Timeline for Non-standard low-pass filter design
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 16, 2023 at 10:01 | answer | added | Antonio51 | timeline score: 0 | |
| Nov 16, 2023 at 4:00 | comment | added | periblepsis | bkxp, It's technically 4th order. But with the risk of missing something important, you can imagine that R4+C1+R5+C3 form a 2nd order LP to the (+) input. This must be reflected at the (-) input. So there's the same 2nd order LP at the (-) input (ideally.) This must be integrated by C2 to form Vout, because vin*(LP 2nd order)=vout*s*C2, or vout/vin=(LP 2nd order)/(s*C2). That's 3rd order behavior. (Granted R6+C4 forms a 1st order LP of vout into the node joining R1+R2. But it's ignorable here because vout is still the integral of a 2nd order LP.) Can't be inverted. | |
| Nov 15, 2023 at 14:39 | answer | added | Andy aka | timeline score: 2 | |
| Nov 15, 2023 at 13:40 | history | undeleted | bkxp | ||
| Nov 15, 2023 at 13:39 | history | deleted | bkxp | via Vote | |
| Nov 15, 2023 at 13:39 | history | asked | bkxp | CC BY-SA 4.0 |