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

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

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

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

4           Cordic Algorithms

5  Key: Given cos α, sin α, tan α we can derive Cordic Algorithms iαiαi

6  Find  Cordic Algorithms (Example)

7 Cordic Algorithms        

8 Logarithms – Method 1  Find 

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

10 Logarithms – Method 1 (Example)  Find ln(x), x = = _ x _ _ x _

11 Logarithms – Method 1 (Example)  -ln x = (1.-1) + ln(1.01) + ln( )

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

13 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

14 Logarithms – Method 2 (Example) x x __ y 1 = 1 x 2 / x _ y 2 = 1  Find ln 2 (x), x = 1.11 (1.75)

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

16 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 _

17 Exponentiation e x        

18  I.  II.  min:  max:


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