Combinational Logic Circuits

combinational circuits
Digital logic circuits can be classified into
“combinational” and “sequential”.
A combinational logic circuit is one whose output
solely depends on its current inputs.
sequential circuits, on the other hand, are built using
combinational circuits and memory elements called
“flip-flops

Steps for converting a verbal problem
statement to a truthtable
1. Determine the input variables and output variables that are involved.
2. Assign mnemonic/letter symbols to each variable.
3. Determine the size of the truth table; how many input combinations exist:
2x= y
where x = number of input variables and y = number of combinations.
4. Construct a truth table containing all of the input variable combinations.
5. By careful reading of the problem statement determine the combinations of
inputs that cause a given output to be true. For the remaining conditions
output will be false.

Example 1
A combinational logic circuit with three input variables that will produce a logic 1 output when
more than one input variables are logic 1. Derive the truth table.
Solution: Given problem specifies that there are three input variables and one output variable. We
assign A, B and C letter symbols to three input variables and assign Y letter symbol to one output
variable. The relationship between input variables and output variable can be tabulated as shown in
truth table. The truth table specifies the output state (either O or 1) for all possible combinations of
input variables A B C Y
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1

Example 2
Two motors M2 and M1 are controlled by three sensors s3, s2, s1. One motor M2 is to run anytime all
three sensors are on. The other motor is to run, whenever sensors s2 or s1 but not both are on and s3 is
off. For all sensor combination, where M1 is on, M2 is to be off except when all the three sensors are off
and then both motors must remain off, construct the truth table and write the Boolean output equation.
Solution:
 No. of input variables: 3; No. of output variables: 0
S3 S2 S1 M1 M2
0 0 0 0 0
0 0 1 1 0
0 1 0 1 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

Example 4: A NASA system consists of three computers, two of which are on-line
(connected to the system) at any given time. The system uses three computers to
ensure safety in spacecraft operation through redundancy. If one computer
experiences a problem it is taken off-line and another computer is brought on-line.
Self-checking diagnostics determine each computer's operating status and generate
an output in the event of failure. When one computer fails it must be switched off-
line. No more than two computers are to be on-line at any given time. Design the
control logic to connect or disconnect the computers. In the event that two
computers are unavailable, generate a warning and allow the third computer to come
online. If all three computers are unavailable, generate a second warning signal that
invokes emergency procedures

A conveyor system brings raw material in from three different sources. The three
sources converge into a single output conveyor. Sensors mounted adjacent to each
source conveyor indicate the presence of raw material. All four conveyors have
separate motors so they can be individually controlled. Each source conveyor can
have a different speed. The output product flow rate is fixed; it can be turned only on
or off. The output product rate must match the source flow rates. To accomplish this.
the following conditions must be met. If source l has product, then sources 2 and 3
must be turned off; if source 1 is empty, then either 2 or 3 or both can be turned on.
In the event that no product is available from the three sources, the output conveyor
must be turned off. If no product is available, the respective source conveyor must
be turned off.M4 is on .

Sum of Products and Product of Sum Forms
1. Boolean algebra variables have only two possible values, 1 or 0 and it is the
language used for performing binary logic operations.
Sum of Product (SOP) is logical OR of multiple product terms. Ex 𝑎 +
𝑎𝑏̅ + 𝑎𝑏𝑐
ABC +A
2𝐴𝐵𝐶 + 𝐴𝐵̅𝐶̅
𝑃̅𝑄 + 𝑄𝑅 + 𝑅𝑆̅
SOP is also known as disjunctive normal form (independent)
Product of Sum (POS) is logical AND of multiple sum terms. Ex 𝑎. (𝑎 + 𝑏). (𝑎 + 𝑏̅ + 𝑐)
𝐴(𝐵̅ + 𝐴)(𝐵 + 𝐶)
(𝑃 + ̅𝑄)(𝑅 +S)
POS from is also known as conjunctive normal form. (dependent)

Standard SOP and POS Forms
Consider the expression 𝐴𝐵 + 𝐴𝐵𝐶̅ the first product term do not
contain literal C. i.e all the individual terms do not involve all
literals. If each term in SOP form contains all the literals then the
SOP form is known as standard or canonical SOP form.
Each individual term in an SOP is called as minterm. Ex. 𝐴𝐵̅𝐶 + 𝐴𝐵𝐶
+ 𝐴̅𝐵𝐶
Similarly, a canonical POS form example is (𝐴 + 𝐵 + 𝐶)(𝐴 + 𝐵̅ + 𝐶).
Each individual term in POS is called as maxterm.

