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Digital Electronics
NUMBER SYSTEM
BINARY CODES
BOOLEAN ALGEBRA
BINARY LOGIC
LOGIC GATES
BOOLEAN ALGEBRA LAWS
CONSENSUS THEOREM
DIFFERENT FORS OF BOOLEAN EQUATIONS
RELATION B/W MAX & MIN TERMS
16 type of LOGIC FUNCTIONS
AND & OR GATE
Other GATES
XOR & XNOR gates
NEGATED & COMPLIMENTRY GATES
TRISTATE gates & DIP
Illustration of NEGATIVE & POSITIVE LOGIC
Relation B/W XOR & XNOR gates
UNIVERSAL GATES
Implementation of XOR using minimum gates
Implementation of XNOR using minimum gates
Different levels of CIRCUIT
Special Characteristics of an IC
QUESTIONS
Q1 (Timing Diagram)
Q2 (Timing Diagram)
Q3 (Timing Diagram - DIfferent units)
Q4 (Timing Diagram)
Q5 (Output of Series of NOR gate )
Q6 (Output of combination of XOR )
Q7 (Circuit of NAND gates & diff delays)
K MAPS
COMBINATIONAL CKT
SEQUENTIAL CIRCUITS
TIMING CIRCUITS

 

 

Relation between min and max terms(canonical form):

The compliment of the function expressed in terms of sum of min terms can be obtained by taking sum of missing min terms in the original functions

e.g.  f= ∑(4, 5, 7) = m7 + m5 + m4  

Its compliment is              f’ = ∑(0, 1, 2, 3, 6) = m0 + m1 + m2 + m3 + m6

The compliment of the function expressed in terms of product of max terms can be obtained by taking product of max missing terms in the original functions

e.g. f= M4 M5 M2 M0 = π (0, 2, 4, 5)

Its compliment is              f’ = π (1, 3, 6, 7)= M7 M6 M3 M1

The compliment of the function expressed iin any of the canonical terms can be obtained by just interchanging the symbols π & ∑ and keeping the list of numbers same

e.g. f= ∑(4, 5, 7) = m7 + m5 + m4  

Its compliment is              f’ = π (4, 5, 7)= M4 M5 M7

e.g. f= M4 M5 M2 M0 = π (0, 2, 4, 5)

Its compliment is              f’∑(0, 2, 4, 5) = m0 + m2 + m4 + m5

The function expressed in terms of product of max terms can be converted to sum of min terms or vice-versa can be done by interchanging π & ∑ and list the numbers which were missing from the original function.

e.g. f= ∑(4, 5, 7) = m7 + m5 + m4  

then f can also be expressed as  f= ∑(4, 5, 7) = π(0, 1, 2, 3, 6)

e.g. f= π (0, 2, 4, 5)

then f can also be expressed as  f= π (0, 2, 4, 5)=∑(1, 3, 6, 7)

 


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