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Answer Q2+Q1

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+22 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

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

+54 votes

Best answer

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

From the given sequence, we have state table as

$Q_2$ | $Q_1$ | $Q_2^+$ | $Q_1^+$ |

$0$ | $0$ | $1$ | $0$ |

$0$ | $1$ | $0$ | $0$ |

$1$ | $0$ | $1$ | $1$ |

$1$ | $1$ | $0$ | $1$ |

Now we have present state and next state, use excitation table of T flip-flop

$Q_2$ | $Q_1$ | $Q_2^+$ | $Q_1^+$ | $T_2$ | $T_1$ |

$0$ | $0$ | $1$ | $0$ | $1$ | $0$ |

$0$ | $1$ | $0$ | $0$ | $0$ | $1$ |

$1$ | $0$ | $1$ | $1$ | $0$ | $1$ |

$1$ | $1$ | $0$ | $1$ | $1$ | $0$ |

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 $

+5

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:

$T_1=Q_2\oplus Q_1 = 1$

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

$T_2=0 \odot Q_2 = 1$

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

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

+17 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**

$T_1 =Q_1\bigoplus Q_2$

$T_2 =Q_1\bigodot Q_2$

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