1. DIGITAL DESIGN MORRIS MANO UNIT 3 ANSWERS OF THE PROBLEMS 1-29
BURCU DAL

2. 3.1 a) F(x,y,z) =∑(0,2,6,7) 1 1 1 1 b) F(A,B,C) =∑(0,2,3,4,6) y B yz00 01 11 10 z y 1 BC A 00 01 11 10 C B
F(A,B,C)= C’ +A’B
F(x,y,z)= xy +x’z’
d) F(x,y,z) =∑(3,5,6,7)
c) F(a,b,c) =∑(0,1,2,3,7) 1 bc a 00 01 11 10 c b 1 yz x 00 01 11 10 z y
F(a,b,c) = a’ +bc
F(x,y,z)= xy + xz + yz

3. 3.2 1 1 F(x,y,z)= y +x’z a) F(x,y,z) =∑(0,1,5,7)
F(x,y,z)= xz +x’y’ yz 1 x 00 01 11 10 z y y
b) F(x,y,z) =∑(1,2,3,6,7)
F(x,y,z)= y +x’z yz 1 x 00 01 11 10 z

4. 3.3 1 1 1 a) xy + x’y’z’ + x’yz’ y yz x b) x’y’ + yz + x’yz’ y yz x z00 01 11 10 z y
a) xy + x’y’z’ + x’yz’
F(x,y,z)= xy + x’z’ 1 yz x 00 01 11 10 z y
b) x’y’ + yz + x’yz’
F(x,y,z)= x’ + yz 1 BC A 00 01 11 10 C B
c) A’B + BC’ + B’C’
F(A,B,C)= C’ + A’B

5. 3.4 1 1 1 1 b) F(A,B,C,D)= ∑(4,6,7,15) C a) F(x,y,z)= ∑(2,3,6,7) y CD00 01 11 10 A B C D
F(A,B,C,D)= A’BD’ + BCD
b) F(A,B,C,D)= ∑(4,6,7,15)
3.4 1 yz x 00 01 11 10 z y
a) F(x,y,z)= ∑(2,3,6,7)
F(x,y,z)= y 1 AB CD 00 01 11 10 A B C D
F(A,B,C,D) = CD + ABC ABD
c) F(A,B,C,D)= ∑(3,7,11,13,14,15) 1 wx yz 00 01 11 10 w x z
d) F(w,x,y,z)= ∑(2,3,12,13,14,15)
F(w,x,y,z)= wx + w’x’y y

6. 3.5 1 1 1 1 a) F(w,x,y,z)= ∑(1,4,5,6,12,14,15) AB CD A B C D00 01 11 10 A B C D
F(A,B,C,D) = A’C’ + A’B’CD’ + ACD +BCD
b) F(A,B,C,D)= ∑(0,1,2,4,5,7,11,15)
F(w,x,y,z)= xz’ + wxy +w’y’z 1 wx yz 00 01 11 10 w x y y
c) F(w,x,y,z)= ∑(2,3,10,11,12,13,14,15) 1 yz 00 01 11 10 x z
d) F(A,B,C,D)= ∑(0,2,4,5,6,7,8,10,13,15) 1 AB CD 00 01 11 10 A B C D wx w
F(A,B,C,D) = B’D’ + BD + A’B
F(w,x,y,z)= wx + x’y z

7. 3.6 1 1 b) x’z + w’xy’ + w(x’y + xy’) = x’z + w’xy’ + wx’y + wxy’
a) A’B’C’D’ + AC’D’ + B’CD’ + A’BCD + BC’D 1 AB CD 00 01 11 10 B C D
F(A,B,C,D) = B’D’ + ABC’ + A’BD 1 wx yz 10 00 01 11 w x
F(w,x,y,z) = xy’ + x’z + wx’y y z A

8. 3.7 1 1 1 1 a) w’z + xz + x’y +wx’z y wx yz w x
F(w,x,y,z) = z + x’y y 1 wx yz 00 01 11 10 w x
b) B’D + A’BC’ + AB’C +ABC’ 1 AB CD 00 01 11 10 B A D
F(A,B,C,D) = BC’ + B’D + AB’C C
c) AB’C + B’C’D’ +BCD +ACD’ + A’B’C + A’BC’D 1 AB CD 00 01 11 10 B C D
F(A,B,C,D) = BD’ + CD + AC + A’BD
d) wxy + yz + xy’z + x’y
F(w,x,y,z) = xz + x’y + wy y 1 wx yz 00 01 11 10 w x

