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Grading Policy
❖ 15%
❖ Assignments. (CA> 75% =>1; 25% < CA < 75%=> 0.8; CA < 25% =>0.3)
❖ Quizzes.
❖ Tutorial
CA: Correct answers
Boolean Expression Simplification❖ Reduction techniques:
❖ Boolean algebra.
❖ K-map simplification:
Rules of Boolean algebra
4
Additional Boolean Algebra
Rules
AB + A’B’
6
Operator Precedence
★Parentheses ( . . . ) • ( . . .)
★NOT x’ + y
★AND x + x • y
★OR
])([ xwzyx ++
])([
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xwzyx
xwzy
xwz
xw
xw
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Salman Bin Abdul Aziz University
❖ Q1. Demonstrate the validity of the following identities by means of truth table:
❖ a) DeMorgan’s theorem:
• (x + y + z)’ = x’y’z’
• (xyz)’ = x’+y’+z’
x y z (z+y+z)’ x’y’z’ (xyz)’ x’+y’+z’0 0 0 1 1 1 10 0 1 0 0 1 10 1 0 0 0 1 10 1 1 0 0 1 11 0 0 0 0 1 11 0 1 0 0 1 11 1 0 0 0 1 11 1 1 0 0 0 0
❖ b) The distributive law
• x + yz = (x +y)(x +z)
• x(y + z) = xy + xz x y z x+yz (x +y)(x +z) x(y + z) xy + xz0 0 0 0 0 0 00 0 1 0 0 0 00 1 0 0 0 0 00 1 1 1 1 0 01 0 0 1 1 0 01 0 1 1 1 1 11 1 0 1 1 1 11 1 1 1 1 1 1
❖ c) The Associative law,
• x + (y + z) = (x + y) + z
• x(yz) = (xy)zx y z x + (y + z) (x + y) + z x(yz) (xy)z0 0 0 0 0 0 00 0 1 1 1 0 00 1 0 1 1 0 00 1 1 1 1 0 01 0 0 1 1 0 01 0 1 1 1 0 01 1 0 1 1 0 01 1 1 1 1 1 1
Simplifying by Boolean laws❖ Q2. Simplify the following Boolean expressions to a minimum
number of literals:
A. xy + xy’
B. (x+y)(x+y’)
C. xyz + x’y + xyz’
D. (A+B)’ (A’+B’)’
❖ Solution: ❖ a) xy + xy’ = x(y + y’) = x.1 = x
a) xy + xy’ = xx y x.y x.y’ xy + xy’
0 0 0 0 0
0 1 0 0 0
1 0 0 1 1
1 1 1 0 1Truth table is not required in this question, it’s just to make sure your answer is correct
❖ b) (x + y)(x + y’) = x+(y.y’) = x+0 = x
b) (x + y)(x + y’) = xx y (x + y) (x + y’) (x + y)(x + y’)
0 0 0 1 0
0 1 1 0 0
1 0 1 1 1
1 1 1 1 1Truth table is not required in this question, it’s just to make sure your answer is correct
❖ c) F(x,y,z) = xyz + x’y + xyz’ = xy(z + z’) + x’y = xy.1 + x’y = xy + x’y = y(x + x’) = y.1 =y
c) xyz + x’y + xyz’ = yx y z xyz x’y xyz’ F0 0 0 0 0 0 00 0 1 0 0 0 00 1 0 0 1 0 10 1 1 0 1 0 11 0 0 0 0 0 01 0 1 0 0 0 01 1 0 0 0 1 11 1 1 1 0 0 1
Truth table is not required in this question, it’s just to make sure your answer is correct
❖ d) (A + B)’(A’ + B’)’ = (A’.B’).(A.B) = (A.A’.B’.B) = 0
d) (A + B)’(A’ + B’)’ = 0
A B (A’.B’) (A.B) (A’.B’).(A.B)
0 0 1 0 0
0 1 0 0 0
1 0 0 0 0
1 1 0 1 0
From Boolean rules
• (X.Y)Z = X(Y.Z)
• X.X’ = 0
Truth table is not required in this question, it’s just to make sure your answer is correct
logic diagram❖ Q3. Draw logic diagram of the following expressions in
question 2 before and after simplification:
A. xy + xy’
B. (x+y)(x+y’)
C. xyz + x’y + xyz’
D. (A+B)’ (A’+B’)’
Boolean Expressions
Before simplification After simplification
a. xy + xy’ x
b. (x+y)(x+y’) x
c. xyz + x’y + xyz’ y
d. (A+B)’ (A’+B’)’ 0
Before simplification
After simplification
Before simplification
After simplification
❖ Q4. Find the complement of the following expressions:
A. xy’ + x’y
B. (x’ + y + z’) (x + y’) (x + z)
C. (A’B + CD)E’ + E
❖ Solution:
