Tugas 2
Tugas 2
Tugas 2
2.2 Prove the identity of each the following Boolean equations, using algebric manipulation. a. + Y + XY + Y + X Y + Y X +
= Y+ = X Y + = b. +Y B+
+AB+
C + C
= B + A B + = B A + =B + B =1
+ C +
c. Y + XZ + XY X + Y + Z = Y + XY + XZ = Y + XY + Y+ XZ = Y + X + XZ = Y + X + XX + Z =X+Y+Z d. XY YZ XZ XY YZ XY XZ YZ
XY YZ XZ XY ( Z Z ) YZ XY XZ YZ XY YZ XZ XYZ XYZ YZ XY XZ YZ XY YZ XZ (1 Y ) YZ (1 X ) XY XZ YZ XY YZ XZ YZ XY XZ YZ XY YZ ( X X ) XZ YZ XY XZ YZ XY XYZ XYZ XZ YZ XY XZ YZ XY (1 Z ) XZ (1 Y ) YZ XY XZ YZ XY XZ YZ XY XZ YZ
2.3 Provide the identity of each of following Boolean equaition, using algebraic manipulation.
a)
ABC BCD BC CD B CD
(WY WXYZ ) (WXYZ WXYZ ) (WXYZ WXYZ ) (WXYZ WXYZ ) WY WXZ XYZ XYZ (WY WXYZ ) (WXYZ WXYZ ) (WXYZ WXYZ ) (WXYZ WXYZ ) WY WXZ XYZ XYZ WY (1 YZ ) WXZ (Y Y ) XYZ (W W ) XYZ (W W ) WY WXZ XYZ XYZ WY .1 WXZ .1 XYZ .1 XYZ .1 WY WXZ XYZ XYZ WY WXZ XYZ XYZ WY WXZ XYZ XYZ
c)
AD AB CD BC ( A B C D)( A B C D)
AD AB CD BC ( A B C D)( A B C D ) AD. AB.CD.BC ( A B C D)( A B C D ) ( A D)( A B )(C D )( B C ) ( A B C D )( A B C D ) ( AA AB AD BD)( BC CC BD DC ) ( A B C D )( A B C D ) ( AB AD BD)( BC BD DC ) ( A B C D )( A B C D ) ABCD ABCD ( A B C D)( A B C D) ( A B C D)( A B C D) ( A B C D)( A B C D )
2.4 A.B = 0 dan A+B= 1, buktikan (A + C) ( + (A + C) ( + ( ( ( ( ( ( ( (
2.6 simplify the following Boolean expression to the indicated number of literials: a. (
( ( ( b. ( = = = c. A B
+( + +
+ ) +
+AC
+B+ )
= 0+0+0 =0
2.7 Reduce the following Boolean expression to the indicated number of literials: a. ( ( b. X + Y ( Z + =X+Y(Z+XZ) =X+Y(Z+X)(Z+Z) =X+YZ+XY = ( X + X )( X + Y ) + Y Z =X+Y+YZ =X+Y c. W X ( Z + Y Z ) + X (W + W Y Z ) =WXZ+WXYZ+WX+WXYZ =WXZ+WXZ+WX =WX+WX =X d. (AB + AB)(CD + CD) + AC = ABCD + ABCD + ABCD + ABCD + A + C = ABCD + A + C = A + C + A(BCD) = A + C + C(BD) = A + C + BD ) ( ( ( (1)
(A+B+C) + (A + C) + (A + B)
b.
F = ABC + AC + AB = (ABC).(AC).(AB)
2.9 Find the complement of the following expression : a. F = (A + B)(A + B) b. F = ((V + W)X + Y)Z c. F = [W + X + (Y + Z)(Y + Z)][W + X + YZ + YZ] d. F = ABC + (A + B)C + A(B + C)