LOGICAL EXPRESSIONS AND THEIR EQUIVALENT FORMS: (PART 2)
In this tutorial, I am going to provide solved examples with all the steps on the topics given in the outline. These topics are explained in detail in the previous tutorial, I discussed the following topics in detail. Please check out the first part here.
OUTLINE:
Solved examples on the following topics:
Minterms
Maxterms
SOP (Sum of Products)
POS (Product Of Sum)
Evaluation of complement of a function with the help of canonical forms (sum of minterms and product of maxterms)
POS to standard POS
SOP to standard SOP
Conversion between standard forms; From SOP to POS and POS to SOP
EXAMPLE 1: Express the function in sum of minterms and product of maxterms (XY + Z)(Y + XZ)
Step 1: Draw ruth table along with sum of minterms and product of maxterms.
Step 2: Express the given function in sum of minterms as:
F = X’YZ + XY’Z + XYZ’ + XYZ
F = m3 + m5 + m6 + m7F = ∑ (3, 5, 6, 7)
Step 3: Express the given function in product of maxterms as:
F = (X +Y +Z)(X +Y +Z’)(X +Y’ +Z)(X’ +Y +Z)
F = M0 . M1 . M2 . M4
F( X, Y, Z) = π (0, 1, 2, 4)
EXAMPLE 2:
Express the function in sum of minterms and product of maxterms (X’ + Y)(Y’ + Z)
Step 1: Draw ruth table along with sum of minterms and product of maxterms.
Step 2: Express the given function in sum of minterms as:
F = X’Y’Z’ + X’Y’Z + X’YZ + XYZ
F = m0 ,+ m1 + m3 + m7
F = ∑ (0, 1, 3, 7)
Step 3: Express the given function in product of maxterms as:
F = (X + Y’ + Z)(X’ + Y +Z)(X’ + Y +Z’)(X’ + Y’ +Z)
F = M2 M4 M5 M6
F( X, Y, Z) = π (2, 4, 5, 6)
EXAMPLE 3:
Express the function in sum of minterms and product of maxterms Y’Z + WXY’ +WXZ’ +W’X’Z
Step 1: Draw ruth table along with sum of minterms and product of maxterms.
Step 2: Express the given function in sum of minterms as:
F = W’X’Y’Z + W’X’YZ + W’XY’Z + WX’Y’Z + WXY’Z’ + WXY’Z + WXYZ’
F = m1 + m3 + m5 + m9 + m12 + m13 + m14
F = ∑ (1, 3, 5, 9, 12, 13, 14)
Step 3: Express the given function in product of maxterms as:
F = (W + X +Y +Z)(W + X +Y’ +Z)(W + X’ +Y +Z)(W + X’ +Y’ +Z)(W + X’ +Y’ +Z’)(W’ + X +Y +Z)(W’ + X +Y’ +Z)(W’ + X +Y’ +Z’)(W’ + X’ +Y’ +Z’)
F = M0 . M2 . M4 . M6 . M7 . M8 . M10 . M11 . M15
F(W, X, Y, Z) = π (0, 2, 4, 6, 7, 8, 10, 11, 15)
EXAMPLE 4: Evaluate complement of a function from the following table using canonical forms.
The function is evaluated from the truth table as
F = m2 + m3 + m6 + m7
The complement of the above function is calculated as;
F’ = (m2 + m3 + m6 + m7)’
F’ = m2’ + m3’ + m6’ + m7’
Mj’ = Mj
The complement function in maxterms form is written below;
F’ = M2 . M3 . M6 . M7
The complement function in minterms form is written below;
F’ = m0 + m1 + m4 + m5
F’(X, Y, Z) = ∑ ( 0, 1, 4, 5)
EXAMPLE 5: Express complement of the function in sum of minterms. F (A, B, C, D) = ∑ (0, 2, 6, 11, 13, 14)
F = m0 + m2 + m6 + m11 + m13 + m14
As I discussed in my previous tutorial, the complement of the above function contains minterms that are missing in the original function.
