Question

A speech signal, band limited to 4 kHz, is sampled at 1.25 times the Nyquist rate. The speech samples, assumed to be statistically independent and uniformly distributed in the range \( -5 \, \text{V} \) to \( +5 \, \text{V} \), are subsequently quantized in an 8-bit uniform quantizer and then transmitted over a voice-grade AWGN telephone channel. If the ratio of transmitted signal power to channel noise power is 26 dB, the minimum channel bandwidth required to ensure reliable transmission of the signal with arbitrarily small probability of transmission error (rounded off to two decimal places) is kHz.

Hint:

For reliable transmission, the channel bandwidth can be estimated using the bit rate and the SNR in dB.

Answer 1

Correct Answer: 9.24

The Nyquist rate is twice the bandwidth: \[ f_{\text{Nyquist}} = 2 \times 4 \, \text{kHz} = 8 \, \text{kHz} \] The sampling rate is 1.25 times the Nyquist rate: \[ f_{\text{sampling}} = 1.25 \times 8 \, \text{kHz} = 10 \, \text{kHz} \] The signal is quantized with 8 bits, so the number of levels is \( 2^8 = 256 \). The bit rate is: \[ \text{Bit rate} = f_{\text{sampling}} \times \text{Bits per sample} = 10 \, \text{kHz} \times 8 = 80 \, \text{kbps} \] To calculate the channel bandwidth, we use the formula: \[ B_{\text{channel}} = \frac{\text{Bit rate}}{\text{SNR in dB}} \] The SNR is given as 26 dB, and the bandwidth is: \[ B_{\text{channel}} = \frac{80 \, \text{kbps}}{26 \, \text{dB}} \approx 9.24 \, \text{kHz} \] Thus, the minimum channel bandwidth required is \( 9.24 \, \text{kHz} \).