Finding the cutoff frequency for an active low-pass filter?

[Steven O.]

I am taking a distance learning "lab" class in EE, using Multisim to
simulate circuits at home. Unfortunately, e-mail communications with
the professor sometimes does, and sometimes does not, result in clear
answers, so once again I'm going straight to the pros.

We are simulating a 741 OpAmp circuit as a low-pass filter. My
question is, what is the correct formula for the cutoff frequency? If
I understand correctly, the cutoff frequency is the frequency where
the output falls to 1/sqrt(2) = .707 of the input voltage. The
circuit is shown at:

http://www.oppenheimercommunications.com/OpAmpProblem.htm

The measured cut-off frequency is around 300 Hz (1884 radians). And,
using the formula for Vout (which I do have), and plugging in various
values for w_in, I eventually find the cutoff frequency to be almost
exactly 310 Hz (1947 radians).

The lab manual for the circuit does not give the formula. The
textbook has a similar-but-not-identical circuit, for which the
formula for the cutoff frequency is (referring to the diagram)
w_cutoff = 1 / (C2 * R4). However, this formula is not really for the
circuit we are simulating, and in fact does not give the correct
cut-off frequency.

Hunting on the Web, I found what seems to be an identical looking
circuit, and the cutoff frequency is given by:
w_cutoff = 1 /sqrt (R2 * R4 * C1 * C2)

Once again, however, this does not give the actual, measured value for
the cutoff frequency. Working from the formula for the output
voltage, I tried to find the formula for w_cutoff myself, but got lost
in a blizzard of algebra. If someone could tell me the correct
formula -- and better yet, steer me towards a derivation somewhere on
the Web -- that would be much appreciated.

FYI, the formula I arrived at for the output voltage is:
Vout = ( Vin / R2 ) / [ (R4/c1) * (1/R2 + 1/c2) + 1/c1 + 1/R2 ];

where cn = 1 / ( j * w * Cn )

(One other thing that would be useful -- if someone could take the
formulas above, and show me how to use Matlab to get the symbolic
result for w_cutoff. I got stumped on that one as well!)

Thanks in advance for all replies.

Steve O.


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[Ken Smith]

I understand correctly, the cutoff frequency is the frequency where
the output falls to 1/sqrt(2) = .707 of the input voltage. The
circuit is shown at:

Yes, that is how we almost always define it.

http://www.oppenheimercommunications.com/OpAmpProblem.htm

The measured cut-off frequency is around 300 Hz (1884 radians). And,
using the formula for Vout (which I do have), and plugging in various
values for w_in, I eventually find the cutoff frequency to be almost
exactly 310 Hz (1947 radians).

Yes, the cut off frequency you get is about right.

[....]

Hunting on the Web, I found what seems to be an identical looking
circuit, and the cutoff frequency is given by:
w_cutoff = 1 /sqrt (R2 * R4 * C1 * C2)

That isn't really the cutoff frequency. Try running it again with
C2=4700nF and see what you get. Then try 47000nF. If you do, you should
get a strong feeling for what that equation really gives you. Obviously,
it only gives you one number but this filter has two poles in it.

Since this one number is based on multiplying all the values together it
is reasonable that a second number, to describe this circuit, would be
based on the ratio of the values.

Try replacing the op-amp with an ideal voltage controlled voltage source
and making the C2=1000*C1 and then C2=10000*C1. You should imediately
suspect what the ratio does and with a little math be able to prove it.

When you now come back to the question of the 3dB point, you should be
able to quickly get the answer and better yet be prepared for other
circuits that do the same thing.

[...]

(One other thing that would be useful -- if someone could take the
formulas above, and show me how to use Matlab to get the symbolic
result for w_cutoff. I got stumped on that one as well!)

I think it is better if you do it by hand at least once. Learning to use
a program can be useful and also a trap. In doing it by hand, you have to
use and apply the theory of how to do it. Nothing sets something in your
mind quite so well as using it a couple of times.

 

[flank]

Try this matlab code. This example is for an RC circuit -1st order
filter. The "bode" command is very valuable for what you are doing.

good luck,
AG

% Filter Analysis using bode plots
% Given a series RC circuit, find the voltage
% across the capacitor as a function of frequency


clear all; close all; clc;

C=0.01e-6; % capacitor
R=10000; % resistor

% Vc(s) can be described as num/den
% num = 1/RC
% den = s + 1/RC

Vc=tf(1/(R*C),[1 1/(R*C)]); % transfer function of Vc(s)
bode(Vc);
title('Voltage across capacitor');
xlabel('Magnitude in dB');
ylabel('Frequency in radians per second');

% To get plots in absolute magnitude instead of dB

figure;
[mag,phase,w]=bode(Vc); % bode will output mag, phase and freq vector
mag=mag); % magnitude must be turned into a column
plot(w./(2*pi),abs(mag));
title('Voltage across capacitor');
ylabel('Absolute magnitude (V)');
xlabel('Frequency in Hertz');

% To select a specific range of input frequencies

figure;
w=2*pi*[10:100:10000]; %inputs from 10Hz to 10khz
[mag,phase]=bode(Vc,w);
mag=mag);
plot(w./(2*pi),abs(mag));
title('Voltage across capacitor, specified frequencies');
ylabel('Absolute magnitude (V)');
xlabel('Frequency in Hertz');