# Finding Frequency and Bandwidth

#### cking2729

Joined Apr 21, 2008
1
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!

Thanks!

#### Caveman

Joined Apr 15, 2008
471
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.

#### rwmoekoe

Joined Mar 1, 2007
172
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

#### Caveman

Joined Apr 15, 2008
471
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.