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Mobile Communications Netrowks

Mobile Communications Netrowks. Tutorial 8. Prob.1. Construct 4 Walsh (Orthogonal) codes for 4 different users by two methods . Assume that 4 users transmit their data with 22Kbps (before spreading) using these codes such that: User 1 transmits +1 @ 500m User 2 transmits -1 @ 1000m

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Mobile Communications Netrowks

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  1. Mobile Communications Netrowks Tutorial 8

  2. Prob.1 • Construct 4 Walsh (Orthogonal) codes for 4 different users by two methods. Assume that 4 users transmit their data with 22Kbps (before spreading) using these codes such that: • User 1 transmits +1 @ 500m • User 2 transmits -1 @ 1000m • User 3 transmits -1 @ 1200m • User 4 transmits -1 @ 1500m • Carrier Freq.=3GHz • Path Loss Factor=2. a) Construct the transmitted and decoded (de-spread) signals for the 4 users. Assuming users 1,2,3&4 see the following attenuation L1X, L2X, L3X and L4X; where Li is the path loss factor with (n=2) plus the following Noise Level; U1: +0.5p, U2: -1p, U3: +0.5p, U4:+1p. Find the decoded signals and calculate the average bit error rate based on X. Given that 0<X<1. b) X is the attenuation factor due to the shadowing effect with Standard deviation =32dB, Find the bit error rate (BER). c) If X has Rayleigh Fading distribution instead of the shadowing effect, Find the bit error rate (BER). d) Find the Average Duration of fade in Case (c) for 100km/hr vehicle.

  3. L4 L3 L2 L1

  4. 1-a • W1=-1 • W2= • -1 • -1 • -1 • 1 • W4= • -1 • -1 • -1 • -1 W0=-1-1-1-1 W1=-1 1 -1 1 W2=-1 -1 1 1 W3=-1 1 1 -1 • -1 • 1 • -1 • 1 W8= • -1 • -1 • 1 • 1 • -1 • 1 • 1 • -1

  5. Wireless Channel 1-a Transmitter Data (2 -2 -2 -2)*L1*X 1 1 1 1 -1 -1 -1 -1 U1 -1 -1 -1 -1 Transmitted Signal (Air) Data 1 -1 1 -1 -1 -1 -1 -1 (2 -2 -2 -2)*L2*X U2 -1 1 -1 1 2 -2 -2 -2 Data 1 1 -1 -1 -1 -1 -1 -1 (2 -2 -2 -2)*L3*X U3 -1 -1 1 1 -1 -1 -1 -1 1 -1 -1 1 (2 -2 -2 -2)*L4*X U4 -1 1 1 -1

  6. (2 -2 -2 -2)*L1*X +0.5p (-2L1X-0.5 +2L1X-0.5 +2L1X-0.5+2L1X-0.5) (2L1X+0.5 -2L1X+0.5 -2L1X+0.5-2L1X+0.5) -1 -1 -1 -1 -1p (2 -2 -2 -2)*L2*X (-2L2X+1 -2L2X-1 +2L2X+1-2L2X-1) (2L2X-1 -2L2X-1 -2L2X-1-2L2X-1) -1 1 -1 1 (2 -2 -2 -2)*L3*X +0.5p (2L3X+0.5 -2L3X+0.5 -2L3X+0.5-2L3X+0.5) (-2L3X-0.5 +2L3X-0.5 -2L3X+0.5-2L3X+0.5) -1 -1 1 1 +1p (2 -2 -2 -2)*L4*X (2L4X+1 -2L4X+1 -2L4X+1-2L4X+1) (-2L4X-1 -2L4X+1 -2L4X+1+2L4X-1) Added Noise @ Rx -1 1 1 -1

  7. (-2L1X-0.5 +2L1X-0.5 +2L1X-0.5+2L1X-0.5) = +4L1X -2p>0  1 (-2L2X+1 -2L2X-1 +2L2X+1-2L2X-1) = -4L2X < 0 -1 (-2L3X-0.5 +2L3X-0.5 -2L3X+0.5-2L3X+0.5) = -4L3X < 0 -1 (-2L4X-1 -2L4X+1 -2L4X+1+2L4X+1) = -4L4X< 0 -1 • L1= • L2= • L3= • L4=

  8. As 0<X (Attenuation) <1 is always positive; = +4*253.3p X -2p>0  1 = -4*63.3p X < 0 -1 = -4*43.9p X < 0 -1 = -4*28.14p X< 0 -1 X is the attenuation factor due to the shadowing effect with Standard deviation =32dB Probability that X > 1.974*(10^-3) So: Power shadowing attenuation> ( 1.974*(10^-3))^2 So: Attenuation>-54 dB or Attenuation<54

  9. b-

  10. c- • If Rayleigh Fading is Considered instead of the shadowing effect,

  11. D-

  12. Auto correlation

  13. Cross correlation

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