Divisor 4 bits LU Shift right Shift left Control test Remai

Divisor 4 bits LU Shift right_ Shift left + Control test Remainder 8 bits Fig. 2 Quotient Deviser Remainder Description Initial values Step Step 1 Step 2 b) Suppose that each operation (addition, subtraction or shift) takes 2ns. Please calculate the time necessary to perform the above division using the given hardware. Assume the registers have been initialized.

Solution

According to above Hardware, we are not using any other register to store quotient. The remainder is taken 8 bits and it is considered as both quotient and remainder

Steps for the Process:

i.Then Go to step 4

i.Then Go to step 8

i.e., First 4 bits is remainder(R) and next 4 bits is quotient (Q)

REMAINDER

QUOTIENT

DIVISOR

REMAINDER -QUOTIENT

DESCRIPTION

SHIFT

0000

1110

0101

Always < 0

Initial values loaded

Initial

0000

0111

0101

Value < 0

Shift values and R-Q < 0

1

0001

1100

0101

Value < 0

Shift values and R-Q < 0

2

0011

100

0101

Value < 0

Shift values and R-Q < 0

3

0111

0000

0101

Value > 0

Shift values and R-Q > 0

4

0010

0000

0101

Subtract (R-D) to R

4

0100

0001

0101

Shift Q by placing 1 at RMB

4

0010

0001

0101

Final Result

4

REMAINDER:      0010

QUOTIENT:         0001

b) TIME TAKEN:

                Shifts : 4 Shift + 1 subtraction = 5 operations

                Total Time :         5 * 2 = 10 ns (assuming 1 operation=2 ns)

REMAINDER	QUOTIENT	DIVISOR	REMAINDER -QUOTIENT	DESCRIPTION	SHIFT
0000	1110	0101	Always < 0	Initial values loaded	Initial
0000	0111	0101	Value < 0	Shift values and R-Q < 0	1
0001	1100	0101	Value < 0	Shift values and R-Q < 0	2
0011	100	0101	Value < 0	Shift values and R-Q < 0	3
0111	0000	0101	Value > 0	Shift values and R-Q > 0	4
0010	0000	0101		Subtract (R-D) to R	4
0100	0001	0101		Shift Q by placing 1 at RMB	4
0010	0001	0101		Final Result	4