ECE405HW7Su2010Sol

ECE405HW7Su2010Sol - ECE 405 HOMEWORK #7 Solutions SU 2010...

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ECE 405 HOMEWORK #7 Solutions SU 2010 ©DR. JAMES S. KANG 1 (a) 2 8 = 256, 32/256 = 0.125 = S = step size, –8.7 – (–16) = 7.3, 7.3/0.125 = 58.4. Closest integer to 58.4 is 58. Binary representation of 58 is obtained by keep dividing 58 by two and taking the remainders. The result of process is 0 0 1 1 1 0 1 0. Notice that two zeros are prefixed to make the code length of 8 bits. (b) Sign and magnitude. 8.7/0.125 = 69.6. Binary representation of 70 is 1 0 0 0 1 1 0. Thus sign and magnitude representation is 1 1 0 0 0 1 1 0. The first bit is a sign bit. (c) 10111010. (d) 10111001. 2 (a) ln 1 ( ) sgn( ) ln(1 ) p m m Qm m μ ⎛⎞ + ⎜⎟ ⎝⎠ = + 8.7 ln 1 255 16 0.890717041174 ln(1 255) + = + 0.890717041174 0.890717041174 16 14.2514726588 p m ×= × = 1.7485273412 13.988219 0.125 = where 0.125 is the step size S obtained by S = 32/256. Binary representation of 14 is 00001110. (b) 14.2514726588 114.01178 0.125 = where 0.125 is the step size S obtained by S = 16/128. Binary representation of 114 in 7 bits is 1110010. Including the sign bit 1, the sign and magnitude code is 11110010. (c) 10001110. (d) 10001101. 3 (a) (SNR) o = 6n dB = 6 × 11 dB = 66 dB. (b) μ = 255, L = 2048. () [] 2 2 2 2 0 2 22 3 2 ) 1 m p m pp m L SNR m mm σ = + ++ Where 1
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2 2 2 2 1 ,| | , 32 pp m m p mm m m σ == 3 = Thus () [] 2 2 0 2 1 3 2048 3 404437.0507 56.07 12 1 ln(1 255) 3 255 2 255 SNR dB + ++ × = 4 (a) (i) For Gaussian, 2 m m π = 2 2 22 0 2 2 3 3 2048 0.000391 2 2 ln(1 ) ln(1 255) 1 2 0.1 1 0.000391 255 16 255 m p m m L SNR m μ μμ × × = 301856.5627 = 54.798 dB (ii) For Laplacian, 2 m m = 2 2 0 2 2 3 3 2048 0.000391 1 2 ) ln(1 255) 1 20 . 1 1 2 0.000391 255 16 255 m p m m L SNR m × × = 310078.2 = 54.9147 dB (b) 2 2 0 () 3 2 0 .1 4915.2 36.9154 16 12 n p mt SNR dB S m ×× = = = 5 ln 1
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ECE405HW7Su2010Sol - ECE 405 HOMEWORK #7 Solutions SU 2010...

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