Electronics and Communication Engineering - Digital Electronics - Discussion

Here we connected a ex-or gate out to input of the register.

In intial state 0110,

1st Clock pulse : i/p of ex-or is 10. So, output is 1 is given to register. Now the register is 1011.

2nd Clock pulse : i/p of exor is 11. So, the output is 0 is given to register. Now the register is 0101.

Finally 3rd clk pulse : i/p of exor is 01. So, the output is 1 is given to register. Now register is 1010.

Hence after 3 clock pulses upline the content of shift register will be 1010.

Just shift only one bit to the right and get the answer and that bit will be your xor's output. Do it thrice as given is 3 clock pulse.

I am not getting it. Please someone explain me clearly.

Initial state 0110.

Last two digit input s xor function that will give input of the first digit.
0110 in that 10 are last two digit xor function 1 = 1011.

Similarly 11 as d input of xor gate o/p is 0 = 0101.
Similarly 3 rd pulse o/p of xor is 1 = 1010.

So final answer is 1010.