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: Review the JK Flip-Flop Truth Table
For a JK flip-flop, the next state depends on inputs J and K as follows: \[ \begin{array}{c|c|c} J & K & \text{Next State} \\ \hline 0 & 0 & \text{No change} \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & \text{Toggle} \end{array} \] 
Here, \( J = 1 \) and \( K = 1 \). Therefore, the flip-flop operates in toggle mode.  

Step 2: Analyze the Toggle Operation
In toggle mode, the output switches state at every clock pulse. If the initial output is 0, the sequence becomes: \[ 0 \rightarrow 1 \rightarrow 0 \rightarrow 1 \rightarrow \dots \] A complete output cycle (0 → 1 → 0) requires two clock pulses. 

Step 3: Compute the Output Frequency
Since the output changes state on every clock edge but completes one full cycle in two clock pulses: \[ f_{out} = \frac{f_{clk}}{2} \] 
Final Result:
\[ \boxed{ f_{out} = \frac{f_{clk}}{2} } \] 
Hence, a JK flip-flop with \( J = K = 1 \) functions as a divide-by-2 frequency divider.