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For Second-Order RC filter the cutoff frequency is: $$fc = \frac{1}{2\pi \sqrt {R_1C_1R_2C_2}}$$

I have taken R1 = R2 = R = 1K Ohm and C1 = C2 = C= 0.1591 micro Farad. So the cutoff frequency comes out to be 999.717 Hz or approximately 1KHz.

Now according to this article the gain of the n-order filter at its cutoff frequency is given by: $$\left(\frac{1}{\sqrt 2}\right)^n$$

It is a second order filter(n=2) so the gain should be -6dB (and similarly for 3rd order filter it should be -9dB). But when simulating the filter in LTspice it shows differnt result.

Filter Circuit: enter image description here

LTSpice Bode Plot: enter image description here

Here in the plot, it can be seen that at the cutoff frequency, at v(p001) the gain is -6dB and at V(p002) the gain is -9dB. But according to the theory, the gain at V(p002) should have been -6dB.

Also, according to its transfer function $$H(s)=\frac{1}{(sRC)^2+s(3RC)+1}$$ the gain at cutoff frequency comes out to be -19.08dB

Three different results

  1. -6dB according to the formula

  2. -9dB in LTspice simulation

  3. -19dB according to the transfer function.

Why is it so? What am I doing wrong?

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    \$\begingroup\$ Your linked article has clear point where its quality collapses. The theory of one R one C lowpass filter is ok, but the handling of cascaded filters is lousy - do not believe it! It presents true formulas which can be used somewhere, but are useless in your case. There's a nice excuse: "But there is a downside too cascading together RC filter stages. Although there is no limit to the order of the filter that can be formed, as the order increases, the gain and accuracy of the final filter declines." User Bimpelrekkie seemingly already has spotted one useless formula. \$\endgroup\$ – user287001 Sep 18 '18 at 8:34
  • \$\begingroup\$ .AC analysis in LTspice is very fast, and there's no reason for not using more points per decade/octave/etc. For your case, use at least 100, 1000 recommended for better resolution. You can also use the arrow keys to move the cursor from point to point, which will give you a precise 1kHz reading. Also, I don't know how you got the 19dB for the TF, I get 0.33323898 (9.5448dB), on point with LTspice with 1000 points (333.239mV/9.5448dB). \$\endgroup\$ – a concerned citizen Sep 18 '18 at 11:23
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Why is it so? What am I doing wrong?

I would like you to explain where you got the formula:

$$fc = \frac{1}{2\pi \sqrt {R_1C_1R_2C_2}}$$

That formula applies to a Sallen-Key second order filter but that's not what you made. You simply cascaded (connected one after the other) two first order RC lowpass filters. The way you did that (with no buffering in between) means that you cannot simply apply the formulas for a first order filter and square it.

the gain at cutoff frequency comes out to be -19.08dB

You cannot define the cutoff frequency like that, for a low pass filter the gain at the cutoff point is by definiton -3 dB. So you find the -3 dB gain point and the corresponding frequency, that frequency is the cutoff point. That also means that if you cascade two first order lowpass filters (with a buffer in between so the transfer functions can simply be multiplied) that the cutoff point will move to a lower frequency.

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  • \$\begingroup\$ Ok, I understand that the formula is wrong and cutoff frequency is wrongly calculated in this case. But, when calculating the gain of the filter using the transfer function at w = 1/(RC) ( = 1KHz) comes out to be -19dB but in LTSpice it shows -9dB. Why so? \$\endgroup\$ – Keestu Sep 20 '18 at 3:57
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This type of passive circuit can be easily solved and expressed in a so-called low-entropy format using the fast analytical circuits techniques or FACTs. Without writing a single line of algebra, you can "inspect" the circuit and determine the transfer function. In this approach, you determine the natural time constants of the circuit by reducing the stimulus (\$V_{in}\$ to 0 V. When you do that, the left terminal of \$R_1\$ is grounded. In this configuration, remove the capacitors and "look" at the resistance from their terminals. That resistance multiplied by the capacitance forms the time constant \$\tau\$ we need. Here, we have two energy-storing elements (with independent state variables) so this is a 2nd-order circuit obeying the following expression for the denominator \$D(s)\$:

\$D(s)=1+s(\tau_1+\tau_2)+s^2\tau_1\tau_{12}\$

We start with \$s=0\$ for which you open all caps. The transfer function is simply:

\$H_0=1\$

If you apply these techniques, you should find:

\$\tau_1=C_1R_1\$

\$\tau_2=C_2(R_1+R_2)\$

\$b_1=\tau_1+\tau_2=C_1R_1+C_2(R_1+R_2)\$

Then, consider shorting \$C_1\$ while you look at the resistance offered by \$C_2\$ terminals in this mode. You have

\$\tau_{12}=C_2R_2\$

\$b_2=\tau_1\tau_{12}=C_1R_1C_2R_2\$

Assembling these expressions, we have the complete transfer function as there is no zero in this network.

\$H(s)=H_0\frac{1}{1+s(C_1R_1+C_2(R_1+R_2))+s^2(C_1R_1C_2R_2)}\$

This is a second-order polynomial form obeying:

\$H(s)=H_0\frac{1}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}\$

in which \$Q=\frac{\sqrt{b_2}}{b_1}\$ and \$\omega_0=\frac{1}{\sqrt{b_2}}\$

If \$Q\$ is sufficiently low (low-\$Q\$ approximation) you can replace the second-order polynomial form by two cascaded poles.

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