In my previous question, I've asked something about acut-off frequency (Finding the cut-off frequency).

But in my plot I don't see those, I see the one that I find when I pick the zero and the pole ones. Is there a way to proof that we don't have a cut-off frequency in my question?

Thanks in advance

  • \$\begingroup\$ You have one zero and one pole, so you know that as \$f\to\infty\$ the frequency response will be flat (as far as this model is concerned). \$\endgroup\$
    – The Photon
    Dec 2, 2015 at 18:29
  • \$\begingroup\$ And also as \$f\to{}0\$. \$\endgroup\$
    – The Photon
    Dec 2, 2015 at 18:31

1 Answer 1


I don`t know exactly what your problem is. However, perhaps the following clarifies something:

With your previous question you gave the transfer function which, however, was not yet shown inthe so called "normal form", which means that the denominator of the transfer function should be D(jw)=1+jw*b+(jw)²c.....

In your case, we simply get by dividing the whole function with (R1+R2) :

H(jw)=N(jw)/D(jw) with

N(jw)=[R2/(R1+R2)][1+jwL/R2) and


From these expressions you immediately can derive that there is one zero at wo=R2/Land a pole at wp=(R1+R2)/L.

This pole is not really identical to the 3dB cut-off but very close to the value as given elsewhere in an answer to this thread (fp=175kHz).


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