Timeline for Time constant of inverting amplifier, filter
Current License: CC BY-SA 4.0
11 events
when toggle format | what | by | license | comment | |
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Apr 23, 2019 at 4:46 | vote | accept | Electro by night | ||
Apr 21, 2019 at 23:53 | comment | added | Chu | The TF has no poles, it has one zero. | |
Apr 21, 2019 at 15:56 | answer | added | Verbal Kint | timeline score: 1 | |
Apr 21, 2019 at 15:37 | comment | added | Electro by night | In this problem the opamp is considered active. How does this change things? | |
Apr 21, 2019 at 10:03 | comment | added | LvW | Verbal Klint-I think, in your answer it would be helpful to mention if the opamp is to be considered as active or not. | |
Apr 21, 2019 at 7:47 | comment | added | Verbal Kint | If you want the time constant of this circuit, reduce the input voltage to 0 V (replace the source by a short circuit) then disconnect capacitor \$C_1\$ and determine the resistance "seen" from its connections. That resistance multiplied by \$C_1\$ is your time constant while the inverse gives the pole. More details in this APEC seminar: cbasso.pagesperso-orange.fr/Downloads/PPTs/… | |
Apr 21, 2019 at 7:41 | comment | added | Electro by night | I am familiar with the laplace transformation, but it seems like the transferfunction has no zeros. I end up with the following: R2 * ( C * s * R1 + 1)/R1 | |
Apr 21, 2019 at 7:23 | comment | added | Chu | Nodal analysis at the inverting input will give the TF. | |
Apr 21, 2019 at 6:55 | comment | added | joribama | If you are familiar with using the Laplace transform to represent impedances, you can calculate the transfer function by dividing R2 by the parallel equivalent of R1 and 1/sC1. You will realize that the gain will be relatively flat till a certain cutoff frequency (I'll let you determine it) and then it will start growing at 20dB/dec after that. | |
Apr 21, 2019 at 6:45 | review | First posts | |||
Apr 21, 2019 at 7:20 | |||||
Apr 21, 2019 at 6:40 | history | asked | Electro by night | CC BY-SA 4.0 |