Not too sure why, but I can’t seem to figure out the transfer function for the filter in question below. Can’t isolate Vout/Vin fully, there’s a straggling Vin that I can’t move over. Any thoughts?
8
-
2\$\begingroup\$ RS produces a VIN-dependent current. What's your goal here? \$\endgroup\$– jonkCommented Mar 11, 2022 at 6:00
-
2\$\begingroup\$ Look at the transfer function of a non-inverting amplifier and consider impedances made of the parallel associations of \$R_3C_2\$ and \$R_sR_1\$. The -5-V line is 0 V in ac (hence the parallel combination with \$R_1\$) if that is what is causing you problem. \$\endgroup\$– Verbal KintCommented Mar 11, 2022 at 6:34
-
3\$\begingroup\$ Use the rule: sum of currents AWAY from node = 0, otherwise it’s easy to make sign errors. You have at least one such error. Don’t be in a hurry to use numerical values; use symbols until the end. \$\endgroup\$– ChuCommented Mar 11, 2022 at 7:57
-
3\$\begingroup\$ Do you really intend to start with KCL? Such a circuit is a basic one and can be analyzed using the well-known gain formula for a simple non-inverting gain stage. \$\endgroup\$– LvWCommented Mar 11, 2022 at 8:00
-
2\$\begingroup\$ Transfer functions assume zero initial conditions, so there will be a problem with the -5V source if it's an initial condition. Superposition is one way out of that problem. \$\endgroup\$– ChuCommented Mar 11, 2022 at 8:52
|
Show 3 more comments
1 Answer
\$\begingroup\$
\$\endgroup\$
3
For help, you should also simulate this.
See the behavior of "voltage shift" ... One can see the "DC" and "AC" parts.
Alternatively, the "AC" part can be shown also like this ... which shows also the "AC frequency transfer function".
-
\$\begingroup\$ A picture is worth a thousand words ? \$\endgroup\$ Commented Mar 11, 2022 at 12:07
-
1\$\begingroup\$ I'm aware of what the Bode plot will look like via simulation - I've already done that. I just need to figure out the numeric transfer function equation at this point. \$\endgroup\$– ZashionyCommented Mar 11, 2022 at 13:18
-
\$\begingroup\$ Ok, Good point. Then you have the two "corners" of the function ... \$\endgroup\$ Commented Mar 11, 2022 at 15:44