Problem

Solutions For Digital Design Chapter 5 Problem 4P

Solution 1

Step #1 of 10

Consider the following data:

A flip-flop has the following operations:

“Clear to 0” when the inputs are

“No change” when the inputs are

“Complement” when the inputs are

“Set to 1” when the inputs are

Step #2 of 10

The flip-flop operation “clear to 0” clears the next state of the flip-flop to “0” irrespective of the present state. That is, for both and.

The flip-flop operation “no change”, makes the next state of the flip-flop as. That is, next state follows the present state and is expressed as, .

The flip-flop operation “complement” makes the next state of the flip-flop as. That is, next state follows the complement of the present state and is expressed as, .

The flip-flop operation “set to 1” sets the next state of the flip-flop to “1” irrespective of the present state. That is, for both and.

Step #3 of 10

(a)

A characteristic table defines the logical operations of a flip-flop by describing its

operation in a tabular form.

The Characteristic table for the givenflip-flop is shown in Table 1:

Table 1

Where,

Present state of the flip-flop

Next state of the flip-flop

Thus, the Characteristic table for the givenflip-flop is obtained.

Step #4 of 10

(b)

The characteristic table for the given flip-flop in Table 1 with the possible present

state and next state outputs is shown in Table 2:

Table 2

Step #5 of 10

The characteristic equation for the given flip-flop can be found by using the Karnaugh map as shown in Figure 1:

Figure 1

The next state output can be obtained by combining the rectangles which consists of “1” as shown in Figure 1.

Thus, the required characteristic equation for the givenflip-flop is,

.

Step #6 of 10

(c)

The excitation table lists the required inputs of the flip-flop from the known present state and next state outputs of the flip-flop.

For the given flip-flop, next state and the present state, are “0” when

the inputsare. Hence, in the excitation table the input P is taken as “0”

and the input N is taken as don’t care.

For theflip-flop, the next state is “1” and the present state, is “0” , only when the inputsare . Hence, in the excitation table the input P is taken as “1” and the input N is taken as don’t care.

For the flip-flop, the next state is “0” and the present state, is “1” , when the inputsare . Hence, in the excitation table the input P is taken as don’t careand the input N is taken as “0”.

For the given flip-flop, the next state and the present state, are “1” when the inputsare . Hence, in the excitation table the input P is taken as don’t careand the input N is taken as “1”.

Thus, the excitation table for the givenflip-flop is shown in Table 3:

Table 3

In Table 3, the symbol is represents the don’t care condition of the flip-flop.

Step #7 of 10

(d)

The schematic diagram for the given flip-flop is shown in Figure 2:

Figure 2

Step #8 of 10

The characteristic table for the flip-flop is shown in Table 4:

Table 4

Step #9 of 10

Construct a conversion table to convert flip-flop into D flip flop using Table 3 and Table 4.

Table 5

Find the Boolean expression for the P and N in terms of D and using k-map.

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Write the Boolean expression for P and N.

Step #10 of 10

Therefore, the given

flip-flop can be converted into D flip flop by connecting

inputs together.

The converted D flip-flop is shown in Figure 3:

Figure 4: equivalent D flip-flop for the givenflip-flop

Solution 2