# Bandwidth confusion Let's imagine that I took a Fourier analysis of a random voice signal that I want to sample and plotted it's frequency components in frequency domain (frequency vs amplitude). Now I want to sample it. Based on Nyquist–Shannon sampling theorem my sampling frequency should be > than 2*B (B-bandwidth). Based on a symbols that I've drawn here can you tell me what a bandwidth (B) of a signal equals to: BW or BW/2=(upper frequency-lower frequency)/2 and why? Thanks!

• That's not actually true. You can sample this sufficiently at 6.2 kHz, but it would require you to understand the concept of aliasing – Marcus Müller Sep 19 '17 at 21:22
• I understand something a bit about sampling and aliasing, but I'm a bit more confused in answering the question: "Based on my picture, is bandwidth equal 3.4kHz-300Hz (which is bandwidth of a human voice) or 3.4kHz-300Hz/2?" – Krushe Sep 19 '17 at 21:26
• why should it be (3400 Hz - 300 Hz)/2 ? I don't really see where you get that from? – Marcus Müller Sep 19 '17 at 21:34
• Please take a look at first image on this link: en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem This is why I'm confused I don't know what is the difference... – Krushe Sep 19 '17 at 21:56
• sorry, still don't understand your confusion. That image shows a two-sided spectrum from -F to +F. Your picture only shows positive frequencies. – Marcus Müller Sep 19 '17 at 22:24

## 1 Answer

In the Nyquist sampling theorem, the bandwidth is usually defined as the largest frequency in the signal; in other words, what the sampling theorem says is that, if you sample this signal at 6.8 kilosamples per second, you will be able to reconstruct the analog signal perfectly.

Now, under some conditions, you can do bandpass sampling, where the bandwidth is considered as $3400 - 300 = 3100\text{ Hz}$, and the signal can be sampled at 6200 samples per second. You can't always do that, though; see here for the Wikipedia explanation. For the actual scholarly source, see

R. G. Vaughan, N. L. Scott, and D. R. White, “The theory of bandpass sampling,” IEEE Transactions on Signal Processing, vol. 39, no. 9, pp. 1973–1984, Sep. 1991.

• If you sample at 6.8 kHz the samples are real and in the frequency domain you will have the conjugate symmetry. If you sample at 3100 Hz the samples will be complex and you lost the conjugate symmetry. The bandwidth in Shannon-Nyquist usually means from 0 Hz to the max frequency. If you are only interested a portion of the frequency range (as in the OP) then you can get away with a lower sample rate. – David Sep 20 '17 at 14:00
• @David If you sample at 3100 Hz the samples will be complex That is incorrect -- there is no way to sample a real signal and end up with a complex signal. – MBaz Sep 20 '17 at 14:02
• The webpage you link to refers to Undersampling which is different than Bandpass sampling. In undersampling you have to ensure there is no other frequency content that will alias into your sampling range. Bandpass sampling you would apply a frequency shift to the data, lowpass filter and then sample - in that case there are very few restrictions. – David Sep 20 '17 at 14:04
• @David I'm not sure what your point is. In any case, you can do bandpass sampling without frequency shifting; in fact, in many cases that's the whole point, you can undersample and avoid the use of a mixer. Also, bandpass sampling is a special case of undersampling. – MBaz Sep 20 '17 at 14:08
• Seems like there is more than one definition of Bandpass sampling. Even in the case Undersampling I think the lowest sample rate would be 6200 Hz not the 3100 Hz - you have to accommodate the negative frequencies band also being shifted down - at least that's what I see from the web page you've linked to. Please explain if I'm wrong or missed something. – David Sep 20 '17 at 14:13