1. Tutorial 8 Mobile Communications Netrowks

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. L1 L2 L3 L4

4. 1-a W1=-1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 -1 1 1 1 -1 1 1 -1 -1 1 -1 -1 1 W0=-1-1-1-1 W1=-1 1 -1 1 W2=-1 -1 1 1 W3=-1 1 1 -1 W8= W2= 1 W4= 1 1 1 11 1

5. 1-a Wireless Channel U1 U2 U3 Data U4 Transmitted Signal (Air) -1 -1 -1 1 -1 -1 1 1 1 1 -1 -1 1 -1 1 1 -1 -1 1 -1 -1 1 2 -2 -2 -2 Transmitter (2 -2 -2 -2)*L 1 *X -1 1 1 -1 (2 -2 -2 -2)*L 2 *X (2 -2 -2 -2)*L 3 *X (2 -2 -2 -2)*L 4 *X

6. +0.5p -1p +0.5p +1p Added Noise @ Rx (2 -2 -2 -2)*L 1 *X (2 -2 -2 -2)*L 2 *X (2 -2 -2 -2)*L 3 *X (2 -2 -2 -2)*L 4 *X (2L 1 X+0.5 -2L1X+0.5 -2L1X+0.5 -2L1X+0.5) (2L2X-1 -2L2X-1 -2L2X-1 -2L2X-1) -1 -1 -1 -1 -1 1 -1 1 -1 -1 1 1 -1 1 1 -1 (2L 3 X+0.5 -2L3X+0.5 -2L3X+0.5 -2L3X+0.5 ) (2L4X+1 -2L4X+1 -2L4X+1 -2L4X+1) (-2L 1 X-0.5 +2L1X-0.5 +2L1X-0.5 +2L1X-0.5) (-2L2X+1 -2L2X-1 +2L2X+1 -2L2X-1) (-2L 3 X-0.5 +2L3X-0.5 -2L3X+0.5 -2L3X+0.5) (-2L4X-1 -2L4X+1 -2L4X+1 +2L4X-1)

7. = +4L 1 X -2p>0  1 = -4L 2 X < 0  -1 = -4L 3 X < 0  -1 = -4L 4 X< 0  -1 L1= L2= L3= L4= (-2L 1 X-0.5 +2L1X-0.5 +2L1X-0.5 +2L1X-0.5) (-2L2X+1 -2L2X-1 +2L2X+1 -2L2X-1) (-2L 3 X-0.5 +2L3X-0.5 -2L3X+0.5 -2L3X+0.5) (-2L4X-1 -2L4X+1 -2L4X+1 +2L4X+1)

8. X is the attenuation factor due to the shadowing effect with Standard deviation =32dB = +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 As 0<X (Attenuation) <1 is always positive; 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-

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

12. D-

13. Auto correlation

15. Cross correlation