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 – Sensit...

Ngày tải lên: 17/09/2012, 09:13

... 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 n...

Ngày tải lên: 17/09/2012, 09:13

... = 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–...

Ngày tải lên: 13/09/2012, 10:20

... 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 bandwidt...

Ngày tải lên: 17/09/2012, 09:13

... 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...

Ngày tải lên: 17/09/2012, 09:13

... 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...

Ngày tải lên: 17/09/2012, 09:13

... 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...

Ngày tải lên: 17/09/2012, 09:13

... 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 Wh...

Ngày tải lên: 20/08/2012, 11:57

... dBm) Link budget : Margin = 20 dBm + 2, 2 dBi + 2, 2 dBi - (-85 dBm) - Ab with Margin = 0 equation becomes : Ab = 20 dBm + 2, 2 dBi + 2, 2 dBi - (-85 dBm) A b = 20 + 2, 2 + 2, 2 + 85 Tx and Rx at the ... log 10 (100 mW) + 2, 2 dBi = 20 dBm + 2, 2 dBi attenuation : Ab = 20 log 10 (f) + γ log 10 (d) + A sol (n) – 28 We get: gain (dBi) - loss (dB) - sensitivity (dBm) = 2...

Ngày tải lên: 17/09/2012, 09:13