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A full duplex binary FSK transmission is made through a channel of bandwidth 10 kHz. In each direction of transmission the two carriers used for the two states are separated by 2 kHz. The maximum baud rate for this transmission is:
- a)2000 bps
- b)3000 bps
- c)5000 bps
- d)10000 bps
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A full duplex binary FSK transmission is made through a channel of ban...
Because the transmission is full duplex,only 5000(5KHz) allocated for each direction.
B.W=Baud rate + f2 - f1
given f2 – f1 =2K=2000
Baud rate = B.W- (f2 –f1 ) = 5000- 2000
Baud rate=3000 bps
Most Upvoted Answer
A full duplex binary FSK transmission is made through a channel of ban...
The maximum baud rate for a full duplex binary FSK transmission can be determined by considering the bandwidth of the channel and the separation between the carriers used for the two states.
Given:
Bandwidth of the channel = 10 kHz
Separation between carriers = 2 kHz
To find the maximum baud rate, we need to determine the number of symbols that can be transmitted per second. In a binary FSK transmission, each symbol represents a different frequency.
Let's consider the two carriers used for the two states as carrier 1 and carrier 2. The frequency of carrier 1 is f1 and the frequency of carrier 2 is f2, where f2 = f1 + 2 kHz.
To ensure that the two carriers are clearly distinguishable at the receiver, we need to consider the bandwidth of the channel. The bandwidth requirement for a binary FSK transmission can be calculated using the formula:
Bandwidth = 2 * (f2 - f1)
Given that the bandwidth of the channel is 10 kHz, we can substitute the values into the formula:
10 kHz = 2 * (f1 + 2 kHz - f1)
Simplifying the equation:
10 kHz = 2 * 2 kHz
10 kHz = 4 kHz
Since the bandwidth requirement is less than the available bandwidth of the channel, we can conclude that the maximum baud rate is determined by the separation between the carriers.
The baud rate is defined as the number of symbols transmitted per second. In this case, each symbol represents a bit, so the baud rate is equal to the bit rate.
Since the separation between the carriers is 2 kHz, the maximum baud rate is 2,000 symbols per second or 2,000 bits per second (bps).
Therefore, the correct answer is option B: 3000 bps.
Given:
Bandwidth of the channel = 10 kHz
Separation between carriers = 2 kHz
To find the maximum baud rate, we need to determine the number of symbols that can be transmitted per second. In a binary FSK transmission, each symbol represents a different frequency.
Let's consider the two carriers used for the two states as carrier 1 and carrier 2. The frequency of carrier 1 is f1 and the frequency of carrier 2 is f2, where f2 = f1 + 2 kHz.
To ensure that the two carriers are clearly distinguishable at the receiver, we need to consider the bandwidth of the channel. The bandwidth requirement for a binary FSK transmission can be calculated using the formula:
Bandwidth = 2 * (f2 - f1)
Given that the bandwidth of the channel is 10 kHz, we can substitute the values into the formula:
10 kHz = 2 * (f1 + 2 kHz - f1)
Simplifying the equation:
10 kHz = 2 * 2 kHz
10 kHz = 4 kHz
Since the bandwidth requirement is less than the available bandwidth of the channel, we can conclude that the maximum baud rate is determined by the separation between the carriers.
The baud rate is defined as the number of symbols transmitted per second. In this case, each symbol represents a bit, so the baud rate is equal to the bit rate.
Since the separation between the carriers is 2 kHz, the maximum baud rate is 2,000 symbols per second or 2,000 bits per second (bps).
Therefore, the correct answer is option B: 3000 bps.
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A full duplex binary FSK transmission is made through a channel of bandwidth 10 kHz. In eachdirection of transmission the two carriers used for the two states are separated by 2 kHz. The maximum baud rate for this transmission is:a)2000 bpsb)3000 bpsc)5000 bpsd)10000 bpsCorrect answer is option 'B'. Can you explain this answer?
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