9. 3.8 1 1 1 a) xy + yz + xy’z y yz x b) C’D + ABC’ + ABD’ + A’B’D C CD00 01 11 10 z
F(x,y,z)= ∑(3,5,6,7)
b) C’D + ABC’ + ABD’ + A’B’D 1 AB CD 00 01 11 10 B C D
F(A,B,C,D) = ∑(1,3,5,8,9,10,12,13) A
c) wxy + x’z’ + w’xz yz
F(w,x,y,z) = ∑ (0,2,5,7,8,10,14,15) y 1 wx 00 01 11 10 w x z

10. 3.9
a) F(w,x,y,z)=∑(0,2,4,5,6,7,8,10,13,15) y 1 wx 00 01 11 10 w x z
F(w,x,y,z) = x’z’ + w’x + xz
F(w,x,y,z) = x’z’ + w’z’ + xz
Essential B’D’, AC, A’BD
Nonessential AB, A’B’D, A’C’D, B’CD
b) F(A,B,C,D) = ∑(0,2,3,5,7,8,10,11,14,15) C CD AB 00 01 11 10 00 1 1 1 01 1 1 B 11 1 1 A 10 1 1 1 D
F(A,B,C,D) = B’D’ + AC + A’BD + CD
F(A,B,C,D) = B’D’ + AC + A’BD + B’C Essential B’D’, AC, A’BD
Nonessential CD, B’C AB CD
c) F(A,B,C,D) = ∑(1,3,4,5,10,11,12,13,14,15) 1 00 01 11 10 B D A
F(A,B,C,D) = AC + BC’ + A’B’D
Essential AC, BC’
Nonessential AB, A’B’D, A’C’D, B’CD C 00 01 11 10

11. 3.10
a) F(w,x,y,z)=∑(0,2,4,5,6,7,8,10,13,15) y 1 wx 00 01 11 10 w x z
F(w,x,y,z) = x’z’ + w’x + xz
F(w,x,y,z) = x’z’ + w’z’ + xz
Essential B’D’, AC, A’BD
Nonessential AB, A’B’D, A’C’D, B’CD
b) F(A,B,C,D) = ∑(0,2,3,5,7,8,10,11,14,15) 1 AB CD 00 01 11 10 B C D A
F(A,B,C,D) = B’D’ + AC + A’BD + CD
F(A,B,C,D) = B’D’ + AC + A’BD + B’C Essential B’D’, AC, A’BD
Nonessential CD, B’C
c) F(A,B,C,D) = ∑(1,3,4,5,10,11,12,13,14,15) 1 00 01 11 10 B D A
F(A,B,C,D) = AC + BC’ + A’B’D
Essential AC, BC’
Nonessential AB, A’B’D, A’C’D, B’CD AB C CD

12. 3.11
a) F(A,B,C,D,E) =∑(0,1,4,5,16,17,21,25,29) 00 01 11 10
F(A,B,C,D,E)= B’C’D’ + A’B’D’ +AD’E 00 1 1
A=0 01 1 1
2. Gösterim: 11 1 00 01 11 10
A=0
A=1 DE BC 10 BC 00 01 11 10 00 1 1
A=1 01 1 11 1 10 1
b) F = A’B’CE’ + A’B’C’D’ + B’D’E’ + B’CD’ + CDE’ + BDE’ 1 00 01 11 10 DE BC
A=1
A=0 1 00 01 11 10 DE BC
F(A,B,C,D,E) = BDE’ +CDE’ +B’D’E’ + B’CD’ + A’B’D’

