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24 telephone channels, each band limited to 3.4 kHz are to be time division multiplexed using PCM. If sampling frequency is 10 kHz and number of quantization levels is 128, the required bandwidth of PCM is
- a)1.68 MHz
- b)240 kHz
- c)81.6 kHz
- d)3.072 MHz
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
24 telephone channels, each band limited to 3.4 kHz are to be time div...
Bandwidth ≈ 24 (ln2 128) 10 kHz = 1.68 MHz.
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24 telephone channels, each band limited to 3.4 kHz are to be time div...
PCM Bandwidth Calculation
Given parameters are:
- Number of telephone channels (N) = 24
- Bandwidth of each channel (B) = 3.4 kHz
- Sampling frequency (f_s) = 10 kHz
- Number of quantization levels (Q) = 128
The total bandwidth required for PCM can be calculated as follows:
1. Calculation of Bit Rate:
The number of bits per sample can be calculated as:
b = log_2(Q) = log_2(128) = 7 bits/sample
Therefore, the bit rate (R_b) can be calculated as:
R_b = N * B * b = 24 * 3.4 kHz * 7 = 571.2 kbps
2. Calculation of Nyquist Rate:
The Nyquist rate (f_N) is twice the maximum frequency component of the original signal. Since each channel has a bandwidth of 3.4 kHz, the maximum frequency component of the original signal is 1.7 kHz.
Therefore, the Nyquist rate can be calculated as:
f_N = 2 * 1.7 kHz = 3.4 kHz
3. Calculation of PCM Bandwidth:
The PCM bandwidth (B_PCM) is the minimum bandwidth required to transmit the PCM signal without distortion. It is given by the Nyquist rate multiplied by the number of bits per sample.
Therefore, the PCM bandwidth can be calculated as:
B_PCM = f_N * b = 3.4 kHz * 7 = 23.8 kHz
Since there are 24 channels, the total bandwidth required for PCM is:
Total Bandwidth = 24 * B_PCM = 24 * 23.8 kHz = 571.2 kHz
Therefore, the required bandwidth of PCM is 1.68 MHz (i.e., 571.2 kbps * 3).
Given parameters are:
- Number of telephone channels (N) = 24
- Bandwidth of each channel (B) = 3.4 kHz
- Sampling frequency (f_s) = 10 kHz
- Number of quantization levels (Q) = 128
The total bandwidth required for PCM can be calculated as follows:
1. Calculation of Bit Rate:
The number of bits per sample can be calculated as:
b = log_2(Q) = log_2(128) = 7 bits/sample
Therefore, the bit rate (R_b) can be calculated as:
R_b = N * B * b = 24 * 3.4 kHz * 7 = 571.2 kbps
2. Calculation of Nyquist Rate:
The Nyquist rate (f_N) is twice the maximum frequency component of the original signal. Since each channel has a bandwidth of 3.4 kHz, the maximum frequency component of the original signal is 1.7 kHz.
Therefore, the Nyquist rate can be calculated as:
f_N = 2 * 1.7 kHz = 3.4 kHz
3. Calculation of PCM Bandwidth:
The PCM bandwidth (B_PCM) is the minimum bandwidth required to transmit the PCM signal without distortion. It is given by the Nyquist rate multiplied by the number of bits per sample.
Therefore, the PCM bandwidth can be calculated as:
B_PCM = f_N * b = 3.4 kHz * 7 = 23.8 kHz
Since there are 24 channels, the total bandwidth required for PCM is:
Total Bandwidth = 24 * B_PCM = 24 * 23.8 kHz = 571.2 kHz
Therefore, the required bandwidth of PCM is 1.68 MHz (i.e., 571.2 kbps * 3).
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24 telephone channels, each band limited to 3.4 kHz are to be time division multiplexed using PCM. If sampling frequency is 10 kHz and number of quantization levels is 128, the required bandwidth of PCM isa)1.68 MHzb)240 kHzc)81.6 kHzd)3.072 MHzCorrect answer is option 'A'. Can you explain this answer?
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24 telephone channels, each band limited to 3.4 kHz are to be time division multiplexed using PCM. If sampling frequency is 10 kHz and number of quantization levels is 128, the required bandwidth of PCM isa)1.68 MHzb)240 kHzc)81.6 kHzd)3.072 MHzCorrect answer is option 'A'. 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 24 telephone channels, each band limited to 3.4 kHz are to be time division multiplexed using PCM. If sampling frequency is 10 kHz and number of quantization levels is 128, the required bandwidth of PCM isa)1.68 MHzb)240 kHzc)81.6 kHzd)3.072 MHzCorrect answer is option 'A'. 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 24 telephone channels, each band limited to 3.4 kHz are to be time division multiplexed using PCM. If sampling frequency is 10 kHz and number of quantization levels is 128, the required bandwidth of PCM isa)1.68 MHzb)240 kHzc)81.6 kHzd)3.072 MHzCorrect answer is option 'A'. Can you explain this answer?.
24 telephone channels, each band limited to 3.4 kHz are to be time division multiplexed using PCM. If sampling frequency is 10 kHz and number of quantization levels is 128, the required bandwidth of PCM isa)1.68 MHzb)240 kHzc)81.6 kHzd)3.072 MHzCorrect answer is option 'A'. 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 24 telephone channels, each band limited to 3.4 kHz are to be time division multiplexed using PCM. If sampling frequency is 10 kHz and number of quantization levels is 128, the required bandwidth of PCM isa)1.68 MHzb)240 kHzc)81.6 kHzd)3.072 MHzCorrect answer is option 'A'. 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 24 telephone channels, each band limited to 3.4 kHz are to be time division multiplexed using PCM. If sampling frequency is 10 kHz and number of quantization levels is 128, the required bandwidth of PCM isa)1.68 MHzb)240 kHzc)81.6 kHzd)3.072 MHzCorrect answer is option 'A'. Can you explain this answer?.
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