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What does the Rate field in TCPIC0 do?

Tim Jackson
Intellectual 480 points

As the data input to TCP2 is presented as 1/5 rate, how does the value of the Rate field change the decoding operation?

Background: I'm using a 6455 and I want to use rates other than the standard 1/5, /14, 1/3, 1/2 & 3/4.  Without understanding what the rate field does I do not know what value to set it to when I use a non-standard rate.

  • 0
    TI__Mastermind 21295 points

    Setting the code rate in TCP2 means the following:
     
    1/5 – mother code rate of the encoder defined for 3GPP2
    1/4 – mother code with rate 1/5 is used with standard defined puncturing which results in code rate 1/4
    1/3 - mother code rate of the encoder defined for 3GPP
    1/2 – mother code with rate 1/3 is used with standard defined puncturing which results in code rate 1/2
    3/4 – mother code with rate 1/3 is used with standard defined puncturing which results in code rate 3/4

    However, as it is stated in the feature list of TCP2, any code rates greater than or equal to 1/5 may be achieved. That would require appropriate puncturing (rate matching) block outside the TCP2. So, one could set the TCP2 for mother code 1/5 or 1/3 and then do his/her own puncturing pattern.

    Regards,

    John Herrington

    WBI HW Apps

  • 0 in reply to JohnH
    Intellectual 480 points

    John,

    Many thanks for your response - it confirms my understanding. 

    However, I'm still left a bit puzzled as to how the internal workings of the TCP2 change depending on the Rate Field.  As the data is presented to the TCP2 as 1/5 rate (regardless of 'real' rate) why do I need to tell the TCP2 what I've loaded it with, just decode it as though it were 1/5 rate.  Or to put it another way, if I have 1/4 rate data which I pad to 1/5 rate but set the Rate field to 1/3 will it produce different results?  If it produces the same results then the Rate field is redundant, which I very much doubt.  If the results are different, then what has the TCP2 done differently given that it was loaded with 1/5 rate and runs at 1/5 rate?

    I'm sure I'm missing something REALLY obvious but I just don't see it yet!  Many thanks for your time, it's really appreciated.

    Regards,   Tim