1. CLD-II, Chapter 2, problem 2.2, parts b, and c.
2.2.a Draw Schematic for XY+XZ using AND,OR, and
NOT gates
b. Draw Schematic for (X(Y+Z))’
2. CLD-II, Chapter 2, problem 2.4, parts a, and b.
2.4.a Draw a Schematic for (X(Y’Z’))’
using NAND gates and inverters only
b. Draw a Schematic for XY+XZ using NAND gates and inverters only
3. CLD-II, Chapter 2, problem 2.6, parts c, and d.
2.6.c Prove (X+Y’)Y = XY
(X+Y’)Y = (XY)+(Y’Y) (6b)
XY+Y’Y = XY+0 (7b)
XY+0 = XY (5a)
d. Prove (X+Y)(X’+Z) = X’Y + XZ
Expand:
(X+Y)(X'+Z) = (X(X'+Z)+Y(X'+Z)) (6b)
(X(X'+Z)+Y(X'+Z) = XX'+XZ+YX'+YZ (6b)
XX'+XZ+YX'+YZ = 0+XZ+YX'+YZ (7b)
0+XZ+YX'+YZ = X'Y+XZ+YZ (5a)
Intuitively
the YZ term is unneeded because the only way YZ
will add
a 1 is if both Y and Z are 1. If Y and Z are both
replaced
with 1, the equation becomes
X'(1)+X(1) = X'+X = 1 (7a)
Therefore
YZ should be eliminated. To do this we expand YZ
by ANDing
it with a 1 in the form of (X+X').
X'Y+XZ+YZ
= X'Y+XZ+YZ(X+X') (5b,7a)
X'Y+XZ+YZ(X+X') = X'Y+XZ+XYZ+X'YZ (6b)
X'Y+X'YZ+XZ+XYZ = X'Y(1+Z)+XZ(1+Y) (6b)
X'Y(1+Z)+XZ(1+Y) = X'Y(1)+XZ(1) = X'Y+XZ
4. CLD-II, Chapter 2, problem 2.8, parts b, and d.
2.8.b Use truth tables to prove (A+B’)B =
AB
|
A B |
B’ |
A+B’ |
(A+B’)B |
AB |
|
0 0 |
1 |
1 |
0 |
0 |
|
0 1 |
0 |
0 |
0 |
0 |
|
1 0 |
1 |
1 |
0 |
0 |
|
1 1 |
0 |
1 |
1 |
1 |
d. Use truth tables to show that
ABC+A’BC+A’B’C+A’BC’+A’B’C’ =
BC+A’B’+A’C’
|
A B C |
ABC |
A’BC |
A’B’C |
A’BC’ |
A’B’C’ |
ALL |
BC |
A’B’ |
A’C’ |
|
0 0 0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
|
0 0 1 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
|
0 1 0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
|
0 1 1 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
|
1 0 0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
1 0 1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
1 1 0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
1 1 1 |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
5. CLD-II, Chapter 2, problem 2.10, parts b, and d.
2.10.b Find the Complement of ABC+B(C’+D’)
(ABC+B(C’+D’))’ = (ABC)’(B(C’+D’))’
(ABC)’(B(C’+D’))’ = (A’+B’+C’)(B’+(C’+D’)’)
(A’+B’+C’)(B’+(C’+D’)’) = (A’+B’+C’)(B’+(C’’D’’))
(A’+B’+C’)(B’+(C’’D’’))
= (A’+B’+C’)(B’+(CD))
(A’+B’+C’)(B’+(CD))
d. Find the Complement of X+YZ’
(X+YZ’)’ = X’(YZ’)’
X’(YZ’)’ = X’(Y’+Z’’)
X’(Y’+Z’’) = X’(Y’+Z)
X’(Y’+Z)
6. CLD-II, Chapter 2, problem 2.12.
2.12 Verify that the following diagram implements an XOR function
Realized
function is: (((XY)’Y)’(X(XY)’)’)’
