TOPIC
: DIGITAL
LOGIC
Assalamualaikum and hi everyone. ^^
Before I moving on further, just a simple introduce about
my self.
The names given is Nur Umira Binti Mustafa. Well,you can just call me Nur. ^^
I studying in Technical University Malaysia Melaka (UTEM) , and taking software development as my course.
Alright,lets quick chatting and start with our main topic for today’s discussion. .. ^^
The names given is Nur Umira Binti Mustafa. Well,you can just call me Nur. ^^
I studying in Technical University Malaysia Melaka (UTEM) , and taking software development as my course.
Alright,lets quick chatting and start with our main topic for today’s discussion. .. ^^
LEARNING
OUTCOMES
In order for you to fully understand and take a full
control of this topic, you need to dig out a few information about the basic
component in this topic.
Lets we analyst our learning outcomes of this discussion
1) understand and apply basic logic gates
2) understand and apply law and rules of Boolean,algebra
and DeMorgan’s theorems to Boolean expressions
3) Able to design a combinational logic circuit for given
Boolean output expression.
// The learning outcomes seems though isn’t it..?
But don’t worry, neither you are beginner nor expert, I will share all
information about this topic with you, to enable you to get more detail
information about this topic.
WHAT IS LOGIC GATE ?
So,friends, have you ever wondering what is meant by LOGIC GATE ?
Do you still remember what is BOOLEAN ALGEBRA ?
(if you can answer 2 simple warming up questions above, you are ready to
explore this topic ^^ )
Boolean algebra
ALTERNATIVE NAME : Boolean logic
WHAT IS IT ?
but with the
numeric operations of >
-multiplication xy
-addition x + y
-The Boolean operations are these and all other
operations that can be built from these, such as x∧(y∨z). These turn out to coincide with the set of all operations on the
set {0,1} that take only finitely many arguments; there are 2^2^n such
operations when there are n arguments.
Basic logic gates
Boolean algebra uses variables and operation to represent
the logic circuit. The variables and the function have the only and value, 0
and 1 .
The compliment of the value is shown by a bar over the letter or a single quotes on the right side above the letter.
Example : A’ / B’
The compliment of the value is shown by a bar over the letter or a single quotes on the right side above the letter.
Example : A’ / B’
Type
|
Distinctive shape
|
Rectangular shape
|
Boolean algebra between A & B
|
Truth table
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|||||||||||||||||||
AND
|
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A.B
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||||||||||||||||||||
OR
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A+B
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|||||||||||||||||||||
NOT
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||||||||||||||||||||||
NAND
|
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||||||||||||||||||||||
NOR
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|||||||||||||||||||||
XOR
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||||||||||||||||||||||
XNOR
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||||||||||||||||||||||
DECORDER :
} Use decoder to build
larger components
} The most common type of decoder
has an n-bit input and 2n outputs, where only one
output is asserted for each input combination.
} This decoder translates the n-bit
input into a signal that corresponds to the binary value of the n-bit
input.
} The outputs are shows a 3-bit
decoder and the truth table. This decoder is called a 3-to-8 decoder
} since there are 3 inputs and 8
(23) outputs. There is also a logic element called an
} encoder that performs the inverse
function of a decoder, taking n inputs and producing
} an n-bit output.
} This decoder is called a 3-to-8
decoder means 3 inputs and 8 (23) outputs. There is also
a logic element called an encoder that performs the inverse function of
a decoder, taking n inputs and producing an n-bit output.
Multiplexor
(Selector)
} The output is one of the inputs
that is selected by a CONTROL
} The left side shows this
multiplexor has three inputs: two data values and a selector (or control)
value. The selector value determines which of the inputs becomes the
output. We can represent the logic function computed by a two-input
multiplexor, shown in gate form on the right side as C = (A & S’ )
+ (B & S)
} Multiplexors can be created with
an arbitrary number of data inputs.
} If there are only two inputs, the
selector is a single signal that selects one of the inputs
} if it is true (1) and the other
if it is false (0). If there are n data inputs, there will need to be ⎡log2n⎤ selector inputs.
} In this case, the multiplexor
basically consists of 3 parts:
} 1. A decoder that generates n signals,
each indicating a different input value
} 2. An array of n AND
gates, each combining one of the inputs with a signal from the decoder
}
3.
A single large OR gate that incorporates the outputs of the AND gates
}
To
associate the inputs with selector values, we often label the data inputs
numerically
How to deal with COMBINATIONAL CIRCUITS ?
~ For your information , a logic
block contains no memory and computes the output given the current inputs.
~ combinational circuits can be
defined in three ways.
1) truth table – the truth table shows many possible combination
of input values, in tabular from between the input values and the result of a
specific Boolean operator or combination on the input variables.
IMPORTANT !
For n inputs,there are 2n of combination
2) Graphical symbol – the
layout of connected gates that represent the logic circuit
examples :
3) Boolean equations-
Boolean function that consist possible combination of inputs that produce an
output signal.
Boolean Equations Forms.
A
Boolean algebra is the combination of variables and operators. Typically , it
has one or more input and produces an output in the range of 0 or 1. The
compliment of a variable is shown by bar or a single quotes on the right side above the letter.
Example : A’ / B’
Example : A’ / B’
All
Boolean equation can be represent in two forms :
Sum of Products Expansion(SOP )
Definition
-
Combination
of input values that produce
1s is convert into
equivalent variables, AND ed, together then OR ed with other combination
variables with same output.
-
SOP
is easier to derive from the truth
Table.
IMPORTANT !
