Thanks:  8

1. ## Important: Relation between Bandwidth and Bitrate and Spreading Factor

As a rule: High BW >>> High Bit Rate >>>More power consumption>>>Less Number of users >>> Low SF (Spreading Factor)

If I take code from the tree with SF= 8 my data rate will be= 384Kbps, and if I take code from the tree with SF=16 my data rate will be=128 Kbps, then the rate is affected with spreading factor,

My question: what is the effect of Bandwidth on the rate; I mean that Carrier BW is 5 Mega what will happen if it will be 10 M or 15 M also what will be happened if it will be 3,84M like chip rate or less for ex. 2 Mega ?

2.

3. ## Re: Important: Relation between Bandwidth and Bitrate and Spreading Factor

The more bandwidth, the higher bit rate. This is the case for the evolutions of 3G:
- Dual Carrier (using 2 adjacent 5 MHz carriers) enable to double the HSPA throughput,
- LTE is reaching higher bit rates using OFDM and higher bandwidth up to 20MHz.

4. ## Re: Important: Relation between Bandwidth and Bitrate and Spreading Factor

Originally Posted by islam wagih
As a rule: High BW >>> High Bit Rate >>>More power consumption>>>Less Number of users >>> Low SF (Spreading Factor)

If I take code from the tree with SF= 8 my data rate will be= 384Kbps, and if I take code from the tree with SF=16 my data rate will be=128 Kbps, then the rate is affected with spreading factor,

My question: what is the effect of Bandwidth on the rate; I mean that Carrier BW is 5 Mega what will happen if it will be 10 M or 15 M also what will be happened if it will be 3,84M like chip rate or less for ex. 2 Mega ?
The answer is embedded in the "good old" Claude Shannon's formula:

Bit Rate = Bandwidth x log2(1+C/I),
where C/I is the carrier-to-interference ratio.

Keeping C/I constant, it is fairly obvious that the bit rate is linearly proportional to the bandwidth.

5. ## Re: Important: Relation between Bandwidth and Bitrate and Spreading Factor

Could you Philip provide more explaination for how transmitted power, bit rate will be changed regarding to spreading bandwidth ?
Originally Posted by stilloboy
The answer is embedded in the "good old" Claude Shannon's formula:

Bit Rate = Bandwidth x log2(1+C/I),
where C/I is the carrier-to-interference ratio.

Keeping C/I constant, it is fairly obvious that the bit rate is linearly proportional to the bandwidth.

6. ## Re: Important: Relation between Bandwidth and Bitrate and Spreading Factor

Originally Posted by islam wagih
Could you Philip provide more explaination for how transmitted power, bit rate will be changed regarding to spreading bandwidth ?
Well, yet again, the answer lies in the classical Shannon-Hartley law given earlier: The bit rate has a logarithmic relationship to the transmitter power (or carrier power), assuming the bandwidth and interference (or noise power) remain constant.

7. ## Re: Important: Relation between Bandwidth and Bitrate and Spreading Factor

Originally Posted by islam wagih
As a rule: High BW >>> High Bit Rate >>>More power consumption>>>Less Number of users >>> Low SF (Spreading Factor)

If I take code from the tree with SF= 8 my data rate will be= 384Kbps, and if I take code from the tree with SF=16 my data rate will be=128 Kbps, then the rate is affected with spreading factor,

My question: what is the effect of Bandwidth on the rate; I mean that Carrier BW is 5 Mega what will happen if it will be 10 M or 15 M also what will be happened if it will be 3,84M like chip rate or less for ex. 2 Mega ?
Couple of basic Queries...

For CS Voice/Video Call, it will be SF=16 ?

When a HSDPA call is established what will be the SF ?
To achieve 14.2 Mbps what should be the SF on single 5M career ?

8. ## Re: Important: Relation between Bandwidth and Bitrate and Spreading Factor

@islam wagih
In a single carrier mode of operation in WCDMA, all services are transmitted over the air occupying the same 5 MHz bandwidth. The relation between WCDMA bandwidth and User Bitrate is the spreading factor:
Symbol Rate per one physical channel (Rs) = 3,84 Mcps / SF
Symbol Rate in HSDPA (Hs) = Rs x Number of Codes
Bit Rate in HSDPA = Hs x bit per symbol
QPSK = 2 bits per symbol
16QAM = 4 bits per symbol
64QAM = 6 bits per symbol

For Instance:
SF=16 (this is the spreading factor for the HSDPA channel)
Rs = 3,84 / 16 = 0,24 Msps
In case of 64QAM (6 bits per symbol),
HSDPA Bit Rate 15 codes = 0,24 x 15 x 6 = 21,6 Mbps
So we can say that in order to reach 21 Mbps, 15 codes and 64QAM are required.

Each service requires a specific bandwidth and then a specific SF:
Voice AMR 12,2kbps - SF=64 in UL and SF=128 in DL
Voice AMR 4,75kbps - SF=128 in UL and SF=256 in DL
If Videocall is carrier out over CS57 - SF=16 in UL and SF=32 in DL

As Philippe says, the use of more than 1 carrier in WCDMA is linear. If you get with one carrier, for instance, 7Mbps, you get 14Mbps with 2 carriers in the same situation.

@scorpion
14Mbps can be achieved with 10 codes and 64QAM or 15 codes and 16QAM. This can be checked easily with the formulas above in a excel sheet.

I hope it helps
NYQUIST

9. ## Re: Important: Relation between Bandwidth and Bitrate and Spreading Factor

Originally Posted by scorpion
Couple of basic Queries...
Hi,

For CS Voice/Video Call, it will be SF=16 ? -- voice is using SF 128 signalling is SF256

When a HSDPA call is established what will be the SF ? --- HSDPA is always using SF 16
To achieve 14.2 Mbps what should be the SF on single 5M career ?
--- to achieve 14.4 Mbps you a higher number of codes in the code tree. Normally we reserve 2-5 codes per cell. If there are no users in the cell except you then you can get up to 14 codes for you to have 14.4 Mbps given that proper license is loaded.

Br,
Sen

10. ## Re: Important: Relation between Bandwidth and Bitrate and Spreading Factor

Dear,

CS (Voice & Video) are carried over R99, and SF is used to assign bit rate, while in HSDPA, HS Channels are introduced, along with higher modulation rate 16-QAM, lower TTI, Fast Scheduling. user can reach 14.4 Mbps if 15 HS-PDSCH code are assigned to him. (release 5)

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)

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

Reps and thanks are welcomed if you find it useful

Originally Posted by scorpion
Couple of basic Queries...

For CS Voice/Video Call, it will be SF=16 ?

When a HSDPA call is established what will be the SF ?
To achieve 14.2 Mbps what should be the SF on single 5M career ?

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