I would like to discuss this question which is frequently asked for both fresher and experienced people in any of the telecom domain interview. Everybody talks about the uplink and downlink speed which we can achieve in WCDMA (3G) but how we can achieve it; is the main thing which everybody should know.
First we should know the frame structure and its basic unit calculation which is explained below.
1 Frame = 10 ms = 5 sub frames
1 Sub frame = 2 ms = 3 slots
1 slot = 2560 symbols
So, 1 frame carries 5*3*2560 = 38400 chips/10ms. (ms = millisecond)
Ques. – How do we achieve 14.4 Mbps max data rate in HSDPA?
Ans. – The UMTS supports 3.84 MCPS (Mega chips / sec)
So 3840000 chips /s
- 3840000/100 = 38400 chips / 10ms = 38400/frame
- 38400/5 = 7680 chips/2ms = 7680 chips/sub frame
- 7680/3 = 2560 chips/slots
That means in a UMTS system, in every slot; 2560 chips must be sent.
To achieve maximum data rate we use 16 QAM with spreading factor 16 and assuming that no redundancy bit is added because of ideal condition.
The below calculation is done for 1 pdsch code.
- 2560 chips / 16 = 160 symbols / slot (de-spreading)
- 160 X 4 = 640 bits /slots (demodulation)
- 640 X 3 = 1920 bits/sub frame = 1920 bits per 2ms
- 1920 X 5 = 9600 bits / frame = 9600 bits per 10 ms
- 9600 X 100 = 960000 bits per second
Hence, per pdsch code we can achieve 960000 bits per second
We have 15 such pdsch codes available.
So, If all the 15 codes are assigned to a single UE then max DL data rate per UE is (960000X15) = 14400000 bits = 14.4 Mbps. đŸ˜€
Ques – How do we achieve 5.76 Mbps max data rate in HSUPA?
Ans – To achieve max data rate HSUPA supports two SF2 code and two SF 4 code to a single UE. Also we use BPSK (only used mod Scheme for HSUPA) modulation scheme. Assuming there is only one UE and no redundancy bit is added because of ideal condition.
- (2560 chips / 2) + (2560 chips / 4) = (1280 + 640) = 1920 symbols / slot (de-spreading)
- 1920 X 1 = 1920 bits /slots (demodulation)
- 1920 X 3 = 5760 bits/sub frame = 5760 bits per 2ms
- 5760 X 5 = 28800 bits / frame = 28800 bits per 10 ms
- 28800 X 100 = 2880000 bits per second = 2.88 Mbps for one SF2 and one SF4 codes.
Hence, max data rate achieved per UE is 2.88 Mbps X 2 = 5.76 Mbps đŸ™‚
With the LTE-release8 (4G) technology we have achieved 300 Mbps in DL and 75 Mbps in UL. How do we achieve such a huge data rate, I would discuss in my next blog. If anybody has doubt in the above calculation please let me know.
Note: Also let me know in case if you have any doubt question related to 3G and 4G technology.
