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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. $Q_1 \oplus Q_2$
asked in Digital Logic by Veteran (59.7k points) | 3.3k views
–1

A

Answer Q2+Q1

0
prateek not getting  ur answer .
0
@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 .....
0
how u reach ur answer ?

don't say by putting all options.
0
if u know to solve it by using truth table then ...again post it mann. It will helpful for us.

3 Answers

+52 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 $

answered by Veteran (55.4k points)
edited by
+1

how did you get next state of Qand Q1?

+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
@praveen

steps are not clear in ur answer how u draw first table ?
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:
$T_1=Q_2\oplus Q_1 = 1$
$1T Q_1 (1) = 0=Q_1^+$
$T_2=0 \odot Q_2 = 1$
$1 T Q_2 (0)=1=Q_2^+$
$\therefore Q_2^+Q_1^+ = 10$
+1
$Q_2=0,Q_1=0$

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

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

$00\rightarrow 10$

$Q_2=1,Q_1=0$

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

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

$00\rightarrow 10\rightarrow 11$
0
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.
+14 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

answered by Boss (22.5k points)
edited by
0
@pC Best explaination!!!
0
I think the k map for t2 is wrong could you explain if it was correct
+4 votes
T1 = XQ1' + X'Q1

T2 = (Q2+Q1)'

sequence is 00,10,11,01,00
X = Q1Q2' + Q1'Q2
answered by Veteran (55.6k points)
0
How did you Get T1 ?
Anyone pls help me to understand concepts.
0
will you please elaborate that how x comes?
+1
@akhilnadhpc did u get it now. or should i help.
0
Can some one explain this a bit plz... !!
Answer:

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