A. F = xy’ + x’y F’=??
F’ = [xy’ + x’y]’ = [(xy’)’ . (x’y)’] = (x’ + y).(x + y’) = x.x’ + x’.y’ + x.y + y.y’ = 0 + x’y’ + xy + 0 = x’y’ + xy
Hint, [xy’ + x’y] is (x XOR y) [x’y’ + xy] is (x XNOR y)
This question could be “Demonstrate the validity that
XOR inverse XNOR”
F= (xy’ + x’y) = x⊕y
So, F’=(x⊕y)’ = x’y’ + xy
B. F = (x’ + y + z’) (x + y’) (x + z) F’=??
F’ = [(x’+y+z’)(x+y’)(x+z)]’ = (x’+y+z’)’ + (x+y’)’ + (x+z)’ = xy’z + x’y + x’z’
C. F = (A’B + CD)E’ + E F’=??
F’ = [(A’B + CD)E’ + E]’
= [(A’B + CD)E’]’.E’
= [(A’B + CD)’ + E].E’
= [(A + B’)(C’ + D’) + E] . E’
= (A + B’)(C’ + D’).E’ + E.E’
= (A + B’)(C’ + D’).E’
First Way
C. F = (A’B + CD)E’ + E F’=??
= (E’+E).(A’B+CD + E)
= A’B + CD + E
F’ = [A’B + CD + E]’
= [(A’B)’ . (CD)’ . E’]
= [(A+B’) . (C’+D’) . E’]
Second Way
Converting between SoP and PoS❖ Any function can be express as a sum of minterms or as
product of maxterms.
❖ Two ways to write sum of minterms:
• Numerical form: F(a,b,c) = Σ( 1,3,6)
• Literals form: F(a,b,c) = a’b’c + a’bc + abc’
• Two ways to write product of maxterms:
• Numerical form: F(a,b,c) = ∏(0,2,4,5).
• Literals form: F(a,b,c) = (a+b+c) (a+b’+c) (a’+b+c) (a’+b+c’)
Hint, in SoP,(0)10 —> (000)2 which is a’b’c’
Hint, in PoS,(0)10 —> (000)2
which is (a+b+c)
Minterms, Sum of Products (SoP)
❖ F1(x,y,z) = Σ (0,2,4,6,7)
❖ F1(x,y,z) =x’y’z’ + x’yz' + xy’z' + xyz’ + xyz
minterms x y z F1
0 0 0 0 1
1 0 0 1 0
2 0 1 0 1
3 0 1 1 0
4 1 0 0 1
5 1 0 1 0
6 1 1 0 1
7 1 1 1 1
If you have one of these, you can get the rest:1.Boolean Expression Numerical form2.Boolean Expression literal form3.Truth table4.Logic diagram
Maxterms, Product of Sums (PoS)❖ F2(x,y,z) = ∏ (1,3,5)
❖ F2(x,y,z) = (x+y+z’) (x+y’+z’) (x’+y+z’)
Maxterms x y z F2
0 0 0 0 1
1 0 0 1 0
2 0 1 0 1
3 0 1 1 0
4 1 0 0 1
5 1 0 1 0
6 1 1 0 1
7 1 1 1 1
F Complement in SoP/PoS
SoP (minterms)
PoS (maxterms)
complement of SoP
complement of PoS
F(x,y) = Σ() F(x,y) = ∏() F’(x,y) = Σ() F’(x,y)= ∏()
F(x,y)= Σ(0,1) F(x,y)= ∏(2,3) F’(x,y)= Σ(2,3) F’(x,y)= ∏(0,1)
m/M
x y F F’
0 0 0 1 0
1 0 1 1 0
2 1 0 0 1
3 1 1 0 1
Hint, When you change from column to another you change just 2 of the 3 components (F, Σ/∏ , numbers)
❖ Q6. Convert each of the following to the other canonical form:
A. F(x,y,z) = Σ(2,5,6) ——> SoPsolution ,, F(x,y,z) = ∏ (0,1,3,4,7) ——> PoS
❖ Q.6
B. F(A,B,C,D) = ∏(0,1,2,4,7,9,12) solution ,, F(A,B,C,D) = Σ(3,5,6,8,10,11,13,14,15)
❖ Q7. Express the complement of the following function in SoP (minterms) form:
A. F(x,y,z,w) = Σ(3,5,9,11,15) solution ,, F’(x,y,z,w) = Σ(0,1,2,4,6,7,8,10,12,13,14)
