Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

PAPR reduction using SLM

Status
Not open for further replies.

Marina90

Newbie level 4
Joined
Dec 5, 2012
Messages
6
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,324
I tried reducing the PAPR by using SLM. but there is not even a change in PAPR can any one help me?
 

show me your code . may be i can help you
 

    V

    Points: 2
    Helpful Answer Positive Rating
from your code, i can see that there is no difference between original and slm because you don't choose the lowest papr
why don't you see this example
 

Attachments

  • slm.zip
    1.5 KB · Views: 85

i wanted to see how much the papr reduce before choosing the minimum one.so you mean only when we choose the lowest the papr will reduce?
 

according to your script, we can say that 'numRun'. as number of symbols
the problem is we run with a lot of symbol, and every symbol must choose the lowest papr, every symbol has their own phase rotation
 

The whole idea of SLM method is to generate a set of alternative data blocks and select the one with the lowest PAPR to transmit. Say, your original data has a PAPR of 12 dB and you multiply your original data with 4 phase sequences which yields 4 set of different data blocks. Let say one of the data blocks has a PAPR of 9 dB which is the minimum among 4 of them. So, the entire process reduce the PAPR from 12 dB to 9 dB.

Hope this helps.
 

thanks both of you.now i clear about it.really appreciate your help:razz:
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top