A

Answer Q2+Q1

29 votes

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

- $Q_2^c$
- $Q_2 + Q_1$
- $\left(Q_1 + Q_2\right)^c$
- $Q_1 \oplus Q_2$

75 votes

Best answer

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: $D$

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: $D$

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 $0$

0

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?

0

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.

0

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:

$T_1=Q_2\oplus Q_1 = 1$

$1T Q_1 (1) = 0=Q_1^+$

$T_2=0 \odot Q_2 = 1$

$1 T Q_2 (0)=1=Q_2^+$

$\therefore Q_2^+Q_1^+ = 10$

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:

$T_1=Q_2\oplus Q_1 = 1$

$1T Q_1 (1) = 0=Q_1^+$

$T_2=0 \odot Q_2 = 1$

$1 T Q_2 (0)=1=Q_2^+$

$\therefore Q_2^+Q_1^+ = 10$

23 votes

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

4 votes

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**

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.

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.

0 votes

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**