TUGAS KELOMPOK PSD KELAS S.205 JUM’AT

1. AHMAD SALAAM MIRFANANDA

2. ANGGA HILMAN HIZRIAN

3. MUHAMMAD GIAN GIFFAR

4. UBAY MUHAMMAD NOOR

5. YOHAN BINSAR GULTOM

Jawaban nomor 2.2

a) X’Y’ + X’Y + XY = X’ + Y

= (X’Y + X’Y’) + (X’Y + XY)= X’(Y + Y’) + Y(X + X’)= X’ + Y

b) A’B + B’C’ + AB + B’C = 1

= (A’B + AB) + (B’C’ + B’C)= B(A’ + A) + B’(C’ + C)= B . 1 + B’.1= B + B’= 1

c) Y + X’Z + XY’ = X + Y + Z

= Y + XY’ + X’Z= (Y + X)(Y + Y’) + X’Z= Y + X + X’Z= Y + (X + X’)(X + Z)= X + Y + Z

d) X’Y’ + Y’Z + XZ + XY + YZ’ = X’Y’ + XZ + YZ’

= X’Y’ + Y’Z(X + X’) + XZ + XY + YZ’= X’Y’ + XY’Z + X’Y’Z + XZ + XY + YZ’= X’Y’(1 + Z) + XY’Z + XZ + XY + YZ’= X’Y’ + XZ(1 + Y’) + XY + YZ’= X’Y’ + XZ + XY(Z + Z’) + YZ’=X’Y’ + XZ + XYZ + YZ’(1 + X)=X’Y’ + XZ(1 + Y) + YZ’

= X’Y’ + XZ + YZ’

Jawaban nomor 2.3

a) ABC’ + BC’D’ + BC + C’D = B + C’D

= ABC’ + ABC + BC + BC’D’ + BC’D + C’D= AB(C’ + C) + BC’(D’ + D) + BC + C’D= AB + BC’ + BC + C’D= B + AB + C’D= B + C’D

b) WY + W’YZ’ + WXZ + W’XY’ = WY + W’XZ’ + X’YZ’ + XY’Z

= (WY + WX’YZ’) + (W’XYZ’ + W’X’YZ’) + (WXYZ + WXY’Z) + (W’XY’Z + W’XY’Z’)= (WY + WXYZ) + (W’XYZ’ + W’XY’Z’) + (W’X’YZ’ + WX’YZ’) + (WXY’Z + W’XY’Z)= WY + W’XZ’(Y + Y’) + X’YZ’(W’ + W) + XY’Z(W + W’)= WY + W’XZ’ + X’YZ’ + XY’Z

c) AD’ + A’B + C’D + B’C = (A’ + B’ + C’ + D’)(A + B + C + D)

= AD’ + A’B + C’D + B’C= (A’ + D)(A + B’)(C + D’)(B + C’) =( A’B’ + AD + B’D)(BC + BD’ + C’D’) = A’B’C’D’ + ABCD= (A + B + C + D)(A’ + B’ + C’ + D’)= (A’ + B’ + C’ + D’)(A + B + C + D)

Nomor 2-4.A.B = 0 , A+B = 1

Buktikan

(A+C) ( +B)(B+C) = BCA̅�

=(AB+ C+BC) (B+C)A̅�=ABB+ABC+ BC+ CC+BBC+BCCA̅� A̅�=AB+BC(A+ )+ C+BC+BCA̅� A̅�=AB+BC+ CA̅�=0+C( +B)(1)A̅�=C( +B)(A+B)A̅�=C(0+ B+AB+BB)A̅�=C(B( +A+B))A̅�=CB

Nomor 2-5

Langkah awal yang dilakukan adalah menentukan semua elemen dari aljabar sebagai 4 bit seperti A, B, C

A =(A3, A2, A1, A0)

B =(B3, B2, B1, B0)

C =(C3, C2, C1, C0)

Selanjutnya buat OR1, AND1, dan NOT1 sesuai dengan aturan AND, OR, NOT

Maka :

a. A + B = C sedemikian hingga dibuat agar untuk semua i , i = 0, ... , 3, Ci sama dengan OR1 dari Ai dan Bi.

b. A B = C sedemikan hingga dibuat agar untuk semua i, i = 0, ... ,3, Ci sama dengan AND1 dari Ai dan Bi

c. 0 didefinisikan sedemikian hingga agar A=”0’ untuk semua i, i = 0, ... ,3, Ai sama dengan 0

d. 1 didefinisikan sedemikian hingga agar A=”1” untuk semua I, I = 0, ….. 2, Ai sama dengan 1

e. Untuk Elemen A apa saja, A didefinisikan sedemikian hingga untuk semua I, I = 0,…..,3, Ai sama dengan NOT1 dari Ai