(((XY)’Y)’(X(XY)’)’)’ = XY’+X’Y
((XY)’Y)’’+(X(XY)’)’’ =
XY’+X’Y
((XY)’Y)+(X(XY)’) = XY’+X’Y
((X’+Y’)Y)+(X(X’+Y’)) = XY’+X’Y
(X’Y+Y’Y)+(XX’+XY’) = XY’+X’Y
(X’Y+0)+(0+XY’) = XY’+X’Y
X’Y+XY’ = XY’+X’Y
7. CLD-II, Chapter 2, problem 2.17, parts c, and d.
2.17.c Simplify f(X,Y,Z)=Y'Z+X'YZ+XYZ
Y'Z+X'YZ+XYZ = Y'Z+YZ(X'+X) (8d)
Y'Z+YZ(X'+X) =
Y'Z+YZ(1) (5)
Y'Z+YZ(1) = Y'Z+YZ (1d)
Y'Z+YZ = Z(Y'+Y) (8d)
Z(Y'+Y) = Z(1) (5)
Z(1) = Z (1d)
f(X,Y,Z) = Z
d. Simplify f(X,Y,Z)=(X+Y)(X'+Y+Z)(X'+Y+Z')
For
the sake of clarity let A=(X'+Y+Z')
(X+Y)(X'+Y+Z)(X'+Y+Z') = (X+Y)(X'+Y+Z)A
(X+Y)(X'+Y+Z)A = (X+Y)(X'A+YA+ZA) (8d)
Expand
X'A:
X'(X'+Y+Z') =
X'X'+X'Y+X'Z' (8)
X'X'+X'Y+X'Z'
= X'+X'Y+X'Z' (3d)
X'+X'Y+X'Z' =
X'(1+Y+Z') (8)
X'(1+Y+Z') = X'(1) = X' (2),(1d)
Expand
YA:
Y(X'+Y+Z') = X'Y+YY+YZ' (8)
YY+YX'+YZ' = Y+YX'+YZ' (3d)
Y+YX'+YZ' = Y(1+X'+Z') (8)
Y(1+X'+Z') = Y(1) = Y (2),(1d)
Expand
ZA:
Z(X'+Y+Z') = X'Z+YZ+ZZ' (8)
X'Z+YZ+ZZ' = X'Z+YZ+0 =
X'Z+YZ (5d),(1)
X'Z+YZ = Z(X'+Y)
Recombine,
keeping only one copy of each term:
(X+Y)(X'+Y+Z(X'+Y)) = (X+Y)((X'+Y)(1+Z)) (8)
(X+Y)((X'+Y)(1+Z)) = (X+Y)((X'+Y)(1)) (2)
(X+Y)((X'+Y)(1)) = (X+Y)(X'+Y) (1d)
(X+Y)(X'+Y) = (X(X'+Y)+Y(X'+Y)) (8)
X(X'+Y)+Y(X'+Y) =
XX'+XY+YX'+YY (8)
XX'+XY+YX'+YY = 0+YX+YX'+Y (3d,5d)
YX+YX'+Y = Y(1+X+X') (8)
Y(1+X+X') = Y(1) = Y (2,1d)
8. CLD-II, Chapter 2, problem 2.20, all parts
2.20.a Write f(A,B,C,D)=∑m(1,2,3,5,8,13) in
canonical minterm form
A’B’C’D+A’B’CD’+A’B’CD+A’BC’D+AB’C’D’+ABC’D
b. Write f in canonical maxterm form
(A+B+C+D)(A+B’+C+D)(A+B’+C’+D)(A+B’+C’+D’)(A’+B+C+D’)(A’+B+C’+D)(A’+B+C’+D’)(A’B’+C+D’)(A’+B’+C’+D)(A’+B’+C’+D’)
c. Write the complement of f in “little m” notation and as a canonical minterm expression
f’(A,B,C,D)=∑m(0,4,6,7,9,10,11,12,14,15)
A’B’C’D’+A’BC’D’+A’BCD’+A’BCD’+AB’C’D+AB’CD’+AB’CD+ABC’D’+ABCD’+ABCD
d. Write the complement of f in “big M” notation and as a canonical maxterm expression
f’(A,B,C,D)=∏M(1,2,3,5,8,13)
(A+B+C+D’)(A+B+C’+D)(A+B+C’+D’)(A+B’+C+D’)(A'+B+C+D)(A’+B’+C+D’)
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