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The group of bits 11001 is serially shifted (right-most bit first) into a 5-bit parallel output shift register with an initial state 01110. After three clock pulses, the register contains ________
- a)01110
- b)00001
- c)00101
- d)00110
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The group of bits 11001 is serially shifted (right-most bit first) int...
Given:
- Group of bits: 11001
- Serial shift (right-most bit first)
- 5-bit parallel output shift register
- Initial state: 01110
To Find:
- State of the register after three clock pulses
Solution:
Step 1: Initial State
The initial state of the register is given as 01110.
Step 2: Clock Pulse 1
During the first clock pulse, the right-most bit of the group of bits is shifted into the right-most bit of the register. The remaining bits in the register are shifted to the right, and the left-most bit is discarded.
- Right-most bit of the group of bits: 1
- Register after clock pulse 1: 10111
Step 3: Clock Pulse 2
During the second clock pulse, the right-most bit of the group of bits is shifted into the right-most bit of the register. The remaining bits in the register are shifted to the right, and the left-most bit is discarded.
- Right-most bit of the group of bits: 0
- Register after clock pulse 2: 01011
Step 4: Clock Pulse 3
During the third clock pulse, the right-most bit of the group of bits is shifted into the right-most bit of the register. The remaining bits in the register are shifted to the right, and the left-most bit is discarded.
- Right-most bit of the group of bits: 0
- Register after clock pulse 3: 00101
Final Answer:
After three clock pulses, the register contains the bits 00101. Therefore, the correct answer is option C.
- Group of bits: 11001
- Serial shift (right-most bit first)
- 5-bit parallel output shift register
- Initial state: 01110
To Find:
- State of the register after three clock pulses
Solution:
Step 1: Initial State
The initial state of the register is given as 01110.
Step 2: Clock Pulse 1
During the first clock pulse, the right-most bit of the group of bits is shifted into the right-most bit of the register. The remaining bits in the register are shifted to the right, and the left-most bit is discarded.
- Right-most bit of the group of bits: 1
- Register after clock pulse 1: 10111
Step 3: Clock Pulse 2
During the second clock pulse, the right-most bit of the group of bits is shifted into the right-most bit of the register. The remaining bits in the register are shifted to the right, and the left-most bit is discarded.
- Right-most bit of the group of bits: 0
- Register after clock pulse 2: 01011
Step 4: Clock Pulse 3
During the third clock pulse, the right-most bit of the group of bits is shifted into the right-most bit of the register. The remaining bits in the register are shifted to the right, and the left-most bit is discarded.
- Right-most bit of the group of bits: 0
- Register after clock pulse 3: 00101
Final Answer:
After three clock pulses, the register contains the bits 00101. Therefore, the correct answer is option C.
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Community Answer
The group of bits 11001 is serially shifted (right-most bit first) int...
LSB bit is inverted and feed back to MSB:
01110 → initial
10111 → first clock pulse
01011 → second
00101 → third.
01110 → initial
10111 → first clock pulse
01011 → second
00101 → third.
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The group of bits 11001 is serially shifted (right-most bit first) into a 5-bit parallel output shift register with an initial state 01110. After three clock pulses, the register contains ________a)01110b)00001c)00101d)00110Correct answer is option 'C'. Can you explain this answer?
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