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hw2_solution

# hw2_solution - 2(1 SNR We have W = 300 Hz(SNR dB = 3...

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3.13 a. (30 pictures/s) (480 × 500 pixels/picture) = 7.2 × 10 6 pixels/s Each pixel can take on one of 32 values and can therefore be represented by 5 bits: R = 7.2 × 10 6 pixels/s × 5 bits/pixel = 36 Mbps b. We use the formula: C = B log 2 (1 + SNR) B = 4.5 × 10 6 MHz = bandwidth, and SNR dB = 35 = 10 log 10 (SNR), hence SNR = 10 35/10 = 10 3.5 , and therefore C = 4.5 × 10 6 log 2 (1 + 10 3.5 ) = 4.5 × 10 6 × log 2 (3163) C = (4.5 × 10 6 × 11.63) = 52.335 × 10 6 bps 3.15 Using Shannon's equation: C = B log
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Unformatted text preview: 2 (1 + SNR) We have W = 300 Hz (SNR) dB = 3 Therefore, SNR = 10 0.3 C = 300 log 2 (1 + 10 0.3 ) = 300 log 2 (2.995) = 474 bps 3.16 Using Nyquist's equation: C = 2B log2M We have C = 9600 bps a. log 2 M = 4, because a signal element encodes a 4-bit word Therefore, C = 9600 = 2B × 4, and B = 1200 Hz b. 9600 = 2B × 8, and B = 600 Hz Additional Problem: a.-+ -00- + 000+ -+ -00- + b.-+ -00- +00+ -00-...
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