Steps to Convert SOP
to Standard SOP
• Identify the missing variable(s) in each AND term.
• AND the missing term and its complement with the original AND
term, xy (z + z').
• Because (z + z') = 1, the original AND term value is not changed.
• Expand the term by application of the property of distribution,
xyz + xyz'.
• Reduce the expression by omitting repeated product terms if
any. Because A+ A= A.

Steps to Convert POS to Standard POS
• Identify the missing variable(s) in each OR term.
• OR the missing term(s) and its complement with the original OR
term, x + y' +zz'
• Because zz' = 0, the original OR term value is not changed.
• Expand the term by application of distributive property, (x + y' +
z) (x + y' + z').
• Reduce the expression by omitting repeated sum terms if any.
Because A . A = A.

Example 3
f (a,b,c) = (a+b`)(b`+c)

Lower case m is used to denotes a “minterm” and uppercase M is
used to denote “maxterm”.
For minterm 1 represents normal form, 0 represents complemented
form
For maxterm 0 represents normal form, 1 represents complemented form

Practice Problems
• Express the following function in standard SOP.
𝐹1 = 𝐴𝐵 + 𝐶̅𝐷 + 𝐴𝐵̅𝐶
• Determine the canonical SOP form of
f (x,y,z) = (𝑥𝑦 + 𝑧̅)(𝑦 + 𝑥𝑧)̅
• Convert the give expression in standard POS form
𝑓(𝑃, 𝑄, 𝑅) = (𝑃 + 𝑄)̅ (𝑃 + 𝑅)
• Convert the given expression in standard POS form
𝑌 = 𝐴(𝐴 + 𝐵 + 𝑐)

Generation of Switching Equations From Truth Tables
Example 1: Construct the truth table for the following Boolean
Expression
F (a, b, c) = ∑m (0,1,2,5,6,7)
a b c f
0 0 0 1
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 1

Example 2: F(a,b,c,d) = πM (0,1,5,6,7,11,12,14,15)
a b c d f
0 0 0 0 0
0 0 0 1 0
0 0 1 0 1
0 0 1 1 1
0 1 0 0 1
0 1 0 1 0
0 1 1 0 0
0 1 1 1 0
1 0 0 0 1
1 0 0 1 1
1 0 1 0 1
1 0 1 1 0
1 1 0 0 0
1 1 0 1 1
1 1 1 0 0
1 1 1 1 0

Example 3: Write the canonical minterm and maxterm expressions for the following
table. M is the output variable. a, b, m, s are the input variables
The minterm expression for the output variable, M, is
M = f(a, b, m, s) = a'bms + ab’ms + abms = ∑(7, 11, 15)
a b m s f
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 1
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
1 1 0 0 0
1 1 0 1 0
1 1 1 0 0
1 1 1 1 1

Example 4 : Express the following SOP equations in
a minterm list (shorthand decimalnotation) form:

Example 5 : Express the following POS equations in a maxterm list
(decimal notation) form:

Example 6: Convert the following Boolean function into:
R= f (a,b,c) = (a` + b) (b + c`) minterm Canonical form
R=f(x,y,z) = x + x`z`(y + z`) maxterm Canonical form
Solution:
1. R= f (a,b,c) = (a` + b) (b + c`)
= (a` + b + c.c`) (a.a` + b + c`)
= (a` + b + c) (a` +b + c`) (a + b + c`) (a` + b + c`)
= (a` +b +c) (a` + b + c`) (a + b + c`)
= πM(4, 5, 1)
Minterm Canonical form =∑m(0,2,3,6,7,)

R=f(x,y,z) = x + x`z`(y + z`)
= x + x`z`y + x`z`z`
=x(y + y`) + x`z`y + x`z`
= xy + xy` + x`z`y + x`z`(y + y`)
= xy(z + z`) +xy`(z + z`) +x`z`y + x`z`y +x`z`y`
= xyz + xyz` + xy`z + xy`z` + x`z`y + x`z`y + x`z`y`
= xyz + xyz` + xy`z + xy`z` + x`z`y + x`z`y`
= ∑m(0, 2, 4, 5, 6, 7)
Maxterm Canonical form = πM(1, 3)

Note:
It is not necessary to first write switching equations in a variable
name format and then convert to a minterm or maxterm list (decimal
format). It is easier to write the canonical output equations directly
from the truth table in a minterm or maxterm numerical list than to
write the equations using input variable names. For example,
consider the canonical equations in following Table

