WCDMA chip, symbol, bit rate

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Can anybody clarify me the speed calculations of 3G/UMTS (European

The European 3G/UMTS WCDMA so called "chip rate" is 3,84 Megachips a
second. This means that 3,84 million so called CHIPS are ALWAYS sent
on every active, unidirectional, "physical channel" in the system.
(Several physical channels are of course used simultaneously in the
air, by multiple terminals and base stations, uplink and downlink.) I
would think that the electric/radio wave modulation, which happens to
be QPSK (quadrature phase shift keying) in the downlink direction,
would not matter, if the chip rate is given to be 3,84 Mchip/s.

Now every "chip" is part of a "coded" user bit, and the "spreading
factor" determines, how many CHIPS there are per information BIT. I
will not go into the basic idea of "code division" here, but "coding"
is needed to separate the different "physical channels" from each
other at the receiver. The shortest "spreading factor" the UMTS
standard uses is 4. Thus the highest achievable raw information speed
(with no formatting or forward error correction or checksums or any
other added bit overhead) on one physical channel would be 3,84 / 4 =
960 kb/s.

But somehow for example the Holma-Toskala "WCDMA for UMTS" book on my
desk calculates double my rates. It talks about spreading factor 4,
"symbol rate" of 960 and "channel bit rate" of 1920.

I would understand that modulation related "wave symbols" would always
carry EXACTLY TWO CHIPS and be sent at the speed of 1920 ksymbol/s,
the fixed chip rate of 3, 84 Mchip/s divided by two. (The QPSK
modulation carries two chips per phase shift.) And if we want to name
the 4 chip unit, that forms one information bit, an "information
symbol", the "information symbol" rate here would be 960 ksymbol/s,
but in that case the "information symbol" would correspond exactly to
one channel bit = raw user bit = information bit.

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