edited by
11,045 views
32 votes
32 votes

The next state table of a $2-$bit saturating up-counter is given below.

$\begin{array}{cc|cc}  Q_1 & Q_0 & Q_1^+ & Q_0^+ \\ \hline  0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{array}$

The counter is built as a synchronous sequential circuit using $T$ flip-flops. The expressions for $T_1$ and $T_0$ are

  1. $T_1 = Q_1Q_0, \quad T_0= \bar{Q_1} \bar{Q_0}$
  2. $T_1 = \bar{Q_1}Q_0, \quad T_0= \bar{Q_1} + \bar{Q_0}$
  3. $T_1 = Q_1+Q_0, \quad T_0= \bar{Q_1} \bar{Q_0}$
  4. $T_1 = \bar{Q_1}Q_0, \quad T_0= Q_1 + Q_0$
edited by

4 Answers

Best answer
28 votes
28 votes

Answer is B)

$$\begin{array}{|cc|cc|c|c|}\hline
\bf{Q_1}&    \bf{Q_0}&  \bf{{Q_1}^+}&\bf{ {Q_0}^+}& \bf{T_1}& \bf{T_2} \\\hline
0&0&0&1&0 &1 \\ 0&1&1&0&1&1\\    1& 0&1&1&0&1 \\   1&1&1&1&0&0 \\ \hline  
 \end{array}$$

By using above excitation table,

$T_1={Q_1}'{Q_0},$

$T_2={(Q_1Q_0)}'={Q_1}'+{Q_0}'$

edited by
7 votes
7 votes
Q1 Q0 Q1+ Q0+ T1 T0
0 0 0 1 0 1
0 1 1 0 1 1
1 0 1 1 0 1
1 1 1 1 0 0

 

T1(Q1,Q0) = Q1’ Q0

T0(Q1,Q0) = Q1’Q0’ + Q1’Q0 + Q1 Q0’

                  = Q1’ + Q1 Q0’

                  = Q1’ + Q0’

option B

5 votes
5 votes
My answer is  B

Expected toggling -- Q1'Qo=0100  & Q1'+Qo'=1110 gives the required next state!

Correct me if i am wrong.
edited by
2 votes
2 votes
Option B --

Expected Toggling Q1'Qo=0100 and  Q1' + Qo'=1110 gives  expected next state.
Answer:

Related questions

39 votes
39 votes
4 answers
1
24 votes
24 votes
7 answers
4
Arjun asked Feb 14, 2017
6,743 views
The representation of the value of a $16\text{-bit}$ unsigned integer $X$ in hexadecimal number system is $\textsf{BCA9}$. The representation of the value of $X$ in octal...