Ex : For a PCM system with the following parameters, determine (a)minimum sample rate (b) minimum number of bits used in the PCM code (c)resolution (d) quantization errorMaximum analog input frequency = 4 kHzMaximum decode voltage at the receiver = ±2.55VMinimum dynamic range = 46 dB

To determine the minimum sample rate, we need to consider the maximum analog input frequency. In this case, the maximum analog input frequency is given as 4 kHz. According to the Nyquist-Shannon sampling theorem, the minimum sample rate should be at least twice the maximum analog input frequency. Therefore, the minimum sample rate would be 2 * 4 kHz = 8 kHz.

To determine the minimum number of bits used in the PCM code, we need to consider the minimum dynamic range. The dynamic range is given as 46 dB. The number of bits required in the PCM code can be calculated using the formula:

Number of bits = dynamic range / (6.02 * resolution)

Here, the resolution is the smallest change in the analog input that can be represented by the PCM code. To calculate the resolution, we can use the formula:

Resolution = (maximum decode voltage) / (2^n - 1)

Where n is the number of bits used in the PCM code. In this case, the maximum decode voltage is given as ±2.55V.

By substituting the values into the formula, we can calculate the resolution. Once we have the resolution, we can then calculate the minimum number of bits required in the PCM code.

To calculate the quantization error, we need to consider the resolution. The quantization error is the difference between the actual analog input value and the quantized value represented by the PCM code. It can be calculated as half of the resolution.

I hope this helps! Let me know if you have any further questions.