Timeline for If a square wave has infinite bandwidth, how can we see it on an oscilloscope?
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21 events
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| Nov 18, 2020 at 19:42 | answer | added | gep wemple | timeline score: 0 | |
| Nov 17, 2020 at 21:50 | comment | added | Sam | @Acccumulation Oh yes, absolutely, it doesn't really matter what you call them, but there is always a finite limit to the screen's ability to define an image (a sub-pixel is still a singular item). Even the old analog scopes of yore had a resolution in the sense that each "pixel" represented each grain of phosphor on the display and a pixel that is half on and half off blurs into one that is half as bright because of light bleed, otherwise we'd just have one continuous film but even then we'd still have a finite resolution, it's just that each pixel is now one atom in size... | |
| Nov 17, 2020 at 20:33 | answer | added | Fat32 | timeline score: 0 | |
| Nov 17, 2020 at 20:25 | answer | added | Acccumulation | timeline score: 0 | |
| Nov 17, 2020 at 20:23 | comment | added | Acccumulation | @Sam Do oscilloscopes have pixels? youtube.com/watch?v=Ea6tw-gulnQ | |
| Nov 17, 2020 at 19:30 | comment | added | EvilSnack | Old fart here from the analog oscilloscope days. When the transition times for a square wave are very short, the edges of the square wave get fainter. The maximum slope (delta voltage over delta time) sets an upper limit on how faint the trace can be. | |
| Nov 16, 2020 at 21:57 | comment | added | Sam | Another fundamental reason no-one's mentioned is that you would need infinitely small pixels to fully define an infinitely sharp squarewave (which our eyes couldn't resolve anyway), but because most modern scopes have pretty chunky pixels (e.g. compared to phones), edges that only slope a fraction get rounded up to a single, straight line of pixels so you lose a bit of the finer details (analog scopes had more effective "pixels" but then had focus issues to worry about). As a wise man once said "There are lies, damned lies, and what's on the screen of the oscilloscope" | |
| Nov 16, 2020 at 21:22 | answer | added | hacktastical | timeline score: 5 | |
| Nov 16, 2020 at 21:00 | history | tweeted | twitter.com/StackElectronix/status/1328442748056313857 | ||
| Nov 16, 2020 at 19:05 | comment | added | user57037 | If you zoom in you will see that the rise time and fall time are finite. Therefore, what you observe on the oscilloscope is not a perfect square wave with infinite bandwidth. Instead it is an imperfect square wave with finite bandwidth. | |
| Nov 16, 2020 at 16:16 | history | became hot network question | |||
| Nov 16, 2020 at 12:07 | answer | added | Martin | timeline score: 3 | |
| Nov 16, 2020 at 11:10 | answer | added | analogsystemsrf | timeline score: 1 | |
| Nov 16, 2020 at 10:49 | history | edited | JRE | CC BY-SA 4.0 |
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| Nov 16, 2020 at 8:38 | answer | added | Transistor | timeline score: 18 | |
| Nov 16, 2020 at 8:21 | answer | added | Bimpelrekkie | timeline score: 26 | |
| Nov 16, 2020 at 8:17 | answer | added | Marcus Müller | timeline score: 57 | |
| Nov 16, 2020 at 8:16 | review | First posts | |||
| Nov 30, 2020 at 8:16 | |||||
| Nov 16, 2020 at 8:15 | history | edited | Bimpelrekkie | CC BY-SA 4.0 |
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| Nov 16, 2020 at 8:15 | history | edited | Marcus Müller | CC BY-SA 4.0 |
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| Nov 16, 2020 at 8:13 | history | asked | OVIYA N | CC BY-SA 4.0 |