GATE CSE 2004 | Question: 61

Question

Consider the partial implementation of a $2-bit$ counter using $T$ flip-flops following the sequence $0-2-3-1-0,$ as shown below.

To complete the circuit, the input $X$ should be

• A. $Q_2^c$
• B. $Q_2 + Q_1$
• C. $\left(Q_1 + Q_2\right)^c$
• D. $Q_1 \oplus Q_2$

Comments

@wanted for this question there is slightly contradiction btw two options according to me

it could be either (D) Q1 XOR Q2    OR

                          (B) Q2 + Q1

will see what was the official answer given

what I feel that in option D they mentioned XOR instead of XNOR .....

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how u reach ur answer ?

don't say by putting all options.

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if u know to solve it by using truth table then ...again post it mann. It will helpful for us.

Answer 1

Sequence is $0-2-3-1-0$

From the given sequence, we have state table as
$${\begin{array}{c|c|c|c}
\bf{Q_2}&    \bf{Q_1}&  \bf{Q_2^+}&\bf{ Q_1^+}\\\hline
0&0&1&0 \\\hline 0&1&0&0 \\ \hline    1&0&1&1 \\ \hline   1&1&0&1 \\  
\end{array}}$$
Now we have present state and next state, use excitation table of $T$ flip-flop
$${\begin{array}{c|c|c|c}
\bf{Q_2}&    \bf{Q_1}&  \bf{Q_2^+}&\bf{ Q_1^+} & \bf{T_2} & \bf{T_1}\\\hline
0&0&1&0 & 1&0\\\hline 0&1&0&0&0&1 \\ \hline    1&0&1&1&0&1 \\ \hline   1&1&0&1&1&0 \\    
\end{array}}$$

From state table, $T_2=Q_2 \odot Q_1$, and $T_1 = Q_2 \oplus Q_1 $

$X=T_1 = Q_2 \oplus Q_1 $

Correct Answer: $D$

Comments

I'm doing exactly that, but All I'm getting is 00, 10, 01, 10, 01, 10...

Here's my trace:
(T defines the T flip flop, $aTb$: $a=$input, $b=$current state (value))
SECOND CLOCK:
$Q_2Q_1$: 10
$T_{1} = Q_{2}\oplus Q_{1} = 1$
$1 T Q_{1} (0)=1={Q_{1}^+}$
$T_2=1 \odot Q_2 = 1$
$1 T Q_2 (1)=0={Q_{2}^{+}}$
$\therefore {Q_{2}^{+}}Q_{1}^{+}=01$

THIRD CLOCK:
$T_1=Q_2\oplus Q_1 = 1$
$1T Q_1 (1) = 0=Q_1^+$
$T_2=0 \odot Q_2 = 1$
$1 T Q_2 (0)=1=Q_2^+$
$\therefore Q_2^+Q_1^+ = 10$

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$Q_2=0,Q_1=0$

$T_2=Q_2\odot Q_1= 1,T_1=Q_2\oplus Q_1= 0$

$Q_2^+=T_2\oplus Q_2= 1,Q_1^+=T_1\oplus Q_1= 0$

$00\rightarrow 10$

$Q_2=1,Q_1=0$

$T_2=Q_2\odot Q_1= 0,T_1=Q_2\oplus Q_1= 1$

$Q_2^+=T_2\oplus Q_2= 1,Q_1^+=T_1\oplus Q_1= 1$

$00\rightarrow 10\rightarrow 11$

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I see. ${T_2, T_1}$ are computed simultaneously, then ${Q_2^+, Q_1^+}$ are computed. I thought that LSB flip flop will generate the result first, then pass the value to MSB flip flop. Thanks sir for the explanation.

Answer 2

Design a 2-bit counter using T - Flip Flop Which will follow the sequence $0-2-3-1-0$

STEP-1: Find the Minimum Number of Flip-Flops Required

Here , we are having $4$ distinct states so we need minimum $2$ Flip-Flops

STEP-2: Obtain the State Diagram

STEP-3: Type of Flip-Flop and Excitation Table

Present State is $Q_n$ and Next State is $Q_{n+1}$

STEP-4: Minimization

STEP-5:Minimized Expression

$T_1 =Q_1\bigoplus Q_2$

$T_2 =Q_1\bigodot Q_2$

Similar Question

• https://gateoverflow.in/42222/design-counter-following-binary-sequence-repeat-hyderbad

Comments

@pC Best explaination!!!

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I think the k map for t2 is wrong could you explain if it was correct

Answer 3

T1 = XQ1' + X'Q1

T2 = (Q2+Q1)'

sequence is 00,10,11,01,00
X = Q1Q2' + Q1'Q2

Comments

How did you Get T1 ?
Anyone pls help me to understand concepts.

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will you please elaborate that how x comes?

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@akhilnadhpc did u get it now. or should i help.

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Can some one explain this a bit plz... !!

Answer 4

Counter produces:

Q2 Q1

0    0

1    0

1    1

0    1

and so on

T2 or T1 FF toggles depending on the previous outputs produced by both T2 and T1.

If you analyze how counter value changes, you will find T2 toggles when previous outputs are equal else T1 toggles.

T2 = Q2 XNOR Q1

T1 = Q2 XOR Q1

Comments

This is the more intuitive answer, I was going to post the same thing.

If previous state had an even number of 1s (including zero 1s) then the current state MSB will be complement/toggle of previous state MSB.

If previous state had an odd number of 1s, then current state LSB will be complement/toggle of previous state LSB.

From this one can easily deduce that ans is D.