digital logic enel 111. digital systems a digital system is a system whose inputs and outputs fall...
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Digital Logic
ENEL 111
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Digital systems
A digital system is a system whose inputs and outputs fall within a discrete, finite set of values
Two main types Combinational
Outputs dependent only on current input Sequential
Outputs dependent on both past and present inputs
1
7
3
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Combinational Logic Circuits
Aims To express the inputs and outputs of a system in
binary form To develop the relationships between these inputs and
outputs as a truth table To simplify the Boolean expression using algebra or
Karnaugh maps To select suitable electronic devices to implement the
required function
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Example
Consider a buzzer which sounds when : The lights are on and The door is open and No key is in the ignition
Variable Value SituationA 1 Lights are on
0 Lights are offB 1 Door is open
0 Door is closedC 1 Key is in ignition
0 Key is out of ignitionP 1 Buzzer is on
0 Buzzer is off
Alarm system ActiveBA
CP
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Example
Truth Table A Truth Table can be used
to show the relationships between : the 3 inputs and the single output
Implementation as a circuit using logic gates
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 0
A B C P
A
B P
C
lights
door
keys
buzzer
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Summary
Inputs and Outputs are expressed in Binary Form A truth table showing relationships between
inputs and outputs is constructed A circuit is built to implement the circuit
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This lecture
Truth tables for primitive functions Boolean notation Sum of products Boolean algebra
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Truth Tables and Boolean Notation
Circuits with one input
Buffer P = A
Not P = A
A P0 01 1
A P0 11 0
A P
A P
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Basic AND / OR
Circuits with two Inputs
AND P = A.B
ORP=A+B
A B P0 0 00 1 01 0 01 1 1
AB P
A B P0 0 00 1 11 0 11 1 1
AB P
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Basic NAND / NOR
Problems with two Inputs
NAND P = A.B
NOR P=A+B
A B P0 0 10 1 11 0 11 1 0
AB P
A B P0 0 10 1 01 0 01 1 0
AB P
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Basic XOR / XNOR
Circuits with two Inputs:
XOR P = A B
XNOR P = A B
A B P0 0 00 1 11 0 11 1 0
A B P0 0 10 1 01 0 01 1 1
AB P
AB P
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Primitive gates
All circuits can actually be made using AND, OR and NOT gates if required.
In terms of components used, it is generally easier to build inverting functions. They typically require less transistors and also work faster than their non-inverting cousins.
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Exercise
Complete the truth table for this circuit and name the equivalent primitive function/gate.
0
1
1
1
A.B
01111
10101
10110
00000
PA.BA+BBA
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Not Symbol
You should be aware that
A . Bnot A and not B
A . Bnot (A and B) equivalent to NAND
and
are different.
011
001
010
100
BA A.B
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Combinational Logic Circuits
Reminder from our overview To express the inputs and outputs of a system in
binary form To develop the relationships between these inputs and
outputs as a truth table To simplify the Boolean expression using algebra or
Karnaugh maps To select suitable electronic devices to implement the
required function
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Our Example
Consider a buzzer which sounds when :
The lights are on and
The door is open and
No key is in the ignition
Variable Value SituationA 1 Lights are on
0 Lights are offB 1 Door is open
0 Door is closedC 1 Key is in ignition
0 Key is out of ignitionP 1 Buzzer is on
0 Buzzer is off
Alarm system ActiveBA
CP
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Very simple!
The truth table
The buzzer sounds only under this condition
A.B.C0111
1011
0101
0001
0110
0010
0100
0000
PCBA
Alarm system ActiveBA
CP
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Slightly more complexConsider my car which complains by sounding a buzzer when I
have left the lights on or left the car in gear (not in Park) and taken the keys out of the ignition:
0
1
0
1
0
1
0
0
P
(buzzer)
111
both011
101
left lights on001
110
left in gear010
100
000
what I’ve doneC
(keys out = 0)
B
(in gear = 1)
A
(lights on = 1)
A.B.C + A.B.C + A.B.C
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Minimization
The expression can be simplified in one of two ways:
via algebra via Karnaugh maps
to A.C + B.C
as the following truth table shows:
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Truth table shows the same result
0
1
0
1
0
0
0
0
A.C
0
1
0
0
0
1
0
0
B.C
0
1
0
1
0
1
0
0
P
0111
1011
0101
1001
0110
1010
0100
0000
A.C + B.CCB A
A.B.C + A.B.C + A.B.C = A.C + B.C = (A + B).C
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Means fewer logic gates are required
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Minterms and Maxterms
Notice the truth table has all possible combinations of A,B and C included:
The minterm is obtained from the “product” of A,B and C by AND-ing them
A.B.C
The maxterm is obtained from the “sum” of A,B and C by OR-ing them and inverting inputs
A + B + C 0111
1011
0101
0001
0110
0010
0100
0000
PCBA
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Sum of Products/Product of Sums For all combinations of inputs for which the output is a
logical true:
Combining the minterms with OR gives the sum-of-products
For all combinations of inputs for which the output is a logical false:
Combining the maxterms with AND gives the product-of sums.
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From our example:
0
1
0
1
0
1
0
0
P
111
011
101
001
110
010
100
000
CB A sum-of-products:
A.B.C + A.B.C + A.B.C
product-of-sums:
A+B+C . A+B+C . A+B+C . A+B+C . A+B+C
Normally the expression is derived using sum-of-products although product-of-sums yields fewer terms when there are more 1 outputs than 0 outputs.
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Exercise
0
0
1
1
0
0
1
1
P
111
011
101
001
110
010
100
000
CB AWrite out the sum-of-products expression for the truth table :
A.B.C + A.B.C + A.B.C + A.B.C
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Summary
A circuits desired outputs can be specified in terms
An boolean (logical) expression can be derived from the truth table.
The boolean expression can then be simplified
now we see how…
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Algebraic Laws
DeMorgan’s Laws The AND and OR functions can be shown to be
related to each other through the following equations:
A B A B
A B A B
A B A B
A B A B
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DeMorgan
DeMorgan’s Laws Example: Implement the expression A.B + C.D using
only NAND gates
NOT the individual terms Change the sign NOT the lot
.
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Boolean Algebraic Laws
(A.B).C = A.B.C = A.(B.C)Associative
A.(B + C) = A.B + A.C
A + (B.C) = (A + B).(A + C) Distributive
A . B = B . A
A + B = B + ACommutative
A . 0 = 0 A . 1 = A
A + 0 = A A + 1 = 1Operating with logic 0
and logic 1
A . A = 0
A + A = 1Complementary
A . A = A
A + A = ATautology (Idempotent)
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Basic rules of Boolean Algebra
Example: Simplify the following Expression
PA.(B C) A(C B)
A
A.B + A.C+ A.C + A.B distributive
A.(B + B) + A.(C + C) re-distribute
A.1 + A.1 complementary
A + A op with logic 1
A idempotent
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Exercises
You should be able to: Construct truth tables given boolean expressions Compare expressions using truth tables Produce a sum-of-products form from a truth table by
combining minterms Simplify the resulting expression algebraically Represent the expression as a circuit using logic gates