We send the data in bytes to the RS Encoder block. The RS Encoder block encodes the data with the Reed Solomon (225,223) code. It waits for 223 bytes and after this encode them to make a codeword of 255 bytes.

What happens if I can send only 219 bytes instead of 223 bytes? Does the length of the virtual fill increase only?

  • $\begingroup$ To my mind , what will happen is that you will perform virtual filling before applying you RS encoder. The process is described below : > src (buffer of 219 bytes : data) > add virtual filling to src (buffer of 223 bytes : virtual fill + data) > ApplyRS encoding ( buffer of 225 bytes : virtual fill + data + parity) > remove virtual fill (buffer of 251 bytes : data + partity) On the receiver size you must precise the expected length of the useful data. Therefore you will be able to perform the exact inverse operations. $\endgroup$ – Nathan Huchon Nov 23 '20 at 13:14
  • $\begingroup$ @NathanHuchon mmm I will perform virtual filling two times, before RS and after. What should I do with the first one? Should I remove it as well? I am thinking the first virtual fill could give me at the receiver side problem ( performance decreasing). $\endgroup$ – Anna Koroleva Nov 24 '20 at 8:17
  • $\begingroup$ @NathanHuchon what to do if it is not possible to precise the expected length of the useful data $\endgroup$ – Anna Koroleva Nov 24 '20 at 8:27
  • $\begingroup$ I wonder how you can decode your RS block(s) without knowing any prior informations about their lengths. I suppose you can estimate the current length of you RS blocks with the synchronisation word placed ahead your RS block(s). With a single RS block (ie without interleave) the structrure of the received data flow should be as follow: [SyncWord][Usefull Data][Parity][SyncWord][...] Therefore you'll be able to estimate the number of bytes between two consecutives SyncWord. $\endgroup$ – Nathan Huchon Nov 24 '20 at 12:21

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