Rabu, 10 Juni 2009

ujian tengah semester

1. A = 3F

B = 5C

Y1 = A + B

Y2 = A B

Y3 = A . B

JAWAB:

A = 3F

= 3x161 + F(15)x 160

= 48 + 15

= 6310 = 1 1 1 1 1 12

B= 5C

= 5x161 + C(12)x160

=80 + 12

= 9210 = 1 0 1 1 1 0 02

A

B

Y1 = A + B

Y2 = A B

Y3 = A . B

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Y1=1x26+1x25+1x24+1x23+1x22+1x21+1x20

=64 + 32 + 16 + 8 + 4 + 2 +1

=12710 = 7F16

Y2 = 1x26+1x25+0x24+0x23+0x22+1x21+1x20

= 64 + 32 +0 +0 +0 +2+ 1

=9910 = 6316

Y3= 0x26+0x25+1x24+1x23+1x22+0x21+0x20

= 0+0+16+8+4+0+0

=2810 =1C16

2. 2.A= 3C

B = 5F

X1=A+B

X2=A – B

JAWAB:

A= 3x161+C(12)x160

= 48 +12

=6010 = 1 1 1 1 0 02

B=5x161+F(15)x160

=80 +15

=9510 = 10111112

X1=A + B

A = 0 1 1 1 1 0 0

B = 1 0 1 1 1 1 1 +

S = 1 0 0 1 1 0 1 12

= 15510

= 9B16

X2 =A – B

B’ = 0 1 0 0 0 0 0

1 +

= 0 1 0 0 0 0 1

A = 0 1 1 1 1 0 0

B’= 0 1 0 0 0 0 1 +

S = 1 0 1 1 1 0 1

3. A = 516 = 5 x 16 = 8010 = 1 0 1 0 0 0 02

B = 616 = 6 x 16 = 9610 = 1 1 0 0 0 0 02

C = 716 = 7 x 16 = 11210 =1 1 1 0 0 0 02

D = 816 = 8 x 16 = 12810 =1 0 0 0 0 0 0 02

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Q = A’B’C’D + ABCD’ + A’BCD’ + AB’CD’ + A’B’C’D’

= A’B’C’(D+D’) + BCD’(A+A’) +AB’CD’

= A’B’C’+BCD’ +AB’CD’






Nama : nirmadiati

Bp : 0801081015

Ujian

1. A=3F = 315=001100010101

B=5C =512=010100010010

Y1=A+B

Y2=A B

Y3=A*B

2.

A=3C=312=001100010010

B=5F=515=010100010101

X1=A+B

X2=A-B

X1=A+B= 0011 0001 0010

0101 0001 0101 +

1101 1111 1101

X2=A-B =A+(B + 1)

B=0101 1110 1010

1+Q

1010 1110 1011

A-B= 0011 0001 0010

0101 1110 1011+

1101 1111 1101

3. Q=0

Jika input (A,B,C,D)=5H,6H,7H,8H gambarkan rangkaiannya

Misalkan input A=1,B=1,C=1,dan D=1 maka Q akan bernilai 0,begitu juga kalau A=0,B=0,C=0,dan D=0 maka Q=0.begitu juga A=0,B=1,C=0,dan D=1 maka Q=0.