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An 8-bit serial in/serial out shift register is used with a clock frequency of 2 MHz to achieve a time delay (td) of ________
- a)16 us
- b)8 us
- c)4 us
- d)2 us
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
An 8-bit serial in/serial out shift register is used with a clock freq...
Calculation of Time Delay (td) in a Serial In/Serial Out Shift Register
Serial In/Serial Out Shift Register: A serial in/serial out shift register is a digital circuit that can hold and shift data bits in a serial manner. It has a single input and a single output, and the data is shifted through the register one bit at a time.
Clock Frequency: The clock frequency is the number of clock cycles per second. It determines the rate at which data is transferred through the shift register.
Calculation of Time Delay (td): The time delay (td) is the time taken by the shift register to transfer one bit of data from the input to the output. It can be calculated using the following formula:
td = 1 / (clock frequency x number of bits)
In this case, we have an 8-bit serial in/serial out shift register and a clock frequency of 2 MHz. Therefore, the time delay (td) can be calculated as follows:
td = 1 / (2 MHz x 8)
td = 1 / 16 us
td = 0.0625 us
Rounding off to the nearest microsecond, we get:
td ≈ 4 us
Therefore, the time delay (td) of the 8-bit serial in/serial out shift register with a clock frequency of 2 MHz is approximately 4 microseconds. Option C is the correct answer.
Serial In/Serial Out Shift Register: A serial in/serial out shift register is a digital circuit that can hold and shift data bits in a serial manner. It has a single input and a single output, and the data is shifted through the register one bit at a time.
Clock Frequency: The clock frequency is the number of clock cycles per second. It determines the rate at which data is transferred through the shift register.
Calculation of Time Delay (td): The time delay (td) is the time taken by the shift register to transfer one bit of data from the input to the output. It can be calculated using the following formula:
td = 1 / (clock frequency x number of bits)
In this case, we have an 8-bit serial in/serial out shift register and a clock frequency of 2 MHz. Therefore, the time delay (td) can be calculated as follows:
td = 1 / (2 MHz x 8)
td = 1 / 16 us
td = 0.0625 us
Rounding off to the nearest microsecond, we get:
td ≈ 4 us
Therefore, the time delay (td) of the 8-bit serial in/serial out shift register with a clock frequency of 2 MHz is approximately 4 microseconds. Option C is the correct answer.
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Community Answer
An 8-bit serial in/serial out shift register is used with a clock freq...
One clock period is = (1⁄2) micro-s = 0.5 microseconds. In serial transmission, data enters one bit at a time. So, the total delay = 0.5*8 = 4 micro seconds time is required to transmit information of 8 bits.
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