NATIONAL COUNCIL FOR TECHNICAL VOCATIONAL AND EDUCATION Republic of Sierra Leone PROGRAMME: HIGHER DIPLOMA EXAMINATIONS- 2026 SUBJECT: TELECOMMUNICATION SYSTEMS INSTRUCTIONS: Answer Any FIVE (5) Questions

The bandwidth of the audio signal is $B = 20 \text{ kHz}$. For an audio signal, this bandwidth represents the maximum frequency component, $f_m$.
According to the Nyquist sampling theorem, the minimum sampling rate ($f_s$) must be at least twice the maximum frequency component.

$$f_s = 2 \times f_m$$ $$f_s = 2 \times 20 \text{ kHz}$$ $$f_s = 40 \text{ kHz}$$

The minimum Nyquist sampling rate is $\boxed{\text{40 kHz}}$.

The samples are quantized using a uniform PCM system with $n = 12$ bits per sample.
Using the minimum Nyquist sampling rate $f_s = 40 \text{ kHz}$ from Step 1, the bit rate ($R_b$) is calculated as:

$$R_b = n \times f_s$$ $$R_b = 12 \text{ bits/sample} \times 40 \text{ ksamples/s}$$ $$R_b = 480 \text{ kbits/s}$$

The resulting bit rate is $\boxed{\text{480 kbits/s}}$.

For a uniform PCM system with $n = 12$ bits per sample, the theoretical maximum SQNR in decibels is given by the formula:

$$SQNR_{dB} = (6.02n + 1.76) \text{ dB}$$ $$SQNR_{dB} = (6.02 \times 12 + 1.76) \text{ dB}$$ $$SQNR_{dB} = (72.24 + 1.76) \text{ dB}$$ $$SQNR_{dB} = 74 \text{ dB}$$

The theoretical maximum SQNR is $\boxed{\text{74 dB}}$.