Finding Frequency and Bandwidth

Discussion in 'Homework Help' started by cking2729, Apr 21, 2008.

  1. cking2729

    Thread Starter New Member

    Apr 21, 2008
    Hi guys,

    I am in my first experiences with analog filters and am in need of serious help with the following problem:

    I need to find the frequency and bandwidth of the above circuit all by hand. If anyone could provide step by step help, it'd be GREATLY appreciated!

  2. Caveman

    Active Member

    Apr 15, 2008
    First things first. Simplify. Since U2 has a low impedance output, you can calculate that without all that junk on the left. However, note that the two circuits have the same topology with only a couple of value changes.

    Second, let's see what you know. Let's call U2 output, V1_out.
    1. U1 non-inverting input is 1/2 V1_out.
    2. We will assume that U1 inverting input is the same.
    3. Since the inverting input of U2 and U1 are the same, the non-inverting input of U2 is also 1/2 V1_out.

    Now you can calculate the currents through all of the other parts using the impedance formulas for capacitors and resistors.
  3. rwmoekoe

    Active Member

    Mar 1, 2007
    the cutoff frequency should be around the -3db cutoff of the 374k and 0.01uf lowpass filter. the upper opamp is bootstrapping the input, giving a steeper cutoff. but still with a 6db/octave slope. the second stage doubles the effect, further making the cutoff steeper, and doubles the slope to 12 db/octave.

    so, how do we do it by hand? go down to the s? it'll be a mess though
  4. Caveman

    Active Member

    Apr 15, 2008
    First calculate the output of U1 in terms of V1_out. You know that the current through the 5.36k and 0.01uF are the same, and you know that the inverting input of U1 is V1_out/2.

    Then using the output of U1, do a KCL on the non-inverting node of U2 to relate V1_out to V1. Solve.

    Then just run the numbers again for the second stage with the new resistor values. Multiply and done.