shift register

Question

A 8-bit left-shift register and a D-flip flop are connected together as shown in the figure below and they are synchronized with the same clock.

If the D-flip-flop is initially cleared, then the circuit will act as

It will work like b7xor 0    b6 xorb5      b4 xorb3    b2xorb1 ???????

please give explain 

Answer 1

The output of XOR gate is z=b_i+1⊕b_i and this shift the register to left, initially z=0.
after 1^st clock z = b₇⊕ 0 = b₇
after 2^nd clock z = b₇⊕b₆
after 3^rd clock z = b₆⊕b₅
after 4^th clock z = b₅⊕b₄
So it will work as Binary to Gray Converter.

Comments

But how? Aren't same bits are getting XORed?

Answer 2

Answer is  Binary to Gray code.

The D flip flop creates delay of one clock cycle, since the D flip flop is clear initially the 0 XOR b7 will take place of b1 with left shift at same time.

Now the one cycle is completed and D flip flop will give result of previous operation that is b7 (coz in D --> Qn+1 = D) and at this point b6 will be entering D flip flop and one of the input of XOR gate and we have b7 XOR b6 . After that it is easy to understand why it is binary to Gray code.