For a PCM system with the following parameters Maximum analog input frequency = 4 kHz, Maximum decoded voltage at the receiver=±2.55V, Minimum dynamic range=46 dB, Determine (a)Minimum sample rate (b)Minimum number of bits used in the PCM code (c)Resolution (d)Quantization error (e)Coding efficiency

The given data -

Maximum Analog Input Frequency = $f_a$ = 4 kHz

Maximum Decoded Voltage at the Receiver = $V{max}$ = ± 2.55 V

Minimum Dynamic Range = $DR$ = 46 dB

To find -

a] Minimum Sample Rate = $f_s$ = ?

b] Minimum number of bits used in the PCM code = n = ?

c] Resolution = $V_{min}$ = ?

d] Quantization Error = $Q_e$ = ?

e] Coding Efficiency = ƞ = ?

Formulae -

Minimum Sample Rate - $$f_s = 2f_a$$

Minimum number of bits used in the PCM code - $$2^n - 1 \ge DR$$

Resolution - $$V_{min} = \frac {V_{max}}{DR} = \frac {V_{max}}{2^n - 1}$$

Quantization Error - $$ Q_e = \frac {Resolution}{2}$$

Coding Efficiency - $$ ƞ = \frac{Minimum \ Number\ of\ Bits (including \ sign \ bit) }{Actual\ Number \ of\ Bits (including \ sign \ bit)} \times 100 \ \%$$

Solution -

a] Minimum Sample Rate = $f_s$

$$f_s = 2f_a = 2 (4 \ kHz)$$ $$f_s = 8 \ kHz$$

b] Minimum number of bits used in the PCM code = n

The number of bits used in the PCM system depends on the Dynamic Range (DR).

Dynamic Range (DR) in terms of dB -

$$DR (dB) = 20\ log (2^n - 1)$$ $$ 46 = 20\ log (2^n - 1)$$ $$ \frac {46}{20} = log (2^n - 1)$$ $$ 2.3 = log (2^n - 1)$$ $$ 10^{2.3} = 2^n - 1$$ $$ 199.53 = 2^n - 1$$ $$ 2^n = 199.53 + 1$$ $$ n = \frac{log \ 200.53}{log\ 2} = 7.64$$ $$ n \ge 7.64 \ (Excluding \ Sign \ bit)$$

• Since the maximum decoded voltage at the receiver side is ± 2.55V.
• Therefore, one more bit is needed to represent the + ve and – ve voltage.
• Hence, a minimum of 8.64 bits must be used for the magnitude along with 1 additional sign bit representation.
  • Minimum Number of Bits = n = 7.64 (Excluding Sign Bit)
  • Minimum Number of Bits = n = 8.64 (Including Sign Bit)
  • Actual Number of Bits = 8.64 $\approx$ 9 (Including Sign Bit).

c] Resolution = $V_{min}$

$$V_{min} = \frac {V_{max}}{DR} = \frac {V_{max}}{2^n - 1} = \frac {2.55}{2^8 - 1} = 0.01 V$$

Where n = Minimum number of Bits (Excluding Sign Bit)

Therefore, $n = 7.64 \approx 8$

d] Quantization Error = $Q_e$

$$ Q_e = \frac {Resolution}{2} = \frac {0.01}{2} = 0.005 V $$

e] Coding Efficiency = ƞ

$$ƞ = \frac{Minimum \ Number\ of\ Bits (including \ sign \ bit)}{Actual\ Number \ of\ Bits (including \ sign \ bit)} \times 100 \ \%$$ $$ ƞ = \frac{8.64}{9} \times 100 \ \% = 96 \ \%$$

Final Summarization of the above Findings -

a] Minimum Sample Rate = $f_s$ = 8 kHz

b] Minimum number of bits used in the PCM code = n = 7.64 $\approx$ 8 (Excluding Sign Bit)

c] Resolution = $V_{min}$ = 0.01 V

d] Quantization Error = $Q_e$ = 0.005 V

e] Coding Efficiency = ƞ = 96 %