Boolean Functions and Reduced Forms

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Exercise
Question 1
1. Determine the values of A, B, C and D that make the product term A’BC’D equal to 1.
i)
A = 0, B = 1, C = 0, D = 1 ✔
ii)
A = 0, B = 0, C = 0, D = 1
iii)
A = 1, B = 1, C = 1, D = 1
iv)
A = 0, B = 0, C = 1, D = 0
Answer 1
i) A = 0, B = 1, C = 0, D = 1. This makes A'=1, B=1, C'=1 and D=1, so the product term becomes 1.
Question 2
2. The binary value of 1010 is converted to the product term A’B’C’D.
i)
True
ii)
False ✔
Answer 2
ii) False. For 1010, literals are AB'CD', not A'B'C'D.
Question 3
3. Which of the following expression is in SOP form?
i)
(A+B)(C+D)
ii)
AB(CD)
iii)
(A)B(CD)
iv)
AB+CD ✔
Answer 3
iv) AB+CD. It is an OR of product terms, which is exactly SOP form.
Question 4
4. A truth table for SOP expression ABC’+ AB’C + A’B’C has how many input combinations?
i)
1
ii)
2
iii)
4
iv)
8 ✔
Answer 4
iv) 8. With 3 variables, total input combinations are 23 = 8.
Question 5
5. POS equivalent of ABC+ AB’C’+AB’C+ABC’+A’B’C will be:
i)
(A’+B’+C’)(A’+B+C’)(A’+B+C)
ii)
(A’+B’+C’)(A+B’+C)(A+B’+C)
iii)
(A+B+C)(A+B’+C)(A+B’+C’) ✔
iv)
(A+B+C)(A’+B+C’)(A+B’+C)
Answer 5
iii) (A+B+C)(A+B'+C)(A+B'+C'). This product of sums matches the same output rows as the given SOP expression.
Question 6
6. Converting the Boolean expression LM+M(NO+PQ) to SOP form we get ______
i)
LM +MNOPQ
ii)
L+MNQ+MPQ
iii)
LM+ M+NO+MPQ
iv)
LM+MNO+MPQ ✔
Answer 6
iv) LM+MNO+MPQ. Distribute M over the bracket: M(NO+PQ)=MNO+MPQ.
Question 7
7. State whether AC + ABC = AC is
i)
True ✔ or
ii)
False.
Answer 7
i) True. By absorption law, AC+ABC = AC(1+B)=AC.
Question 8
8. A student makes a mistake somewhere in the process of simplifying the following Boolean expression. Determine where the mistake was made, and what proper sequence of steps should be used to simplify the original expression.
i)
ab + a(b+c) = ab+ab+c = ab+c
ii)
(x’y+z)’ = (x’)’+y’.z’ = x+y’.z’
Answer 8
i) Correct sequence:
ab + a(b+c)
= ab + ab + ac
= ab + ac
= a(b+c)
The mistake was treating a(b + c) as ab + c instead of ab + ac.
ii) Correct sequence:
(x'y+z)'
= (x'y)'.z'
= (x+y').z'
DeMorgan must be applied to the whole sum term correctly.
Question 9
9. When grouping cells within k-map, the cells must be combined in groups of
i)
2s
ii)
1, 2, 4, 8, etc ✔
iii)
4s
iv)
3s
Answer 9
ii) 1, 2, 4, 8, .... K-map groups must be in powers of 2 for valid simplification.
Question 10
10. Mapping the SOP expression A'B'C' + A'BC' + A'BC + ABC' we get
i)
C’
C
AB
A’B’
1
1
A’B
1
AB’
1
ii)
A
B
AB
1
A’B’
1
A’B
1
AB’
1
1
iii)
C’
C
AB
1
A’B’
A’B
1
1
AB’
1
1
iv)
C’
C
AB
1
1
A’B’
A’B
1
AB’
1
v)
00
01
11
10
00
0
0
0
0
01
0
1
1
0
11
0
1
1
0
10
0
0
0
0
vi)
C’D’
C’D
CD
CD’
A’B’
0
0
0
0
A’B
0
0
0
0
AB’
1
1
1
1
AB’
0
0
0
0
Answer 10
The given options do not exactly match the corrected mapping pattern. The required K-map is the option-(i) style 3-variable table with all cells blank under C' and all cells as 1 under C.
