Penyelesian tugas 3 sistem digital ALJABAR BOOLEAN [www.tofa90.blogspot.com]

NAMA : MOH TOFARIYANTO

NPM    : 2011.02.01.0.0113

KELAS  : IIID

SoalLatihan I :

Sederhanakan ekspresi logika dibawah dengan

Aljabar Boolean :

1. AB’+BC + C’A

2. A’(BC + AB + BA’)

3. ABC + AB +A

4. (A’+AB ) (A’B)

5. BC + AD + ABCD +ADC +A’

SoalLatihan II :

BUATLAH TABEL KEBENARAN DARI PERSAMAAN

LOGIKA DIBAWAH:                                                                      

(a) X . Y + X’. Y + X’ . Y’ = X’+ Y

(b) A . B . C + A . C + B . C = A + B + C ganti A.B.C + A.C + B.C = (A+B).C

(c) ( X’. Y + Y’ . X ) + X . Y = ( X . Y’ )

(d) A . B . D + A’. B’ . D + A . B’.D’=A. ( B’.D’+B.D ) ganti (X’.Y + Y’.X) + X.Y = (X+Y)

PENYELESAIAN

 SoalLatihan I :

1. AB’+BC+C’A

AB’+AC’+BC = A(B’+C’)+BC

                   = A(B,C)’ + (BC)

                   = A D’ + D

                   = (BC)’ A + (BC)

                   = (BC) + (BC)’ A

                   = BC + A                                                       

2. A’(BC + AB + BA’) = A’ (B(C+A+A’)

                               = A’ (B(C + 1)

                               = A’ (B(C + 1)

                               = A’ (B(1))

                               =A’ (B)

                               = A’ B

3. ABC(+A(B+1)

= ABC + A

= A (BC + 1)

= A(1)

= A

4. (A’+AB) (A’B)

= A’A’B + (AB) (A’B)

= A’B + (AB) (A’B)

= A’B (1+AB)

= A’B (1)

= A’B

5. BC+AD+ABCD+ADC+A’

= A’+BC+(AD+ADC)+ABCD

= A’+BC+AD+ABCD

= A’+BC+AD(1+BC)

= A’+AD+(BC+ABCD)

= A’+AD+BC(1+AD)

= A’+AD+BC

SoalLatihan II

TABEL KEBENARAN DARI PERSAMAAN

aa)    X.Y + X’.Y + X’.Y’ = X’Y

X	Y	X’	Y’	XY	X’Y	X’Y’	XY+X'Y+X'Y'	X'+Y
0	0	1	1	0	0	1	1	1
0	1	1	0	0	1	0	1	1
1	0	0	1	0	0	0	0	0
1	1	0	0	1	0	0	1	1

bb)      A.B.C + A.C + B.C = (A+B).C

A	B	C	A+B	A.B	A.C	B.C	(A.B).C	(A.B.C)+(A.C)	(A.B.C)+(A.C)+(B.C)	(A+B).C
1	1	1	1	1	1	1	1	1	1	1
1	1	0	1	1	0	0	0	0	0	0
1	0	1	1	0	1	0	0	1	1	1
0	0	0	0	0	0	0	0	0	0	0
0	1	1	1	0	0	1	0	0	1	1
0	1	0	1	0	0	0	0	0	0	0

cc)     (X’.Y + Y’.X) + X.Y = (X+Y)

X	Y	X’	Y’	X.Y	X’.Y	X.Y’	(X’.Y)+( X.Y’)	(X’.Y+X.Y’)+ (X.Y)	X+Y
1	1	0	0	1	0	0	0	1	1
1	0	0	1	0	0	1	1	1	1
0	1	1	0	0	1	0	1	1	1
0	0	1	1	0	0	0	0	0	0

 

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