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CSE 246: Computer Arithmetic Algorithms and Hardware Design

CSE 246: Computer Arithmetic Algorithms and Hardware Design. Fall 2006 Lecture 11 Cordic, Log, Square, Exponential Functions. Instructor: Prof. Chung-Kuan Cheng. Cordic Algorithm Idea. Coordinate Rotations Digital Computer Rotate vector (x,y) to (x’,y’). (x’,y’). α. (x,y).

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CSE 246: Computer Arithmetic Algorithms and Hardware Design

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

  2. Cordic Algorithm Idea • Coordinate Rotations Digital Computer • Rotate vector (x,y) to (x’,y’) (x’,y’) α (x,y)

  3. Cordic Algorithm Equations Main equations Derived equations

  4. Cordic Algorithms • Key: Given cos α, sin α, tan α we can derive

  5. Cordic Algorithms (Example) • Find

  6. Cordic Algorithm Example Continued

  7. Logarithms – Method 1 • Find

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

  9. 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 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

  10. 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

  11. Logarithms – Method 2 • Let define • Initially x<2, ie. y0=0 • If

  12. Logarithms – Method 2 for i = 1 to l do x = x2 if x ≥ 2 then yi = 1 x = x/2 else yi = 0

  13. Logarithms – Method 2 (Example) • Find ln2(x), x = 1.11 (1.75) x2 1.1 1 x 1.1 1 1 1 1 1 1 1 + 1 1 1 __ 1 1 0 0 0 1 y1 = 1 x2/2 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 1 0 0 0 1 _ 1 0.0 1 0 1 1 0 0 0 0 1 y2 = 1

  14. Logarithms – Method 2 (Example) (x2/2)2/2 = 1.00101100001 y3 = 0 ln2 1.11 ≈ 0.110

  15. 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 _

  16. Exponentiation ex

  17. Exponentiation ex • I. • II. • min: • max:

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