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For first order systems, it's easy to get the time constant by using the Thevenin concept.

Suppose I have a second-order system, for example a two-stage cascaded low-pads filter with identical resistance and capacitor values.

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I am sure that this system will have more than 1 time constant.

However, my guess is that the transfer function of the desired output with respect to the input voltage will have 2 poles and the reciprocal of these poles will give the time constants.

I am confused whether depending on the output of interest the poles will change. I think they won't, but I am not able to convince myself.

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  • \$\begingroup\$ please help me with how you evaluate the time constants in second order case \$\endgroup\$
    – Ashik Anuvar
    Commented Nov 30, 2015 at 15:45
  • \$\begingroup\$ If all components were impedances Z(f) would you be able to make an expression for this circuit ? Now fill in Z(f) = R for the resistors and Z(f) = 1/jwC for the capacitors. Then group the equation such that you have RC products. Since both RC influence each other you might not get "nice" timeconstants. For that you need to isolate them from each other. \$\endgroup\$
    – Bimpelrekkie
    Commented Nov 30, 2015 at 15:52

2 Answers 2

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The poles of a system do not depend on the selected output node and, more than that, the also do not depend on the selected input node - as long as you select a node that was at ground potential before.

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Start with naming the middle node Vx, then you can say that: -

\$V_{OUT} = V_X \dfrac{1}{1+sCR}\$

You then have to find Vx in terms of Vin - this is the start I'd consider making but, be aware, the algebra becomes more complex because in finding Vx the component to ground that Vx develops across is C in parallel with R and C in series.

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  • \$\begingroup\$ my question is whether depending on the output of interest will the time constants change because you are prone to get a different transfer function with respect to input \$\endgroup\$
    – Ashik Anuvar
    Commented Nov 30, 2015 at 16:49
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    \$\begingroup\$ Apart from 0V there are three nodes and Vx is one of them. If the input were swapped with the node on C1, C1 would fall out of the reckoning for poles so that's dealt with that one because placing the output at any other relevant node is a trivial calculation. Placing Vin where Vx currently is makes a single order system and, again, this is trivial. The only difficult-to-decipher setup is where Vout is across the righthand capacitor and Vin is to the far left. Use maths to make comparisons. That's what I'm trying to encourage. \$\endgroup\$
    – Andy aka
    Commented Nov 30, 2015 at 20:48
  • \$\begingroup\$ @Andyaka I did the math in this more intricate case where Vx is the middle node and indeed it was a bit more complicated. If my math is correct, H(s=0) = 0.5 and H(s=inf) = 0, which does look like a low pass filter as we expected from two cascaded LPF. HOwever I have completely forgotten how to examine what happens in the middle given my two calculated poles which look ugly. Is it as simple as having a 20dB/dec drop after the first and a 40dB/decade after the second? I used to be able to remember drawing bode plots...sigh \$\endgroup\$
    – Geo
    Commented Dec 15, 2020 at 2:07
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    \$\begingroup\$ @George above the "middle point" i.e. in the roll-off area, the response will be 40 dB/decade. \$\endgroup\$
    – Andy aka
    Commented Dec 15, 2020 at 9:44

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