Question

consider a PCM encoder that accepts a signal with full scale voltage of 25V. The maximum actual quantized voltage is 24.99923706 for uniform quantization. Do not round the numbers. Go up to 9 decimal places.

a. Calculate the number of bits used for quantization

b. Calculate the number of quantized levels

c. Calculate the normalized step size

d. Calculate the actual step size

e. Calculate the maximum normalized quantized voltage

f. Calculate the normalized resolution

g. If this encoder operates at a data rate of 750 kbps, what is the required channel bandwidth

Answer 1

Given data :

PCM encoder uses uniform quantization .

Full scale voltage = 25 volts.

Maximum actual quantized voltage = 24.99923706 volts

a) We have to calculate no of bits (n) used for quantization.

Quantization error = Actual value - quantized value = 25 - 24.99923706 = 0.00076294 volts

Quantization error ranges from . where is the step size.

=2*0.00076294= 0.00152588

Step size , = Full scale voltage / no of quantization levels =

bits

b) No of quantized levels =

c) Normalized Step size =

d) Actual step size =

e) Maximum normalized quantized voltage = normalized step size * no of quantization levels= 0.000061035 * 16384 = 0.99999744

f) Normalized resolution = Normalized step size = 0.000061035

g) Data rate / Bit rate of the encoder, R_b= 750 Kbps = sampling rate * no of bits = (Samples / sec) * (bits /samples) = 750 kilo bits / sec

Sampling rate = 1/ T_S

Data rate = n * Sampling rate = n / T_s = 1/ T_b

T_b= bit duration

Bandwidth of PCM signal = 1/ Tb Hz = 750 kHz