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i have a signal of 120Mhz and sampling at 800Mhz. we look at the region of fs/2=400MHz so obvios we have harmonics of the fundamental which is 240 360. But in the document i see that there is another harmonic -4*120M+800M I cant see why we have negative harmonics and their copies. Upon what theory its based?

they say the -120M is the only one copied to the 0-400M region but i thought we could take also

-6*120M+fs=-720M+800M=80MHz

I cant see what why we have negative harmonics and which one is copied to the 0-fs/2 region.

Thanks.

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  • $\begingroup$ What is "the document"? Is this a homework-like question? Some tutorial? My answer wouldn't change, but it would be helpful to understand the context better. I might try and answer though, since it might be a little unwieldy for a comment... $\endgroup$ – Nigel Redmon Oct 1 '20 at 21:05
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First, about "why we have negative frequencies and which one is copied"...

In a real signal, there is no difference between a negative frequency and a positive frequency in this sampling context. Consider that you have captured 1000 samples that you know contains a single sine wave. You analyze it and find it contains exactly ten cycles of a sine wave. Knowing the sample rate, you determine that it's a 100 Hz sine wave. Now analyze the capture from the other direction—end to start. You would also find that it's 100 Hz, but since we moved backwards through time, it would be -100 Hz.

That is, if you choose to plot negative frequencies, they will always be a mirror of the positive frequencies. You will never find -80 Hz without 80 Hz, for instance. There is never a choice of "which on is copied". If you find somehow that there is a -80 Hz component, you can be sure there is an 80 Hz component.

Sampling is a type of amplitude modulation (AM). Pulse Amplitude Modulation (PAM) in the analog domain, Pulse Code Modulation (PCM) in the digital domain. In the analog domain, it's a multiplication of the signal with an impulse train at the sampling frequency. Looking at just the first harmonic of the impulse train, it's a sinusoid (cosine in particular) at the sampling frequency. In your case this means you are multiplying your signal by 800 MHz (as well as its pulse harmonics, but we need only consider the closest one here).

Amplitude modulation produces sum and difference frequencies, but for this problem we only care about the difference frequencies at the moment. That means if your signal has a 120 MHz first harmonic, sampling will also produce 800-120 HMz = 780 Mhz. And the fourth harmonic of your signal, 480 MHzHz, would produce a new component of 800-480 = 320 MHz. Note that this is exactly the same answer as your "-4*120M+800M". Except that I came up with my answer without considering negative frequencies (800M minus 4x120M).

To summarize, negative frequencies always have positive counterparts. And there is no particular need to consider negative frequencies with what you are looking at—it seems it just adds confusion, but may make more sense in context of the document you refer to.

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  • $\begingroup$ Hello So why we are not considering -720M mirrores as 80Mhz? -6*120M+fs=-720M+800M=80MHz Thanks $\endgroup$ – rocko445 Oct 2 '20 at 9:27
  • $\begingroup$ I'm not sure how to answer that—I'd say it's really the other way around. Below half the sample rate lies our original signal, before sampling. Sampling creates sidebands—the mirrors—around every multiple of the sample rate. Coincidentally, I published an AM widget for experimentation. Scroll down to the "Digital audio" and give it a try. These mirrors ("images") are an unavoidable consequence of sampling. earlevel.com/main/2020/10/02/am-widget $\endgroup$ – Nigel Redmon Oct 3 '20 at 0:01
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Note the trig identity:

-sin(x) == sin(-x)

This means that a negative aliased frequency just means the aliased samples happen to be inverted in phase from that of the fundamental signal. You can't tell, just by looking at these aliased samples, whether the harmonic was an inverted sinewave, or a negative frequency sinewave. So you can call that set of aliased samples either (or both).

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  • $\begingroup$ Good point in the general case with positive and negative frequencies, but when discussion sampling I think it's even more to the point to note that cos(x) == cos(-x). In other words, positive and negative frequencies yield the same thing. $\endgroup$ – Nigel Redmon Oct 21 '20 at 22:38

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