I have a question about a problem in a textbook I can't exactly solve. It goes as follow:

Given a message m(t) as input to a DM system $$m(t) = 3cos(890\pi t) - 0.7sin(1000\sqrt{3}t)$$

• Determine the minimum step size E necessary to avoid DM slope overload
• Calculate the average quantization noise power

Now, for the step size, I do: $$\left | \frac{dm}{dt} \right|_{max} = \left | -2670\pi \cdot sin(890\pi t) - 700\sqrt{3} \pi \cdot cos(1000 \sqrt{3} \pi t) \right|_{max} = (2670 + 700\sqrt{3})\pi$$ The maximal slope of the sampled DM signal will be

$$a_{max} = E/T_s \geq (2670 + 700\sqrt{3})\pi$$

$$E \geq T_s (2670 + 700\sqrt{3})\pi$$

Now I suppose to get a numeric value I would consider the fact the bandwidth B of m(t) being $$\500\sqrt{3}\$$ Hz, the minimum $$\f_s\$$ (Nyquist rate) would be 2B and thus $$\f_{s_{min}} = 2B = 1000\sqrt{3}\$$ Hz, and $$\T_s = 1/f_{s}\$$ etc.

First, I'd like to ask if so far what I did makes sense.

Second, for the $$\ SNR = S_o/N_o \$$, I would have: $$S_o = P_{signal} = \sqrt{ \left (\frac{3}{\sqrt{2}} \right) ^2 + \left (\frac{0.7}{\sqrt{2}} \right) ^2 } = \sqrt{\frac{949}{200}}$$

And this is where I get a bit lost. My textbook says that the PSD of the noise power will be $$P_{noise}(f) = \frac{E^2}{6f_s}$$ May I ask where does the hell that come from, how can I calculate it?

Anyway, taking it for cash I get: $$P_{noise} = \int^B_{-B} P_{noise}(f) df = \frac{E^2 B}{3 f_s} = \frac{[(2670 + 700\sqrt{3})\pi T_s]}{3/T_s} = \frac{(2670 + 700\sqrt{3})\pi B}{3 f_s^2}$$

So now $$SNR = P_{signal}/P_{noise} = \sqrt{\frac{949}{200}} \frac{3 f_s^2}{(2670 + 700\sqrt{3})\pi B}$$ with $$\ B = 2\pi 500\sqrt{3} \$$ Hz.

So, if someone could tell me if I made mistakes or what I did makes sense, and perhaps tell me where $$\ P_{noise}(f) = \frac{E^2}{6f_s}\$$ comes from, I'd be very grateful.