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CSE 246: Computer Arithmetic Algorithms and Hardware Design Instructor: Prof. Chung-Kuan Cheng Fall 2006 Lecture 11 Cordic, Log, Square, Exponential Functions

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Project IEEE Computer Society Author Kit (≤5Pages) Introduction Statement of Problem Approaches Examples Experiments Conclusion 30 Minutes

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Cordic Algorithms Coordinate Rotations Digital Computer Rotate vector (x,y) to (x’,y’) α (x’,y’) (x,y)

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Cordic Algorithms

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Key: Given cos α, sin α, tan α we can derive Cordic Algorithms iαiαi 045 126.6 214 37.1 43.6 51.8 60.9 70.4 80.2 90.1

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Find Cordic Algorithms (Example)

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Cordic Algorithms

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Logarithms – Method 1 Find

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Logarithms – Method 1 I. II. III. A table of

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Logarithms – Method 1 (Example) Find ln(x), x = 1.625 1+0.5+0.125=1.625 1. 1 0 1 1.-1 _ -1 -1 0 -1 x 1 1 0 1 _ 0. 1 1 0 1 1.0 1 _ 0 1 1 0 1 x 0 1 1 0 1 _ 1.0 0 0 0 1

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Logarithms – Method 1 (Example) -ln x = (1.-1) + ln(1.01) + ln(1.0000-1) 1. 0 0 0 0 0 1 1. 0 0 0 0 0-1 1. 0 0 0 0 0 0 0 0 0 0

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Logarithms – Method 2 Let define Initially x<2, ie. y 0 =0 If

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Logarithms – Method 2 for i = 1 to l do x = x 2 if x ≥ 2 then y i = 1 x = x/2 else y i = 0

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Logarithms – Method 2 (Example) x 2 1.1 1 x 1.1 1 1 1 1 + 1 1 1 __ 1 1 0 0 0 1 y 1 = 1 x 2 /2 1.1 0 0 0 1 x 1.1 0 0 0 1 1 1 0 0 0 1 + 1 1 0 0 0 1 _ 1 0.0 1 0 1 1 0 0 0 0 1 y 2 = 1 Find ln 2 (x), x = 1.11 (1.75)

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Logarithms – Method 2 (Example) (x 2 /2) 2 /2 = 1.00101100001 y 3 = 0 ln 2 1.11 ≈ 0.110

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Squarer x3 x2 x1 x0 X x3 x2 x1 x0 x3x0 x2x0 x1x0 x0x0 x3x1 x2x1 x1x1 x0x1 x3x2 x2x2 x1x2 x0x2 + x3x3 x2x3 x1x3 x0x3 _ x3x2 x3x1 x3x0 x2x0 x1x0 x0 x3 x2x1 x1 + x2 _

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Exponentiation e x

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I. II. min: max:

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June 2007 Computer Arithmetic, Function EvaluationSlide 1 VI Function Evaluation Topics in This Part Chapter 21 Square-Rooting Methods Chapter 22 The CORDIC.

June 2007 Computer Arithmetic, Function EvaluationSlide 1 VI Function Evaluation Topics in This Part Chapter 21 Square-Rooting Methods Chapter 22 The CORDIC.

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