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Consider the circuit in the diagram. The $\oplus$ operator represents Ex-OR. The D flip-flops are initialized to zeroes (cleared).

The following data: $100110000$ is supplied to the “data” terminal in nine clock cycles. After that the values of $q_{2}q_{1}q_{0}$ are:

  1. $000$
  2. $001$
  3. $010$
  4. $101$
asked in Digital Logic by Active (3.7k points)
edited by | 1.9k views
+5

Option (C) is correct!

4 Answers

+20 votes
Best answer
Data Q0 Q1 Q2
1 1 Q0start XOR Q2start = 0 XOR 0 = 0 Q1start = 0
0 0 1 XOR 0 = 1 0
0 0 0 XOR 0 = 0
1 1 0 XOR 1 = 1 0
1 1 1 XOR 0 = 1 1
0 0 1 XOR 1 = 0
0 0 0 XOR 1 = 1 0
0 0 0 XOR 0 = 0
0 0 0 XOR 1 = 1 0

So, option C. 

answered by Veteran (355k points)
selected by
0
Please tell the approach used to calculate this... Q0 can be calculated easily since it depends on D0 only which is given initially. How to calculate Q2 and Q1.. they depend on D2 and D1 which in turn depend on old values od Q1 and Q2 which are not available initially
0
in the XOR operation why is Q0 taking the old value?

new value of Q0 is already available by the time XOR is calculated?
0
@Prerna+Chauhan That's what we do in synchronous circuits, when a clock is applied output by the application of this clock cycle should be present at each $FF$, which is then used as input in next clock.
0
yes i totally agree, but when the first clock cycle is applied the output of Q0 is 1. so, when we have this new output why is older one used?
+1
Initially, $Q_{2}Q_{1}Q_{0} = 000$ (reset), and when first clock is applied $D_{0} = data = 1$, $Q_{0n} = 1$, $Q_{1n} = Q_{0}$xor$Q_{2} = 0$xor$0 = 0$ and $Q_{2n} = Q{1} = 0$
0
my doubt is here input is data, so it's consider as 1??

o/p of q0 will 1 then perform exor on q0 xor q2 then it will give 1???

plz correct me??
+2
@Arjun sir data is given from LSB to MSB or MSB to LSB?? how to guess it?
0
I too got the same dout.i started from lsb while giving data and my answer is different .how do we guess whether to start from lsb or msb?
0
can you please tell me how you have calculated Q2
0 votes
The question is simple, but just because of delays of XOR gate q1 & q2 are working old datas of q0..  At a particular instance, value present at q0 is from data but q1 is getting from XOR gate, so the value calculated from previous bit. Thats it.
Just follow the data given to q0,
q0 value is the data bit.
q1,q2 will be evaluated from previous state.
C is the ans.
answered by Boss (10.7k points)
–1 vote
answer - A

q0 = data

q1 = q0 exor q2

q2 = q1
answered by Loyal (9k points)
+1
No.. it's (C)

explaination:

q2 q1 q0

0 0 0

0 0 1

0 1 0

1 0 0

0 1 1

1 1 1

1 0 0

0 1 0

1 0 0

0 1 0  <-- answer....
0

q2 q1 q

0 0 0

 0 0 1

0 1 0

1 0 0

0 1 1

1 1 1

1 0 0

0 1 0 how you got this above you got 011 after 100

1 0 0

0 1 0

0
@rahulkr Sahi pakde bhai
0

@mcjoshi How q2=q1start ?? why not q2=q1

–2 votes
c is the correct answer
answered by Active (4.8k points)


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