MOB Subject 2 – "Link budget " practical

... EIRP – 100 + Gr – Margin – Sensitivity⇒ d = 10 ^ (EIRP – 100 –1 0 – Margin – Sensitivity) /20 where Margin = 0We get d = 10 ^ (20 - 110 – Sensitivity) /20 ⇒ d = 10 ^ (– 90 – Sensitivity) / 20 (km )2. ... maximum value of the EIRP.d = 10 ^ (EIRP – 100 + Gr – Margin – Sensitivity) /20 ⇒d = 10 ^ (20 – 100 + Gr – 10 – Sensitivity) /20 ⇒d = 10 ^ (– 90 + Gr – Sensitivity) /20 (km)b) Trace for each flow, the curve ... Alibre + Gr – Marginand Px –L + Ge = EIRP = 20 dBmand Gr = 0 dBi.Thus: 20 log10 (4π / 0,000 125 ) = 100Substitutes:Sensitivity = EIRP – [20 log10(4π / 0, 125 ) + 20 log10 (d)] + Gr – Margin⇒ 20 log10...

MOB Subject 8 - WiFi Practical Traces Analyzes

... 1MOB Subject 8 - WiFi PracticalTraces Analyzes1. Configuration of a WiFi cardYou will have to find the answers ... cards ?2. Traces AnalyzeThe traces has to analyze are to download from the folder:/Infos/lmd /20 04/master/ue /mob2 005fev/Traces_sujet4/.Trace 11. Which information can you obtain from this trace on the ... named"TPMOB" in ad hoc mode or infrastructure, on a network having a shared key WEP "toto".4. What are the parameters that one can modify during the configuration of the WiFi cards ?2. ...

LINK BUDGET

... = EIRP – (100 + 20 log10d) + Gr– Margin⇒ 20 log10d = EIRP – 100 + Gr– Margin –SensitivityHay = 10EIRP – 100 + Gr– Margin – SensitivityDo Gr= 0, EIRP = 20 dBm và Margin = 10dB nên= 1 0– 90 – Sensitivity(km)Với ... baonhiêu?Giải:Afree= 20 log10(4d / λ)= 20 log10(4  / λ) + 20 log10d= 20 log10(4  / 0, 125 ) + 20 log10d= 100 + 20 log10dSensitivity = Pt– L + Gt– Afree+ Gr– Marginvà Pt– L + Gt= EIRP = 20 dBmvà Gr= 0 dBiBài toánThay ... suất, ta có:10log10(P2/P1) = 3 P2/P1= 1,9953  2 lầnAttenuation – độ suy hao• Free-space path loss (FSPL) là mất mátcông suất của tín hiệu điện từ xung quanhline-of-sight (LOS) – nó không bao gồmgain...

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MOB Subject 3 Transmission Techniques

... 1/Tbbit/sAnd Ts = log2M  Tb (with M = 2n)So: R =sT1 =bTM 2log1 =M2log1bT1 =M2log1 D.Thus, the relation between bit rate (D) and modulation rate (R) is:D = R  log2M = R  log22n = R  nDoes the ... = 2B = 6000 baudBecause a signal transmitting with four signal levels, so n = log24 = 2, then:D = R × 2 = 120 00 bps7. We need to send 26 5 kbps over a noiseless channel with a bandwidth of 20 ... - A cos (2 f0t +0) + A cos (2 f0t +0 +  /2) for a binary couple "01"; s(t) = - A cos (2 f0t +0) - A cos (2 f0t +0 +  /2) for a binary couple "00"; s(t) = A cos (2 f0t +0)...

MOB Subject 5 –Access methods Static and dynamic access techniques

... values of Kcorrespond to?i j K1 0 11 1 32 0 42 1 73 0 92 2 123 1 134 0 163 2 194 1 21 5 0 25 The first K values are: 1, 3, 4, 7, 9, 12, 13, 16, 19, 21 , 25 , ... These values correspondto the sizes ... Master of Computer Science 1 - MOB Mobile Internet and Surrounding1/ 12 Baey, Fladenmuller – Subject 5MOB Subject 5 –Access methodsStatic and dynamic ... decreases;Master of Computer Science 1 - MOB Mobile Internet and Surrounding2/ 12 Baey, Fladenmuller – Subject 5- if a cell is malfunctioning or...

