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From what I thought I learned about sampling is that when you draw a sampled signal in the frequency domain, you can show copies or "images" of that signal at integer multiples of the sampling frequency. Is this true?

If so, I'm having a hard time understanding this conceptually. What does it really mean to make a copy at a different location on the frequency line?

For example, if I have a signal which when drawn in the frequency domain looks like a box centered at zero extending from $f_{low}$ to $f_{high}$, can I now say I can draw this box centered at integer multiples of the sampling frequency and so long as my box when redrawn does not touch another box I avoid aliasing? Why do I do this at integer multiples? For example if my sampling frequency is $20 \ \mathrm{Hz}$, what does the next integer multiple at $40 \ \mathrm{Hz}$ tell me? Does it mean I'll see something interesting if the signal were to run at $40 \ \mathrm{Hz}$?

I'm trying to get a better understanding of this so I can then understand bandpass sampling and what it means to select a location at $k$, $k-1$, etc.

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Below is an example of two sine waves at 11 Hz and 1 Hz, both sampled with a 10 Hz sampling clock. The sampled locations are drawn as circles. If you only had the samples, and had no knowledge of any filtering prior to the sampling process, you would not be able to know for certain if the samples came from a 1 Hz analog signal or an 11 Hz analog signal. In fact, any signal at N fs ± 1 Hz for any integer N would produce the same result. (9 Hz, 11 Hz, 19 Hz, 21 Hz, etc).

enter image description here

Thus, with appropriate analog bandpass filtering prior the sampling, we can directly sample ("undersample") a bandpass signal, as long as we follow Nyquist's criteria that the sampling rate be > 2x the signal bandwidth. Do not confuse bandwidth with highest frequency content! The sine wave shown for example has 0 bandwidth in that it occupies an infinitely thin line in the spectrum. Actual bandpass waveforms of interest would have some bandwidth based on their modulation content.

What this picture also helps clarify is the increased sensitivity to any sampling clock jitter when performing bandpass sampling. Note specifically the slope of the red lower frequency signal as compared to that of the blue higher frequency signal in vicinity of the sample locations, and imagine small variations due to clock jitter and how these variations would transform to amplitude error.

Please see this answer for specific details related to a conceptual understanding of having multiple frequency "images": Aliasing after downsampling

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  • $\begingroup$ Thank you for this - it is helpful, but I think I'm still confused with what it means to have a duplicate image centered at some integer multiple of the sampling frequency. When I think of aliasing, I usually do think of the typical example you have shown above. My confusion is with what does that duplicated image centered on the frequency line really mean? If I'm sampling at 20 Hz, and I have an image at 100 Hz, does it mean the sampled value at the same relative location in the other mage will be exactly the same as the original, but just happens to come from a higher frequency signal? $\endgroup$ – user7509231 Feb 28 '17 at 2:12
  • $\begingroup$ The images are the alternate frequency locations of which if energy is present, it will "fold" into your primary Nyquist band (first Nyquist Zone) that is from -fs/2 to fs/2 where fs is the sampling rate. Of course a signal would need to be there for this to happen, but once it does you would not know if it came from the image location or was already in your first Nyquist Zone-- unless you had a prefilter that would exclude the image locaitons. So the picture I showed you demonstrates how a sine wave either at 11 Hz OR at 1 Hz would result in the same digital signal. $\endgroup$ – Dan Boschen Feb 28 '17 at 2:15
  • $\begingroup$ Perhaps in your confusion with this you are imagining a duplicate signal present in the image band-- if so, it is the otherway around-- any signal in the image bands will replicate itself to your first Nyquist zone (all you really see once it is digital), but these images can be completely different from the signal of interest. The fact that my picture shows identical digital signals for the two different analog signals is to show how the aliasing can happen, but the higher frequency signal can be completely different of course. $\endgroup$ – Dan Boschen Feb 28 '17 at 2:19
  • $\begingroup$ I think that's the answer I was looking for...confirming that these duplicate "images" represent the signal frequencies that can be perfect aliases of the base signal (the signal between -fs/2 and fs/2. Now, you just used a word which I think is significant - folded. When the aliasing occurs, are the two aliased points between the two signals at the exact same position in the images, or is it reversed? For example, if I fold a sheet of paper in half, the left most edge becomes the right most edge and now your points look reversed, albeit symmetric. $\endgroup$ – user7509231 Mar 1 '17 at 11:03
  • $\begingroup$ It is actually cyclical: not sure how familiar you are with complex signal notation, but when the signal is complex the spectrum from -fs/2 to 0 can be completely different from the spectrum from 0 to fs/2. When the signal is real these two positive and negative frequency spectrums are conjugate symmetric (same amplitude opposite phase). That said images are always cyclical: what is 1 Hz past +fs/2 rolls around and appears 1 Hz beyond -fs/2, in your Nyquist band. This always happens but when the signal is real it will have a complex conjugate copy which then appears as the file you describe. $\endgroup$ – Dan Boschen Mar 1 '17 at 11:08

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