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The bit rate of a digital communication system is R Kbits/sec. The modulation used is 32-QAM. The minimum band width for intersymbol interference free transmission is –
- a)(R/10)Hz
- b)(R/10)KHz
- c)(R/5)Hz
- d)(R/5)KHz
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
The bit rate of a digital communication system is R Kbits/sec. The mod...
For ISI free transmission according to Nyquist criterion for Baseband signal is:
But for Bandpass signal, the amplitude spectrum get shifted to some high frequency and B.W ≥ Rb
For M-ary signaling B.W
Calculation:
Bit Rate Rb = R k bits/sec
32-ary signaling is used:
But for Bandpass signal, the amplitude spectrum get shifted to some high frequency and B.W ≥ Rb
For M-ary signaling B.W
Calculation:
Bit Rate Rb = R k bits/sec
32-ary signaling is used:
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The bit rate of a digital communication system is R Kbits/sec. The mod...
To determine the minimum bandwidth for intersymbol interference (ISI)-free transmission, we need to consider the modulation scheme used and the bit rate of the digital communication system.
In 32-QAM modulation, each symbol represents 5 bits (log2(32) = 5). Therefore, the symbol rate (Rs) can be calculated by dividing the bit rate (R) by the number of bits per symbol:
Rs = R / log2(M)
Where M is the number of symbols in the modulation scheme.
For 32-QAM, M = 32, so the symbol rate is given by:
Rs = R / log2(32) = R / 5
To avoid ISI, the Nyquist criterion states that the bandwidth (B) should be equal to or greater than the symbol rate (Rs). Therefore, the minimum bandwidth for ISI-free transmission is:
B = Rs = R / 5
So, the minimum bandwidth required for ISI-free transmission in this digital communication system is R/5 Kbits/sec.
In 32-QAM modulation, each symbol represents 5 bits (log2(32) = 5). Therefore, the symbol rate (Rs) can be calculated by dividing the bit rate (R) by the number of bits per symbol:
Rs = R / log2(M)
Where M is the number of symbols in the modulation scheme.
For 32-QAM, M = 32, so the symbol rate is given by:
Rs = R / log2(32) = R / 5
To avoid ISI, the Nyquist criterion states that the bandwidth (B) should be equal to or greater than the symbol rate (Rs). Therefore, the minimum bandwidth for ISI-free transmission is:
B = Rs = R / 5
So, the minimum bandwidth required for ISI-free transmission in this digital communication system is R/5 Kbits/sec.
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The bit rate of a digital communication system is R Kbits/sec. The modulation used is 32-QAM. The minimum band width for intersymbol interference free transmission is –a)(R/10)Hzb)(R/10)KHzc)(R/5)Hzd)(R/5)KHzCorrect answer is option 'D'. Can you explain this answer?
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