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Aug 7 at 16:58 comment added ai_ch @mathkeepsmebusy yes exactly
Aug 7 at 16:40 history became hot network question
S Aug 7 at 16:37 vote accept ai_ch
Aug 7 at 14:46 comment added Math Keeps Me Busy Are R5 and C5 supposed to be the Thevenin equivalent impedance for R4 and C4?
Aug 7 at 13:58 history edited ai_ch CC BY-SA 4.0
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Aug 7 at 12:41 answer added Math Keeps Me Busy timeline score: 3
Aug 7 at 11:20 vote accept ai_ch
S Aug 7 at 16:37
Aug 7 at 10:29 answer added Andy aka timeline score: 2
Aug 7 at 9:04 comment added periblepsis I'm just speaking very generally. If the AC signal has a very low frequency then C4 won't short it out and C5 won't bypass R5, either. Then you just have R4 in one circuit and R5 in the other. So near DC they are about the same. It's only when the frequency increases that these two circuits are no longer similar. As far as your questions go, I can't say I'm getting into your mind very well. When I can put myself into someone else's thinking, then I can re-read their writing and find a match and a segue to help out. But I'm not able to put myself into yours. I'd know, if so.
Aug 7 at 8:53 comment added ai_ch But very high frequencies are no longer present in the measurement of VF4 (which is used as an input on the second source), since the RC filter has already removed them prior to measuring. Or am I missing something?
Aug 7 at 8:47 comment added periblepsis Very high frequencies would be fully present at VF5 because C5 would pass them along (bypassing R5.). But in the case of VF4 it would be close to 0 because C4 would short them to ground. So the two aren't equivalent. (Not sure if you think they are or if there's some other question I've missed.)
S Aug 7 at 8:39 review First questions
Aug 7 at 9:39
S Aug 7 at 8:39 history asked ai_ch CC BY-SA 4.0