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The bit rate of a digital communication system is 34 Mbps. The modulation scheme is QPSK. The baud rate of the system is
- a)68 Mbps
- b)34 Mbps
- c)17 Mbps
- d)85 Mbps
Correct answer is option 'C'. Can you explain this answer?
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The bit rate of a digital communication system is 34 Mbps. The modulat...
In BPSK, baud rate is half the bit rate due to which there is more effective utilization of the available bandwidth of the transmission channel. Given, bit rate = 34 Mbps
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The bit rate of a digital communication system is 34 Mbps. The modulat...
The bit rate of a digital communication system is determined by the combination of the modulation scheme and the baud rate. In this case, the modulation scheme is QPSK (Quadrature Phase Shift Keying) and the bit rate is given as 34 Mbps. We need to find the baud rate of the system.
- Modulation Scheme:
QPSK is a modulation scheme that uses four different phase shifts to represent four different symbols. Each symbol represents two bits of information. Therefore, the modulation scheme has a symbol rate of half the bit rate. In this case, since the bit rate is 34 Mbps, the symbol rate would be 17 Msymbols/s.
- Baud Rate:
The baud rate is the number of symbols (or pulses) transmitted per second. In the case of QPSK, each symbol represents 2 bits, so the baud rate is equal to the symbol rate divided by 2. In this case, the baud rate would be 17 Msymbols/s divided by 2, which is equal to 8.5 Mbps.
Therefore, the baud rate of the system is 8.5 Mbps. However, none of the given options match this value. It seems there might be an error in the question or the options provided.
- Modulation Scheme:
QPSK is a modulation scheme that uses four different phase shifts to represent four different symbols. Each symbol represents two bits of information. Therefore, the modulation scheme has a symbol rate of half the bit rate. In this case, since the bit rate is 34 Mbps, the symbol rate would be 17 Msymbols/s.
- Baud Rate:
The baud rate is the number of symbols (or pulses) transmitted per second. In the case of QPSK, each symbol represents 2 bits, so the baud rate is equal to the symbol rate divided by 2. In this case, the baud rate would be 17 Msymbols/s divided by 2, which is equal to 8.5 Mbps.
Therefore, the baud rate of the system is 8.5 Mbps. However, none of the given options match this value. It seems there might be an error in the question or the options provided.
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The bit rate of a digital communication system is 34 Mbps. The modulation scheme is QPSK. The baud rate of the system isa)68 Mbpsb)34 Mbpsc)17 Mbpsd)85 MbpsCorrect answer is option 'C'. Can you explain this answer?
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