F’ = m1 + m3 + m4 + m5 +m7 + m8 + m9 + m10 + m12 + m15
F’ (A, B, C, D) = ∑ (1, 3, 4, 5, 7, 8, 9, 10, 12, 15)
Or
It can evaluate from the following method as well;
F’ = (m0 + m2 + m6 + m11 + m13 + m14)’
F’ = m0’ . m2’ . m6’ . m11’ . m13’ . m14’
mj’ = Mj
The complement function in product of maxterms;
F’ = M0 . M2 . M6 . M11 . M13 . M14
The complement function in sum of minterms;
F’ = m1 + m3 + m4 + m5 +m7 + m8 + m9 + m10 + m12 + m15
F’ (A, B, C, D) = ∑ (1, 3, 4, 5, 7, 8, 9, 10, 12, 15)
EXAMPLE 6: For each of the following function, determine domain of the function, SOP, standard SOP, and then convert standard SOP into standard POS
(A + B)(C + B’)
Domain: A, B, C
SOP:
Apply distributive law:
= AC + AB’ + BC + B.B’
B.B’ = 0
= AC + AB’ + BC
Standadrd SOP:
Look for non-standard term, and then look for missing variable
= AC (B + B’) + AB’(C + C’) + BC(A +A’)
= ABC + AB’C + AB’C + AB’C’ + ABC + A’BC
Remove those terms that appear twice.
= AB’C + AB’C’ + ABC + A’BC
Standard SOP To Standard POS:
There is 23 = 8 possible combinations. The standard SOP contains 4 combinations, the rest of the combinations are part of the standard POS.
You can write directly as well, but for ease, I draw a truth table and then write standard POS expression.
The standard POS expression is
= (A + B + C)(A + B + C’)(A + B’ +C)(A’ + B’ +C)
(A + B’C)C
Domain: A, B, C
SOP:
= AC + B’CC = AC + B’C
C.C = C
Standadrd SOP:
Look for non-standard term, and then look for missing variable
= AC (B + B’) + B’C (A + A’)
= ABC + AB’C + AB’C + A’B’C
Remove those terms that appear twice.
= ABC + AB’C + A’B’C
Standard SOP To Standard POS:
There is 23 = 8 possible combinations. The standard SOP contains 3 combinations, the rest of the combinations are part of the standard POS.
You can write directly as well, but for ease, I draw a truth table and then write standard POS expression.
The standard POS expression is
= (A + B + C)(A + B’ + C)(A + B’ +C’)(A’ + B +C)(A’ + B’ + C)
(A + C)(AB + AC)
Domain: A, B, C
SOP:
= AAB + AAC + ABC + AAC
A.A = A
= AB + AC + ABC + AC
Remove those terms that appear twice.
= AB + AC + ABC
Standard SOP:
Look for non-standard term, and then look for missing variable
= AB (C + C’) + AC (B + B’) + ABC
= ABC + ABC’ + ABC + AB’C + ABC
Remove those terms that appear twice.
= ABC’ + ABC + AB’C
Standard SOP To Standard POS:
There is 23 = 8 possible combinations. The standard SOP contains 3 combinations, the rest of the combinations are part of the standard POS.
You can write directly as well, but for ease, I draw a truth table and then write standard POS expression.
The POS expression is
= (A + B + C)(A + B + C’)(A + B’ + C)(A + B’ + C’)(A’ + B + C)
AB + CD(AB’ + CD)
Domain: A, B, C, D
SOP:
= AB + AB’CD + CD.CD
= AB + AB’CD + CD
Standard SOP:
Look for non-standard term, and then look for missing variable
= AB (C + C’) + AB’CD + (A + A’)CD
= ABC + ABC’ + AB’CD + ACD + A’CD
Again, look for non-standard terms, and then look for missing variables.
= ABC(D + D’) + ABC’(D + D’) + AB’CD + ACD (B + B’) A’CD (B + B’)
= ABCD + ABCD’ + ABC’D + ABC’D’ + AB’CD + ABCD + AB’CD + A’BCD + A’B’CD
Remove those terms that appear twice.
= ABCD + ABCD’ + ABC’D + ABC’D’ + AB’CD + A’BCD + A’B’CD
The POS expression is
= (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)
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