Question

Design a 4-bit bidirectional serial-in-serial-out shift register using S-R flip flops that trigger on the negative–edge transition. EXPLAIN its operation if binary input 0101 is applied to the register which initially holds binary data 1101. DRAW the timing-diagram for serial-in operation in right-shift mode only.

Answer 1

In this diagram

when Mode Control = 1, the circuit will behave like Right shift serial in serial out register. When M=1(High). AND Gate no 1, 3, 5 and 7 will allow to pass the data to OR Gate whereas AND Gate no 2, 4, 6 and 8 will give output 0[Low] irresptive of any input. So circuit wll allow to pass the data in the right direction

i.e.

Similarly when M= 0 , the circuit will act as a Left shift register. Because only AND Gate no 2, 4, 6 and 8 will allow the data to pass through OR Gate.

So it will transfer the data as

When initial data is 1101

for M=1

for input = 0101

after 1st clock output will be 1110

after 2nd clock output will be 0111

after 3rd clock output will be 1011

after 4th clock output will be 0101

Similarly

for M=0 (Left Shift)

for input = 0101

after 1st clock output will be 1010

after 2nd clock output will be 0101

after 3rd clock output will be 1010

after 4th clock output will be 0101

Here is the timing diagram