Question

A JK flip-flop has inputs $J = 1$ and $K = 1$. 
The clock input is applied as shown. Find the output clock cycles per second (output frequency). 
 

Hint:

A JK flip-flop with $J=K=1$ always works as a divide-by-2 circuit because it toggles at every clock edge.

Answer 1

Step 1: Recall JK Flip-Flop Operation.
For a JK flip-flop:
\[ \begin{array}{c|c|c} J & K & \text{Next State} \\ \hline 0 & 0 & \text{No change} \\ 0 & 1 & 0\\ 1 & 0 & 1 \\ 1 & 1 & \text{Toggle} \end{array} \]
Given:
\[ J = 1, \quad K = 1. \] Hence the flip-flop toggles at every clock pulse.
Step 2: Understand Toggle Behavior.
When toggle mode is active, the output changes state at each clock edge:
If initial state is 0:
\[ 0 \rightarrow 1 \rightarrow 0 \rightarrow 1 \rightarrow \dots \]
Thus, one complete output cycle (0 → 1 → 0) requires two clock pulses.
Step 3: Determine Output Frequency.
Since output changes state at every clock pulse, but completes one full cycle in two pulses:
\[ f_{out} = \frac{f_{clk}}{2}. \]
Final Answer:
\[ \boxed{f_{out} = \frac{f_{clk}}{2}}. \]
Thus, the JK flip-flop with $J=K=1$ acts as a frequency divider by 2.