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.
Best Answer
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:
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:
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.
simulate this circuit