Question

Problem 2 (10 points): Consider the serial adder shown in Figure below. It uses two 4-bit serial shift registers A and B. Initially, register A holds the binary number 0110 and register B holds 0011, while the carry flip-flop is reset to 0. Note that the serial input for shift register B is connected to the logic value zero Use the given table to list the binary values in register A, register B, the S signal, and the carry flip-flop after each of the four shifts Shift SI Shift register A Augend) sO CLK FA Serial input so SI Shift register B Addend) Clear Figure problem 2. Shifts Signals Content of A register Content of 0 (Initial values) 2 4 B register Carry

Answer 1

	0	1	2	3	4
Content of A register	0110	0011	1001	0100	0010
Content of B register	0011	0001	0000	0000	0000
S	0	1	0	0	1
Carry	0	0	1	1	0

Explanation:

x	y	z	S	C
0	0	0	0	0
0	0	1	1	0
0	1	0	1	0
0	1	1	0	1
1	0	0	1	0
1	0	1	0	1
1	1	0	0	1
1	1	1	1	1

All binary additions(x+y+z) done in the below solution are referred from above table.

Initial values: Given A register 0110

B register : 0011

Since all are reset the S and C values are also 0

After first shift:

LSB of A and B are passed to x and y respectively and since C is zero, the Q of D flip flop is also 0, hence z= 0;

x+y+z = 0+1+0 results in S= 1 and C= 0;

Now the registers are right shifted by 1 bit. The MSB of A becomes the previous S while that of B becomes 0 as Serial input 0 is always appended after every shift.

Hence A = 0011, B= 0001

After second shift:

LSBs of the contents present in A and B after first shift becomes x and y while z will be the previous C which is 0

x+y+z = 1+1+0 results in S=0 and C=1;

while A becomes 1001 (MSB from previous S) and B = 0000

After third shift:

LSBs of the contents present in A and B after second shift becomes x and y and z will be the previous C which is 1;

x+y+z = 1+0+1 results in S= 0 and C=1

while A becomes 0100 and B = 0000

After fourth shift:

LSBs of the contents present in A and B after third shift becomes x and y and z will be the previous C which is 1;

x+y+z = 0+0+1 results in S=1 and C=0

while A becomes 0010 and B = 0000