Correct table:
C’
C
AB
1
A’B’
1
A’B
1
AB’
1
⚠️ Note: None of the given options are valid. The correct response should be the one given in answer.
Question 11
11. If you look at the following k-map, you should notice that only two of the input variables- A, B, C, D change their state, in the marked group. The other two variables hold the same value ‘1’. Identify which variable change, and which stay the same:
i)
00
01
11
10
00
0
0
0
0
01
0
1
1
0
11
0
1
1
0
10
0
0
0
0
ii)
C’D’
C’D
CD
CD’
A’B’
0
0
0
0
A’B
0
0
0
0
AB’
1
1
1
1
AB’
0
0
0
0
Answer 11
For each marked group, the variables that remain fixed form the reduced term, and the variables that vary are eliminated.
i)
Rows are 01, 11 and columns are 01, 11. So B=1 and D=1 stay constant, while A and C change. Reduced term: BD.
ii)
Marked row is AB' across all four columns. So A=1 and B'=1 stay constant, while C and D change. Reduced term: AB'.
Question 12
12. Give truth table for
i)
Z = x’+y’+z
ii)
Z = x( y+xz+x’)
Answer 12
Truth values using 4-column tables (x, y, z, Z):
i) Truth table for Z = x' + y' + z
x
y
z
Z
0
0
0
1
0
0
1
1
0
1
0
1
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
0
1
1
1
1
ii) Truth table for Z = x(y + xz + x')
x
y
z
Z
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
0
1
0
1
1
1
1
0
1
1
1
1
1
Question 13
13. Use Boolean algebra to find the most simplified SOP expression for F=ABD+CD+ ACD+ABC+ABCD
i)
F = ABD+ABC+CD ✔
ii)
F= CD+AD
iii)
F = BC+AB
iv)
F= AC+AD
Answer 13
i) F = ABD + ABC + CD. This is the reduced SOP obtained after removing redundant terms.
Question 14
14. From the truth table below, determine SOP and POS expression.
A
B
C
Output
X
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
0
1
0
1
0
1
1
0
1
1
1
1
0
Answer 14
Rows with X=1
001, 011, 110
SOP (canonical)
X = A'B'C + A'BC + ABC' = Σm(1,3,6)
SOP (simplified)
X = A'C + ABC'
Rows with X=0
000, 010, 100, 101, 111
POS (canonical)
X = ΠM(0,2,4,5,7)
POS (simplified)
X = (A + C)(A' + B)(A' + C')
Question 15
15. Specify which axiom(s)/theorem(s) are being used in the following Boolean reductions:
i)
x’y’ + x’y’z = x’y’
ii)
1+A = 1
iii)
D+CD =D
iv)
a’.a’=a’
v)
(bc)’ + bc = 1
vi)
xyz+zx = xz
vii)
ca’b’ + ab = ab + c
Answer 15
i)
Absorption: x'y' + x'y'z = x'y'(1+z) = x'y'. Here X + XY = X with X=x'y'.
ii)
Domination (Null law): 1 + A = 1. OR with 1 always gives 1.
iii)
Absorption: D + CD = D(1 + C) = D. So extra factor C is absorbed.
iv)
Idempotent law: a'·a' = a' because repeating the same literal in AND does not change value.
v)
Complement law: let X = bc, then (bc)' + bc = X' + X = 1.
vi)
Absorption: xyz + zx = zx(y+1) = zx (equivalently X + XY = X with X = zx).
vii)
Redundancy / consensus-style simplification: this step is treated as removal of a redundant term in the chapter’s reduction pattern, giving the shorter form.
