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The clock frequency applied to the digital circuit shown in the figure below is 1 kHz. If the initial state of the output Q of the flip-flop is ‘0', then the frequency of the output wave orm Q in kHz is X
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- 0.25
- 0.5
- 1
- 2
Correct Option: B
From above fig.
X = [(Q ⊕ Q).(Q ⊕ Q)]
X = 1 because
Q ⊕ Q = 1 always ⇒ Q ⊕ Q = 0 always
∴ (q ⊕ Q) . (Q ⊕ Q = (1 . 0) = 0 = 1
∴ ‘ T ’ input = 1 always for ‘τ’ flip flop of input = 1 then O/P will be implemented at the time of triggering.
∴ ƒ1 = 0.5(ƒ) = 0.5(1) = 0.5kHz
