The output Y should be that of a JK flip-flop. That is: Y=JQ'+K'Q

I tried to solve the following way:

1. Y=K'J'Q+K'J+KJQ'

2. =K'(J'Q+J)+KJQ'

3. =JK'+K'Q+JKQ'

4. =J(K'+KQ')+K'Q

5. =J(K'+Q')+K'Q

6. =JK'+JQ'+K'Q

It appears as if the term JK' is 0. Please help.

  • \$\begingroup\$ Explain the last step, please? \$\endgroup\$
    – Eugene Sh.
    Oct 31, 2018 at 15:45
  • \$\begingroup\$ @EugeneSh. Done \$\endgroup\$
    – Soumee
    Oct 31, 2018 at 15:49
  • 1
    \$\begingroup\$ Try writing down the truth tables of the original function and the target one. If these are different, then you have missed something (or the question author did) \$\endgroup\$
    – Eugene Sh.
    Oct 31, 2018 at 15:54
  • \$\begingroup\$ Also draw out the waveforms, with arrows to show what signals (some are anded) affect the outputs, at what times. \$\endgroup\$ Oct 31, 2018 at 16:23

1 Answer 1


Start with the original equation:

Y = K'J'Q + K'J + KJQ'

The trick is to create two terms that are equivalent to the middle term, using the fact that X + X' = 1:

Y = J'K'Q + JK'(Q + Q') + JKQ'

Y = J'K'Q + JK'Q + JK'Q' + JKQ'

Now factor the pairs of terms:

Y = K'Q(J + J') + JQ'(K + K')

Simplify, again using the fact that X + X' = 1:

Y = K'Q + JQ'

This is really obvious if you draw the Karnaugh map for the function. The JK' term is redundant, given the other two terms.


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