I calculate the amplitude of output signal by using the formula shown in figure. where R1 = 1 k ohm and Cf = 1 μF and Vin is sine signal. I got my answer is 5 V. Is it right? I want to know that why Rf (feed back resistor) did not contribute here while calculating V(out).

Sorry, previously I post the wrong circuit in which I showed positive feedback instead of negative feedback.

  • \$\begingroup\$ Try deriving the formula instead \$\endgroup\$ – Pyxzure Jan 30 '14 at 5:41
  • \$\begingroup\$ Are you trying to do a transient time-domain analysis (from time = 0 with initial conditions), as the equation implies, or a steady-state analysis? Hint... the latter is much easier. \$\endgroup\$ – Spehro Pefhany Jan 30 '14 at 13:09
  • \$\begingroup\$ no not time domain analysis ... just the input sin and circuit is given and i have to calculate magnitude of V(out).that formula shown above i used to calculate. \$\endgroup\$ – SW. Jan 31 '14 at 5:23

From the topology you can see this is a low pass filter. I would start with getting some intuition of how this filter is going to behave, rather than trying to see what exactly happens at your frequency. For frequencies well below the rolloff, you can approximate the capacitor as a open, and well above the rolloff as a short. By computing the gain at the limits and looking at where the rolloff frequency is, you can draw out a Bode plot from inspection.

Then you can look at where your frequency of interest fits on that plot and get a quick idea of the gain.

That brings up another issue, which is that it's unclear what your frequency is since you haven't defined the units of T. If T is in seconds, then the frequency is 1000 / (2π s) = 160 Hz.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.