K THUT S - 5Cu 1: (2) Cho hm Bool f(A,B,C,D)= a. Biu din hm f di dng bng s tht. b. Thc hin hm f bng ba K v rt gn. c. Thc hin hm f bng cng NAND 2 ng vo. Cu 2: (3)a. Thit k MUX 4 1 c s khi cho hnh 1 sau:

+ d(1,2,3,6,7,14)

b. Dng MUX 4 1 thit k cu a,thc hin mch cng ton phn FA c s

khi nh hnh 2 trn (ch cn dng cc khi hnh 1 minh ha cc kt ni). Cu 3: (2) Xy dng 1 decoder (mch gii m ) 3 8 ch dng cc decoder 1 2 c s khi nh hnh 3 (khng c dng cc thit b khc k c cng o). nh nhn cc ng vo l A0,A1,A2. Cc ng ra l D0, D1, D2, D3, D4, D5, D6, D7. (V d nu A0A1A2 = 110 th D6 = 1 v tt c cc cng khc bng 0). Cu 4: (2) Thit k mch m ln Mod 10 dng JK FF c xung CK kch cnh xung, s dng phng php dng ng Clear. (Cc bc thit k gm: Xc nh FF, bng s tht, s mch, gin xung v dy m).

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Bi lmCu 1: Biu din hm f di dng bng s tht:

Cu 3: Xy dng Decoder 3 8 t Decoder 1 2

2

Cu 2: a. Thit k MUX 4 1 Ta c bng s tht nh sau:

b. Dng MUX 4 1 thc hin cng ton phn FA Bng s tht ca FA3

Cu 4: - Bc 1: Xc nh FF : Ta c 2n-1 < M < 2n vi M=10 n =4. - Bc 2: Bng s tht Xung CK Q3 Q2 Q1 Q0 K

4

0 1 2 3 4 5 6 7 8 9 10

0 0 0 0 0 0 0 0 1 1 1/0

0 0 0 0 1 1 1 1 0 0 0

0 0 1 1 0 0 1 1 0 0 1/0

0 1 0 1 0 1 0 1 0 1 0

1 1 1 1 1 1 1 1 1 1 0

T bng s tht ta c:

-

Bc 3: S mch:

- Bc 4: Gin xung

5

- Bc 5: Dy m 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 0000 . . .

6