Tài liệu Interpolation and Extrapolation part 2 doc
... extreme right. For example, with N =4, x 1 : y 1 =P 1 P 12 x 2 : y 2 = P 2 P 123 P 23 P 123 4 x 3 : y 3 = P 3 P 23 4 P 34 x 4 : y 4 = P 4 (3.1 .2) Neville’s algorithm is a recursive way of filling in ... x N ) (x 2 − x 1 )(x 2 − x 3 ) (x 2 − x N ) y 2 + ···+ (x−x 1 )(x − x 2 ) (x − x N−1 ) (x N − x 1 )(x N − x 2 ) (x N − x N−1 ) y N (3.1.1) There are N terms, each a...
Ngày tải lên: 26/01/2014, 15:20
... extremely useful (§3 .2) . Trigonometric functions, sines and cosines, give rise to trigonometric interpolation and related Fourier methods, which we defer to Chapters 12 and 13. There is an extensive ... function f(x)=3x 2 + 1 π 4 ln (π − x) 2 +1 (3.0.1) 105 106 Chapter 3. Interpolation and Extrapolation Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC CO...
Ngày tải lên: 21/01/2014, 18:20
... floods of applications filed by their UK counterparts to organize tourist groups to visit Japan. (A) to handle (B) handles (C) handling (D) handle 107. Product _______ should have a bachelor's ... (D) was found Please let me know the arrangements for handing back equipment, company car, etc, and handing over outstanding work and responsibilities. Yours sincerely...
Ngày tải lên: 13/12/2013, 18:15
Tài liệu 50 harvard essays part 2 docx
... know I will do and be something. And the determination and success of my great-grandmother will help me reach this something. She is more than a memory or a story, she has become a part of me: ... left. She has brown hair that falls to her shoulders and deep blue eyes. She is wearing a white blouse and blue jeans and in her right hand is a pair of binoculars. The stage is da...
Ngày tải lên: 14/12/2013, 18:15
Tài liệu Modules and Ports part 1 docx
... understand the components of a module shown above, let us consider a simple example of an SR latch, as shown in Figure 4 -2 . Figure 4 -2. SR Latch The SR latch has S and R as the input ports and ... primitive nand gates // Note, how the wires are connected in a cross-coupled fashion. nand n1(Q, Sbar, Qbar); nand n2(Qbar, Rbar, Q); // endmodule statement endmodule // Modul...
Ngày tải lên: 15/12/2013, 03:15
Tài liệu Modules and Ports part 2 pptx
... b, and c_in and produces an output on ports sum and c_out. Thus, module fulladd4 performs an addition for its environment. The module Top is a top-level module in the simulation and does not ... does not have a list of ports. The module names and port lists for both module declarations in Verilog are as shown in Example 4 -2 . Example 4 -2 List of Ports module fulladd4(sum,...
Ngày tải lên: 15/12/2013, 03:15
Tài liệu Timing and Delay part 2 ppt
... ({c,d} == 2& apos;b01) (c,d *> out) = 11; if ({c,d} != 2& apos;b01) (c,d *> out) = 13; endspecify and a1(e, a, b); and a2(f, c, d); and a3(out, e, f); endmodule Rise, fall, and turn-off ... endspecify and a1(e, a, b); and a2(f, c, d); and a3(out, e, f); endmodule The full connection is particularly useful for specifying a delay between each bit of an...
Ngày tải lên: 15/12/2013, 03:15
Tài liệu Timing and Delay part 3 doc
... input timing constraints can also be specified. 10.3.1 $setup and $hold Checks $setup and $hold tasks are used to check the setup and hold constraints for a sequential element in the design. ... IEEE Standard Verilog Hardware Description Language document. The various steps in the flow that use delay back-annotation are as follows: 1. The designer writes the RTL description and...
Ngày tải lên: 15/12/2013, 03:15
Tài liệu Modeling of Data part 2 doc
... 1964, Mathematical Theory of Probability and Statistics (New York: Academic Press), Chapter X. Korn, G.A., and Korn, T.M. 1968, Mathematical Handbook for Scientists and Engineers , 2nd ed. (New York: McGraw-Hill), ... logarithm, namely, N i=1 [y i − y(x i )] 2 2σ 2 − N log ∆y (15.1.4) Since N, σ ,and yare all constants, minimizing this equation is equivalent to minimizing (15....
Ngày tải lên: 15/12/2013, 04:15
Tài liệu Evaluation of Functions part 2 doc
... familiar examples are the sine function and the Bessel function of the first kind, sin x = ∞ k=0 (−1) k (2k +1)! x 2k+1 (5.1 .2) J n (x)= x 2 n ∞ k=0 (− 1 4 x 2 ) k k!(k + n)! (5.1.3) Both of ... u 0 − u 1 + u 2 −u n−1 + ∞ s=0 (−1) s 2 s+1 [∆ s u n ](5.1.5) Here ∆ is the forward difference operator, i.e., ∆u n ≡ u n+1 − u n ∆ 2 u n ≡ u n +2 − 2u n+1 + u n ∆ 3 u n ≡ u n+...
Ngày tải lên: 15/12/2013, 04:15