Sum of products for Boolean function
is a Boolean expression constructed by:
-Where each OUTPUT result is a 1
-form product of all variables
-sum each product
Definition 1
Sum-of-products expansion
is the sum of minterms.
Disjunctive normal
form
is the same as
sum-of-products.
|
Definition 2
Literal
is a Boolean variable or its complement.
Minterm
of the Boolean variables x1, x2,
..., xn is a
Boolean product y1, y2,
..., yn where yi=xi or yi=xi
D
|
Example 1:
F and G of Table 1 are expressed as sum of
minterms.
F(x,y,z) =
xyz
The sum of one minterm.
G(x,y,z) = xyz + xyz The
sum of two minterms.
Example 2 :
Minterm that is 1 for x1 = 0,
x2 = 0, x3 = 1,
x4 = 1
x1x2x3x4 = 1
Example 3
Find Boolean EQUATION based on TRUTH Table 1.
H =
X’Y’ + XYZ’
TABLE 1 :
X
|
Y
|
Z
|
H
|
0
|
0
|
0
|
1
|
0
|
0
|
1
|
1
|
0
|
1
|
0
|
0
|
0
|
1
|
1
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
1
|
0
|
1
|
1
|
0
|
0
|
1
|
1
|
1
|
0t
|
SOLUTION
X
|
Y
|
Z
|
X’Y’
|
XYZ’
|
H
|
0
|
0
|
0
|
1
|
0
|
1
X’Y’Z’
|
0
|
0
|
1
|
1
|
0
|
1
X’Y’Z
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
1
|
0
|
1
|
0
|
0
|
0
|
1
|
1
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
H = (X’Y’Z’) + (X’Y’Z)
By: Nur Umira
By: Nur Umira
Product-of-sum( POS )
Definition
-
Input
combination that produce 0 in the sum term ( OR’ed variables ) are AND’ed
together.
-
Convert
input values that produce 0s into equivalent variables, OR’ed the variables,
the AND’ed with other OR’ED forms
-
Usually
use if more 1s produce in output function.
same output.
IMPORTANT !
PRODUCT of SUM for Boolean function
is a Boolean expression constructed by:
-Where each OUTPUT result is a 0
-form SUM of all variables
-MULTIPLY each SUM OF VARIABLE
EXAMPLE 4:
Find Boolean EQUATION based on TRUTH Table 2.
H = (A+B+C)(A+B+C’)(A+B’+C)(A’+B+C’)
PRODUCT TERM
(1)A+B+C
(2)A+B+C’
(3)A+B’+C
(4)A’+B+C’
A
|
B
|
C
|
H
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
0
|
0
|
1
|
1
|
0
|
1
|
0
|
1
|
1
|
0
|
1
|
1
|
1
|
1
|
1
|
(1)
(2)
(3)
(4)
POS Expressions :
H = (A+B+C)(A+B+C’)(A+B’+C)
-note : this is not simplified versions
SIMPLIFICATION
OF BOOLEAN EQUATIONS
Rules of
Boolean Algebra
Table
4-1 lists 12 basic rules that are useful in manipulating and simplifying
Boolean expressions. Rules 1 through 9 will be viewed in terms of their
application to logic gates. Rules 10 through 12 will be derived in terms of the
simpler rules and the laws previously discussed.
Table
4-1 Basic rules of Boolean algebra.
Rule 1. A + 0 = A
A
variable ORed with 0 is always equal to the variable. If the input variable A
is 1, the output variable X is 1, which is equal to A. If A is 0, the output is
0, which is also equal to A. This rule is illustrated in Fig.(4-6), where the
lower input is fixed at 0.
Fig.(4-6)
Rule 2. A + 1 = 1
A
variable ORed with 1 is always equal to 1. A 1 on an input to an OR gate
produces a 1 on the output, regardless of the value of the variable on the
other input. This rule is illustrated in Fig.(4-7), where the lower input is
fixed at 1.
Fig.(4-7)
Rule 3. A . 0 = 0
A
variable ANDed with 0 is always equal to 0. Any time one input to an AND gate
is 0, the output is 0, regardless of the value of the variable on the other
input. This rule is illustrated in Fig.(4-8), where the lower input is fixed at
0.
Fig.(4-8)
Rule 4. A . 1 = A
A
variable ANDed with 1 is always equal to the variable. If A is 0 the output of
the AND gate is 0. If A is 1, the output of the AND gate is 1 because both
inputs are now 1s. This rule is shown in Fig.(4-9), where the lower input is
fixed at 1.
Fig.(4-9)
Rule 5. A + A = A
A
variable ORed with itself is always equal to the variable. If A is 0, then 0 +
0 = 0; and if A is 1, then 1 + 1 = 1. This is shown in Fig.(4-10), where both
inputs are the same variable.
Fig.(4-10)
Rule 6. A + A = 1
A
variable ORed with its complement is always equal to 1. If A is 0, then 0 +
0
= 0 + 1 = 1. If A is l, then 1 + 1 = 1+ 0 = 1. See Fig.(4-11), where one input
is the complement of the other.
Fig.(4-11)
Rule 7. A . A = A
A variable ANDed with itself is
always equal to the variable. If A = 0, then 0.0 = 0; and if A = 1. then 1.1 =
1. Fig.(4-12) illustrates this rule.
Fig.(4-12)
Table
4-2
Rule 11. A + AB = A + B
This
rule can be proved as follows:
A
+ AB = (A + AB) + AB
= (AA + AB) + AB =AA +AB +AA +AB
=
(A + A)(A + B) = 1. (A + B)
=
A + B
By:Amirul Ramzani



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