s4 - 16 ECE 253a Digital Image Processing Pamela Cosman...

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# 16 ECE 253a Digital Image Processing Pamela Cosman 11/11/10 SOLUTIONS FOR HOMEWORK 4 1. Scalar Quantization: Comparing pdfs and distortions of binary quantizers: You are given an input signal X to quantize. It has a pdf given by f X ( x ) = A - x, 0 x 1 A + x, - 1 x 0 0 , otherwise A 2-level quantizer for this input signal has output levels y 1 = - 1 2 and y 2 = 1 2 . (a) To Fnd the constant A: i 1 - 1 f X ( x ) dx = i 0 - 1 ( A + x ) dx + i 1 0 ( A - x ) dx = 1 p Ax + 1 2 x 2 Pv v v v 0 - 1 + p Ax - 1 2 x 2 Pv v v v 1 0 = 2 A - 1 = 1 and A = 1 (b) By symmetry, the distribution of the error should be uniformly distributed from - 1 2 to 1 2 . Mathematically, we start from Fnding the CD± of the error: Pr [ ε α ] = Pr b - 1 < X < 0 , ( - 1 2 - X ) < α B + Pr b 1 > X > 0 , ( 1 2 - X ) < α B = i 0 - 1 2 - α (1 + x ) dx + i 1 1 2 - α (1 - x ) dx = α + 1 2 for - 1 2 < α < 1 2 , and the PD± is: f ε ( α ) = d Pr[ ε < α ] = 1 for - 1 2 < α < 1 2 . (c) The distortion is: D = E ( ε 2 X ) = i - 1 2 - 1 2 x 2 × 1 dx = 1 12 (d) If the quantization levels are - 1 3 and 1 3 , the distortion is: D = E ( ε 2 X ) = i 0 - 1 p x + 1 3 P 2 (1 + x ) dx + i 1 0 p x - 1 3 P 2 (1 - x ) dx = 1 18 which is smaller than the distortion in part (c). This is because the probability density of X is concentrated near 0, and so it is better if we design the absolute value of our quantization level smaller than 1 2 . 1

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(e) Since the distortion is defned as the square oF the quantizer error, the sign oF the error wouldn’t matter, and the distortion will be the same. To fnd the new PD± oF the new quantizer error: Pr [ ε α ] = Pr b - 1 < Z < 0 , ( - 1 2 - Z ) < α B + Pr b 1 > Z > 0 , ( 1 2 - Z ) < α B = i 0 - 1 2 - α (1 + z ) dz + i 1 1 2 - α zdz = 3 4 + α - α 2
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s4 - 16 ECE 253a Digital Image Processing Pamela Cosman...

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