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A signal has a bandwidth of 1 MHz. It is sampled at a rate 50% higher than the Nyquist rate and  quantized into 256 level using a  -low quantizer with μ = 225.
Que: It was found that a sampling rate 20% above therate would be adequate. So the maximum SNR, thatcan be realized without increasing the transmission bandwidth, would be
  • a)
    60.4 dB
  • b)
    70.3 dB
  • c)
    50.1 dB
  • d)
    None of the above
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A signal has a bandwidth of 1 MHz. It is sampled at a rate 50% higher ...
Nyquist Rate = 2 MHz
This question is part of UPSC exam. View all Electronics and Communication Engineering (ECE) courses
Most Upvoted Answer
A signal has a bandwidth of 1 MHz. It is sampled at a rate 50% higher ...
Given Information
- Bandwidth of the signal = 1 MHz
- Sampling rate = 50% higher than Nyquist rate
- Quantization level = 256
- Quantizer step size = 225
- Sampling rate should be 20% above the Nyquist rate

To find
- Maximum SNR that can be realized without increasing the transmission bandwidth

Solution
Nyquist Rate:
The Nyquist rate is defined as the minimum sampling rate required to reconstruct a signal without any loss of information. It is given by the formula:
Nyquist rate = 2 x Bandwidth
Nyquist rate = 2 x 1 MHz = 2 MHz

Sampling Rate:
The given sampling rate is 50% higher than the Nyquist rate. Therefore, the sampling rate is:
Sampling rate = 1.5 x 2 MHz = 3 MHz

Quantization:
The given quantization level is 256, which means that the quantizer step size is:
Quantizer step size = (maximum signal amplitude) / (number of quantization levels - 1)
Quantizer step size = (2 x maximum signal amplitude) / 255
Quantizer step size = 225

From the above two equations, we get:
Maximum signal amplitude = (255 x Quantizer step size) / 2
Maximum signal amplitude = (255 x 225) / 2
Maximum signal amplitude = 28687.5

Sampling Rate for Maximum SNR:
The sampling rate should be 20% above the Nyquist rate. Therefore, the sampling rate for maximum SNR is:
Sampling rate = 1.2 x 2 MHz = 2.4 MHz

Signal-to-Quantization-Noise Ratio (SQNR):
The SQNR is given by the formula:
SQNR = (6.02 x number of bits) + 1.76 dB
SQNR = (6.02 x log2(number of quantization levels)) + 1.76 dB
SQNR = (6.02 x log2(256)) + 1.76 dB
SQNR = 49.92 dB

Noise Power:
The quantization noise power is given by the formula:
Noise power = (quantizer step size)² / 12
Noise power = (225)² / 12
Noise power = 4218.75

Signal Power:
The maximum signal amplitude is 28687.5. Therefore, the signal power is:
Signal power = (maximum signal amplitude)² / 2
Signal power = (28687.5)² / 2
Signal power = 412,748,046.875

SNR:
The SNR is given by the formula:
SNR = Signal power / Noise power
SNR = 412,748,046.875 / 4218.75
SNR = 97,856.96

Actual Sampling Rate:
The actual sampling rate is 3 MHz, which is greater than the required sampling rate of 2.4 MHz. Therefore, the SNR will be reduced due to aliasing. The aliasing noise is given by the formula:
Aliasing noise power = (sampling rate - required sampling rate) / sampling rate x Signal power
Aliasing noise power = (3 MHz - 2.4 MHz) / 3 MHz x 412,748,046.875
Aliasing noise power = 37,025
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A signal has a bandwidth of 1 MHz. It is sampled at a rate 50% higher than the Nyquist rate and quantized into 256 level using a -low quantizer with μ = 225.Que: It was found that a sampling rate 20% above therate would be adequate. So the maximum SNR, thatcan be realized without increasing the transmission bandwidth, would bea)60.4 dBb)70.3 dBc)50.1 dBd)None of the aboveCorrect answer is option 'C'. Can you explain this answer?
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