A baseband analog message signal $x(t)$ with bandwidth 20 kHz, power $10^{-3}$, and $|x(t)| \leq 1$ is waveform encoded and transmitted through a channel of bandwidth 160 kHz. Besides, let the sampled data of $x(t)$ be denoted by $x_n$ and suppose that $$ | x_n - x_{n-1}| > \leq 10^{-0.55 f_s / f_N}, $$ where $f_s$ denotes the sampling rate and $f_N$ denotes the Nyquist rate, for all $n$ when $f_s \geq f_N$. If we require SQNR $\geq$ 30 dB, please answer the followng questions.
(a) Determine the maximum bit rate, $R_{b\_max}$, of the coded binary data.
(b) If PCM was used for encoding, determine the range of bits, $v_{PCM}$, can be so that SQNR required is fullfilled.
(c) If DPCM was used for encoding, determine the range of bits, $v_{DPCM}$, can be so that SQNR required is fulfilled according to $R_{b\_max}$ determined in (a).
(d) Determine the condition of $f_s$ so that DM could fulfill the SQNR required according to $R_{b\_max}$ determined in (a).
(e) According to the above results, determine which encoding method(s), among PCM, DPCM, and DM, can be suitable to transmit through the channel.
The following is my attempt to solve the above questions:
(a) According to the Shannon capacity formula, the maximum bit rate is the channel capacity: $$R_{b_\max} = C = W \log_2 (1 + SQNR) = 160 \times 10^3 \times \log_2(1 + 10^{30/10}) \approx 1.5948 \times 10^6 \text{ bits/s}.$$
(b) From Eq. (7.4.4) in [1] we know that $$\left. \text{SQNR} \right|_{\text{dB}} \approx 10 \log_{10} \frac{P_X}{x^2_{\max}} + 6 v_{PCM} + 4.8,$$ where $P_X$ is the power in each sample and $x_{\max}$ is the maximum possible value for $x(t)$. Now $P_X = 10^{-3}$ and $x_{\max} = 1$, so $$ 10 \log_{10} \frac{10^{-3}}{1^2} + 6 v_{PCM} + 4.8 \geq 30.$$ Solving the above equation, we get $$ v_{PCM} \geq \frac{46}{5} = 9.2 \text{ bits}.$$ Because the number of bits must be a positive integer, the range of bits is $$v_{PCM} \geq 10 \text{ bits}.$$
(c) The Nyquist rate is $$ f_N = 2 \times 20 \text{ kHz} = 40 \text{ kHz}.$$ According to Example 7.4.2 in [1], for DPCM, the bit rate is $$ R = v_{DPCM} f_s \geq v_{DPCM} f_N = v_{DPCM} \times 40 \text{ kHz}. $$ On the other hand, $$ R \leq R_{b\_max}. $$ Therefore, $$ v_{DPCM} \times 40 \text{ kHz} \leq 1.5948 \times 10^6 \text{ bits/s}. $$ Then $$ v_{DPCM} \leq \frac{1.5948 \times 10^6}{40 \times 10^3} = 39.87 \text{ bits}. $$ Again, the number of bits must be a positive integer. So $$ v_{DPCM} \leq 39 \text{ bits}. $$
(d) In DM only one bit per sample is employed, so $$ f_s = R \leq R_{b\_max}.$$ This implies that $$ f_s \leq 1.5948 \times 10^6 \text{ Hz} = 159.48 \text{ kHz}.$$ On the other hand, $$ f_s \geq f_N = 40 \text{ kHz}. $$ So $$ 40 \text{ kHz} \leq f_s \leq 159.48 \text{ kHz}.$$
(e) I think PCM, DPCM, and DM can be all suitable to transmit through the channel. But I am not so sure. $\blacksquare$
Reference:
[1] J. G. Proakis and M. Salehi, Fundamentals of Communication Systems, Second Edition, Pearson, 2014.
