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A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V are sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. What is the number of quantization levels required to reduce the quantization noise by a factor of 4? (Answer up to the nearest integer)
Correct answer is '512'. Can you explain this answer?
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Most Upvoted Answer
A speed signal, band limited to 4 kHz and peak voltage varying betwee...
As Noise ∝ 1/L2
To reduce quantization noise by factor 4, quantization level must be two times i.e. 2L.
Now L = 2n = 28 = 256
Thus 2L = 512
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A speed signal, band limited to 4 kHz and peak voltage varying betwee...
Problem Analysis:
To reduce quantization noise by a factor of 4, we need to increase the number of quantization levels by a factor of 4. This can be achieved by increasing the number of bits used to represent each sample.
Given:
- Bandwidth of the speed signal (B) = 4 kHz
- Peak voltage (Vp) = 5 V
- Number of bits used for quantization (n) = 8
Calculating Nyquist Rate:
The Nyquist rate is twice the bandwidth of the signal. Therefore, the Nyquist rate (Fs) can be calculated as:
Fs = 2 * B = 2 * 4 kHz = 8 kHz
Calculating Maximum Frequency Component:
According to the Nyquist-Shannon sampling theorem, the maximum frequency component of the signal should be less than or equal to half the sampling rate. Therefore, the maximum frequency component (fmax) can be calculated as:
fmax = Fs / 2 = 8 kHz / 2 = 4 kHz
Calculating Quantization Step Size:
The quantization step size (Δ) can be calculated using the peak voltage and the number of quantization levels. Since the quantization range is symmetrical around zero, we can consider the peak voltage as the maximum positive value. The quantization step size can be calculated as:
Δ = 2 * Vp / (2^n) = 2 * 5 V / (2^8) = 10 V / 256 = 0.039 V
Calculating Number of Quantization Levels:
To reduce the quantization noise by a factor of 4, we need to increase the number of quantization levels by a factor of 4. Since the quantization step size is fixed, the number of quantization levels (L) can be calculated as:
L = 4 * (2^n) = 4 * (2^8) = 4 * 256 = 1024
However, since the question asks for the number of quantization levels required to reduce the quantization noise by a factor of 4, we need to round up to the nearest power of 2. The nearest power of 2 greater than 1024 is 2048. Therefore, we need 2048 quantization levels.
Answer:
The number of quantization levels required to reduce the quantization noise by a factor of 4 is 2048.
To reduce quantization noise by a factor of 4, we need to increase the number of quantization levels by a factor of 4. This can be achieved by increasing the number of bits used to represent each sample.
Given:
- Bandwidth of the speed signal (B) = 4 kHz
- Peak voltage (Vp) = 5 V
- Number of bits used for quantization (n) = 8
Calculating Nyquist Rate:
The Nyquist rate is twice the bandwidth of the signal. Therefore, the Nyquist rate (Fs) can be calculated as:
Fs = 2 * B = 2 * 4 kHz = 8 kHz
Calculating Maximum Frequency Component:
According to the Nyquist-Shannon sampling theorem, the maximum frequency component of the signal should be less than or equal to half the sampling rate. Therefore, the maximum frequency component (fmax) can be calculated as:
fmax = Fs / 2 = 8 kHz / 2 = 4 kHz
Calculating Quantization Step Size:
The quantization step size (Δ) can be calculated using the peak voltage and the number of quantization levels. Since the quantization range is symmetrical around zero, we can consider the peak voltage as the maximum positive value. The quantization step size can be calculated as:
Δ = 2 * Vp / (2^n) = 2 * 5 V / (2^8) = 10 V / 256 = 0.039 V
Calculating Number of Quantization Levels:
To reduce the quantization noise by a factor of 4, we need to increase the number of quantization levels by a factor of 4. Since the quantization step size is fixed, the number of quantization levels (L) can be calculated as:
L = 4 * (2^n) = 4 * (2^8) = 4 * 256 = 1024
However, since the question asks for the number of quantization levels required to reduce the quantization noise by a factor of 4, we need to round up to the nearest power of 2. The nearest power of 2 greater than 1024 is 2048. Therefore, we need 2048 quantization levels.
Answer:
The number of quantization levels required to reduce the quantization noise by a factor of 4 is 2048.
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A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V are sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. What is the number of quantization levels required to reduce the quantization noise by a factor of 4? (Answer up to the nearest integer)Correct answer is '512'. Can you explain this answer?
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A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V are sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. What is the number of quantization levels required to reduce the quantization noise by a factor of 4? (Answer up to the nearest integer)Correct answer is '512'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V are sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. What is the number of quantization levels required to reduce the quantization noise by a factor of 4? (Answer up to the nearest integer)Correct answer is '512'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V are sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. What is the number of quantization levels required to reduce the quantization noise by a factor of 4? (Answer up to the nearest integer)Correct answer is '512'. Can you explain this answer?.
A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V are sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. What is the number of quantization levels required to reduce the quantization noise by a factor of 4? (Answer up to the nearest integer)Correct answer is '512'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V are sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. What is the number of quantization levels required to reduce the quantization noise by a factor of 4? (Answer up to the nearest integer)Correct answer is '512'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V are sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. What is the number of quantization levels required to reduce the quantization noise by a factor of 4? (Answer up to the nearest integer)Correct answer is '512'. Can you explain this answer?.
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