Technische Grundlagen der Informatik 5. Musterlösung – Automaten ...

) G(A, B, C, D) = A(B + ¯C D) + C D(A + B)(C + Ā)
= AB + A¯C D + (AC D + BC D)(C + Ā)
= AB + A¯C D + AC D + AĀC D + BC D + ĀBC D
= AB + AD + BC D
AB
C D
00
A
00 01 11 10
0 0 1 0
C
01
11
10
0 0 1 1
0 1 1 1
0 0 1 0
D
B
G(A, B, C, D) = AB + AD + BC D
c) H(A, B) = (A|A)|(B|B) = Ā|¯B
= Ā¯B
= A + B
d) I(A, B) = (A ⊕ B) ⊕ A = (A¯B + ĀB) ⊕ A
= (A¯B + ĀB)Ā + (A¯B + ĀB)A
= ĀB + (Ā + B)(A + ¯B)A
= ĀB + (Ā¯B + AB)A
= ĀB + AB
= B
e) J(A, B, C, D) = ((A + B + ¯D)(A + C))⊕((B + ¯C)(A + B + ¯C))
= (Ā¯BD + Ā¯C) ⊕ (¯BC + Ā¯BC)
= (Ā¯BD + Ā¯C) ⊕ ¯BC
= (Ā¯BD + Ā¯C)¯BC + (Ā¯BD + Ā¯C)¯BC
= (Ā¯BD + Ā¯C)(B + ¯C) + (A + B + ¯D)(A + C)¯BC
= Ā¯B ¯C D + ĀB ¯C + Ā¯C + (A + AC + AB + BC + A¯D + C ¯D)¯BC
= Ā¯B ¯C D + ĀB ¯C + Ā¯C + A¯BC + A¯BC + A¯BC ¯D + ¯BC ¯D
= Ā¯C + A¯BC + ¯BC ¯D
AB
C D
00
A
00 01 11 10
1 1 0 0
C
01
11
10
1 1 0 0
0 0 0 1
1 0 0 1
D
B
J(A, B, C, D) = Ā¯C + A¯BC + ¯BC ¯D + Ā¯B ¯D
4 Alexander Biedermann biedermann@iss.tu-darmstadt.de · Marc Stöttinger stoettinger@iss.tu-darmstadt.de · (0 61 51) 16-70470