GATE EE 2003 Digital Electronics Q 4 - Grad Plus
GATE EE 2003 Digital Electronics Q 4

GATE EE 2003 Digital Electronics Q 4

An X-Y flip flop, whose Characteristic Table is given below is to be implemented using a J-K flip flop.

XYQn+1
001
01Qn
10{\overline Q}_n
111

This can be done by making

a) J=X,\;K=\overline Y

b) J=\overline X,\;K=Y

c) J=Y,\;K=\overline X

d) J=\overline Y,\;K=X

Ans: (d)

Explanation

XYQn+1
001
01Qn
10{\overline Q}_n
110
X-Y truth table
JKQn+1
00Qn
010
101
11{\overline Q}_n
J-K truth table
Q(t)Q(t+1)JKXY
000xx1
011xx0
10x11x
11x00x
Excitation table

From the excitation table, we have, (J) should be (\overline Y) and (K) should be (X).

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