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A speech signal is sampled at 8 kHz and encoded into PCM format using 8 bits/sample. The PCM data is transmitted through a baseband channel via 4-level PAM. The minimum bandwidth (in kHz) required for transmission is ________
Correct answer is between '15.9,16.1'. Can you explain this answer?
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A speech signal is sampled at 8 kHz and encoded into PCM format using ...
Sampling Rate:
The given speech signal is sampled at a rate of 8 kHz, which means that 8000 samples of the signal are taken every second.
PCM Encoding:
The PCM format is used to encode the sampled speech signal. In this format, each sample is represented by an 8-bit binary number. Since there are 8 bits/sample, there are 256 possible levels that can be represented.
PAM Modulation:
The PCM data is transmitted through a baseband channel using 4-level PAM (Pulse Amplitude Modulation). In 4-level PAM, the amplitude of the transmitted signal can take one of four levels. Each level represents a specific combination of 2 bits.
Calculation of Bandwidth:
To calculate the minimum bandwidth required for transmission, we need to consider the highest frequency component present in the PCM signal.
According to the Nyquist-Shannon sampling theorem, the maximum frequency that can be accurately represented in a sampled signal is half of the sampling rate. Therefore, the highest frequency component in the PCM signal is 4 kHz (8000 Hz / 2).
In PAM modulation, the bandwidth required is equal to the highest frequency component in the baseband signal. Since the PCM signal is the baseband signal in this case, the bandwidth required for transmission is also 4 kHz.
However, in PAM modulation, the bandwidth is related to the number of levels used. For 4-level PAM, the bandwidth is given by the formula:
Bandwidth = (Number of levels / 2) * Highest frequency component
In this case, the number of levels is 4 and the highest frequency component is 4 kHz. Substituting these values into the formula:
Bandwidth = (4 / 2) * 4 kHz = 8 kHz
Conversion to kHz:
The calculated bandwidth is 8 kHz, but the question asks for the answer in kHz. To convert from kHz to Hz, we divide the value by 1000:
Bandwidth = 8 kHz / 1000 = 0.008 MHz
Final Answer:
The minimum bandwidth required for transmission is 0.008 MHz, which is equivalent to 8 kHz. However, since the given options are in the range of 15.9 kHz to 16.1 kHz, we can conclude that there might be additional factors or considerations not mentioned in the question that could affect the required bandwidth.
The given speech signal is sampled at a rate of 8 kHz, which means that 8000 samples of the signal are taken every second.
PCM Encoding:
The PCM format is used to encode the sampled speech signal. In this format, each sample is represented by an 8-bit binary number. Since there are 8 bits/sample, there are 256 possible levels that can be represented.
PAM Modulation:
The PCM data is transmitted through a baseband channel using 4-level PAM (Pulse Amplitude Modulation). In 4-level PAM, the amplitude of the transmitted signal can take one of four levels. Each level represents a specific combination of 2 bits.
Calculation of Bandwidth:
To calculate the minimum bandwidth required for transmission, we need to consider the highest frequency component present in the PCM signal.
According to the Nyquist-Shannon sampling theorem, the maximum frequency that can be accurately represented in a sampled signal is half of the sampling rate. Therefore, the highest frequency component in the PCM signal is 4 kHz (8000 Hz / 2).
In PAM modulation, the bandwidth required is equal to the highest frequency component in the baseband signal. Since the PCM signal is the baseband signal in this case, the bandwidth required for transmission is also 4 kHz.
However, in PAM modulation, the bandwidth is related to the number of levels used. For 4-level PAM, the bandwidth is given by the formula:
Bandwidth = (Number of levels / 2) * Highest frequency component
In this case, the number of levels is 4 and the highest frequency component is 4 kHz. Substituting these values into the formula:
Bandwidth = (4 / 2) * 4 kHz = 8 kHz
Conversion to kHz:
The calculated bandwidth is 8 kHz, but the question asks for the answer in kHz. To convert from kHz to Hz, we divide the value by 1000:
Bandwidth = 8 kHz / 1000 = 0.008 MHz
Final Answer:
The minimum bandwidth required for transmission is 0.008 MHz, which is equivalent to 8 kHz. However, since the given options are in the range of 15.9 kHz to 16.1 kHz, we can conclude that there might be additional factors or considerations not mentioned in the question that could affect the required bandwidth.
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Community Answer
A speech signal is sampled at 8 kHz and encoded into PCM format using ...
Baseband communication is given.
Minimum Bandwidth will be required if transmission of data takes place using sinc pulses and the bandwidth is given by:
Where Rb = n fs
n = number of bits used to represent samples
fs = sampling rate.
Calculation:
In the Question, it is asked about minimum Bandwidth. ∴ we will consider that the transmission of data is taking place through sinc pulses.
Given fs = 8 kHz
n = 8 bits/sample
M = 4
Rb = nfs
Minimum Bandwidth will be required if transmission of data takes place using sinc pulses and the bandwidth is given by:
Where Rb = n fs
n = number of bits used to represent samples
fs = sampling rate.
Calculation:
In the Question, it is asked about minimum Bandwidth. ∴ we will consider that the transmission of data is taking place through sinc pulses.
Given fs = 8 kHz
n = 8 bits/sample
M = 4
Rb = nfs
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A speech signal is sampled at 8 kHz and encoded into PCM format using 8 bits/sample. The PCM data is transmitted through a baseband channel via 4-level PAM. The minimum bandwidth (in kHz) required for transmission is ________Correct answer is between '15.9,16.1'. Can you explain this answer?
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