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In a three stage counter, using RS flip flops what will be the value of the counter after giving $9$ pulses to its input? Assume that the value of counter before giving any pulses is 1.

  1. 1
  2. 2
  3. 9
  4. 10
in Digital Logic by Boss (30.8k points) | 2.8k views

2 Answers

+13 votes
Best answer

3 stage counter with RS FF is 3 bit counter, so after every 8 clock pulse, it will return to initial state.

Initial state is 1, state after 8 clock pulse will be 1, So, after 9th clock pulse , state will be 2. 

by Veteran (57k points)
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0
I think question is incomplete... its not given whether the counter is asynchronous or synchronus ?.. coz in both the cases answer will be different.
0
No, sequence of states doesn't depend on whether counter is synchronous or not
0
but no of states depends ..

for n bit conuter

1.for synchronous , total no of states possible = n  (Ring counter)

2. for synchronous , total no of states possible = n  (Johnson counter)

3. for asynchronous , total no of states possible = 2 Power n

ur solution is possible only by assuming case 3rd
0
1. n-bit synchronous counter, no of  states = 2^n

2. n-bit asynchronous counter, no of  states = 2^n

3. Ring counter and Johnson counter are special counters.
0
M getting confused here, i read that we can have mod N or mod 2N synchronous counter or mod 2^N asynchronous conuter using n bits.
0
the counter can represent from 0 to 7 right .. ? because its 3 bit ..

but since its starts with 1 , wont the states go like

1->2->3->4->5->6->7->1 (7 clock pulses)
+1
1->2->3->4->5->6->7->0->1
0 votes

A three stage counter can count only to 8 (0-7)

Initially, count = 1.

After 9 clock pulses

$1\rightarrow 2\rightarrow 3\rightarrow 4\rightarrow 5\rightarrow 6\rightarrow 7\rightarrow 0\rightarrow 1\rightarrow 2$

Answer is 2

by Loyal (6.7k points)
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