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Misunderstanting of Carson's rule in datasheet

Enrico S.
Intellectual 290 points
Other Parts Discussed in Thread: CC1020

Hi,

In datasheet of CC1020 I read  "In NRZ mode the maximum modulating signal occurs when transmitting a 0-1-0 sequence. In both Manchester and NRZ mode 2∙fm is then equal to the programmed baud rate."

I don't get it for the NRZ, while it make sense for Manchester (because the programmed baudrate is referred to RF side, therefore not on the bitrate itself which is halved). My explanation for the NRZ is because the modulating signal have a doubled period compared to bitrate used, due to the double transition needed. Therefore tha maximum modulating frequency is always half the bitrate.

Am I correct?

Thanks!

  • 0
    TI__Guru**** 317180 points

    Not sure what your question is. For NRZ baudrate and bitrate is the same, for manchester the bitrate is half of the baudrate. Ref 4.4 "transmitt baudrate", fig 5.6, 5.7

  • 0 in reply to TER
    Intellectual 290 points
    Actually, my question is another. I don't see why, for example setting a baudrate to 9600bps with NRZ, the "fm" paramer sohuld corresponds to half the baudrate, i.e. 4800 (Hz). So I tried to provide my explanation about that.

    With NRZ the bit duration is 1/9600 s, while symbol duration is the same. But since Carson is used when considering a "wave with a period", are considerred two symbols of opposite value (i.e 0-1 and ending with 0, 0-1-0), the total duration considered is assumed to have a doubled period, from which "fm" is assumed to be 4800Hz.

    While using the MAnchester is quite clear, with 9600baudps the bitrate is 4800bps. The bit duration is 1/4800 s while symbol duration is 1/9600 s, (considering the minimum duration of digital levels, not the number of transitions). "fm" can be assumed to be the modulating signal, which is 4800 (Hz).

    Is this make any sense? :)
  • 0 in reply to Enrico S.
    TI__Guru**** 317180 points
    Not really...

    As stated in the datasheet, fm in carson's rule is equal to the baudrate. Eq for NRZ 9600 baud equals 9600 bps equals fm. Not sure where you get the divide by two factor.
  • 0 in reply to TER
    Intellectual 290 points

    TER said:
    Not really...

    As stated in the datasheet, fm in carson's rule is equal to the baudrate. Eq for NRZ 9600 baud equals 9600 bps equals fm. Not sure where you get the divide by two factor.

    From datasheet: "The signal bandwidth (SBW)

    can be approximated by (Carson’s rule):

    SBW = 2 ∙ fm + 2 ∙ frequency deviation

    where fm is the modulating signal. In

    Manchester mode the maximum

    modulating signal occurs when

    transmitting a continuous sequence of 0’s

    (or 1’s). In NRZ mode the maximum

    modulating signal occurs when

    transmitting a 0-1-0 sequence. In both

    Manchester and NRZ mode 2∙fm is then

    equal to the programmed baud rate."

    It seems to me that in NRZ the baudrate is 2*fm, not fm.

  • 0 in reply to Enrico S.
    TI__Mastermind 39860 points

    Carson's rule = 2 x fm + 2 x deviaton

    NRZ: fm is not the same as the baud (bit) rate. Assuming X kbps data rate, fm is X/2 (this will be if you are sending alternating 0's and 1's). 2 x fm is then equal to the Baud (bit) rate

    Manchester: fm is not the same as the baud rate. Assuming X kBaud data rate, the bit rate is X/2. In this case fm is X/2 (this will be if you are a continuous string of 0's or 1's). 2 x fm is then equal to the Baud rate 

    In the above two cases Carson's rule can be re-written as Baud rate + 2 x deviaton

  • 0 in reply to Sverre
    Intellectual 290 points

    Sverre said:

    NRZ: fm is not the same as the baud (bit) rate. Assuming X kbps data rate, fm is X/2 (this will be if you are sending alternating 0's and 1's). 2 x fm is then equal to the Baud (bit) rate

    And this is what the datasheet confirm.

    Now, I was trying to search for a solid explanation: what you are saying is that the fm correspond to a frequency with a given period. This period is determined on the duration of a periodic wave, i.e. 1-0 alternating, with a given period by definition. Obtaining this alternation requires to send 2 symbols (bits), therefore the duration is doubled with respect to the single bit transmitted. This apply as the same way to Manchester, when considering that the data rate is halved. Is this the (quite naive) explanation?

    Thanks to all and sorry for the obviousness of questions.

  • 0 in reply to Enrico S.
    TI__Mastermind 39860 points
    fm is the highest modulating signal frequency, and for NRZ this will be if you transmit a 0-1-0 sequence.

    For Manchester the highest modulating signal frequency is if you transmit a continuous string of 0's or 1's.