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Apr 10, 2023 at 8:54 history edited ocrdu CC BY-SA 4.0
added 29 characters in body; edited title
Apr 10, 2023 at 1:00 comment added periblepsis @karlkarlsen Ernesto just told you how to see it. You know the gain equation for the (-) input and the way it's tied to the output and ground: \$A=1+\frac{R_2}{R_3}\$. That's going to be multiplied by the high-pass CR filter at the (+) input, where \$\omega_{_0}^2=\frac1{R_1\,C_1}\$ and only depends upon \$R_1\$ and \$C_1\$. Learn to sweep away the details for a moment and see larger pictures which will then help you navigate the math forest, later on.
Apr 10, 2023 at 0:55 history became hot network question
Apr 9, 2023 at 18:34 comment added Spehro 'speff' Pefhany Aside from missing the effects of the capacitor, your H(s) units make no sense. It may seem like a small thing but if you get into more complex algebra one tiny error can cause a lot of wasted work.
Apr 9, 2023 at 18:28 answer added Jan Eerland timeline score: 5
Apr 9, 2023 at 16:49 comment added karl karlsen No - but I just saw it. I will read the stackexchange codex ASAP
Apr 9, 2023 at 16:31 answer added Designalog timeline score: 7
Apr 9, 2023 at 16:23 comment added Andy aka Did you read my comment on your other question?
Apr 9, 2023 at 16:22 comment added karl karlsen Hi @Andyaka won't they be zero in this case (because its an non-inverting om-amop)? and what is then Z1 and Z2 ?
Apr 9, 2023 at 16:15 comment added Andy aka You forgot to include C1 and R1 effects. That makes all the difference.
S Apr 9, 2023 at 16:09 review First questions
Apr 9, 2023 at 16:22
S Apr 9, 2023 at 16:09 history asked karl karlsen CC BY-SA 4.0