in Digital Logic retagged by
877 views
1 vote
1 vote

How to solve in brief and answer?

in Digital Logic retagged by
by
877 views

1 Answer

2 votes
2 votes
Best answer

Initially  $\rightarrow 000$

  • After $1$st clock  $\rightarrow 111$ . How ?

The first FF will always toggle on every input .Now, we goto $3$rd FF and take feedback from $Q_{2}' = 1$. This $1$ reaches $2$nd FF and $Q_{0}$= clock input = 1 makes the $2$nd FF toggle giving $Q_{1} = 1$ = input for $3$rd FF with clock being high makes it toggle too.

This gives the next state $111$.

  • After $2$nd clock  $\rightarrow 011$
  • After $3$rd clock  $\rightarrow 110$
  • After $4$th clock  $\rightarrow 010$ . How ?

The first FF will always toggle on every input .Now, we goto $3$rd FF and take feedback from $Q_{2}' = 1$. This $1$ reaches $2$nd FF and $Q_{0}$ = clock input = 0 makes the $2$nd FF not toggle giving $Q_{1} = 1$ = input for $3$rd FF with clock being $0$ makes it not toggle too.

  • After $5$th clock  $\rightarrow 100$
  • After $6$th clock  $\rightarrow 000$
selected by
by

4 Comments

method ?? any other ??? 

2
2
@DD. Good enough
1
1
@Debashish

Right !!
1
1

Related questions