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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

1. $Q_2^c$
2. $Q_2 + Q_1$
3. $\left(Q_1 + Q_2\right)^c$
4. $Q_1 \oplus Q_2$
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0

A

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prateek not getting  ur answer .
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@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.

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: $D$
by Veteran (57k points)
edited
+1

how did you get next state of Qand Q1?

+7

sequence of states given in question

as $0-2-3-1-0$

means next state of $0$ is $2$, next state of $1$ is $0$, next state of $2$ is $3$ and next state of $1$ is $0$

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@praveen

steps are not clear in ur answer how u draw first table ?
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It's quite weird that I'm getting 01, 10, 01, 10... as $Q_2Q_1$ if I place T1 as $Q_2 \oplus Q_1$ and trace the output one by one. How can I get 11?
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from the sequence of states $0-2-3-1-0$, draw table as present state$Q_2Q_1$, say$00$, nextstate $Q_2^{+}Q_1^{+}$ as $10$ and so on.
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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:
$T_1=Q_2\oplus Q_1 = 1$
$1T Q_1 (1) = 0=Q_1^+$
$T_2=0 \odot Q_2 = 1$
$1 T Q_2 (0)=1=Q_2^+$
$\therefore Q_2^+Q_1^+ = 10$
+2
$Q_2=0,Q_1=0$

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

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

$00\rightarrow 10$

$Q_2=1,Q_1=0$

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

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

$00\rightarrow 10\rightarrow 11$
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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.

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

by Boss (21.5k points)
edited by
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@pC Best explaination!!!
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I think the k map for t2 is wrong could you explain if it was correct
T1 = XQ1' + X'Q1

T2 = (Q2+Q1)'

sequence is 00,10,11,01,00
X = Q1Q2' + Q1'Q2
by Veteran (60.8k points)
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How did you Get T1 ?
Anyone pls help me to understand concepts.
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will you please elaborate that how x comes?
+1
@akhilnadhpc did u get it now. or should i help.
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Can some one explain this a bit plz... !!
+1 vote

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

by Active (2.1k points)
0
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.

Normally, we're given equations and we have to figure out the values from them. Here, we're given the values and we need to figure out the equation.

Let's write the values down:-

 $Q_2$ $Q_1$ 0 0 1 0 1 1 0 1

Repeat.

$T_1$ corresponds to $Q_1$ If you notice, $T_1$'s values are: 0,1,0,1...

ie, $T_1$ didn't toggle $Q_1$ in the first cycle, $T_1$ toggled $Q_1$ in the second cycle... so on.

Now check the options. Which one gives $T_1$ the values: 0,1,0,1...

It is Option D

by Loyal (6.3k points)