GATE CSE 1990 | Question: 5-c

Question

For the synchronous counter shown in Fig$.3,$ write the truth table of $Q_{0}, Q_{1}$, and $Q_{2}$ after each pulse, starting from $Q_{0}=Q_{1}=Q_{2}=0$ and determine the counting sequence and also the modulus of the counter.

Comments

Detailed Video Solution, with Two Methods to Solve: https://youtu.be/x7nDZzEtdjo 

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J0 = Q1 + Q2   ,      K0 = 1
J1 = Q2   ,    K1 = Q0'
J2 = Q1' Q0'   ,   K2 = 1
 

Q0   Q1  Q2
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0      0      0
0      0      1
1      1      0
0      1      0
1      0      0
0      0      0

0 -> 1 -> 6 -> 2 -> 4   
5 different  states so MOD 5 counter

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Answer 1

$$\begin{array}{|ccc|ccc|} \hline \textbf{$Q_0$} & \textbf {$Q_1$} &\textbf {$Q_2$} & \textbf {$Q_{0N}$} & \textbf{$Q_{1N }$}&\textbf{$Q_{2N}$} \\\hline \text{0}& \text{0} & \text{0}  & \text{0} & \text{0} &\text{1}\\\hline \text{0}& \text{0} & \text{1}  & \text{1} & \text{1} &\text{0}\\\hline \text{0}& \text{1} & \text{0}  & \text{1} & \text{0} &\text{0} \\\hline\text{0}& \text{1} & \text{1}  & \text{1} & \text{0} &\text{0} \\\hline\text{1}& \text{0} & \text{0}  & \text{0} & \text{0} &\text{0} \\\hline \text{1}& \text{0} & \text{1}  & \text{0} & \text{1} &\text{0} \\\hline \text{1}& \text{1} & \text{0}  & \text{0} & \text{1} &\text{0} \\\hline \text{1}& \text{1} & \text{1}  & \text{0} & \text{1} &\text{0} \\\hline \end{array}$$
$Q_{0N} = Q_0 \implies J_0 =  Q_1 + Q_2, K_0 = 1$

$Q_{1N} = Q_1 \implies J_1 =   Q_2, K_1 = \bar{Q_0}$

$Q_{2N} = Q_2 \implies J_2 =  \bar{Q_1}.\bar{Q_0}, K_2 = 1$

$$0 - 1 - 6 - 2- 4-0$$

So, MOD $5$ counter.

Comments

@rajatmyname YES this is correct convention. If MSB is not given by default highest subscript is the MSB.

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why we didint took those circle(dot) in diagram as not operator

I dont understand sometime we take the dot circle as not sometimes we dont in some other pyq i have see we took the joint as not operator here we took the same as joint only why? I could have solved it correctly but taken dot as not in exam how will we understand this???? @Deepak Poonia sir

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i guess when the dot is connected to any hardware then we have to take it as NOT

Answer 2

${\begin{array}{|c|c|c|}\hline
\bf{Q_0}&    \bf{Q_1}&  \bf{Q_2} \\\hline
0&0&0\\ 0&0&1 \\    1&1&0  \\   0&1&0\\   1&0&0  \\   0&0&0 \\ \hline
\end{array}}$

Counting sequence: $ 0-1-6-2-4$

As there are $5$ different states, so $5$ modulus

Comments

@Rishav Kumar Singh ,yeah I am getting same

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is it not $0\rightarrow 4\rightarrow 3\rightarrow 2\rightarrow 1$

if we interchange MSB and LSB?

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yes, srestha u are correct...

Answer 3

Mod "5" 

Answer 4

Ans. is 0-4-3-2-1- repeat 

Comments

Same I getting

@prashant Dubey

Your table is correct but transition diagram wrong

Answer 5

here if we want to solve such type of question then just  write q0 q1 q2 and also like input as previous output expression and just look at previous state and see how input are becoming for next state. like if  you see here if you want to make easy further then  we can make q0 q1 q3 as postion 0 1 2 now only try to identify  from which previous state postiion output coming to the next state

 

Answer 6