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A speech signal is sampled a rate of 20% above the Nyquist rate. The signal has a bandwidth of 10 kHz. The sample is quantized into 1024 levels and then transmitted through 8-level PAM over an AWGN baseband channel. The bandwidth required for transmission is ______
  • a)
    80 kHz
  • b)
    44 kHz
  • c)
    40 kHz
  • d)
    34 kHz
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A speech signal is sampled a rate of 20% above the Nyquist rate. The s...
Sampling rate

The Nyquist rate is the minimum sampling rate required to accurately represent a signal without any loss of information. According to the Nyquist-Shannon sampling theorem, the sampling rate should be at least twice the bandwidth of the signal.

Given that the signal has a bandwidth of 10 kHz, the Nyquist rate would be 2 * 10 kHz = 20 kHz.

Sampling rate above the Nyquist rate

In this case, the signal is sampled at a rate 20% above the Nyquist rate. Therefore, the sampling rate would be 1.2 * 20 kHz = 24 kHz.

Quantization levels

After sampling, the signal is quantized into 1024 levels. Quantization is the process of reducing the number of amplitude levels in a signal. It allows for the digital representation of the signal.

Pulse Amplitude Modulation (PAM)

The quantized samples are then transmitted using 8-level PAM over an AWGN (Additive White Gaussian Noise) baseband channel. PAM is a modulation technique where the amplitude of a series of pulses is varied to represent the information.

Bandwidth required for transmission

The bandwidth required for transmission can be calculated based on the highest frequency component in the signal. In this case, the bandwidth of the original signal is given as 10 kHz.

However, during the PAM modulation process, additional frequency components are introduced. The bandwidth required for PAM can be estimated using Carson's rule, which states that the bandwidth is equal to the sum of the highest frequency component and twice the modulation frequency.

In this case, the highest frequency component is 10 kHz, and the modulation frequency is equal to the sampling rate of 24 kHz. Therefore, the bandwidth required for transmission would be 10 kHz + 2 * 24 kHz = 58 kHz.

However, it is important to note that the bandwidth required for transmission cannot be less than the Nyquist rate. Since the sampling rate is already 20% above the Nyquist rate (24 kHz), the required bandwidth for transmission would be equal to the sampling rate, i.e., 24 kHz.

Therefore, the correct answer is option A) 80 kHz.
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Community Answer
A speech signal is sampled a rate of 20% above the Nyquist rate. The s...
Concept:

Where,
Rb = nfs
n = no. of bits
M = no. of levels of PAM
Calculation:
fm = 10 kHz
Nyquist sampling rate = 2f­m = 20 kHz.
The signal is sampled at 20% above Nyquist rate:
∴ fs = (1.2 × 20) = 24 kHz.
No. of Quantization levels
= 1024
No. of bits required = log2 (1024)
= log2 210 = 10 bits.



= 80 kHz.
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A speech signal is sampled a rate of 20% above the Nyquist rate. The signal has a bandwidth of 10 kHz. The sample is quantized into 1024 levels and then transmitted through 8-level PAM over an AWGN baseband channel. The bandwidth required for transmission is ______a)80 kHzb)44 kHzc)40 kHzd)34 kHzCorrect answer is option 'A'. Can you explain this answer?
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