❖ Q7
B. .F(x,y,z) = ∏ (0,1,3,4,7) solution ,, F’(x,y,z) = Σ(0,1,3,4,7)
❖ Q.8a Express the following function as a sum of minterms and as product of maxterms (standard forms):
a. F(a,b,c,d) = a’d + b’d + bd
❖ First, Finding F in SoP (standard forms):
❖ Solution,, F(a,b,c,d) = a’b’c’d + a’b’cd + a’bc’d + a’bcd + a’b’c’d + a’b’cd + ab’c’d + ab’cd + a’bc’d + a’bcd + abc’d + abcd
Hint, [a’d + b’d + bd] —> incomplete input in every term so, it’s not SoP [a’bc’d + a’bcd + abc’d + abcd] —> every input in every term so, it’s SoP
❖ Second, Finding F in PoS (standard forms):
❖ Since, F(a,b,c,d) = a’b’c’d + a’b’cd + a’bc’d + a’bcd + a’b’c’d + a’b’cd + ab’c’d + ab’cd + a’bc’d + a’bcd + abc’d + abcd
❖ So, F(a,b,c,d) = ∑(1,3,5,7,9,11,13,15)
❖ And we can find F in PoS (numerical form) F(a,b,c,d) = ∏(0,2,4,6,8,10,12,14)
❖ Thus, F(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)
First way: SoP Literal form —> SoP numerical form —> PoS numerical form —> PoS Literal form
F(a,b,c,d) = a’d + b’d + bd
❖ Second, Convert the Boolean expression to product of maxterms:
❖ We need here extra step, create truth table from SoP.
❖ F(a,b,c,d) = ∏(0,2,4,6,8,10,12,14)
❖ F(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)
m/M a b c d F0 0 0 0 0 01 0 0 0 1 12 0 0 1 0 03 0 0 1 1 14 0 1 0 0 05 0 1 0 1 16 0 1 1 0 07 0 1 1 1 18 1 0 0 0 09 1 0 0 1 1
10 1 0 1 0 011 1 0 1 1 112 1 1 0 0 013 1 1 0 1 114 1 1 1 0 015 1 1 1 1 1
Second way: SoP Literal form —> Truth Table —> PoS numerical form & PoS Literal form
❖ Q8b Express the following function as a sum of minterms and as product of maxterms:
b. (AB+C)(B+C’D)
❖ Solution,,, F(A,B,C,D) = (AB+C)(B+C’D) = ABB + ABC’D + BC +CC’D = AB + ABC’D + BC +0.D = AB + ABC’D + BC
❖ The Boolean expression [F(A,B,C,D) = AB + ABC’D + BC] now is easy to deal with and solve the question as (Q8.a)
❖ Q.8c Express the following function as a sum of minterms and as product of maxterms:
c. x’ + x(x+y’)(y+z’)
❖ Solution,,, F(x,y,z) = x’ + x(x+y’)(y+z’) = x’+ {x[(xy)+(x.z’)+(y’y)+(y’z’)]} = x’ +{x[(xy)+(x.z’)+(y’z’)]} = x’ +[x(xy)+x(x.z’)+x(y’z’)] = x’ +[xxy+xxz’+xy’z’] = x’ + xy+ xz’+ xy’z’
❖ The Boolean expression [F(x,y,z) = x’ + xy+ xz’+ xy’z’] now is easy to deal with and solve the question as (Q8.a)
❖ Q. 9 Write Boolean expression and construct the truth table describing the outputs of the circuits described by the following logic diagram:
❖ y(a,b,c,d,e) = {[a(bcd)’e]’}’ = a(bcd)’e = a(b’+c’+d’)e = ab’e + ac’e + ad’e