Nomor 2-6

2.6.a. BC + B �C = A � C � + A�BC + (A � B �C + B�C)= A � C � + (A �BC + A � B �C) + B �C= A � + C + CC̅� A̅� B̅�= + CA̅� B̅�

2.6.b. (A̅� )( ) = ( )( + + ) ̅�+̅� ̅�B̅� ̅�+̅� ̅�C̅� A̅� .̅�B̅� .̅�C̅� A̅� B̅� C̅� A̅� B̅� C̅�

= + + A̅� B̅� C̅� A̅� B̅� C̅� A̅� B̅� C̅�= A̅� B̅� C̅�

2.6.c. AB + AC = A(B + C)C̅� C̅�

= A(B + C)= AB + AC

2.6.d. ( + B)( + )(A C) = (A )(AC)( + B + )A̅� A̅� C̅� B̅� B̅� A̅� C̅�

= A C( + B + )B̅� A̅� C̅�= (A ) C + A(B )C + A (C )A̅� B̅� B̅� B̅� C̅�= (0) C + A(0)C + A (0)B̅� B̅�= 0

Nomor 2-7

2.7.a. + XYZ + YX̅� Y̅� X̅� = ( + Y) + XYZX̅� Y̅�

= +XYZX̅�= ( + X)( + YZ)X̅� X̅�= + YZX̅�

2.7.b. X + Y(Z + )X̅� ̅�+̅� ̅�Z̅� = X + Y(Z + ) = X + Y(Z + )(Z + )X̅� Z̅� Z̅� X̅�

= X + Y(Z + ) = X + Y + YZX̅� X̅�= (X + Y)(X + ) + YZ = X + Y + YZX̅�= X + Y(1 +Z)= X + Y

2.7.c. X( + Z) + X(W + YZ) = X + X Z + WX + XYZW̅� Z̅� Y̅� W̅� W̅� Z̅� W̅� Y̅� W̅�

= X + WX + XZ( + Y)W̅� Z̅� W̅� Y̅�= X( + Z) + WXW̅� Z̅�= X( + W)W̅�= X

2.7.d. (AB + )( + CD) + = AB + ABCD + + CD + + A̅� B̅� C̅� D̅� A̅� .̅�C̅� C̅� D̅� A̅� B̅� C̅� D̅� A̅� B̅� A̅� C̅�

= (1 + AB ) + (1 + CD) + ABCD + C̅� D̅� A̅� B̅� A̅� B̅� C̅� D̅� = (1 + ) + + A(BCD)C̅� A̅� B̅� D̅� A̅� = + + C(BD)A̅� C̅� = + + BDA̅� C̅�

Number 2-8

2-8 . Using DeMorgan’s theorem, express the function F= A �BC+AC�+A �Ba.) With only OR and complement operation

BC+A + B= (A+ + )+( +C)+(A+ )A̅� C̅� A̅� B̅� C̅� A̅� B̅�b.) With only AND and complement operation

BC+A + B= ( BC)(A )( B)A̅� C̅� A̅� A̅� C̅� A̅�

Number 2-9

2-9 . Find the complement of the following expression:a) ᾹB + AB

ᾹB + AB = (A+ )( + )B̅� A̅� B̅�(AA�+AB�+A �B �+B )�B̅�(0+ (A+ )+ )B̅� A̅� B̅�( + )B̅� B̅�( )B̅�

Complement of ᾹB+AB is B̅�

b) ( + )ZX̅� Y̅�

( + )Z =XY+X̅� Y̅� Z̅�Complement = XY+ Z̅�

c) W + (Y+ +YZ)+ X+( +Z)(Y+ )Z̅� W̅� Y̅� Z̅�

W + (Y+ +YZ)+ X+( +Z)(Y+ )= ( Z( + ))(W+ )(Y + Z)Z̅� W̅� Y̅� Z̅� W̅� Y̅� Y̅� Z̅� X̅� Z̅� Y̅�( Z)(Y + Z)(W+ )W̅� Y̅� Z̅� Y̅� X̅�(0+ Z)(W+ )W̅� Y̅� X̅�

Z(W+ )W̅� Y̅� X̅�ZW̅� X̅�Y̅�

Complement = ZW̅� X̅�Y̅�

d) (A+B+C�)(A�B+C)(A+BC�)(A+B+C �)(AA�B+A�BBC �+AC+BCC�)(A+B+C �)(0+A �BC�+AC+0)AA�BC�+AAC+A�BBC �+ABC+A �BC +AC� C̅� C̅�0+AC+A �BC�+ABC+A �BC�+0B(AC+A � C�)+ACAC(B+1)+BA �C �AC+BA �C �

(A+B+C �)(A�B+C)(A+BC�) = AC+BA �C � ( + )( +A+C)A̅� C̅� B̅�

The Complement is ( + )( +A+C)A̅� C̅� B̅