4 bit Up and Down Counter Using D F/F

M

mindhacker

New Member

• Feb 25, 2011

• #1

see my k map there. Is it right?


1da= A[(c xor d)+ (c xor x) ] + [(x xnor c xnor d)[(A xor b) xnor x]
2db =b[(c xor D) + (c xor x)] + b'( c xnor d xnor x)
3dc= ad' + x'a'(c xnor d) + xa'(c xor d)
4dd=d'
where:
a=qa
b=qb
c=qc
d=qd

 

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Last edited: Feb 25, 2011

R

Ratchit

Well-Known Member

• Feb 25, 2011

• #2

mindhacker,

What does "q" mean? What does 1da,2db,3dc,4dd mean? You can easily check your own work by making a truth table, and verifying whether your reduction gives the same output as your original expression.

Ratch

 

M

mindhacker

New Member

• Feb 25, 2011

• #3

q is the output of the d f/f. qa for first f/f, qb for the 2nd. . . . 1da is the first f/f, db is the 2ndd f/f. . . .

 

R

Ratchit

Well-Known Member

• Feb 25, 2011

• #4

mindhacker,

What is "x" in your logic equations? It does not show up in your K-maps. What is "U" is your K-maps? It does not show up in your logic equations. Where is your truth table?

Ratch