Question 6.4: For a PCM system with the following parameters, determine (a......

a.  Substituting into Equation 1, the minimum sample rate is

f_{s}≥2f_{a}                         (1)

f_{s}=2f_{a}= 2(4 kHz) = 8 kHz

b.  To determine the absolute value for dynamic range, substitute into Equation 4:

DR  =  20 log\frac{V_{max}}{V_{min}}               (4)

46 dB  =  20 log\frac{V_{max}}{V_{min}}

2.3 = log\frac{V_{max}}{V_{min}}

199.5 = DR

The minimum number of bits is determined by rearranging Equation 5b and solving for n:

2^{n}-1 = DR               (5b)

n = \frac{log(199.5\,+\,1)}{log\,2}= 7.63

The closest whole number greater than 7.63 is 8; therefore, eight bits must be used for the magnitude.
Because the input amplitude range is ±2.55, one additional bit, the sign bit, is required. Therefore, the total number of CM bits is nine, and the total number of PCM codes is 2^{9} = 512. (There are 255 positive codes, 255 negative codes, and 2 zero codes.)

To determine the actual dynamic range, substitute into Equation 6:

DR_{(dB)} = 20 log(2^{n} – 1)                (6)

= 20(log 256 – 1)

= 48.13 dB

c.  The resolution is determined by dividing the maximum positive or maximum negative voltage by the number of positive or negative nonzero PCM codes:

resolution = \frac{V_{max}}{2^{n} – 1} = \frac{2.55}{2^{8} – 1}=\frac{2.55}{256 – 1} = 0.01\,V

The maximum quantization error is