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Exercise-5.sol

# Exercise-5.sol - Q 29 2 = x x 3 10-= 2 N E Q b 88 2 = ⇒ x...

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THE UNIVERSITY OF WESTERN ONTARIO FACULTY OF ENGINEERING DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING ECE-433b-COMMUNICATION SYSTEMS Exercise # 5 (Solutions) 1. Given: dt e x erfc x t - = 2 2 ) ( π and - = x t dt e x Q 2 / 2 2 1 ) ( π Let z t = 2 , then dz dt 2 = and when x t = , 2 x z = and when = t , = z . Substituting these, we get, - = 2 2 2 1 ) ( 2 x z dz e x Q π = 2 2 1 x erfc 2. For coherent binary PSK, the error probability is given by: = 0 2 1 N E erfc P b e For error probability 4 10 - = e P , 0 N E b works out to 7.0 , which implies 5 . 3 = b E x 10 10 - . Therefore the average power of the carrier is 0.35mWatts. 3. The energy of ) ( 1 t s =the energy of ) ( 2 t s = ( 29 ( 29 J E b 00125 . 0 01 . 0 2 5 . 0 2 = = = T b dt t s t s E 0 2 1 ) ( ) ( 1 ρ ( 29 ( 29 dt t t E T b = 0 1010 2 cos 5 . 0 1000 2 cos 5 . 0 1 π π ρ ( 29 ( 29 [ ] dt t t + = 01 . 0 0 2010 2 cos 10 2 cos 2 1 00125 . 0 25 . 0 π π ρ ( 29 ( 29 01 . 0 0 4020 2010 2 sin 20 10 2 sin 100 + = π π π π t t

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94 . 0 005 . 0 935 . 0 = + = ρ - = 0 ) 1 ( N E Q P b e ρ ( 29 27 . 0 612 . 0 0002 . 0 ) 06 . 0 ( 00125 . 0 = = =
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Unformatted text preview: Q ( 29 . 2 = x x 3 10-= 2 N E Q b 88 . 2 = ⇒ x 1472 . 4 = ⇒ N E b With phase error, the error probability is Q ( 29 5 . 2 = y x 3 10-= -) 1 ( 2 N E Q b 81 . 2 = ⇒ y 81 . 2 ) 1 ( =-⇒ N E b θ cos 90396 . =-= ⇒ 1 7 . 154 cos = =-. Therefore, the angle between ) ( 1 t s and ) ( 2 t s is 154.7 degrees. The optimum antipodal signaling is for 2 ) 1 ( =-or =180 degrees. Therefore, phase error 3 . 25 7 . 154 180 =-= φ ....
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Exercise-5.sol - Q 29 2 = x x 3 10-= 2 N E Q b 88 2 = ⇒ x...

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