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+12 votes

Find the contents of the flip-flop $Q_2, Q_1$ and $Q_0$ in the circuit of figure, after giving four clock pulses to the clock terminal. Assume $Q_2Q_1Q_0=000$ initially.

asked in Digital Logic by Veteran (52k points)
edited by | 976 views
The sequence is 000 - 100 - 010 - 101 - 011 - 000

It's Mod 5 counter ..and it's self starting ..but not free running as state 111 and 110 is not in counting

2 Answers

+19 votes
Best answer

Initial $Q_2=0, Q_1=0, Q_0=0$

Clock 1:

  • $Q_2=1$  $[J =$ ( old $Q_0)'=1, K=1,$ New $Q2=$ Complement of old $Q2=1]$
  • $Q_1 =0$  $[D =$old $Q_2=0,$ new $Q1= D =0]$
  • $Q_0=0$  $[T=$ old $Q1=0,$ New $Q_0 = $old $Q_0 =0]$

Clock 2:

  • $Q_2=0$   $[J =$ ( old $Q_0)'=1, K=1,$ New $Q2=$ Complement of old $Q2=0]$
  • $Q_1 =1$   $[  D =$old $Q_2=1,$ new $Q1= D =1]$
  • $Q_0=0$  $[ T=$ old $Q_1=0,$ New $Q_0 = $old $Q_0 =0]$

Clock 3:

  • $Q_2=1$  $[J =$ ( old $Q_0)'=1, K=1,$ New $Q_2=$ Complement of old $Q2=1]$     
  • $Q_1 =0$  $[D =$ old $Q_2=0,$ New $Q_1= D =0]$
  • $Q_0=1$  $[T= $ old $Q1=1,$ New $Q_0 =$ complement of old $Q_0 =1]$

Clock 4:

  • $Q2=0$  $[J =($old $Q_0)'=0, K=1$, new $Q_2=$ Reset$=0]$
  • $Q1 =1$  $[D =$old $Q_2=1$, new $Q_1= D =1]$
  • $Q0=1$  $[T= $old $Q_1=0$, new $Q_0 = $old $Q_0 =1]$

After $4$ clock pulses $Q2Q1Q0$ is $011$

Note : for JK flipflops, $Q_{(t+1)} = JQ' +K'Q$, for D flipflops, $Q_{(t+1)} = D$, and for T flipflops $Q_{(t+1)}= T\oplus Q $ Where $Q_{(t+1)}$ represent new value of $Q$

answered by Veteran (55.8k points)
edited by
Yes, you are right.
have we considered that the propagation delay of each flipflop is exactly equal to the clock period ???Then we can say at the end of each clock cycle the previous old value transmits to the next flipflop... all flipflops are behaving like a positive edge triggered then..
+3 votes

The below table clearly demonstrates how after 4th cycle it will be 011 state. 

answered by Loyal (9k points)

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