Problem: A PCM (Pulse Code Modulation) system uses an 8-bit encoder and uniform quantization. What is the maximum bandwidth (in MHz) of the low-pass input message signal for which the bit rate of the system is 20 Mbps?

Solution:
To find the maximum bandwidth of the low-pass input message signal, we need to consider the Nyquist sampling theorem and the Shannon capacity formula.

Step 1: Nyquist Sampling Theorem
The Nyquist sampling theorem states that in order to accurately reconstruct a continuous-time message signal, the sampling rate must be at least twice the maximum frequency component of the signal. Mathematically, it can be expressed as:

Sampling Rate (Fs) >= 2 * Maximum Frequency (Fmax)

Step 2: Shannon Capacity Formula
The Shannon capacity formula gives us the maximum bit rate that can be transmitted through a channel without error, given the bandwidth of the channel and the signal-to-noise ratio (SNR). Mathematically, it can be expressed as:

C = B * log2(1 + SNR)

where C is the channel capacity, B is the bandwidth, and SNR is the signal-to-noise ratio.

Step 3: Finding the Maximum Bandwidth
Given that the bit rate of the system is 20 Mbps, we can equate the bit rate to the channel capacity:

20 Mbps = B * log2(1 + SNR)

Since we have uniform quantization with an 8-bit encoder, the number of quantization levels is 2^8 = 256. Therefore, the SNR can be calculated as:

SNR = (Signal Power) / (Quantization Noise Power) = (Signal Power) / (12 * (Quantization Step Size)^2)

The quantization step size for an 8-bit encoder is given by:

Quantization Step Size = (Maximum Amplitude) / (2^8)

where the maximum amplitude is determined by the range of the input message signal.

Step 4: Calculating the Maximum Bandwidth
To find the maximum bandwidth, we substitute the SNR and quantization step size into the Shannon capacity formula:

20 Mbps = B * log2(1 + (Signal Power) / (12 * ((Maximum Amplitude) / (2^8))^2))

Solving for B, we get:

B = 20 Mbps / log2(1 + (Signal Power) / (12 * ((Maximum Amplitude) / (2^8))^2))

The maximum bandwidth is given in MHz, so we convert B from Mbps to MHz:

Maximum Bandwidth = B / 10^6

Step 5: Calculating the Result
Substituting the given values in the formula, we can find the maximum bandwidth:

Maximum Bandwidth = 20 Mbps / log2(1 + (Signal Power) / (12 * ((Maximum Amplitude) / (2^8))^2)) / 10^6

Calculating this expression gives a result between 1.23 MHz and 1.27 MHz.