HW3S - ECE 402 Communications Engineering Homework 3...

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ECE 402: Communications Engineering Homework 3 Solutions (Fall 2009) 1. The Nyquist rate is 2 W = 200 Hz. The sampling rate is 50 % larger: f s = 1 . 5 · 200 = 300 Hz. The samples are limited to the range ± V = ± 1, so the number of bits k needed in the quantizer satisﬁes 4 . 8 + 6 k + 10 log 10 ( P x /V 2 ) 60 , or k 10 . 7. Thus we need at least k = 11 bits. The resulting minimum data rate is R = kf s = 3 . 3 kbps for each channel, or a total of 6.6 kbps for both channels. 2. The combined data rate of the two tracks is at most 10 Mbps, so each track must contribute at most 5 Mbps. To avoid aliasing, each stereo track must be sampled at a rate not less than f s > 2 W = 40 kHz. The data rate of each track is R = kf s 5 Mbps, so k 5 × 10 6 bps 4 × 10 4 Hz = 125 bits . The resulting peak SNR is therefore (SNR) peak = 4 . 8 + 6 k 754 . 8 dB . A very big number! 3. (a) The distribution function is given by F X ( x ) = Z x -∞ f X ( ξ ) = ( R x 0 2 ξ exp( - ξ

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This note was uploaded on 01/28/2012 for the course ECE 402 taught by Professor Townsend during the Fall '08 term at N.C. State.

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HW3S - ECE 402 Communications Engineering Homework 3...

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