GATE CS 2017 Set 2 Q.42 (CS Q.42): Sequential Circuits, Counters and Registers

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

$Q_1$	$Q_0$	$Q_1^+$	$Q_0^+$
0	0	0	1
0	1	1	0
1	0	1	1
1	1	1	1

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

$T_1 = Q_1 Q_0, \quad T_0 = \bar{Q}_1 \bar{Q}_0$

$T_1 = \bar{Q}_1 Q_0, \quad T_0 = \bar{Q}_1 + \bar{Q}_0$

$T_1 = Q_1 + Q_0, \quad T_0 = \bar{Q}_1 + \bar{Q}_0$

$T_1 = \bar{Q}_1 Q_0, \quad T_0 = Q_1 + Q_0$

Official answer and solution

Official answer: B
From: Official GATE 2017 answer key

Solution

Construct the excitation table for the T flip-flops from the given next-state table using $T = Q \oplus Q^+$ , then derive the Boolean expressions for $T_1$ and $T_0$ .

For a T flip-flop, the excitation equation is $T = Q \oplus Q^+$ .
Compute $T_1$ and $T_0$ for each row of the next state table:

$Q_1$	$Q_0$	$Q_1^+$	$Q_0^+$	$T_1 = Q_1 \oplus Q_1^+$	$T_0 = Q_0 \oplus Q_0^+$
0	0	0	1	$0 \oplus 0 = 0$	$0 \oplus 1 = 1$
0	1	1	0	$0 \oplus 1 = 1$	$1 \oplus 0 = 1$
1	0	1	1	$1 \oplus 1 = 0$	$0 \oplus 1 = 1$
1	1	1	1	$1 \oplus 1 = 0$	$1 \oplus 1 = 0$

Find minimal sum-of-products expressions:

For $T_1$ : $T_1 = 1$ only when $Q_1 = 0$ and $Q_0 = 1$ . Thus:

T_1 = \bar{Q}_1 Q_0

For $T_0$ : $T_0 = 1$ for minterms 0, 1, 2 (all except $Q_1 = 1, Q_0 = 1$ ). By De Morgan's Law:

T_0 = \overline{Q_1 Q_0} = \bar{Q}_1 + \bar{Q}_0

The resulting expressions are $T_1 = \bar{Q}_1 Q_0$ and $T_0 = \bar{Q}_1 + \bar{Q}_0$ , which corresponds to option B.

Final answer

Common trap

Using D flip-flop excitation ( $D = Q^+$ ) instead of T flip-flop excitation ( $T = Q \oplus Q^+$ ).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2017 Computer Science and Information Technology (CS) Session 2
Question: 42 of 65, printed as CS Q.42
Section: Computer Science and Information Technology

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