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.


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