by Wong Poh Ling B031210033
Showing posts with label Digital Logic. Show all posts
Showing posts with label Digital Logic. Show all posts
Sunday, 21 October 2012
Combinational Circuit
Combinational
Circuits :
1.
No memory – output
depends on present input only (contrast to sequential logic).
2.
The output is a pure
function of the present input only.
3.
Can be defined in
three ways :
i.
Truth table
ii.
Graphical symbols
iii.
Boolean equations
Truth
table :
|
A
|
B
|
C
|
F(SOP)
|
F(POS)
|
|
0
|
0
|
0
|
0
|
0
|
|
0
|
0
|
1
|
0
|
1
|
|
0
|
1
|
0
|
0
|
1
|
|
0
|
1
|
1
|
0
|
1
|
|
1
|
0
|
0
|
0
|
1
|
|
1
|
0
|
1
|
0
|
1
|
|
1
|
1
|
0
|
0
|
1
|
|
1
|
1
|
1
|
1
|
1
|
Graphical
symbols :
Boolean
equation :
Equation that consist of possible
combination of inputs that produce an output signal.
Boolean
Equation Forms :
1.
Represented in 2
forms
i.
Sum-of-products (SOP)
1 = A , 0 = A’
ii.
Products-of-sum
(POR)
1 = A’ ,0 = A
Example
of SOP :
F = A’BC + A’B’C + ABC
For
SOP, output 1 will be taken.
SOP
expression :
F
= A’BC + A’B’C + ABC
NOTE
: This is not simplified version
Example
of POR :
F = (A + A’)(AB + ABC)
|
A
|
B
|
C
|
A’
|
AB
|
ABC
|
F
|
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
|
0
|
0
|
1
|
1
|
0
|
0
|
0
|
|
0
|
1
|
0
|
1
|
0
|
0
|
0
|
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
|
1
|
0
|
1
|
0
|
0
|
0
|
0
|
|
1
|
1
|
0
|
0
|
1
|
0
|
1
|
|
1
|
1
|
1
|
0
|
1
|
1
|
1
|
For POS, output 0 will be taken.
POS expression :
F = (A + B + C)(A + B + C’)(A + B’
+C)(A +B’ +C’)(A’ + B + C)(A’ + B + C’)
Simplification
Of Boolean Equation :
F = A’BC + A’BC’ + AC
F = A’(BC + BC’) + AC-------------Distributive
Law
F = A’( B(C + C’) ) + AC-----------Distributive
Law
F = A’( B(1) ) + AC-----------------Inverse
Law
F = A’B + AC------------------------Identity
Law
Law
of Boolean Algerba :
Karnaugh
Map :
Example :
a)
Truth Table :
F = A’BC + AB’C’ + AB’C + ABC’ + ABC
|
A
|
B
|
C
|
F
|
Minterm
|
|
0
|
0
|
0
|
0
|
A’B’C’
|
|
0
|
0
|
1
|
1
|
A’B’C
|
|
0
|
1
|
0
|
1
|
A’BC’
|
|
0
|
1
|
1
|
1
|
A’BC
|
|
1
|
0
|
0
|
1
|
AB’C’
|
|
1
|
0
|
1
|
1
|
AB’C
|
|
1
|
1
|
0
|
1
|
ABC’
|
|
1
|
1
|
1
|
1
|
ABC
|
So, the question become F = A + BC
a)
Boolean Law :
F = A’BC + AB’C’ + AB’C + ABC’ + ABC
F = A’BC + A ( BC + B’C + BC’ + B’C’ )-------------Distributive
Law
F =A’BC + A ( B(C + C’) + B’ (C’ + C)
)--------------Distributive Law
F = A’BC + A ( B(1) + B’ (1) )-------------------------Inverse
Law
F = A’BC + A ( B + B’ )--------------------------------Identity
Law
F = A’BC + A (1)---------------------------------------Inverse
Law
F = A’BC + A-------------------------------------------Identity
Law
F = A + BC---------------------------------------------Absorption
Law
By CHONG LEE MAN - B031210367
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