PCM vs DM SQNR for 6cos(5000πt) with 128 levels

Given: $x(t) = 6\cos(5000\pi t)$, so $A = 6$ V, $f_m = 2500$ Hz. $L = 128$. Nyquist rate $f_s = 2f_m = 5$ kHz.

a) SQNR of PCM

$n = \log_2 128 = 7$ bits. For a full-load sinusoid ($m_p = 6$ V):

$$\begin{aligned} \Delta &= \frac{2 \times 6}{128} = 0.09375\ \text{V}, \quad N_q = \frac{\Delta^2}{12} \\ \text{SQNR} &= \frac{A^2/2}{\Delta^2/12} = 1.5L^2 = 1.5 \times 128^2 = 24\,576 \\ \text{SQNR}_{dB} &= 10\log_{10}24\,576 = 43.91\ \text{dB} \end{aligned}$$

b) Sampling frequency for DM with the same SQNR

For DM with a sinusoid at the slope-overload limit and output filter $f_M = f_m$:

$$\text{SQNR}_{DM} = \frac{3}{8\pi^2}\left(\frac{f_s}{f_m}\right)^3$$

Set equal to 24 576:

$$\begin{aligned} \left(\frac{f_s}{f_m}\right)^3 &= \frac{24\,576 \times 8\pi^2}{3} = 6.468 \times 10^5 \\ \frac{f_s}{f_m} &= 86.48 \\ f_s &= 86.48 \times 2500 = 216.2\ \text{kHz} \end{aligned}$$

c) SQNR degradation of DM at the Nyquist rate

With $f_s = 5$ kHz, $f_s/f_m = 2$:

$$\begin{aligned} \text{SQNR}_{DM} &= \frac{3}{8\pi^2}(2)^3 = \frac{3}{\pi^2} = 0.304 \\ \text{SQNR}_{DM,dB} &= -5.17\ \text{dB} \\ \text{Degradation} &= 43.91 - (-5.17) = 49.08\ \text{dB} \end{aligned}$$

Answer: (a) 43.91 dB; (b) $f_s \approx 216.2$ kHz; (c) DM is about 49.08 dB worse than PCM.