Ive read the chapter based on sequential circuits, but its all based in theory w

Question

Ive read the chapter based on sequential circuits, but its all based in theory which makes it hard for me to follow instead of examples.

The combination requires sequence for 2bit binary codewords

AB are generated using switches, the switches are flipped to establish the correct values of A and B in preparation for Next state transition.

AB=X4aX4b ; X3aX3b; X2a X2b ; X1aX1b ;LET (Y4Y3Y2Y1)10

Each codeword XNaXNb is the least significant two bits of decimal digit YN

The Five Les are to provide an indication that the lock is in the following states

LED1: state a ; LED2: state b, f ; LED3: stage c, g ; LED 4: States d, h ; LED5: state e

This is that part i have trouble with, can someone please explain..

Explanation / Answer

It appears the table is arranged like this:

CURRENT STATE || INPUT || NEXT STATE

Q1 Q2    Q3 || A B || Q1* Q2* Q3*

I believe this is called a "state change table." The way to get from the state diagram to the table is to simply check how many states are present in the machine (in this case 8) and assign a number to each. Since 8<=2^3, a three-bit number can represent any state. Then each state gets a number 0 to 7 (order doesn't matter, but they went alphabetically by state in this case). Based on the inputs A and B and the current state Q, you must determine the next state Q*. To fill out the table, you'll need to enter the number for the appropriate Q* in the three boxes for Q1*Q2*Q3* on each row. For example, the first state (a):

I think the first label for the correct input is 01, but it's hard to see. I'll proceed under the assumption that that I am correct...

First line: You are in state a, and the input is 00 (A=0, B=0). Since that is INCORRECT, you should proceed to state f, which has been assigned the number 101 (according to the table). So for the first line, Q1*=1, Q2*=0, and Q3*=1.

Second line: You are in state a, and the input is 01 (A=0, B=1). Since that is CORRECT, you should proceed to state b, which has been assigned the number 001. So for the second line Q1*=0, Q2*=0, and Q3*=1.

As for the J and K fields, it appears the implementation is going to use JKFF's. You just have to fill in the proper J and K driver values to change each FF to the next state where J1,K1 drive Q1; J2,K2 drive Q2; and J3,K3 drive Q3. The interesting thing about the JK implementation is that there will always be two ways to get each FF to the proper next state, which leaves some room for possible optimization in the final circuit. Anyway, the first couple lines could be:

D-LIKE-EXCITATION (OR)    T-LIKE-EXCITATION

J1 K1 || J2 K2 || J3 K3 (OR) J1 K1 || J2 K2 || J3 K3

1 0 0 1 1 0       1 1 0 0 1 1

0 1 0 1 1 0 0 0 0 0 1 1

You can, of course, mix and match the two excitation methods as you like.

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