Nguyn Trng Lut BM in T - Khoa in-in T - H Bch Khoa TP. HCM

1

BI TP C LI GII PHN 1 MN K THUT S

B mn in t i Hc Bch Khoa TP.HCM

Cu 1 Cho 3 s A, B, v C trong h thng s c s r, c cc gi tr: A = 35, B = 62, C = 141.

Hy xc nh gi tr c s r, nu ta c A + B = C.

Cu 2 S dng tin v nh l: a. Chng minh ng thc: A B + A C + B C + A B C = A C

b. Cho A B = 0 v A + B = 1, chng minh ng thc A C + A B + B C = B + C

nh ngha gi tr: A = 3r + 5, B = 6r +2, C = r2 + 4r + 1 A + B = C (3r + 5) + (6r + 2) = r2 + 4r + 1

PT bc 2: r2 - 5r - 6 = 0 r = 6 v r = - 1 (loi) H thng c s 6 : tuy nhin kt qu cng khng hp l v B = 62: khng phi s c s 6

VT: A C + A B + B C = (A + B) C + A B ; A + B = 1 = C + A B = C + A B + A B ; A B = 0 = C + ( A + A ) B = B + C : VP

VT: A B + A C + B C + A B C = B ( A + A C) + A C + B C = B ( A + C ) + A C + B C ; x + x y = x + y = A B + B C + A C + B C = A B + A C + C ( B + B ) = A B + A C + C = A B + A + C = A ( B + 1) + C = A + C = A C : VP

Nguyn Trng Lut BM in T - Khoa in-in T - H Bch Khoa TP. HCM

2

Cu 3 a. Cho hm F(A, B, C) c s logic nh hnh v. Xc nh biu thc ca hm F(A, B, C).

Chng minh F c th thc hin ch bng 1 cng logic duy nht.

b. Cho 3 hm F (A, B, C), G (A, B, C), v H (A, B, C) c quan h logic vi nhau: F = G H Vi hm F (A, B, C) = (0, 2, 5) v G (A, B, C)= (0, 1, 5, 7).

Hy xc nh dng hoc ca hm H (A, B, C) (1,0 im)

Cu 4 Rt gn cc hm sau bng ba Karnaugh (ch thch cc lin kt) a. F1 (W, X, Y, Z) = (3, 4, 11, 12) theo dng P.O.S (tch cc tng)

B

.

.

F

A

C

F = (A + B) C B C = ((A + B) C) (B C) + ((A + B) C) (B C) = (A + B) B C + ((A + B) + C) (B + C) = A B C + B C + (A B + C) ( B + C) = B C (A + 1) + A B + B C + A BC + C = B C + A B + C (B + A B + 1) = A B + B C + C = A B + B + C = A + B + C : Cng OR

F = G H = G H + G H = G H F = 1 khi G ging H F = 0 khi G khc H

A B C F G H 0 0 0 0 1 0 0 0 1 1 1 1 0 1 0 0 0 1 0 1 1 1 0 0 1 0 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 0 1 1 1 1 1 1

H (A, B, C) = (1, 2, 7) = (0, 3, 4, 5, 6)

00 01 11 10 00

01

11

10

WX YZ F1

0 0

0 0 0 0

0 0

0 0 0 0

(X + Y)

(X + Z)

(Y + Z)

F1 = ( X + Y ) ( X + Z ) ( Y + Z )

Hoc F1 = ( X + Z ) ( Y + Z ) ( X + Y )

Nguyn Trng Lut BM in T - Khoa in-in T - H Bch Khoa TP. HCM

3

b. F2 (A, B, C, D, E) = (1, 3, 5, 6, 7, 8, 12, 17, 18, 19, 21, 22, 24) + d (2, 9, 10, 11, 13, 16, 23, 28, 29)

c. Thc hin hm F2 rt gn cu b ch bng IC Decoder 74138 v 1 cng logic

Cu 5 Ch s dng 3 b MUX 4 1,

hy thc hin b MUX 10 1

c bng hot ng:

A B C D F A B C D F 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0

IN0 IN1 IN2 IN3 IN4

0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1

IN5 IN6 IN7 IN8 IN9

B E

00 01 11 10 00

01

11

10

BC DE 11 01 00 10

A 0 1 F2

1

1

1

1

1 X

X

1

X

X

X

1

1

1

1

X

1

X

1

1 X

X B D E

B D

F2 = B D E + B D + B E

F2 (B, D, E) = B D E + B D + B E = ( 1, 2, 3, 4)

Y4

Y0 Y1 Y2 Y3

Y5 Y6 Y7

C (MSB) B A (LSB)

G1 G2A G2B

IC 74138

B D E

1 0 0

F2

Sp xp li bng hot ng:

Ng vo IN8 v IN9 c chn ch ph thuc vo A v D

A D B C F 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 1 0 0

IN0 IN2 IN4 IN6 IN1 IN3 IN5 IN7 IN8 IN9

D0 D1 D2 D3

S0 (lsb)

Y

S1

MUX 4 1

D0 D1 D2 D3

S0 (lsb)

Y

S1

MUX 4 1

D0 D1 D2 D3

S0 (lsb)

Y

S1

MUX 4 1

IN0 IN2 IN4 IN6

C B

IN1 IN3 IN5 IN7

C B

IN8 IN9

D A

F

Nguyn Trng Lut BM in T - Khoa in-in T - H Bch Khoa TP. HCM

4

Cu 6 Mt hng gh gm 4 chic gh c xp theo s nh hnh v:

Nu chic gh c ngi ngi th Gi = 1, ngc li nu cn trng th bng Gi = 0 (i = 1, 2, 3, 4). Hm F (G1, G2, G3, G4) c gi tr 1 ch khi c t nht 2 gh k nhau cn trng trong hng. Hy thc hin hm F ch bng cc cng NOR 2 ng vo.

G1 G2 G3 G4

G1 G2 G3 G4 F 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1

1 1 1 1 1 0 0 0 1 1 0 0 1 0 0 0

Lp bng hot ng: 00 01 11 10 00

01

11

10

G1G2 F

1 1

1 0 0 1

0 0

0 0 0 1

G3 G4

G3G4

1

1 1

0

G1 G2

G2 G3

F = G1 G2 + G2 G3 + G3 G4

= G1 + G2 + G2 + G3 + G3 + G4

G1

G2

G3

G4

F