GATE 2020 IN Q.40: solution and answer

GATE 2020 · INQ.40 · 2 marks · NAT

Two T-flip flops are interconnected as shown in the figure. The present state of the flip flops are: A = 1, B = 1. The input x is given as 1, 0, 1 in the next three clock cycles. The decimal equivalent of $(A B y)_{2}$ with A being the MSB and y being the LSB, after the 3rd clock cycle is ____

GATE 2020 IN question 40: Two T flip-flops, A (top) and B (bottom), share a common clock line labelled clk. — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

ANSWER

After the third clock, $A = 1$ , $B = 1$ and $y = A + B = 1$ , so $(A B y)_{2} = (111)_{2} = 7$ .

Official answer: 7Source: Official answer key, GATE 2020 IN (IIT Delhi)

Read the circuit (complement written with ): the NAND gate takes $x$ and $B$ , so $T_{A} = \overline{x B}$ ; the wire from the $x$ junction goes straight to flip-flop B, so $T_{B} = x$ ; the OR gate takes $A$ and $B$ , so $y = A + B$ .
T flip-flop rule: $Q^{+} = T \oplus Q$ . The flip-flop toggles when $T = 1$ and holds when $T = 0$ . The T inputs are worked out from the present state, before the clock edge.
Clock 1 ( $x = 1$ , present $A = 1$ , $B = 1$ ): $T_{A} = \overline{1 \cdot 1} = 0$ , so A holds at 1. $T_{B} = 1$ , so B toggles to 0. New state $A B = 10$ .
Clock 2 ( $x = 0$ , present $A = 1$ , $B = 0$ ): $T_{A} = \overline{0 \cdot 0} = 1$ , so A toggles to 0. $T_{B} = 0$ , so B holds at 0. New state $A B = 00$ .
Clock 3 ( $x = 1$ , present $A = 0$ , $B = 0$ ): $T_{A} = \overline{1 \cdot 0} = 1$ , so A toggles to 1. $T_{B} = 1$ , so B toggles to 1. New state $A B = 11$ .
Output after the 3rd clock: $y$ is a purely combinational function of the new state, so $y = A + B = 1 + 1 = 1$ .
$(A B y)_{2} = (111)_{2} = 4 + 2 + 1 = 7$ .

KEY IDEA

Write down the excitation equations from the gates ( $T_{A} = \overline{x B}$ , $T_{B} = x$ ). Then step through a state table one clock at a time, working each T input out from the present state and that cycle's $x$ , and using $Q^{+} = T \oplus Q$ . The output $y = A + B$ is combinational, so read it from the state after the 3rd clock.

WATCH OUT FOR

The usual slips: (i) reading the gate as an AND and forgetting the bubble, which gives $T_{A} = x B$ and a different sequence; (ii) working a cycle's $T_{A}$ out from the value B will have after the edge instead of its present value (at clock 1, for example, $T_{A}$ uses $B = 1$ , not the new $B = 0$ ); (iii) taking $y$ from the state before the 3rd clock, or leaving $y$ out of the 3-bit word.

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 30 Sequential questions, with notes

GATE 2020 IN Q.40: solution and answer

Answer and solution

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