But after this , follows the 'spreading' that will cause that the data rate will increase even more but since the channel has a bandwidth of 5 Mhz , then how is this possible?
( I mean 3.84mcps x SF > 5Mhz )
Thanks to everyone!!
PD:
I'm trying to learn this stuff on my own and its a bit complicated in some points. If Im getting this wrong please let me know it means a lot for me!
Re: WCDMA: How is much is increased the velocity of the chips after spreading?
Your diagram is partly wrong. Only the channelization code causes user data spreading; the scrambling code does not. The chip rate of 3.84Mcps along with room for filter roll-off will occupy a bandwidth of 5 MHz.
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