Timeline for What exactly are harmonics and how do they "appear"?
Current License: CC BY-SA 3.0
24 events
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| May 24, 2012 at 7:21 | comment | added | stevenvh | @oxakhil - You'll only see a 50Hz sine, the 3rd harmonic is at 150Hz and that's already too far away. Remember the square wave doesn't have even harmonics, so the 100Hz harmonic doesn't exist. | |
| May 24, 2012 at 7:19 | comment | added | Kortuk | @AndrejaKo, I think we should talk more in chat if you want to discuss this, this comment chain has become too long. | |
| May 24, 2012 at 7:17 | comment | added | 0xakhil | @stevenvh Make that a 10th order filter. I want to know if I will get that same square at the output or a square wave with its higher harmonics of frequencies greater than 100Hz blocked?? | |
| May 24, 2012 at 4:57 | comment | added | stevenvh | @oxakhil - A first order RC filter will not completely block the higher frequencies, just decrease them. The effect will be that the edges are less steep, and somewhat exponential. That's because of the time required to charge/discharge the capacitor via the resistor. | |
| May 23, 2012 at 19:10 | comment | added | 0xakhil | @stevenvh There is something that relates the electronics to the sinusoidal waves as the basis functions. If I pass a 50Hz square wave through a 100 Hz RC low pass filter, what will I get at the output? Will I get the same square wave at the output? | |
| May 22, 2012 at 13:16 | comment | added | stevenvh | @avakar - Say you have a 100Hz sawtooth wave. That has all harmonics: 100Hz, 200Hz, 300Hz, 400Hz, etc. If you would filter the fundamental out and would listen to the result you would still hear a 100Hz tone! That's because your brain "knows" 200Hz can't be the fundamental if you have also 300Hz and 500Hz, for instance. | |
| May 21, 2012 at 14:14 | comment | added | avakar | @stevenvh, I see, I didn't realize that a frequency would be called fundamental even though its amplitude is zero, though Wikipedia agrees. My apologies. | |
| May 21, 2012 at 14:03 | comment | added | stevenvh | @avakar - Suppose you have a 3Hz sine and a 4Hz sine, so no integer multiples. Then the repeating wave will have a frequency of 1Hz (greatest common divider). That's the fundamental frequency of the sum, and the harmonics will be multiples of that. What do we have? A suppressed fundamental (amplitude zero), no second harmonic, a third harmonic and a fourth. Nothing else. "every repeating signal can be deconstructed into a fundamental and harmonics" still stands. Every harmonic has its own amplitude, and that can be zero. Look at the even harmonics in a square wave. | |
| May 21, 2012 at 13:53 | comment | added | avakar | "every repeating signal can be deconstructed into a fundamental and harmonics" that's false and highly confusing. Surely there are two sines such that neither of them has the frequency that is an integer multiple of the other's, yet their sum will be periodic. | |
| May 21, 2012 at 12:33 | comment | added | stevenvh | @Telaclavo - "We can use square waves as the basis functions". That's correct, and I also commented on that on a previous question. Here I assumed sines, like everybody else would. | |
| May 21, 2012 at 12:21 | comment | added | Telaclavo | "Sinusoidal waves don't have harmonics because it's exactly sine waves which combined can construct other waveforms". Wrong. That's only true if the base functions used in the decomposition are sinusoidal themselves. The base functions need to be orthogonal, but there are infinite functions we can choose from. We can use square waves as the basis functions and, under that choice, pure sinusoidal functions would have inifinite frequency components. Wavelets is just another choice. | |
| May 21, 2012 at 11:13 | comment | added | user | It's also worthwhile to note that your final illustration does not show AM (double sideband full power carrier) but rather double sideband suppressed carrier. In "true" AM, the carrier is at 50% of total (maximum) envelope power and the remaining 50% are distributed equally between the upper and lower sidebands. | |
| May 21, 2012 at 10:02 | comment | added | leftaroundabout | "because it's exactly sine waves which combined can construct other waveforms". Actually, you could use any other complete set of orthonormal waves as well (e.g. wavelets). The reason trigonometric functions / complex exponentials are most popular is that they're eigenvectors of the differential operator, which is why Fourier transform immediately solves linear differential equations. But if those weren't so important, some other transform would probably prevail. | |
| May 21, 2012 at 9:39 | vote | accept | John Quinn | ||
| May 21, 2012 at 9:39 | comment | added | John Quinn | @stevenvh Well I'll be, Spice agrees with you (hopefully not by coincidence). I've come up with a lot to carry on with here, thank you guys. :-) | |
| May 21, 2012 at 9:30 | history | edited | stevenvh | CC BY-SA 3.0 |
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| May 21, 2012 at 9:17 | comment | added | stevenvh | @John - Yes, it may look implausible, but if you switch on and off once per second that signal will have a 1Hz sine in it. And a 3Hz sine, etc. You can isolate each one of them by passing the square wave through a narrow band-pass filter. If you would filter 0.8Hz to 1.2Hz you would clearly see the 1Hz sine! It's all Fourier's fault, really! ;-) | |
| May 21, 2012 at 9:11 | comment | added | John Quinn | @stevenvh, and so viewing as many harmonics on top of eachother as possible (if the signal has them) will make for a clean signal on the scope? If I turn a switch on and off (gnd, DC5V, gnd, ...) will there be actual harmonics to capture (not just decomposing in to) and be clearer than some random square wave that is composed of many harmonics? came out of my mind weird, sort of makes sense. | |
| May 21, 2012 at 8:49 | comment | added | stevenvh | @John - Nobody composes the signal from harmonics, but the math says they're there. The frequency spectrum will be infinitely wide. If you pass such a signal through a low-pass filter it's shape will change because harmonics are cut off. The scope's limited bandwidth works as a low-pass filter. | |
| May 21, 2012 at 8:48 | comment | added | Kortuk | @JohnQuinn, Any signal that exists can be made up of sin waves, this is how we look at what the spectral content is of a signal(ie. The amount of which frequencies exist) and most circuits can be looked at as affecting frequencies differently. When I was acting as a teaching assistant I found most often teaching an understanding of frequency domain to be top 5 on things that allow an electrical engineer to be a great one. | |
| May 21, 2012 at 8:46 | comment | added | Kortuk | @JohnQuinn, In real life a square wave is made up of spectral content as shown. Making a signal instantly change from 0V to 5V takes an infinite amount of power, in reality there is some rise time to the square wave and this determines the amount of spectral content required. High speed digital signals can be the devil for unwanted radiated transmission if allowed because the fast rise time means you are driving some very high frequencies. | |
| May 21, 2012 at 8:42 | comment | added | John Quinn | "[...] on the scope if only up to the fifth harmonic are shown." - Why on earth are harmonics used to construct the form, and not "5V = high up, 0V = down", like an CRO moving the beam? Sure a square can be composed of many harmonics, but in real life a square wave is a square wave, and an oscilloscope will show that, it can be analysed in to harmonics to reconstruct it but that loses resolution if it cannot read all of those harmonics. For the AM: If there are many harmonics, they cannot integrate well (having so many "sines") and create these unwanted frequencies? | |
| May 21, 2012 at 8:37 | history | edited | stevenvh | CC BY-SA 3.0 |
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| May 21, 2012 at 8:19 | history | answered | stevenvh | CC BY-SA 3.0 |