A sequential circuit takes an input stream of $0's$ and $1's$ and produces an output stream of $0's$ and $1's.$ Initially it replicates the input on its output until two consecutive $0's$ are encountered on the input. From then onward, it produces an output stream, which is the bit-wise complement of input stream until it encounters two consecutive 1's, whereupon the process repeats. An example input and output stream is shown below.

$\begin{array}{ll}

\text{The input stream:} & 101100|01001011|011\\

\text{The desired output:}& 101100|10110100|011\\

\end{array}$

$\text{J-K}$ master-slave flip-flops are to be used to design the circuit.

- Give the state transition diagram
- Give the minimized sum-of-product expression for $\text{J}$ and $\text{K}$ inputs of one of its state flip-flops

## 2 Answers

We can design a Mealy Machine as per the requirement given in the question.

From which we will get state table, and we can design sequential circuit using any Flip-flop from the state table (with the help of excitation table) :

As we get $4$ states (renaming state component to binary states), we need two FFs to implement it.

Let $A$ and $B$ be present states, $x$ be the input and $y$ be the output.$$\small \begin{array}{c|c|c|c|cc|cc|}

\text{Present State}&\text{Input}&\text{Next State}&\text{Output}&\rlap{\text{FF}}&&\rlap{\text{FF}}\\

&&&&\rlap{\text{inputs}}&&\rlap{\text{inputs}}\\

\hline

AB&X&A'B'&Y&J_A&K_A&J_B&K_B\\\hline

00&0&01&0&0&X&1&X\\

00&1&00&1&0&X&0&X\\

01&0&10&0&1&X&X&1\\

01&1&00&1&0&X&X&1\\

10&0&10&1&X&0&0&X\\

10&1&11&0&X&0&1&X\\

11&0&10&1&X&0&X&1\\

11&1&00&0&X&1&X&1\\

\end{array}$$ \begin{array}{cc|cc} Q_t&Q_{t+1}&J&K\\\hline 0&0&0&X\\0&1&1&X\\1&0&X&1\\1&1&X&0\end{array}

### 16 Comments

Shaik Masthan Please add your answer. I want to learn to design circuit without Master Slave Filp Flop also

there is no specific representation for Master-slave FF, in this answer the A flipflop and B flipflops are Master slave Flipflop's only ! ( the internal circuit is different for normal and master-slave, but o/p is same. )

Then Why my answer is not Matching ?

without state diagram, i tried and got with 3-FF, it is not minimized !

Just one doubt how could you replace q0 with 00.. why can't we replace it with 11(or any other combination).. can you please help me with this @Shaik Masthan @Praveen Saini

normally in the state diagram, the states are represented the o/p of FF's only.

Here what is the output of FF, we don't know about it right ??

By taking the initial state as 00, we form that entire sequence !

I agree we generally take 00 as initial state unless it is specified to take any other value, but how will the second state be 01 ? :(

@Shaik Masthan Sir

the master slave JK flipflop are a modification of normal JK flipflop to avoid RACE AROUND condition in JK Flipflop when we use level trigerred flipflops.

If we use edge trigerred then no use of master slave JK flipflop.right?