I do not understand the logic behind this restriction. What would go wrong if this condition is not fulfilled?

E.g, in frequency domain \$\dfrac{1}{(s+1)^2}\$ which has an inverse Laplace of \$t \cdot e^{-t}\$ which seems pretty bounded and stable but the above mentioned condition violates in this network function.

  • \$\begingroup\$ Air - your question has nothing to do with stability (don`t mix it with transfer functions). In nearly all cases, the driving point function is identical to the operational input impedance/admittance. Hence, in case the inductive component dominates for rising frequencies, the function can be "unbounded". \$\endgroup\$
    – LvW
    Mar 17, 2016 at 14:59
  • \$\begingroup\$ Air - do you know any circuit with real components having a driving point function as given in your example? \$\endgroup\$
    – LvW
    Mar 17, 2016 at 15:35
  • \$\begingroup\$ Air - are you still interested in the problem? \$\endgroup\$
    – LvW
    Mar 18, 2016 at 7:04
  • \$\begingroup\$ No, I don't think I have come across a circuit like such yet. Maybe there does exist such a circuit. \$\endgroup\$ Mar 18, 2016 at 16:59
  • \$\begingroup\$ No - in any case, for frequencies approaching infinity only one of the three alternatives (capacitice, inductive, resistive) can determine the frequency-dependence(for real circuits with parasitics: always capacitive). Can you imagine any input impedance with a frequency-dependence of second order? No - it is not possible. \$\endgroup\$
    – LvW
    Mar 18, 2016 at 17:28

1 Answer 1


For very high frequencies the driving point admittance/impedance (real and positive) primarily can be only (a) inductive, (b) capacitive or (c) resistive. For this reason, the difference in degree between N(s) and D(s) can be at most "1".


Remember: The degree of a function is determined by the highest exponent (s³ gives third degree). And - on the other hand - for very large frequencies it is the highest degree which dominates. For this reason, the answer is derived from circuit properties for frequencies approaching infinity.

Because for very large frequencies (infinity) the function must be of first order only (capacitive or inductive), the degree of N(s) and D(s) must be differ by "1" only (examples: s³/s² or s²/s³).

  • \$\begingroup\$ This answer deserves more explanation and detailling \$\endgroup\$
    – MaximGi
    Mar 17, 2016 at 12:32
  • \$\begingroup\$ This explanation can be found in many textbooks about system theory - and it is logical. Where is the answer from the downvoting person? If you make up your mind, you will find out that for frequencies approaching infinity there are only two alternatives for the characteristics of input impedances/admittances (capacitice or inductive). The third option (resistive) is only of theoretical interest. \$\endgroup\$
    – LvW
    Mar 17, 2016 at 14:31
  • \$\begingroup\$ You are not fully answering the question, but just giving a vague hint. Nothing personnal about it, no need to be condescending. This site is about taking your time sharing knowledge and provide complete and accurate answers, not rushing to dump incomplete answers ("it's in the textbooks" is not an answer, btw). I wanted to tell you nicely, you decided to be stubborn, that's your choice. \$\endgroup\$
    – MaximGi
    Mar 17, 2016 at 14:49
  • \$\begingroup\$ Maxim- what is YOUR anwer? If you make up your mind you will find out that there are just two (realistic) alternatives: capacitive or inductive. This has nothing to do with "stubborness" - that is the logical consequence of circuit theory. Are you able to give any counterexample? If not - you do not understand the question. \$\endgroup\$
    – LvW
    Mar 17, 2016 at 14:55
  • \$\begingroup\$ May I add: If somebody is able to qualify a contribution as "not fully answering the question" he should be able to give a better (complete) answer, am I wrong? \$\endgroup\$
    – LvW
    Mar 17, 2016 at 15:05

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