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Four J-K flip-flops are cascaded with their J-K inputs tied HIGH. If the input frequency (fin) to the first flip-flop is 32 kHz, the output frequency (fout) is ________
  • a)
    1 kHz
  • b)
    2 kHz
  • c)
    4 kHz
  • d)
    16 kHz
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Four J-K flip-flops are cascaded with their J-K inputs tied HIGH. If t...
32/2 = 16:-first flip-flop,
16/2 = 8:- second flip-flop,
8/2 = 4:- third flip-flop,
4/2 = 2:- fourth flip-flop.
Since the output frequency is determined on basis of the 4th flip-flop.
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Four J-K flip-flops are cascaded with their J-K inputs tied HIGH. If t...
Explanation:

To solve this problem, we need to understand the operation of J-K flip-flops and how they are cascaded together.

J-K Flip-flop:
A J-K flip-flop is a sequential logic device with two inputs: J (set) and K (reset) and two outputs: Q (output) and Q' (complement of the output). The flip-flop has two stable states: '0' and '1'. The output changes based on the inputs and the current state.

When J = 0 and K = 0, the output remains unchanged.
When J = 0 and K = 1, the output is reset to '0'.
When J = 1 and K = 0, the output is set to '1'.
When J = 1 and K = 1, the output toggles between '0' and '1'.

Cascading Flip-flops:
When flip-flops are cascaded together, the output of one flip-flop is connected to the input of the next flip-flop. This allows for the creation of longer shift registers or counters.

In this problem, four J-K flip-flops are cascaded together, with their J-K inputs tied HIGH. This means that the J and K inputs of all the flip-flops are permanently set to '1'.

Input Frequency:
The input frequency (fin) to the first flip-flop is given as 32 kHz.

Output Frequency:
Since the J and K inputs of all the flip-flops are permanently set to '1', the flip-flops will toggle their outputs at half the frequency of the input signal.

Therefore, the output frequency (fout) will be half of the input frequency (fin).

Calculating the output frequency:
fout = fin / 2 = 32 kHz / 2 = 16 kHz.

Therefore, the output frequency (fout) is 16 kHz.

The correct answer is option 'D'
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Comprehension:Read the following paragraph and answer the questions.A digital counter is a set of flipflops whose states change in response to the pulses applied at Input. Counters can be asynchronous counters or synchronous counters. A counter is an example of a state machine; the number of states is called the modulus. Two basic types of state machines are the Moore and the Mealy. In Moore machine, the combinational logic is a gate array with outputs that determine the next state of the flip-flops in the memory. For Mealy machine, the present state affects just as in Moore machine but in addition, the inputs also affect the outputs.In a 4-bit asynchronous binary counter, each D flip-flop is negative edge-triggered and has a propagation delay for 10 nanoseconds. What is the highest frequency allowed for the counter to avoid problems due to propagation delay?

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Four J-K flip-flops are cascaded with their J-K inputs tied HIGH. If the input frequency (fin) to the first flip-flop is 32 kHz, the output frequency (fout) is ________a)1 kHzb)2 kHzc)4 kHzd)16 kHzCorrect answer is option 'B'. Can you explain this answer?
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