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Below you can see the frequency spectrum of a square pulse. But what exactly is the purpose of the negative frequencies?

Lets say I have a square pulse with a width of 10ns (f=100MHz) and I want to pass it through an opamp, I want most of the energy to be preserved, so I will look for an opamp with a good bandwidth . Lets say the op amp has a bandwidth of 100Mhz. Does this mean I get the whole energy between -100MHz to 100MHz or just the energy between 0 and 100MHz?

It is probably a dumb question but I want to be sure.

enter image description here

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It's not a dumb question. In the case of your graph, the frequency is negative with respect to the centre frequency. So if centre frequency is 10MHz, and we talk about the 3dB points being +/-1MHz, they are 9 and 11MHz.

If you want to put a 100MHz square wave through an opamp, you will indeed need a bandwidth much more than 100MHz. Square waves have odd harmonics, so you need to pass frequencies of 300MHz, 500MHz and so on. The more of these you get through, the better the shape of the square wave will be.

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  • \$\begingroup\$ but what if I just want to pass a single square pulse? \$\endgroup\$ – Yoomo May 27 at 9:59
  • \$\begingroup\$ you can imagine that as a brief burst of those harmonics that dies away very quickly. But you still need to get them through the opamp to get a decent pulse shape. \$\endgroup\$ – danmcb May 27 at 10:00
  • \$\begingroup\$ ok thank you. But I still don't get your point with the center frequency. What is the center frequency? Could you give an example \$\endgroup\$ – Yoomo May 27 at 10:02
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    \$\begingroup\$ en.wikipedia.org/wiki/Center_frequency @Yoomo \$\endgroup\$ – Mitu Raj May 27 at 10:04
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    \$\begingroup\$ But anyway, in the real world, outside of abstract mathematics, frequency is a positive quantity. You can't vibrate something -10 times a second. \$\endgroup\$ – danmcb May 27 at 10:16
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Negative frequencies are not intuitive. They appear in Fourier transforms because those transforms present signals as sums of complex exponentials exp(j(2Pi)ft).

Presenting signals as sums of usual sines would be perfectly possible and they would not need negative frequencies, but algebraic manipulations and formula derivations would become much more difficult.

In math we could use any orthogonal base function set, so complex exponentials are in that sense not better nor worse than the common real sine functions. But complex exponentials are easier to manipulate and the connection between them and real sinusoidals is simple - the spectrum is mirrored at negative frequencies.

So: For pure convenience we have taken into use negative frequencies and calculate spectrums assuming the signal is a sum of complex exponentials. The price is that every spectrum of normal time dependent voltages must be mirrored to the negative frequency side and every practical filter frequency response must be mirrored in the same way. That's no problem after one has got some time to adapt himself to the convention.

There's more advantages than easier algebraic manipulations. Some advanced signal processing will become possible if we allow signals to have also the imaginary part. That's, of course only a separate memory word in digital signal processing. Real part is in one word and the imaginary part is in another. Spectrums of those signals are not symmetric around f=0, but they are calculated with the same Fourier transform formulas with complex exponentials.

You may see in the principle explanations of modulators and detectors I and Q channels. Those applications are perfect examples of the advantages of complex valued signals.

Let's assume you have built an amp or filter circuit which handles practical voltage signals say 0 to 1MHz. You can in normal talk say that it covers band 0...1MHz, but when calculating exactly its effect to signal spectrums which have negative frequencies you should automatically keep in mind that in frequency domain it affects symmetrically around f=0.

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