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I am a bit confused with Laplace domain and its equivalent time domain conversion

Consider the s-domain of first order LPF filter which is $$\frac{V_o(s)}{V_i(s)}=\frac{1}{1+sRC}$$

Now for a second order LPF filter in s-domain is simply the multiplication of the transfer function by itself i.e $$\frac{V_o(s)}{V_i(s)}=\frac{1}{(1+sRC)^2}$$ The implmentation of such a transfer function with resistor and capacitor are two RC filters cascaded like shown in the figure

enter image description here

Now for analysis of the above implemented filter in time domain, considering a step input the analysis of this filter is \$V_1(t)/V_{\text{in}}(t)=1-e^{-(t/RC)}\$ and \$V_o(t)/V_1(t)=1-e^{-(t/RC)}\$, and hence \$V_o(t)/V_{\text{in}}(t)=(1-e^{-(t/RC)})^2\$

But in time domain the multiplication of Laplace domain transfer function should be convolution, yet the second order RC filters are implemented as multiplications. Also the Laplace transform of \$V_o(t)/V_{\text{in}}(t)=(1-e^{-(t/RC)})^2\$ is not \$V_o(s)/V_i(s)=1/(1+sRC)^2\$

What am I missing here?

Here is an exercise I tried. Assuming \$V_i(t)=u(t)\$, unit step function, which in the \$s\$ domain is \$1/s\$, the Laplace transfer function for a first order LPF is \$V_o(s)=V_i(s)\times 1/(1+sRC) = V_o(s)=1/s(1+sRC)\$. The inverse Laplace of this function is \$V_o(t)=u(t)\times (1-e^{-t/RC})\$. This checks out which I verified in Matlab in the time and \$s\$ domains.

Now for the second order LPF and step input with a buffer in between like in the circuit by MatteoRM: the Laplace transform \$ V_o(s)=1/s(1+sRC)^2\$ right? If I follow the same exercise as before, the inverse Laplace is \$1 - (te^{(-t/(RC))})/RC) - e^{(-t/(RC))}\$. Now this does not check out in time domain. Again, what am I doing wrong?

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2 Answers 2

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At first: the next formulas

enter image description here

are conceptually crap although in math they can be true when the denominators Vin(t) and V1(t) are exactly =1 and the right sides happen to present the nominators. You should write V1(t)=something, Vo(t)=something.

The circuit error: RC lowpass filters have their well known step responses only in case there's nothing connected to the output, at least everything which takes some current like another RC filter are forbidden OR the whole transfer function should be recalculated from the very beginning for the whole circuit. Some operational amplifier circuits can be cascaded without this problem because they have stiff outputs which do not drop if there's some reasonable loading. With them you can multiply the s-domain transfer functions.

Then the most fatal error: Multiplying step responses to get the step response of a cascaded circuit is your own unique poetry, it's pure nonsense in math which you probably have catched from nowhere because it felt comfortable. Laplace domain transfer function multiplication is meaningful, but only if the circuits do have stiff (=low impedance) outputs so that the transfer functions stay no matter is there a load or not.

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  • \$\begingroup\$ Well even if I write the response in the format Vo(t) = something, Vo(t)= Vin(t)*(1-e^(-t/rc))^2 right? Multiplying step responses to get the step response of a cascaded circuit is your own unique poetry, it's pure nonsense in math which you probably have catched from nowhere because it felt comfortable--> do you think this circuit wont work in reality when a step input is given? Offcourse there will be attenuation at the output \$\endgroup\$
    – RAN
    Commented Apr 2, 2020 at 22:14
  • \$\begingroup\$ NO!!!!!!!!!! The only facts are 1) When a single RC lowpass filter gets a step input, the output voltage = 1-exp(-t/RC) and 2) When a filter which is 2 identical LC lowpass filters cascaded the step response is something else than (1-exp(-t/RC)^2, that something else must be calculated by starting from the basic circuit laws, it cannot be derived from 1-exp(-t/RC) because that step response is valid only for a non-loaded filter. The difference is more complex than pure attenuation. \$\endgroup\$
    – user136077
    Commented Apr 2, 2020 at 22:21
  • \$\begingroup\$ Ok now i understand why this circuit wont work for a step response. But this is still an RC LPF of second order right? Especially for sinusoidal signals which will be attenuated after the cut off frequency? or is that not true either? \$\endgroup\$
    – RAN
    Commented Apr 2, 2020 at 22:55
  • \$\begingroup\$ This is 2nd order LPF although quite ineffective because RC circuits cannot create steep frequency response curves. -3dB cutoff frequency exists, but it is at different frequency than in a single R single C LPF. The frequency response of cascaded version cannot be calculated as 2nd power of the single RC frequency response. \$\endgroup\$
    – user136077
    Commented Apr 2, 2020 at 23:56
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I think you need two suggestions:

1)Pay attention to the fact that a cascade connection of two low pass passive filters is not the multiplication of the transfer functions. It is true just in this case, with a buffer:

schematic

simulate this circuit – Schematic created using CircuitLab

In your circuit, since the output impedance of the first filter is similar to the input impedance of the second filter something happens in the middle of the circuit.

2)As you can check here:

https://www.wolframalpha.com/widgets/gallery/view.jsp?id=1f9f9d8ff75205aa73ec83e543d8b571

the inverse Laplace trasformation of \$ \frac{V_O}{V_{in}}(s)=\frac{1}{(1+sRC)^2}\$ is not what you expect.

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  • \$\begingroup\$ Right, this makes sense, i do have more questions to ask, but it is pretty late here, thank you for your reply \$\endgroup\$
    – RAN
    Commented Apr 2, 2020 at 23:04
  • \$\begingroup\$ I verified the same in Matlab, for some reason the second order transfer function does not work as expected. I have edited the question to include the this exercise \$\endgroup\$
    – RAN
    Commented Apr 3, 2020 at 12:09
  • \$\begingroup\$ Also what is according to you the s domain and time domain transfer function of the circuit you have drawn \$\endgroup\$
    – RAN
    Commented Apr 3, 2020 at 12:10

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