# A voice signal is band-limited to 3.3 kHz. What is the Nyquist frequency?

I'm having a some trouble with an assignment I have to do. I don't come from electrical engineering background and would appreciate any help I can get.

"A voice signal is band-limited to 3.3 kHz. What is the Nyquist frequency? It is sampled using a guard band of 1.4 kHz is applied and converted to digital data with 7 bits per sample. What is the resulting data rate?"

I calculated the Nyquist frequency using the formula $$f_{Ny} = \frac{1}{2}f_s = \frac{1}{2}3300\text{ Hz} = 1650\text{ Hz}.$$

Is this correct? Moving along I have no idea how to do the last part of the question, my lecturer didn't have any examples of this type of question so I'm really lost. Or else if someone could point me to a source that will show me how to do this question, that would be great too.

• The "data rate" part of the question should be fairly straightforward: once you know how many samples per second you're taking, and how many bits you have per sample, you can easily find how many bits per second you're producing. – MBaz Apr 24 '18 at 21:48

If an analog signal is limited in frequency, and $W$ is its highest frequency component, then the minimum sampling frequency to avoid time-aliasing is $f_s = 2W$.
In your case, and to avoid time-aliasing, the Nyquist frequency would be $f_{Nyquist} = 3 300\text{ Hz}$. But, since an additional guard band is considered, then the Nyquist frequency is:
$$f_{Nyquist} = 3 300\text{ Hz} + 1 400\text{ Hz} = 4 700\text{ Hz}$$