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The mathematics result does not match with simulator results.

The transfer function of a first order passive low pass filter is:

$$\frac{V_{out}}{V_{in}} = \frac{1}{(sCR + 1)}$$

If I have two stages and I need to calculate the phase shift of the output. All the resistors in the circuit are 1000 ohm. All capacitors are 100 nF. operating frequency is 1000 Hz (sine wave).

I think the new transfer function would be:

$$\frac{V_{out}}{V_{in}} = \frac{1}{(sCR + 1)^2}$$

Now I substitute the values of the circuit in the transfer function where,

$$s = j \times 2\pi f$$

The result was the phase shift = - 64 deg.

When I simulated the circuit on Protues, The phase shift was about - 72 deg.

This website says the result is -72 deg.

Why mathematical result is different from the result of simulator and the calculator of website? Am I missing something?

I used the arg function of my calculator to get the phase shift form the complex transfer function.

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    \$\begingroup\$ If that SCR is meant as Laplace operator then you better change to sRC ot sT. If you connect two RC filters in series the lapalace function is not multiplied, as they have mutal relationship. You should place an active element like op amp to isolate both elemnts, then you get F(s)^2 \$\endgroup\$ – Marko Buršič Feb 29 '16 at 16:36
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You have come up with an equation without thinking much about the circuit. When you have two RC pairs cascaded, as a passive filter, you will have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The transfer function for this circuit can be obtained using Kirchoff's as follows:

$$\frac{V_x - V_{in}}{R_1} + sC_1 V_x + \frac{V_x - V_{out}}{R_2} = 0$$ $$ \frac{V_{out} - V_x}{R_2} + sC_2 V_{out} = 0$$

Simplification of these equations leads to:

$$V_x = \frac{R_1 R_2}{sR_1 R_2 C_1 + R_1 + R_2}\left(\frac{V_{in}}{R_1}+\frac{V_{out}}{R_2}\right)$$ $$V_{out} = \frac{V_x}{sR_2 C_2 + 1}$$

You can see that Vout behaves regarding to Vx as you would expect with your RC transfer function. But the output of the first RC pair relates to all four components. This reaches the point to be made: the only situation in which your equation for a cascaded second-order filter would apply is if the second RC pair had infinite input impedance. Vx would then "see" the second pair as an open circuit.

That can be achieved by means of active components like an ideal Op-Amp. Check out the following circuit:

schematic

simulate this circuit

Now the Op Amp presents a high impedance to the first RC pair and your proposed equation stands for no output load. It would be dumb to add a second Op Amp to get around this, so in "real-world" applications, second order RC filters are not setup like this, but follow well known topologies like Sallen-Key.

schematic

simulate this circuit

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Your equation is for two ideal low-pass filters in series. The resistor/capacitor realisation is not ideal you have a finite input and output impedance.

To get a better approximation of the ideal behavior you have to make the value of the resistor in the second stage larger compared to the resistor in the first stage. Don't forget to scale the capacitor accordingly.

Alternatively you could add a buffer to the output of the first filter, just to see the difference.

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This is a typical example in which there is no need to write a single line of algebra. Determine the time constants of this 2-energy-storing element network (2nd order) and you are done. The denominator follows the form \$D(s)=1+sb_1+s^2b_2\$. For \$s=0\$, open the caps and determine the gain \$H_0\$. Here it is one as nothing loads the network. Then, reduce the excitation to 0 V (replace the \$V_{in}\$ source by a short circuit) and determine the resistance driving each capacitor. Sum up the these time constants, you have the term \$b_1\$. Then short \$C_1\$ and determine the resistance driving \$C_2\$ in this mode. Finally, assemble the whole thing as

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

The below picture guides you in these simple steps:

enter image description here

You can then apply the low-\$Q\$ approximation considering well-spread poles and you can nicely factor the final expression as shown below:

enter image description here

In particular with passive circuits, complex polynomial expressions can be determined by inspection without writing a single line of algebra: just draw small sketches and determine the \$a_i\$ and \$b_i\$ terms for \$N\$ or \$D\$ individually. Here it was easy, there was no zero. If you see a mistake, just correct the guilty term without re-starting from scratch. It complicates a bit with controlled sources but the spirit remains the same. If you want to know more about FACTs, have a look at the seminar taught at APEC 2016

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

but also the numerous transfer functions derived in the book

http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf

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  • \$\begingroup\$ But from where did you know for example that wp1 = 1/b1 and wp2 = b1/b2 ?? What is the origin of this? \$\endgroup\$ – G36 May 28 '17 at 15:30
  • \$\begingroup\$ Hi G36, this is when you apply what is called "the low-Q approximation". In a second-order network, when Q is high, you have two conjugate roots. Reduce Q to 0.5 and these roots become coincident. Then further reduce Q to a low value and these roots or poles spread apart with one dominating at low freq (1/b1) and the other one being in higher frequency (b1/b2). Actually, you have wp1=W0*Q and Wp2=W0/Q. Replace W0 and Q by their definition in D(s)=1+s.b1+s^2.b2=1+(s/W0.Q)+(s/W0)^2 and you will find my definitions. You can even generalize the formula for higher orders with well-spread poles. \$\endgroup\$ – Verbal Kint May 28 '17 at 15:50

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