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Leno4ka [110]
4 years ago
5

A 4-bit left shift register is initially in the 0000 state, with all the flip flops storing 0s. A group of bits 1011 is serially

left-shifted into the register. What will the register contain after two clock pulses?
Computers and Technology
1 answer:
frutty [35]4 years ago
6 0

Answer:

0010

Explanation:

Serially left shifted means that the left most bit will enter the register first. The left most bit already stored in the register will move out of the sequence. The "bold" bits mentioned below highlight these left most bits:

Initial State of the Register:

0000

Group of bits entering:

1011

<u>First Clock Cycle:</u>

0000 <em>(This bold bit will move out)</em>

1011 <em>(This bold bit will move in from right side, shifting the whole sequence one place to the left).</em>

The resulting Sequence:

0001

<u>Second Clock Cycle:</u>

0001 <em>(This bold bit will move out)</em>

1011 <em>(This bold bit will move in from right side, shifting the whole sequence one place to the left).</em>

The resulting Sequence:

0010 <em>(Final Answer)</em>

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Answer:

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To calculate G if A=3, b=4 and C=5

cobegin

p1: A^3 = 27

P2:C^2 =25

P3: p1 +B =31

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Explanation:

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