So each reduction is justified by a standard Boolean identity and an intermediate factoring step where needed.
Question 16
16. Construct a truth table for the following functions and from the truth table obtain an expression for the inverse function.
i)
F1(A,B,C) = A + BC’
ii)
F2(A,B,C) = AC + BC + AB’
Answer 16
i) Truth table of F1(A,B,C) = A + BC'
A
B
C
F1
0
0
0
0
0
0
1
0
0
1
0
1
0
1
1
0
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1
Inverse output gives F1' = 1,1,0,1,0,0,0,0, so F1' = Σm(0,1,3) = A'(B'+C).
ii) Truth Table of F2(A,B,C) = AC + BC + AB'
A
B
C
F2
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1
Inverse output gives F2' = 1,1,1,0,0,0,1,0, so F2' = Σm(0,1,2,6). Equivalent form by De Morgan: F2' = (A'+C')(B'+C')(A'+B).
Question 17
17. Examine the given truth table and then write both SOP and POS Boolean expressions describing the output.
A
B
C
Output
0
0
0
1
0
0
1
0
0
1
0
1
0
1
1
0
1
0
0
0
1
0
1
1
1
1
0
1
1
1
1
0
Answer 17
Rows with output 1 are 000, 010, 101, 110 i.e. minterms m(0,2,5,6).
SOP (canonical):
F
= A'B'C' + A'BC' + AB'C + ABC'
= Σm(0,2,5,6)
Rows with output 0 are 001, 011, 100, 111 i.e. maxterms M(1,3,4,7).
POS (canonical):
F
= (A + B + C')(A + B' + C')(A' + B + C)(A' + B' + C')
= ΠM(1,3,4,7)
Question 18
18. Write POS Boolean expressions for F(a, b) = ab’+a’b. Show through Boolean algebra reduction that the SOP and POS expressions are indeed equivalent to one another.
Answer 18
Given SOP: F = ab' + a'b.
Required POS form is F = (a+b)(a'+b').
Verification by expansion:
(a+b)(a'+b')
= aa' + ab' + ba' + bb'
= 0 + ab' + a'b + 0
= ab' + a'b
So the POS expression simplifies exactly to the original SOP expression. Therefore, both forms are equivalent representations of the same Boolean function.
Question 19
19. Minimize the following Boolean functions using algebraic method:
i)
Z = f(a, b, c) = a’b’c’ +a’b + abc’ +ac
ii)
Z = f (a,b,c) = a’b +bc’ +bc+ ab’c’
iii)
Z = f (a,b,c) = a’b’c’ + …
Answer 19
i) Z
= a'b'c' + a'b + abc' + ac
= a'b'c' + a'b + a(c+bc')
= a'b'c' + a'b + a(b+c)
= a'b'c' + a'b + ab + ac
= a'b'c' + b + ac (since a'b + ab = b)
= b + a'c' + ac (using b + b'X = b + X)
ii) Z
= a'b + bc' + bc + ab'c'
= a'b + b + ab'c' (since bc' + bc = b)
= b + ab'c' (because b + a'b = b)
= b + ac' (using b + b'X = b + X)
iii)
The third expression is incomplete in the current question text (a'b'c' + ...), so its algebraic minimization cannot be determined until the full expression is provided.