a b c d e ab’e ac’e ad’e Y0 0 0 0 00 0 0 0 10 0 0 1 00 0 0 1 10 0 1 0 00 0 1 0 10 0 1 1 00 0 1 1 10 1 0 0 00 1 0 0 10 1 0 1 00 1 0 1 10 1 1 0 00 1 1 0 10 1 1 1 00 1 1 1 11 0 0 0 01 0 0 0 1 1 1 1 11 0 0 1 01 0 0 1 1 1 1 11 0 1 0 01 0 1 0 1 1 1 11 0 1 1 01 0 1 1 1 1 11 1 0 0 01 1 0 0 1 1 1 11 1 0 1 01 1 0 1 1 1 11 1 1 0 01 1 1 0 1 1 11 1 1 1 01 1 1 1 1
Empty cells are 0
❖ y(a,b,c,d,e) = ab’e + ac’e + ad’e
❖ Terms meaning in SoP:
❖ ab’e —> when a=1&b=0&e=1 =>F=1
❖ ac’e —> when a=1&c=0&e=1 =>F=1
❖ ad’e —> when a=1&d=0&e=1 =>F=1
a b c d e Y0 0 0 0 00 0 0 0 10 0 0 1 00 0 0 1 10 0 1 0 00 0 1 0 10 0 1 1 00 0 1 1 10 1 0 0 00 1 0 0 10 1 0 1 00 1 0 1 10 1 1 0 00 1 1 0 10 1 1 1 00 1 1 1 11 0 0 0 01 0 0 0 1 11 0 0 1 01 0 0 1 1 11 0 1 0 01 0 1 0 1 11 0 1 1 01 0 1 1 1 11 1 0 0 01 1 0 0 1 11 1 0 1 01 1 0 1 1 11 1 1 0 01 1 1 0 1 11 1 1 1 01 1 1 1 1
Empty cells are 0
The Second Way to Find Truth Table
❖ Q. 9 Write Boolean expression and construct the truth table describing the outputs of the circuits described by the following logic diagram:
❖ y1(a,b,c,d,e,f) = a ⊕(c+d+e) = a(c+d+e)’ + a’(c+d+e) = a(c’d’e’) + a’c + a’d + a’e = ac’d’e’ + a’c + a’d + a’e
❖ y2(a,b,c,d,e,f) = {[b’(c+d+e)f]’}’ = b’(c+d+e)f = b’f(c+d+e) = b’cf + b’df + b’ef
Hint, we know that [x ⊕ y = x’.y+x.y’] So,a ⊕ (c+d+e) = a’. (c+d+e) + a. (c+d+e)’
a b c d e f ac’d’e’ a’c a’d a’e y10 0 0 0 0 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 00 0 0 0 1 0 0 0 0 1 10 0 0 0 1 1 0 0 0 1 1
Continue until 111111
a b c d e f b’cf b’df b’ef y20 0 0 0 0 0 0 0 0 00 0 0 0 0 1 0 0 0 00 0 0 0 1 0 0 0 0 00 0 0 0 1 1 0 0 1 1
Continue until 111111
❖ y1= ac’d’e’ + a’c + a’d + a’e
❖ y2= b’cf + b’df + b’ef
❖ ———————-
❖ Terms meaning in SoP for y1:
❖ ac’d’e’ —> when =>F=1
❖ a’c —> when =>F=1
❖ a’d —> when =>F=1
❖ a’e —> when =>F=1
❖ ——————
❖ Terms meaning in SoP for y2:
❖ …
a b c d e f y10 0 0 0 0 0 00 0 0 0 0 1 00 0 0 0 1 0 10 0 0 0 1 1 1
Continue until 111111
a b c d e f y20 0 0 0 0 0 00 0 0 0 0 1 00 0 0 0 1 0 00 0 0 0 1 1 1
Continue until 111111
The Second Way to Find Truth Table
❖ Q. Create the corresponding Boolean expression in sum-of-products form.
❖ X(A,B,C) = A’B’C’ + A’B’C + A’BC’ + AB’C’ = A’B’C’ + A’B’C + A’B’C’ + A’BC’ + A’B’C’ + AB’C’ = A’B’(C’+C) + A’C’(B+B’) + B’C’(A’+A) = A’B’ + A’C’ + B’C’
❖ F(A,B,C) = A’B’C’ + A’B’C + A’BC’ + AB’C’ = A’B’(C’+C) + A’C’(B+B’) + B’C’(A’+A) = A’B’ + A’C’ + B’C’
❖ F(A,B,C,D) = (A+C+D)(A+C+D’)(A+C’+D)(A+B’) = [A+C+(D.D’)] . [(A+D+(C.C’)] . (A+B’) = (A+C) (A+D) (A+B’)
هامش= ACD + ACD’ + AC’D + AB’= AC(D+D’) + AD(C+C’) +AB’
= AC + AD +AB’
Extra Questions to think of
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