pcshw8_soln - THE STATE UNIVERSITY OF NEW JERSEY RUTGERS...

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THE STATE UNIVERSITY OF NEW JERSEY RUTGERS College of Engineering Department of Electrical and Computer Engineering 332:322 Principles of Communications Systems Spring 2004 Problem Set 8 Haykin: 3.7, 3.9-3.15 1. Baby Quantization, Just in Case: This is a simple problem, but it’s worth doing just to make sure everyone is in synch. Suppose you have (for t 0 ) a periodic signal s ( t ) = u - 2 ( t ) + 2 X k =1 ( - 1) k u - 2 ( t - 2 k + 1) for t 0 where u - 2 ( t ) is the unit ramp function. (a) Sketch s ( t ) over one cycle. SOLUTION: refer to figure 1, where two cycles are shown. (b) What might the PDF of s ( t ) look like (use your instincts – you do NOT have the formal machinery to deal with this just yet). SOLUTION: Uniform on ± 1 (c) Q () is a one bit quantizer with x 0 = 0 , q 0 = - 1 , q 1 = 1 . Sketch he quantized signal Q ( s ( t )) and the error signal s ( t ) - Q ( s ( t )) . Assume s () is uniform on ± 1 . Is Q () an optimum one bit quantizer for s ( t ) ? If not, what IS the optimum one bit quantizer for s () ? SOLUTION: Analytically we have Q ( s ( t )) = X k =0 u - 1 ( t - 2 k + 1) for t 0 . The Lloyd-max conditions state that x 0 = ( q 1 + q 0 ) / 2 which is true. They also state that q 0 = E [ X | X < 0] which is NOT true (should be - 1 / 2 ). It’s symmetric so the optimal q 1 = 1 / 2 . So it was NOT an optimal quantizer. (d)
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This homework help was uploaded on 04/03/2008 for the course ECE ECE332 taught by Professor Rose during the Spring '08 term at Rutgers.

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pcshw8_soln - THE STATE UNIVERSITY OF NEW JERSEY RUTGERS...

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