MOB Subject 6 - IEEE 802.11 standards Wifi operation

... cards8 02. 11a and 8 02. 11b?No. This is not the same frequency band and not the same encoding: 5 GHz + OFDM for8 02. 11a, 2. 4 GHz + CCK for 8 02. 11b.8. The same question with 8 02. 11a and 8 02. 11g?Not ... families of protocol8 02. 11?The data modulation: BPSK to 8 02. 11, CCK (Complementary Code Keying) for 8 02. 11b andOFDM for 8 02. 11a and 8 02. 11g.Master of Computer Science 1 - MOB ... Computer Science 1 - MOB Mobile Internet and SurroundingFladenmuller Baey - ”IEEE 8 02. 11standards” 1/ 8MOB Subject 6 - IEEE 8 02. 11 standardsWifi...

MOB Subject 7 –IEEE 802.11 standards Wifi performances

... 1 92 bits and 1 Mbit/s 192overhead MAC 2 + 2 + 6 + 6 + 6 + 2 + 6 + 4 = 34 octets =27 2 bits and 54 Mbit/s5data 1500 octets = 12 000 bits 22 2Preamble 12signal extension 6SIFS 10overhead PHY 192ACK ... RTS/CTS:T’(µs) = (1 – p) 13814 (no retransmission)+ (1 – p) p (TtotalData+ 13814) (1 retransmission)+ (1 – p) p2 (2TtotalData + 13814) (2 retransmission)+ ...= 13814 + 7 92 p by neglecting the terms p2.T = ... = (1 – p) 13138 (no retransmission)+ (1 – p) p (TtotalData+ 13138) (1 retransmission)+ (1 – p) p2 (2TtotalData + 13138) (2 retransmission)+ ...= 13138 + 129 04 p by neglecting the terms p2. With...

C Sharp 2.0 Practical Guide For Programmers

... 23 1A .2. 2 Types 23 1A .2. 3 Variables 23 2A .2. 4 Expressions 23 2A .2. 5 Statements 23 3A .2. 6 Namespaces 23 5A .2. 7 Classes 23 5A .2. 8 Structs 23 7A .2. 9 Arrays 23 7A .2. 10 Interfaces 23 7A .2. 11 Enums 23 8A .2. 12 ... Attributes Using Reﬂection 22 110.3 Where to Go from Here 22 3A C# 2. 0 Grammar 22 7A.1 Lexical Grammar 22 7A.1.1 Line Terminators 22 8A.1 .2 White Space 22 8A.1.3 Comments 22 8A.1.4 Tokens 22 8A.1.5 Unicode Character ... Notation 62 Classes, Objects, and Namespaces 92. 1 Classes and Objects 1 02. 1.1 Declaring Classes 1 02. 1 .2 Creating Objects 1 12. 2 Access Modiﬁers 122 .2. 1 Controlling Access to Classes 122 .2. 2 Controlling...

MOB TME Subject 1 – Signal Transmission

... dBm)Link budget :Margin = 20 dBm + 2, 2 dBi + 2, 2 dBi - (-85 dBm) - Abwith Margin = 0equation becomes : Ab = 20 dBm + 2, 2 dBi + 2, 2 dBi - (-85 dBm)Ab = 20 + 2, 2 + 2, 2 + 85Tx and Rx at the ... log10(100 mW) + 2, 2 dBi = 20 dBm + 2, 2 dBiattenuation :Ab = 20 log10(f) + γ log10(d) + Asol(n) – 28 We get: gain (dBi) - loss (dB) - sensitivity (dBm) = 2, 2 dBi - 0 - (-85 dBm)Link budget :Margin ... = 0.We get: 20 log10(f) + γ log10(d) - 28 = 109,4 dB 20 log10(5800) + 31 log10(d) - 28 = 109,4 dB 75 .26 9 + 31 log10(d) - 28 = 109.4 dB d = 10^[(109,4 – 75 ,26 9 + 28 )/31] d...

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