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Assuming:

Radio frame

  1. Subdivided in 15 slots
  2. Number of chips: 38.400
  3. Duration: 10 ms; each slots = 666.6 μs and has 2.560 chips

Example of channels

DPDCH (Dedicated Physical Data Channel) (UL)

  • Spreading factor varies between 4 and 256
  • Data rates 960, 480, 240, 120, 60, 30 and 15 kbit/s
  • Spreading factor 4, 640 bits/slot, 15 slots/frame, 100 frames/s -> 960 kbit/s
  • Spreading factor 256, 640 bits/slot, 15 slots/frame, 100 frames/s ->15 kbit/s
  • A connection can have at most 6 DPDCH -> 5, 74 Mbit/s

Question

I don't really understand how the bitrate for DPDCH is determined like "960kbit/s" for example. Which formula is used to derive these values?

Here is a picture of the frame structure.

enter image description here

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1 Answer

up vote 5 down vote accepted

Spreading factor 4, 640 bits/slot, 15 slots/frame, 100 frames/s = 960 kbit/s

I don't really understand how the bandwidth for DPDCH is determined like "960kbit/s" for example.

There's no equation. It's just another way of stating the information that's already there. By dimensional analysis:

$$ \require{cancel} \frac{640\, \text{b}}{\cancel{\text{slot}}} \frac{15 \cancel{\text{ slot}}}{\cancel{\text{frame}}} \frac{100 \cancel{\text{frame}}}{s} \frac{1\,\text{k}}{1000} = \frac{960\,\text{kb}}{\text{s}} $$

I also would not call "960 kbit/s" a bandwidth. That may be acceptable in the context of say, a PCI bus, but in the context of a radio protocol, bandwidth also means the width of the signal in the frequency domain. I'd call it a bitrate to avoid that confusion.

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Thank you @Phil you answered my question. Also your emphasis on my usage of "bandwidth" is correct ! Should've been bitrate. –  beginningperl Jan 16 at 18:42
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