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The amplitude of a random signal is uniformly distributed between -5 V and 5 V.
If the signal to quantization noise ratio required in uniformly quantizing the signal is 43.5 dB, the step size (in V) of the quantization is approximately
- a)0.0333
- b)0.05
- c)0.0667
- d)0.10
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The amplitude of a random signal is uniformly distributed between -5 ...
To determine the step size of quantization, we need to first understand the concept of signal to quantization noise ratio (SQNR).
Signal to Quantization Noise Ratio (SQNR):
SQNR is a measure of the quality of quantization in a system. It represents the ratio of the power of the input signal to the power of the quantization noise introduced by the quantization process. Higher SQNR values indicate better quantization quality.
Given:
Amplitude of the random signal = uniformly distributed between -5 V and 5 V
Required SQNR = 43.5 dB
Step 1: Conversion of SQNR to linear scale
The required SQNR is given in decibels, so we need to convert it to a linear scale using the formula:
SQNR_linear = 10^(SQNR/10)
Here, SQNR = 43.5 dB
SQNR_linear = 10^(43.5/10)
SQNR_linear ≈ 3162.2776
Step 2: Calculation of signal power
The power of a signal can be calculated using the formula:
Power = (Amplitude^2)/2
Since the amplitude is uniformly distributed between -5 V and 5 V, the average amplitude is 0 V. Therefore, the signal power can be calculated as:
Signal power = (5^2)/2 = 25/2 = 12.5 V^2
Step 3: Calculation of quantization noise power
Quantization noise power can be calculated using the formula:
Quantization noise power = Signal power / SQNR_linear
Quantization noise power = 12.5 V^2 / 3162.2776 ≈ 0.00395 V^2
Step 4: Calculation of step size
Step size is the difference between two adjacent quantization levels. In a uniform quantizer, the step size can be calculated as:
Step size = (Maximum amplitude - Minimum amplitude) / Number of quantization levels
Here, the maximum amplitude = 5 V
The minimum amplitude = -5 V
Number of quantization levels = (Maximum amplitude - Minimum amplitude) / Step size
Let's assume the step size as S.
Number of quantization levels = (5 V - (-5 V)) / S = 10 V / S
Quantization noise power can also be calculated as:
Quantization noise power = (Step size^2) / 12
By equating the two equations, we can solve for the step size.
0.00395 V^2 = (S^2) / 12
Solving this equation, we get:
S ≈ 0.0667 V
Therefore, the step size of the quantization is approximately 0.0667 V, which corresponds to option C.
Signal to Quantization Noise Ratio (SQNR):
SQNR is a measure of the quality of quantization in a system. It represents the ratio of the power of the input signal to the power of the quantization noise introduced by the quantization process. Higher SQNR values indicate better quantization quality.
Given:
Amplitude of the random signal = uniformly distributed between -5 V and 5 V
Required SQNR = 43.5 dB
Step 1: Conversion of SQNR to linear scale
The required SQNR is given in decibels, so we need to convert it to a linear scale using the formula:
SQNR_linear = 10^(SQNR/10)
Here, SQNR = 43.5 dB
SQNR_linear = 10^(43.5/10)
SQNR_linear ≈ 3162.2776
Step 2: Calculation of signal power
The power of a signal can be calculated using the formula:
Power = (Amplitude^2)/2
Since the amplitude is uniformly distributed between -5 V and 5 V, the average amplitude is 0 V. Therefore, the signal power can be calculated as:
Signal power = (5^2)/2 = 25/2 = 12.5 V^2
Step 3: Calculation of quantization noise power
Quantization noise power can be calculated using the formula:
Quantization noise power = Signal power / SQNR_linear
Quantization noise power = 12.5 V^2 / 3162.2776 ≈ 0.00395 V^2
Step 4: Calculation of step size
Step size is the difference between two adjacent quantization levels. In a uniform quantizer, the step size can be calculated as:
Step size = (Maximum amplitude - Minimum amplitude) / Number of quantization levels
Here, the maximum amplitude = 5 V
The minimum amplitude = -5 V
Number of quantization levels = (Maximum amplitude - Minimum amplitude) / Step size
Let's assume the step size as S.
Number of quantization levels = (5 V - (-5 V)) / S = 10 V / S
Quantization noise power can also be calculated as:
Quantization noise power = (Step size^2) / 12
By equating the two equations, we can solve for the step size.
0.00395 V^2 = (S^2) / 12
Solving this equation, we get:
S ≈ 0.0667 V
Therefore, the step size of the quantization is approximately 0.0667 V, which corresponds to option C.
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Community Answer
The amplitude of a random signal is uniformly distributed between -5 ...
(S/Nq)0dB = 1.76 + 6.02u
⇒43.5 = 17.6 + 6.024
⇒ n ≈ 7
Δ(Step size) = 2A/2′′
=10/27 = 0.07 Volts
Which is close to 0.0667 Volts
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The amplitude of a random signal is uniformly distributed between -5 V and 5 V.If the signal to quantization noise ratio required in uniformly quantizing the signal is 43.5 dB, the step size (in V) of the quantization is approximatelya)0.0333b)0.05c)0.0667d)0.10Correct answer is option 'C'. Can you explain this answer?
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