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A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.
If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (Answer up to the nearest integer)
Correct answer is '32'. 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...
Fm = 4 KHz
fs = 2fm = 8 kHz
Bit Rate Rb = nfs = 8 x 8 = 64 kbps
The minimum transmission bandwidth is
BW = Rb/2 = 32 kHz
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A speed signal, band limited to 4 kHz and peak voltage varying betwee...
To determine the minimum bandwidth required for distortion-free transmission of a speed signal, we need to consider the Nyquist-Shannon sampling theorem, quantization, and the use of bipolar pulses.
Nyquist-Shannon Sampling Theorem:
The Nyquist-Shannon sampling theorem states that in order to accurately reconstruct a continuous signal, it must be sampled at a rate greater than or equal to twice the bandwidth of the signal. In this case, the speed signal is band-limited to 4 kHz, so the Nyquist rate would be 2 * 4 kHz = 8 kHz.
Quantization:
Quantization is the process of converting the continuous amplitude of the signal into discrete levels or steps. In this case, the speed signal is represented by 8 bits. Each bit can have two possible values, either 0 or 1, resulting in 2^8 = 256 discrete levels.
Bipolar Pulses:
Bipolar pulses are used to transmit the quantized signal. Each bit is represented by a pulse that can take on two possible values, either positive or negative. This allows for the representation of both positive and negative values.
Determining the Minimum Bandwidth:
To determine the minimum bandwidth required for distortion-free transmission, we need to consider the maximum rate of change of the signal. In this case, the peak voltage of the speed signal varies between 5 V and -5 V.
The maximum rate of change of the signal can be approximated by the formula:
Rate of change = (2 * Peak voltage) / (Sampling rate)
Substituting the values, we have:
Rate of change = (2 * 5 V) / (8 kHz) = 1.25 kV/s
According to the Nyquist-Shannon sampling theorem, the bandwidth required is equal to the maximum rate of change of the signal. Therefore, the minimum bandwidth required for distortion-free transmission is 1.25 kHz.
However, in this case, the bits are transmitted using bipolar pulses. Each pulse represents a bit and can take on two possible values, either positive or negative. Since there are 8 bits in total, the minimum bandwidth required for distortion-free transmission would be 8 * 1.25 kHz = 10 kHz.
Rounding up to the nearest integer, the answer is 10 kHz, which is equal to 10,000 Hz or 10,000 Hz / 1000 = 10 kHz.
Nyquist-Shannon Sampling Theorem:
The Nyquist-Shannon sampling theorem states that in order to accurately reconstruct a continuous signal, it must be sampled at a rate greater than or equal to twice the bandwidth of the signal. In this case, the speed signal is band-limited to 4 kHz, so the Nyquist rate would be 2 * 4 kHz = 8 kHz.
Quantization:
Quantization is the process of converting the continuous amplitude of the signal into discrete levels or steps. In this case, the speed signal is represented by 8 bits. Each bit can have two possible values, either 0 or 1, resulting in 2^8 = 256 discrete levels.
Bipolar Pulses:
Bipolar pulses are used to transmit the quantized signal. Each bit is represented by a pulse that can take on two possible values, either positive or negative. This allows for the representation of both positive and negative values.
Determining the Minimum Bandwidth:
To determine the minimum bandwidth required for distortion-free transmission, we need to consider the maximum rate of change of the signal. In this case, the peak voltage of the speed signal varies between 5 V and -5 V.
The maximum rate of change of the signal can be approximated by the formula:
Rate of change = (2 * Peak voltage) / (Sampling rate)
Substituting the values, we have:
Rate of change = (2 * 5 V) / (8 kHz) = 1.25 kV/s
According to the Nyquist-Shannon sampling theorem, the bandwidth required is equal to the maximum rate of change of the signal. Therefore, the minimum bandwidth required for distortion-free transmission is 1.25 kHz.
However, in this case, the bits are transmitted using bipolar pulses. Each pulse represents a bit and can take on two possible values, either positive or negative. Since there are 8 bits in total, the minimum bandwidth required for distortion-free transmission would be 8 * 1.25 kHz = 10 kHz.
Rounding up to the nearest integer, the answer is 10 kHz, which is equal to 10,000 Hz or 10,000 Hz / 1000 = 10 kHz.
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A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (Answer up to the nearest integer)Correct answer is '32'. 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, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (Answer up to the nearest integer)Correct answer is '32'. 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, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (Answer up to the nearest integer)Correct answer is '32'. 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, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (Answer up to the nearest integer)Correct answer is '32'. Can you explain this answer?.
A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (Answer up to the nearest integer)Correct answer is '32'. 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, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (Answer up to the nearest integer)Correct answer is '32'. 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, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (Answer up to the nearest integer)Correct answer is '32'. Can you explain this answer?.
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