Timeline for PWM demodulator implementation
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Nov 2 at 17:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 3 at 13:39 | comment | added | Erinç Utku Öztürk | I made several attemps but was unable to solve the transfer function. Could you help me solve the circuit if you are able to? The ratio t/T feels impossible to incorporate the equation | |
| Oct 3 at 13:20 | history | edited | Erinç Utku Öztürk | CC BY-SA 4.0 | deleted 121 characters in body |
| Oct 3 at 12:43 | history | edited | Erinç Utku Öztürk | edited tags | |
| Oct 3 at 9:31 | history | edited | Erinç Utku Öztürk | CC BY-SA 4.0 | deleted 424 characters in body |
| Oct 2 at 23:25 | comment | added | Marcus Müller | no, the sample and hold element samples at the beginning of the on period and holds thereon. So, there's no connection through while the PWM signal is on; otherwise, this would just be an amplfiier. This is explained starting in line 64 of column 2 of the patent document. | |
| Oct 2 at 23:22 | comment | added | Erinç Utku Öztürk | Makes sense. I edited the post again as I was trying to draw the schmatic when the PWM is on. When the PWM is on doesnt the circuit look like as i drew? If so, how would the integrator work? @MarcusMüller | |
| Oct 2 at 23:17 | history | edited | Erinç Utku Öztürk | CC BY-SA 4.0 | deleted 1 character in body |
| Oct 2 at 23:08 | history | edited | Erinç Utku Öztürk | CC BY-SA 4.0 | added 18 characters in body |
| Oct 2 at 23:01 | comment | added | Marcus Müller | yes, that's the same thing. PWM is on (1) or off (0). Multiplying with 1 or 0 is like connecting the reference signal or 0V to R1. This is really simple! | |
| Oct 2 at 23:00 | comment | added | Erinç Utku Öztürk | About the multiplicator (the X sign enclosed with circle), somehow that symbol represents a switch. It is explained in the text of the patent. But about the sample and hold circuit, when it is sampling can’t i model it as a close circuit | |
| Oct 2 at 22:49 | comment | added | Marcus Müller | your blue assumptions are wrong. A sample and hold is not a short circuit. that's the point here! And the assumption that your multiplicator also is just a constant feedthrough of the reference circuit makes no sense either, because now your assumption completely removes the actual input signal completely from the system. | |
| Oct 2 at 22:49 | history | edited | Erinç Utku Öztürk | CC BY-SA 4.0 | added 292 characters in body |
| Oct 2 at 22:30 | comment | added | Erinç Utku Öztürk | Thank you for the edits and the notices. I will keep them in mind next time I post anything. | |
| Oct 2 at 22:03 | history | edited | Marcus Müller | CC BY-SA 4.0 | added 127 characters in body |
| Oct 2 at 21:59 | answer | added | Marcus Müller | timeline score: 1 | |
| Oct 2 at 21:55 | comment | added | Andy aka | Add a circuit image to your post. Do this because if the link dies your question becomes liable to subsidence and, you probably don't want that. In a wider context, questions should be self contained and rely only on information posted by the OP (yourself). | |
| Oct 2 at 20:50 | history | edited | JYelton | CC BY-SA 4.0 | added 1 character in body |
| Oct 2 at 20:28 | history | edited | toolic | CC BY-SA 4.0 | edited body; edited title |
| Oct 2 at 20:27 | history | edited | Erinç Utku Öztürk | CC BY-SA 4.0 | added 142 characters in body |
| S Oct 2 at 20:25 | review | First questions | |||
| Oct 2 at 20:28 | |||||
| S Oct 2 at 20:25 | history | asked | Erinç Utku Öztürk | CC BY-SA 4.0 |