π
<-

## TD_CHOR

### File hierarchy

LicenceLicense : Non spécifiée / IncluseUnspecified / Included

Vote :

### Informations

Catégorie :Category: nCreator TI-Nspire
Auteur Author: dlk
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 2.29 Ko KB
Mis en ligne Uploaded: 24/11/2022 - 16:34:15
Visibilité Visibility: Archive publique

### Description

Fichier Nspire généré sur TI-Planet.org.

Compatible OS 3.0 et ultérieurs.

<<
Problem 1 Consider an AWGN channel with bandwidth 50 MHZ, received signal power 10mW, and noise PSD N0/2 where N0=2.10^-9 W/HZ. How much does capacity increase by doubling the recieved power? How much does capacity increase by doubling the channel bandwidth? Solution Pb1 B = 50 MHz ; P = 10mW ; N0 ; N=N0B; C=6.87 Mbps So Pnew=20mW, C=13.15 Mbps (for x<<1, log(1+x) = x) rappel: C=Blog2(1+S/N) B=100MHz, Notice that both the bandwidth and noise power will increase. So C=7Mbps Problem 2 Consider two users simultaneously transmitting to a single receiver in an AWGN channel. This is a typical scenario in a cellular system with multiple users sending signals to a base station. Assume the users have equal received powr of 10mW and total noise at the receiver in the bandwidth of interest of 0.1 mW. The channel bandwidth for each user is 20MHz. (a) Suppose that the receiver decodes user 1's signal first. In this decoding, user 2's signal acts as noise (assume it has the same statistics as AWGN). What is the capacity of user 1's channel with this additional interference noise? (b) Suppose that, after decoding user 1's signal, the decoder re-encodes it and subtracts it out of the received singal. Now, in the decoding of user 2's signal, there is no interference from user 1's signal. What then is the Shannon capacity of user 2's channel? Solution : Pnoise =0.1mW ; B = 20MHZ rappel : for AWGN   channel capacity   = 1/2log2(1+P/N) Problem 8 " What is the PAPR of an OFDM System with N subcarriers? Solution Problem 8 " Let us assume that the average power of a single subcarrier with M-QAM modulation is S. This is also the average power of the OFDMsignal " The peak power of the OFDM signal occurs when ALL subcarriers areon phase, i.e., they add in amplitude and not in power. In this casethe power of the PFDM signal is N*S " Therefore PAPR=N*S/S=N Made with nCreator - tiplanet.org
>>

-
Search
-
Social TI-Planet
-
Featured topics
1234567
-
For more contests, prizes, reviews, helping us pay the server and domains...

Discover the the advantages of a donor account !

-
Stats.
351 utilisateurs:
>333 invités
>12 membres
>6 robots
Record simultané (sur 6 mois):
6892 utilisateurs (le 07/06/2017)

-
Other interesting websites
Texas Instruments Education
Global | France
(English / Français)
Banque de programmes TI
ticalc.org
(English)
La communauté TI-82
tout82.free.fr
(Français)