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• File src/osmo-bts-trx/sched_lchan_tchh.c:

  • Patch Set #2, Line 415: goto send_burst; /* send garbage */

    Thanks, for the explanation, it's clearer now. […]

    Before answering this question I thought it's just garbage, but actually no. Short answer: half-filled bursts mixed with zeroes. Basically whatever remained in the buffer from a previous invocation of this function: bursts of a previously encoded frame. Long answer below.

    Normally when tch_dl_dequeue() yields a L2 frame (let's say 'b'), the burst buffer already contains 262 bits of the previous frame (let's say 'a') at the odd numbered bit positions (bursts b0 and b1). We fill in 262 bits of the new frame 'b' to the even numbered bit positions of bursts {b0, b1} and the other 262 bits to the odd numbered bit positions of bursts {b2, b3}.

            b0   b1   b2   b3   b4   b5
          +----+----+----+----+----+----+
     ...  | ba | ba | _b | _b | __ | __ | ...
          +----+----+----+----+----+----+
           bid0 bid1 bid0 bid1 bid0 bid1
          ^
          |
          | We're here.

    Now imagine tch_dl_dequeue() yields NULL, and we're not permitted to schedule FACCH/H. We have nothing to fill in to the even numbered bit positions of b0 and b1 (marked with '_'), which were initialized with zeroes thanks to the memset() above.

            b0   b1   b2   b3   b4   b5
          +----+----+----+----+----+----+
     ...  | _a | _a | __ | __ | __ | __ | ...
          +----+----+----+----+----+----+
           bid0 bid1 bid0 bid1 bid0 bid1
          ^
          |
          | We're here.

    By jumping to 'send_burst' we send half-filled burst b0 and then b1, so that the MS is still able to decode the previous frame 'a'. Please let me know if something is not clear.

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