13. 3.12
a) F(w,x,y,z) = ∑(0,2,5,6,7,8,10)
b) F(A,B,C,D) =∏(1,3,5,7,13,15)= ∑(0,2,4,6,8,9,10,11,12,14) yz w y
F = x’z’ + w’xy + w’xz
F’= x’z + wx+ xy’z’
F= (x+z’)(w’+ x’)(x’ + y + z) 1 wx 00 01 11 10 z x C CD AB 00 01 11 10 00 1 1 01 1 1 B 11 1 1 A 10 1 1 1 1 D
F(A,B,C,D) = D’ + AB’ =(A + D’)(B’ + D)

14. 3.13 1 1 1 a) x’z’ + y’z’ + yz’ + xy b) F = AC’ + B’D + A’CD + ABCD yyz x 00 01 11 10 z y
a) x’z’ + y’z’ + yz’ + xy
F = z’ + xy
F’ = x’z + y’z
b) F = AC’ + B’D + A’CD + ABCD 1 AB CD 00 01 11 10 A D
F(A,B,C,D) = AC’ + CD + B’D
F’= CD’ +A’D’ + A’BC’
F(A,B,C,D)= (C’ + D)(A + D)(A + B’ + C) B C
c) (A’ + B’ +D’ )(A +B’+C’)(A’ + B +D’)(B +C’ +D’)
F(A,B,C,D) = AD’ +A’C’ +B’D’
F’= ABD + A’BC + AB’D +B’CD 1 AB CD 00 01 11 10 A D B C

15. 3.14 1 F(A,B,C,D) = A’B’D’ + AB’CD’ + A’BD + ABC’D C CD AB B A D 00 0111 10 A D
F(A,B,C,D) = B’CD’ + A’B’D’ + A’BD + BC’D B C

16. 3.15 X 1 X 1 1 X a) F(x,y,z)=∑(0,1,2,4,5) d(x,y,z)=∑(3,6,7)00 01 11 10 z y
a) F(x,y,z)=∑(0,1,2,4,5) d(x,y,z)=∑(3,6,7)
F(x,y,z)=1=∑(0,1,2,3,4,5,6,7)
b) F(A,B,C,D) =∑(0,6,8,13,14)
d(x,y,z)=∑(3,6,7) X 1 AB CD 00 01 11 10 A D
F(A,B,C,D) = B’D’ + CD’ + ABC’D =∑(0,2,6,8,10,13,14) B C
c) F(A,B,C,D) =∑(1,3,5,7,9,15)
d(x,y,z)=∑(4,6,12,13) 1 X AB CD 00 01 11 10 A D B C
F(A,B,C,D) = A’D + C’D + BD =∑(1,3,5,7,9,12,13)

17. 3.16 1 1 a) AB’ + ABD +ABD’ + A’C’D’ + A’BC’ C F = A + BC’ + C’D’ CD00 01 11 10 A D B C
F = A + BC’ + C’D’
b) BD + BCD’ + AB’C’D’ 1 AB CD 00 01 11 10 A D B C
F = BC + BD + AB’C’D’

18. 3.17
F(A,B,C,D) =∑(0,1,2,3,4,8,9,12) 1 AB CD 00 01 11 10 A D B C
F = A’B’ + B’C’ + C’D’→ F’ = (A+B)(B+C)(C+D)
F’ = AC + BC + BD→ F = (A’+C’)(B’+C’)(B’+D’)

19. 3.18
(AB + A’B’)(CD’ + C’D)

20. 3.19 1 a) F = wx’ + y’z’ + w’yz’ y yz wxwx yz 00 01 11 10 w z x y
F = wx’ + w’z’ + y’z’ F’ = (w’+x)(w+z)(y+z)
F’= w’z + xz + wxy F = (w+z’)(x’+z’)(w’+x’+y’)

21. 3.19
b) F(w,x,y,z) =∑ (5,6,9,10) 1 wx yz 00 01 11 10 w z x y
F = w’xy’z + w’xyz’ + wx’y’z + wx’yz’
= (y’z + yz’)(w’x + wx’)
F’ = wx + w’x’ + yz + y’z’
F = (w’ + x’ )(w + x)(y’ + z’)(y + z)

