f and the requirement of 100 cycles at 433MHz.
  • Another participant clarifies that the period for one cycle at 433MHz is approximately 2.3 ns, leading to a total time of 0.23 microseconds for 100 cycles.
  • This second participant questions the initial claim about symbols and bits, suggesting that perhaps 2 bits form a symbol instead of 2 symbols forming a bit.
  • The second participant estimates a bit rate of approximately 8700 bps based on their interpretation of the timing and symbols.
  • There is a mention that baud rate is defined as the maximum rate at which the signal can change states, and the participant guesses a baud rate of 9600, which contrasts with the earlier calculations.
  • A repeated question from the first post seeks confirmation of the initial calculation and introduces a new guess of 2.165 MHz, which raises further uncertainty.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between symbols and bits, as well as the correct calculation of baud rate. No consensus is reached regarding the baud rate or the definitions involved.

Contextual Notes

There are unresolved assumptions about the definitions of symbols and bits, as well as the calculations leading to different baud rate estimates. The discussion reflects varying interpretations of the relationship between frequency, cycles, and data transmission rates.

Sir_Pogo
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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?
 
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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.

Sir_Pogo said:
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.
 
Sir_Pogo said:
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...
 

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