A binary 3-bit down counter uses \(๐ฝ-๐พ\) flip-flops, \(๐น๐น_๐ \) withโฆ
2017
A binaryย 3-bit down counterย usesย \(๐ฝ-๐พ\)ย flip-flops,ย \(๐น๐น_๐ย \) with inputsย \(๐ฝ_๐,ย ๐พ_๐\)ย and outputsย \(๐_๐,ย ๐ย =ย 0,1,2\)ย respectively. The minimized expression for the input from followingย is :
\(I. \ \ ๐ฝ_0=๐พ_0=0 \\ II. \ \ ๐ฝ_0=๐พ_0=1 \\ III \ \ ๐ฝ_1=๐พ_1=๐0 \\ IV . \ J_{1} = K_{1} = \overline{Q}_{0} \\ V. \ \ ๐ฝ_2=๐พ_2=๐1๐0 \\ VI. \ J_{2} = K_{2} =\overline{Q}_{1} \overline{Q}_{0}\)
- A.
I, III, V
- B.
I, IV, VI
- C.
II, III, V
- D.
II, IV, VI
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Correct answer: D
Derivation: obtain JK inputs by examining the next-state transitions for a 3-bit down counter (N โ Nโ1 mod 8).
Key idea: For a JK flip-flop, when the bit must toggle, set J = K = 1; when it must hold at 0 or 1 use J = K = 0; use the excitation patterns to identify toggling conditions.
Least-significant bit (Q0):
Q0 alternates every count (0โ1, 1โ0) for every clock pulse, so it always toggles.
Therefore J0 = K0 = 1.
Middle bit (Q1):
Observing the transitions shows Q1 toggles exactly when Q0 = 0 (for states where Q0 = 0, Q1 changes; where Q0 = 1, Q1 holds).
Therefore J1 = K1 = ~Q0.
Most-significant bit (Q2):
Q2 toggles only when Q1 = 0 and Q0 = 0 (i.e., on the transition from 000โ111 and 100โ011), so the toggle condition is Q1 = 0 AND Q0 = 0.
Therefore J2 = K2 = ~Q1 AND ~Q0.
Final minimized JK inputs:
J0 = K0 = 1
J1 = K1 = ~Q0
J2 = K2 = ~Q1 AND ~Q0
These are the correct minimized expressions for the inputs of the three JK flip-flops in the 3-bit binary down counter.