22. 3.20
F = (AB’ + CD’)E + BC(A +B)=AB’E + CD’E + ABC + BC = AB’E +CD’E+BC(A+ 1) = AB’E + CD’E + BC

23. 3.21 1 w(x + y + z) + xyz = wx + wy + wz + xyz y yzwx yz 00 01 11 10 w z x y
F’ = w’y’ + w’z’ + wx’y’z’
F = (w + y)(w + z)(w’ + x + y + z)

24. 3.22
BCD to Excess 3 Code Converter: Inputs: A, B, C, D Outputs: w, x, y, z
z=D’ , y = CD + C’D’ , x= B’C + B’D + BC’D’ , w = A + BC +BD

25. 3.23 X 1 F(A,B,C,D) =∑(0,1,2,9,11) d(A,B,C,D) = ∑(8,10,14,15) C CD ABAB CD 00 01 11 10 A D B C
F = B’D’ + B’C’ + AC
F’ = B + A’CD
F = B’(A + C’ + D’)

26. 3.24
F(A,B,C,D) =∑(0,1,2,3,4,8,9,12) 1 AB CD 00 01 11 10 A D B C
F = A’B’ + B’D’ + C’D’ → Sum of products
F’ = (A + B)(B + D)(C + D)
F’= AD + BC + BD
F = (A’ + D’)(B’ + C’)(B’ + D’) → Product of sums
a) NAND - AND

27. 3.24
b) AND - NOR F
c) OR - NAND
d) NOR - OR F’

28. 3.25 Degenerate Two Level Forms: AND-AND AND-NAND NAND-NOR NAND-OR
NOR-AND
NOR-NAND
OR-NOR
OR-OR A B C D
(AB.CD)’
AND-AND
AND-NAND
AB.CD
AB.CD A B C D
NAND-NOR
NAND-OR
(AB)’+(CD)’

29. 3.25 A’B’.C’D’ (A+B+C+D)’ A’B’.C’D’ (A’B’.C’D’)’ A+B+C+D A+B+C+D
NOR-AND
NOR-NAND
OR-NOR
OR-OR

30. 3.26 1 1 1 F = f.g f = wxy’ + y’z + w’yz’ + x’yz’
g = (w + x + y’ + z’)(x’ + y’ +z)( w’ + y + z’) g’ = w’x’yz + xyz’ + wy’z
f için;
g için; 1 wx yz 00 01 11 10 w z x y 1 wx yz 00 01 11 10 w z x y
f(w,x,y,z) = ∑(1,2,5,6,9,10,12,13) g(w,x,y,z) = ∑(0,1,2,,4,5,7,8,10,11,12,15)
F(w,x,y,z) = ∑(1,2,5,10,12)
F için; 1 wx yz 00 01 11 10 w z x y
F = wy’z + x’yz’ + wxy’z’

31. 3.27
XOR : X xor Y=X’Y+XY’
Dual of X xor Y : (X’+Y)(X+Y’) = XX’+X’Y’+XY+YY’
= X’Y’+XY
Complement XOR : (X xor Y)’ = [ (X’+Y)(X+Y’) ]’
= XX’+X’Y’+XY+YY’
= X’Y’+XY = (X xor Y)’
Dual of X xor Y = Complement XOR

32. 3.28 1 P = ∑(0,3,5,6) B BC A P = A’B’C’ + A’BC + AB’C + ABC’1 1 BC A 00 01 11 10 C B
P = A’B’C’ + A’BC + AB’C + ABC’
= A’(B xor C)’ + A (B xor C)
= A xor (B xor C)’

33. 3.28
C = ∑(0,3,5,6,9,10,12,15) 1 AB CP 00 01 11 10 A P B C A B C P Ç 1
C= A’B’(C’P’ +CP) + A’B(C’P + CP’) + AB(C’P’ + CP) + AB’(C’P + CP’)
=(C xor P)’(A xor B)’ + (C xor P)(A xor B) = [(A xor B) xor (C xor D)]’

34. 3.29 Half Adder Circuit: * D = A xor B xor C
* E = A’BC + AB’C= (A xor B)C
* F = ABC’ + (A’+ B’)C= AB xorC
* G = ABC

35. 3.30 F = AB’CD’ + A’BCD’ + AB’C’D + A’BC’D
= AB’(C xor D) + A’B(C xor D)
=(A xor B)(C xor D)