Question

Click Submit to complete this assessment. Question 4 Operation table of a PN flipflop is give below. According to this table, obtain the excitation table of PN flipflop. P N Operation O O Reset 0 1 Complement 1 O No change 1 1 Set . 00 AXI • 01 10 8.0 C 1x D.ox 11 A Click Submit to complete this assessment.

Answer 1

Solu :

The operation table for the PN flip flop is given as :

P	N	Operation
0	0	Resert
0	1	Complement
1	0	No change
1	1	Set

The excitation table for the PN flip flop can be found as :

Characterisistic table for PN flip flop :

P	N	un	$Q_{n+1}$	Operation
0	0	0	0	Reset
0	0	1	0	Reset
0	1	0	1	Complement
0	1	1	0	Complement
1	0	0	0	No change
1	0	1	1	No change
1	1	0	1	Set
1	1	1	1	Set

The excitation table can be found as :

Case A : When $Q_{n}=0$ and

Qn+1 0

This condition happen with either

P=0 and N=0 or

P=1 and N=0

Therefore , the desiered output is obtained when

Qn+1 0

P=x and N=0

Case B : When $Q_{n}=0$ and

Qn+1

This condition happen with either

P=0 and N=1 or

P=1 and N=1

Therefore , the desiered output is obtained when

Qn+1

P=x and N=1

Case C : When $Q_{n}=1$ and

Qn+1 0

This condition happen with either

P=0 and N=0 or

P=0 and N=1

Therefore , the desiered output is obtained when

Qn+1 0

P=0 and N=x

Case D : When $Q_{n}=1$ and

Qn+1

This condition happen with either

P=1 and N=0 or

P=1 and N=1

Therefore , the desiered output is obtained when

Qn+1

P=1 and N=x

Therefore , the excitation table can be given as :

$Q_{n}$	$Q_{n+1}$	P	N
0	0	x	0
0	1	x	1
1	0	0	x
1	1	1	x

Hence , the correct option for the given problem is given as :

0 0 = B ( x 0 )

0 1 = A ( x 1 )

1 0 = D ( 0 x )

1 1 = C ( 1 x )

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