Question 20
20. Reduce the following Boolean expression using k-map
i)
F (a, b, c) = a’b’c’+ a’bc’+ a ‘bc’ + a’bc + ab’c’ + abc’ to SOP form
ii)
F(w,x,y,z) = (w + x) ( x + z’) (w’ + y’ + z) to SOP form
Answer 20
i) F(a,b,c)
= a'b'c' + a'bc' + a'bc' + a'bc + ab'c' + abc'
= a'b'c' + a'bc' + a'bc + ab'c' + abc'(remove duplicate term)
= c'(a'b' + a'b + ab' + ab) + a'bc
= c'(1) + a'bc
= c' + a'bc
= (c' + a'b)(c' + c)
= c' + a'b
Reduced SOP:F = c' + a'b
ii) F(w,x,y,z)
= (w+x)(x+z')(w'+y'+z)
= (x+wz')(w'+y'+z) (using (x + a)(x + b) = x + ab)
= x(w'+y'+z) + wz'(w'+y'+z)
= xw' + xy' + xz + wz'w' + wz'y' + wz'z
= xw' + xy' + xz + wz'y' (since ww' = 0 and z'z = 0)
Reduced SOP:F = xw' + xy' + xz + wy'z'
Question 21
21. Convert the given forms to SOP and POS form (as provided in the question table).
i)
A
B
C
F1
0
0
0
1
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
0
1
0
1
1
1
1
0
0
1
1
1
0
To SOP and POS form
ii)
A
B
C
D
F2
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0
1
1
0
0
1
0
0
0
0
1
0
1
0
0
1
1
0
0
0
1
1
1
0
1
0
0
0
1
1
0
0
1
0
1
0
1
0
1
1
0
1
1
1
1
1
0
0
0
1
1
0
1
0
1
1
1
0
1
1
1
1
1
1
To SOP and POS form
Answer 21
i)
F1=1 at rows 000,001,011,101⇒ Σ m(0,1,3,5).
Canonical SOP:
F1 = A'B'C' + A'B'C + A'BC + AB'C.
F1=0 at rows 010,100,110,111⇒ Π M(2,4,6,7).
Canonical POS:
F1=(A+B'+C)(A'+B+C)(A'+B'+C)(A'+B'+C').
ii)
F2=1 at rows 1000,1010,1011,1110,1111⇒ Σ m(8,10,11,14,15).
Canonical SOP:
F2 = AB'C'D' + AB'CD' + AB'CD + ABCD' + ABCD.
Simplification steps (grouping idea / algebraic merge):
F2
= AB'C'D' + AB'CD' + AB'CD + ABCD' + ABCD
= AB'D'(C'+C) + AB'C(D'+D) + ACD'(B'+B)
= AB'D' + AB'C + ACD'
= A(B'D' + B'C + CD')
= A(C + B'D')
F2 = 0 at rows 0,1,2,3,4,5,6,7,9,12,13⇒ Π M(0,1,2,3,4,5,6,7,9,12,13).
Canonical POS:
F2 = Π M(0,1,2,3,4,5,6,7,9,12,13).
Question 22
22. Obtain the minterm canonical form of the Boolean expression by algebraic method
i)
xyz + xy + x'(yz’ + y’z)
ii)
ab + abc + a’b + ab’c
Answer 22
i)
F
= xyz + xy + x'(yz' + y'z)
= xyz + xy(z+z') + x'yz' + x'y'z
= xyz + xyz + xyz' + x'yz' + x'y'z
= xyz + xyz' + x'yz' + x'y'z
= Σm(7,6,2,1)
= Σm(1,2,6,7).
ii)
F
= ab + abc + a'b + ab'c
= ab + a'b + ab'c (since ab + abc = ab)
= b(a+a') + ab'c
= b + ab'c
= b(c+c') + ab'c
= bc + bc' + ab'c
= abc + a'bc + abc' + a'bc' + ab'c
= Σm(7,3,6,2,5)
= Σm(2,3,5,6,7)
Note i: Minterm index is read from binary variable values in order. For part (i), using order xyz: x'y'z=001→1, x'yz'=010→2, xyz'=110→6, xyz=111→7. So Σm(7,6,2,1) and Σm(1,2,6,7) are the same set; ascending order is standard.
Note ii: Using order abc: a'bc'=010→2, a'bc=011→3, ab'c=101→5, abc'=110→6, abc=111→7. Hence Σm(7,3,6,2,5) and Σm(2,3,5,6,7) represent the same minterm set.