0
\$\begingroup\$

For first order systems, its easy to get the time constant by using thevenin concept.

Suppose i have a second order system ,for example consider a two stage cascaded LOW PASS FILTER with identical resistance and capacitor values.

enter image description here

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

However,

My guess is that the transfer function of desired output with respect to 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, will the poles change, i think it won't but i am not able to convince myself.

\$\endgroup\$
  • 1
    \$\begingroup\$ Are you able to convince yourself that now is a good time to ask a question? \$\endgroup\$ – Andy aka Nov 30 '15 at 15:42
  • \$\begingroup\$ yea :P i am able to \$\endgroup\$ – Ashik Anuvar Nov 30 '15 at 15:43
  • \$\begingroup\$ So, feel free to ask one... \$\endgroup\$ – Brian Drummond Nov 30 '15 at 15:44
  • \$\begingroup\$ please help me with how you evaluate the time constants in second order case \$\endgroup\$ – Ashik Anuvar Nov 30 '15 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 Nov 30 '15 at 15:52
1
\$\begingroup\$

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.

\$\endgroup\$
1
\$\begingroup\$

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.

\$\endgroup\$
  • \$\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 Nov 30 '15 at 16:49
  • 1
    \$\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 Nov 30 '15 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.