eee206_2005_sols - Numerical Solutions to EEE206, 2004-5...

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Numerical Solutions to EEE206, 2004-5 Session, B. Chambers Q1. (c) (i) Carrier power = 10 3 = E C 2 /2 x 50 . E C = 316.2 V Then lower sideband amplitude = mE C /2 = 0.6 x 316.2/2 = 94.9 V since m=0.6 when V in = 5V. (ii) Total sideband power = 2 x 316.2 2 x 0.6 2 /8 x 50 = 180 W Ratio of SB to carrier power = 180 / 1000 = 0.18 (also from m 2 /2) (iii) Since the transfer characteristic of the modulator is linear, if input amplitude reduced from 5V to 3V = 0.6 then m decreases from 0.6 to 0.6 x 0.6 = 0.36 Output power is now 1000(1+0.36 2 /2) = 1065 W Q2. (c)(i) m = 1. Then from tables supplied, J 0 (1) = 0.7652, J 1 (1) = 0.4401. When m = 0, J 0 (0) = 1, all J n (0)=0. Then carrier power developed across 1 Ω resistor = E c 2 J 0 2 (0)/2 = E c 2 /2 When m = 1, power in carrier and first sideband pair = E c 2 J 0 2 (1)/2 + E c 2 J1 2 (1)/4 = 0.7652 2 /2 + 0.4401 2 /4. Then fractional power = 0.341/.5 = 0.682 = 68.2% . (ii) When m = 1, from tables supplied, number of significant sideband pairs = 4. The sideband separation = message frequency = 3 kHz. Hence bandwidth = 4 x 2 x 3 = 24 kHz (iii) noise induced carrier phase shift φ = sin -1 (N/S) = sin -1 (1/3) = 0.3398 rads (19.47 degs).
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This note was uploaded on 09/25/2008 for the course EEE 206 taught by Professor Langley during the Spring '08 term at University of Sheffield.

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eee206_2005_sols - Numerical Solutions to EEE206, 2004-5...

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