Timeline for Design a filter that allows frequencies below 5 kHz to pass freely, but all frequencies above 5.2 kHz must be undetectable
Current License: CC BY-SA 3.0
25 events
| when toggle format | what | by | license | comment | |
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| Mar 21, 2018 at 17:38 | comment | added | richard1941 | In the past, we radio hams have used crystal an mechanical filters for such extreme requirements. You need to define "undectable". | |
| Mar 14, 2018 at 13:10 | vote | accept | Tamir Shklaz | ||
| Mar 12, 2018 at 13:39 | comment | added | J... | The flaws here are in the failure to have a clear specification. What does "freely" mean? What is "undetectable"? What is "minimal"? What is "not too important"? This is engineering, not drama. A spec needs numbers. The first thing an engineer should do in the face of such wooly "specifications" is to figure out what they actually need to design. At this point, you don't know what you need to design, and making assumptions will always get you into trouble. | |
| Mar 11, 2018 at 14:48 | comment | added | Bruce Abbott | "with minimal ripple in the pass band" - 'minimal' is not a specification. | |
| Mar 11, 2018 at 13:07 | answer | added | mathreadler | timeline score: 4 | |
| Mar 11, 2018 at 12:51 | comment | added | mathreadler | Maybe you can use Daubechies wavelet packets. Daubechies are famous for being max flat in pass band for the low-pass filter (for being orthogonal filter pair). You will need to design the sub-band decomposition to fit the 5-5.2 kHz with fine granularity but that will be the major work. Every frequency area you don't have to subdivide will mean some calculations saved. That would basically be all areas "away from" 5-5.2 kHz. | |
| Mar 10, 2018 at 21:44 | answer | added | bobflux | timeline score: 4 | |
| Mar 10, 2018 at 21:30 | history | edited | Tamir Shklaz | CC BY-SA 3.0 |
Fixed some glaring math mistakes and added more specific detail.
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| Mar 10, 2018 at 19:33 | history | tweeted | twitter.com/StackElectronix/status/972556187521765376 | ||
| Mar 10, 2018 at 18:28 | answer | added | Olin Lathrop | timeline score: 21 | |
| Mar 10, 2018 at 18:19 | answer | added | user110971 | timeline score: 21 | |
| Mar 10, 2018 at 17:28 | comment | added | Bruce Abbott | How much passband ripple is acceptable? | |
| Mar 10, 2018 at 17:16 | answer | added | analogsystemsrf | timeline score: 3 | |
| Mar 10, 2018 at 16:48 | comment | added | Harry Svensson | What kind of answer are you expecting Tamir? Analog passive/active? Digital IIR/FIR? Wave digital filter? FFT + Remove unwanted stuff + IFFT? | |
| Mar 10, 2018 at 16:44 | comment | added | Dave Tweed | @WhatRoughBeast ??? I realize that he divided 4500 by 20 to get 225 poles; what I'm challenging is how he got the 4500 in the first place. | |
| Mar 10, 2018 at 16:43 | comment | added | WhatRoughBeast | @DaveTweed - He divided 4500 by 20. | |
| Mar 10, 2018 at 16:23 | history | edited | SamGibson♦ | CC BY-SA 3.0 |
Fixed glaring typo (and incorrect capitalisation of units) in the title.
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| Mar 10, 2018 at 15:31 | comment | added | Dave Tweed | How did you get 4500 dB/decade? \$\log_{10}\frac{5.2}{5} = 0.017\$, which gives a value closer to 1200 dB/decade. | |
| Mar 10, 2018 at 15:23 | comment | added | PlasmaHH | Calling -20dB undetectable is quite a stretch | |
| Mar 10, 2018 at 15:01 | comment | added | Neil_UK | use an elliptic filter | |
| Mar 10, 2018 at 15:00 | history | edited | Neil_UK | CC BY-SA 3.0 |
it's
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| Mar 10, 2018 at 15:00 | answer | added | user136077 | timeline score: 4 | |
| Mar 10, 2018 at 14:38 | history | edited | TonyM | CC BY-SA 3.0 |
Tidied.
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| Mar 10, 2018 at 14:33 | review | First posts | |||
| Mar 10, 2018 at 16:07 | |||||
| Mar 10, 2018 at 14:32 | history | asked | Tamir Shklaz | CC BY-SA 3.0 |