GSM, a common problem of burst pulse sequence, asks experts to answer how to calculate the eight and a half pulses marked in the figure below.

This is certainly not something you can understand the first time you read this book, but you will understand it when you see the back of the book.

First of all, it should be understood as 8 "semi-bursts" instead of 8.5 bursts. There is a slight problem with the interruption of the book. It's no big deal.

Because in the A/D conversion of GSM, a 20 ms speech block is extracted, and this speech block is finally encoded into 456 bits. In GSM, a burst can transmit 156.25bit, which contains 20 ms of speech block information 57 * 2 = 1 14 bit (others are 26-bit training sequence, 2-bit frame stealing flag, and so on). 8.25-bit guard interval), that is to say, an ordinary burst can only transmit 1 14 bits in a 20-ms speech block, so to completely transmit a 20-ms speech module, it takes four complete bursts to complete the transmission (1 14 * 4 = 456 bits). It divides the 456 bits of 20ms into 8 small blocks (456 bits ÷ 57 bits = 8), then interleaves them internally once (it is easy to understand, not to elaborate), and then interleaves the speech blocks again, that is, interleaves two consecutive 20ms speech blocks, such as the first speech block. The interweaving of these two 20ms speech blocks forms (a4, b0) in the first burst, (a5, b 1) in the second burst, and forms (a6, b2), (a7, b3), (b4, a0), (b5, a 1) and (respectively). Eight consecutive bursts are needed to complete the transmission, and any one of these eight small blocks can only occupy the first 57 bits or the last 57 bits in the burst (that is, it can be understood as a "semi-burst"). However, some handover signaling needs to borrow a complete 20ms speech block to transmit, that is, it needs to borrow continuous "8 and a half bursts".

I'm incompetent. I've talked about so many details. I hope you can understand.