KARNAUGH MAP (K-Map)
• The Karnaugh map is a matrix of squares. Each square represents a
minterm or maxterm from a Boolean equation. The arrangement of the
matrix square permits identification of input variable redundancies,
which helps reduce the output equation.
• The Karnaugh map identifies all of the cases for a given set of input
variables where groups of minterms may contain redundant variables of
the form of x + x' = 1.
• When these groups are identified, the redundant variables can be
eliminated, resulting in a simplified output function (abc + abc' = ab).
• If a given switching equation contains a minterm, then a 1 is entered
into the square that represents that term. A maxterm is represented by
a 0.
• One-Variable, Two-Variable, Three-Variable and Four-Variable Maps
• The basis of this method is a graphical chart known as Karnaugh map (K-
map). It contains boxes called cells. Each of the cell represents one of
the 2n possible products that can be formed from n variables.

Representing Standard SOP on K-Map
Example 1: Plot Boolean expression Y = A BC’+ A BC+ A’B’C on the
Karnaugh map

Grouping Cells for Simplification

• Generalized procedure to simplify Boolean expressions as follows:
• Plot the K-map and place ls in those cells corresponding to the ls in the
truth table or sum of product expression. Place 0s in other cells.
• Check the K-map for adjacent ls and encircle those 1s which are not
adjacent to any other l s. These are called isolated ls.
• Check for those ls which are adjacent to only one other 1 and encircle
such pairs.
• Check for quads and octets of adjacent ls even if it contains some ls that
have already been encircled. While doing this make sure that there are
minimum number of groups.
• Combine any pairs necessary to include any l s that have not yet been
grouped.
• Form the simplified expression by summing product terms of all the
groups.

Rules for Simplifying logic function using K-map are:
• Group should not include any cell containing a zero.
• The number of cells in a group must be a power of 2, such as 1, 2, 4, 8 or 16.
• Group may be horizontal, vertical but not diagonal.
• Cell containing 1 must be included in at least one group.
• Groups may overlap.
• Each group should be as large as possible to get maximum simplification.
• Groups may be wrapped around the map. The leftmost cell in a row may be
grouped with the rightmost cell and the top cell in a column may be grouped
with the bottom cell.
• A cell may be grouped more than once. The only condition is that every group
must have at least one cell that does not belong to any other group. Otherwise,
redundant terms will result.
• All above rules are stated considering the SOP simplification. In case of POS
simplification all rules are same except O (zero) takes place of 1 (one).

Example 4: Y=f (A, B, C, D) = ∑ (1,5,6,7,11,12,13,15)

Example 7 Reduce the following function
using Karnaugh map technique
Y=f (A, B, C, D) = ∑ (0,1,4,8,9,10)

Essential Prime Implicants
After grouping the cells, the sum terms which appear in the
K-map are called prime implicant groups. It is observed that
some cells may appear in only one prime implicant group;
while other cells may appear in more than one prime
implicant group. ln the K-map shown above, cells 1, 4, 9 and
10 appear in only one prime implicant group. These cells are
called essential cells and corresponding prime implicants are
called essential prime implicants.
Note: In the following examples placing of MSB and LSB
variables is different

Incompletely Specified Functions (Don't
Care Terms)
P=f (A, B, C, D) = ∑ (1, 2, 4, 7, 8) + ∑ d (10, 11, 12, 13, 14, 15)

Find the reduced SOP form of the following function
f (A, B, C, D) = ∑ (1, 3, 7, 11, 15) + ∑ d (0, 2, 4)
Example 2:

Example 3: Reduce the following function using Karnaugh map
technique. f (A, B, C, D) = ∑ (5, 6, 7, 12, 13) + ∑ d (4, 9, 14, 15)
f (A, B, C, D) = B
Example 2:

Example 2:
f (A, B, C) = ∑ (0, 1, 3, 7) + ∑ d (2, 5)
f (A, B, C) = A′ + C

Example 2:
f (W, X, Y, Z) = ∑ (0, 7, 8, 9, 10, 12) + ∑ d
(2, 5, 13).
W, X, Y, Z) = X′ Z′+ W′X Z+ WY′

Example 2: Using K-map find the minimized expression
for the following function and implement it using basic
gates: f (A, B, C, D) = ∑ (2, 4, 8, 11, 15) +∑ d(l, 10, 12, 13).

