Điện tử số part 1

... 0 0 0 1 0 0 0 1 0 1 0 0 1 0 0 1 0 0 1 1 0 1 0 1 0 0 0 1 0 1 0 1 1 1 0 1 1 0 0 1 0 1 1 1 1 1 1 0 0 0 0 1 1 0 0 1 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 2/24 © ... 1 1 1 1 1 0 0 1 1 1 1 1 0 x y z w F 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 1 0 1 1 0...

Ngày tải lên: 08/05/2014, 14:29

... blocks 3 /12 © R.Lauwereins Imec 20 01 Digital design Combina- torial circuits Sequential circuits FSMD design VHDL Gated SR Latch C S R Q(next) 0 0 0 Q 0 0 1 Q 0 1 0 Q 0 1 1 Q 1 0 0 Q 1 0 1 0 1 1 0 1 1 ... positive edge-triggered Negative edge-triggered S R Q(next) 0 0 Q 0 1 0 1 0 1 1 1 NA Characteristic table (for design of SR flip-flop) Q Q(next) S R 0 0 0 x 0 1...

Ngày tải lên: 08/05/2014, 14:31

... 2 Instruction format RF OE1 RE1RA0RA1WA1R RA2WEWA0WA2SCOECL Register File Read Port 1 Register File Write Port Counter 012 345678 910 111 213 1 415 1 617 18 1920 212 22324252627282930 31 32-bit instruction word For ... Data = Data >> 1 ENDWHILE Outport = OCnt 0 1 2 3 4 5 Comp x00000 Update 010 100 Load 11 1x00 s =1 Temp x00 010 z=0 Out x000 01 z =1 s=0 s Data Data = Inpor...

Ngày tải lên: 08/05/2014, 14:32

... component Comparator; begin Compare1: component Comparator port map (In1,In2,Out1); Compare2: component Comparator port map (In1,In3,Out2); end architecture Struct1; Two instantiations of the same ... like in C /* Dual comparator Test program */ main() { int In1, In2, In3; int Out1, Out2; Out1 = Compare(In1, In2); Out2 = Compare(In1, In3); } Two calls to the function ‘Compare’ with its argume...

Ngày tải lên: 08/05/2014, 14:32

... phân 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0000 00 01 0 010 0 011 010 0 010 1 011 0 011 1 10 00 10 01 1 010 10 11 110 0 11 01 111 0 11 11 00 01 02 03 04 05 06 07 10 11 12 13 14 15 16 17 0 1 2 3 4 5 6 7 8 9 A B C D E F ng 1. 1. Các t hp mã nh phân 4 ... hoc thp c phân hoc bát phân. Ví d 1. 3 : Xét vic biu din s nh phân 10 111 110 111 111...

Ngày tải lên: 15/10/2012, 15:28

... là phng f(x 1 ,x 2 ) x 1 x 2 0 1 0 1 f x 1 x 2 x 3 0 1 00 01 11 10 f x 1 x 2 x 3 x 4 00 01 11 10 00 01 11 10 f x 2 x 3 x 4 x 5 00 01 11 10 00 01 11 10 10 11 01 00 x 1 =0 x 1 =1 Chng 2. i ... f(x 1 ,x 2 ,x 3 ,x 4 ) = x 1 + x 4 00 01 11 10 00 X X 1 X 01 X 0 1 X 11 0 X X 1 10 1 1 X 1 00 01 11 10 00 X X 1 X 01 X 0 1 X 11 0 X X...

Ngày tải lên: 15/10/2012, 15:28

... 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0  bng khai trin trên ta xây dng c bng u vào kích cho JKFF nh sau: Q n Q n +1 S n R n 0 0 0 X 0 1 1 X 1 0 X 1 1 1 X 0  th ... x x ⊕ x = 0 x ⊕ x = 1 Hình 3 .16 e. S dng cng NOR làm cng NAND y = 212 1 21 .1 xxxxxxy =+=+= x 1 x 2 y 1 1 x 2 x x 1 x 2 yy x 1 x 2 y 0 0 0 0 1 1 1...

Ngày tải lên: 15/10/2012, 15:28

... 1 1 1 1 1 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0 1 0 1 1 0 1 1 0 1 0 0 1 1 1 1 1 1 0 0 1 0 1 0 0 0 1 1 0 0 1 1 0 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 0 0 0 0 1 0 0 0 1 1 1 1 1 1 1 1 ... 0 01 0 1 x 0 11 0 0 x x 10 0 1 x x 00 01 11 10 00 0 0 x 0 01 0 0 x 0 11 0 0 x x 10 1 0 x x 00 01 11 10 00 0 1...

Ngày tải lên: 15/10/2012, 15:28

... 1 1 1 1 1 1 1 1 1 1 1 1 1 1 F F F H ROM 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 H 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 F F F H RAM 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 0 0 0 H 0 1 0 1 1 1 1 1 1 1 ... 1 1 1 1 1 1 5 F F F H RAM 2 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 H 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 7 F F F H RAM 3 1 0 0 0 0 0...

Ngày tải lên: 15/10/2012, 15:28

... là phng f(x 1 ,x 2 ) x 1 x 2 0 1 0 1 f x 1 x 2 x 3 0 1 00 01 11 10 f x 1 x 2 x 3 x 4 00 01 11 10 00 01 11 10 f x 2 x 3 x 4 x 5 00 01 11 10 00 01 11 10 10 11 01 00 x 1 =0 x 1 =1 Chng 2. i ... f(x 1 ,x 2 ,x 3 ,x 4 ) = x 1 + x 4 00 01 11 10 00 X X 1 X 01 X 0 1 X 11 0 X X 1 10 1 1 X 1 00 01 11 10 00 X X 1 X 01 X 0 1 X 11 0 X X...

Ngày tải lên: 09/01/2014, 15:48