Simplifying Maxterm Equations
Simplify the function G = f(a, b, c, d) = π (0, 4, 5, 7, 8, 9, 11, 12, 13, 15)
G= ( a′+d′) (b′+d′) (c+d)

Example 2: Simplify the function G = f(a, b, c, d) = π (1,
3, 8, 10, 12,13, 14, 15

Example 4: Simplify the function f(a, b, c, d) = π ( 0, 2, 3, 8, 9,
12, 13, 15)

Practice Problems
Design a logic circuit with inputs P, Q, R so that output S is high whenever P
is zero or whenever Q = R = 1.

Practice Problems
3) (A, B, C, D) = π (4, 5, 6, 7, 8, 12, 13) + d (1, 15)

Practice Problems
Design a logic circuit that has 4 inputs, the output will only be high when
the majority of the inputs are high. Use K-map to simplify

Practice Problems
Using K-map obtain the minimal sum of products and the minimal product
of sums form of the function f (a, b, c, d) = ∑ ( 1,2, 3, 5, 6, 7, 8,13).

Practice Problems
Get the minimized sum of products expression for f (a, b, c, d) = ∑ (0,1,5,6,7,8,9)
with don't cares: ∑ (l0,11,12,13,14,15). Use Karnaugh map for simplification

Practice Problems
f (a, b, c, d) = ∑ (1,2,3,5,6,7,11,12,13,14,15) for the above expression:
i)Draw the logic diagram using AOI logic for minimal sum. Obtain minimal sum
using Kmap.

Simplifying Maxterm Equations
Find minimal sum and minimal product for the following function using K-map f (a,
b, c, d) =∑ (6,7,9,10,13) + ∑ d (1,4,5,11,15)

Identify all prime implicants and essential prime implicants of the following
functions using K-map.
i) f(a,b,c,d) = ∑ (0,1,2,5,6,7,8,9,10,13,14,15) ii) f(a,b,c,d) =π (0,2, 3,8,9,10,12,14)

Limitations of K-Map method
• map method of simplification is convenient as long as the
number of variables does not exceed five or six. As the
number of variables increases it is difficult to make
judgements about which combinations form the minimum
expression.
• In case of complex problem with 7, 8, or even 10 variables it
is almost an impossible task to simplify expression by the
mapping method. Another important point is that the K-
map simplification is manual technique and simplification
process is heavily depends on the human abilities.
• To meet this need, W. V. Quine and E. J. McCluskey
developed an exact tabular method to simplify the Boolean
expression. This method is called the Quine-McCluskey, or
Tabular method.

Example 1: D = f(a, b, c, d) =- ∑(0, 1, 2, 3, 6, 7, 8, 9, 14, 15)

D = b' c' + bc + a'b′
or
D = b' c' + bc + a' c

Quine-McClusky Using Don't CareTerms
Consider the following problem:·
S = f(w. x, y, z) = ∑ (1, 3, 13, 15) + ∑ d (8, 9, 10, 11)
Step 1: Construct a list of minterms and don't care terms classified
according to the number of ls. Indicate the don't care terms by using a *
symbol. Don't care terms are never included as prime implicants by
themselves.

Step 2: Compare terms in group n, including don't care terms, with terms
in group n + 1, looking for a single variable change. Treat don't care
terms as a 1 in finding primeimplicants.

Step 3: Repeat step 2. creating an additional table indicating groups of
four minterm/don't care term groups. Repeat step 3 until no further
grouping can occur.

Step 4: Construct a prime implicant chart and determine essential
prime implicants. Treat any don't care terms not part of a group
containing a minterms as 0s. They do not need to be covered.
Each is a candidate for becoming an EPI. However. set {8*, 9*, 10*, 11 *}
contains only don't care terms and is, therefore, not aPI.
The resultingsimplified equations is
S = x'z + wz

Example 1: F (A, B, C, D) = ∑ (0, 2, 3, 6, 7, 8, 10, 12, 13).

Example 2: F (A, B, C, D) = ∑ (2, 4,5, 9, 12, 13)

Example 3:Y (A,B,C,D) = ∑ (1, 2, 3, 5, 9, 12, 14, 15) + ∑d (4, 8, 11).

Example 4 f (A,B,C,D)= ∑ ( 0, 1, 9, 15, 24, 29, 30) + d (8, 11, 31)

Example 5 f (A,B,C,D,E) = ∑ (4, 5, 9, 11, 12, 14, 15, 27, 30) + ∑ d (1,
17, 25, 26, 31)

Example 6 f (A,B,C,D) = ∑ (0,1,4,5, 9,10,12, 14,15) + ∑d (2, 8, 13)

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