When we analyse electric circuits we often use transfer functions. To calculate the poles and zeros of such a function can be done in different ways. When we look to a transfer function in the Laplace domain, it looks like:


Now, if we want to set it in the complex frequency domain we substitute:


Where $$j^2=-1$$.

Now we substitute that into the Laplace domain transfer function:


To calculate the (absolute value of the) poles and zeros of the original transfer function (in the Laplace domain) we can solve:

 1. When we have one capacitor or inductor (called cutoff frequency): $$\Re\left[\text{H}_{\text{T}}\left(j\omega\right)\right]=\Im\left[\text{H}_{\text{T}}\left(j\omega\right]\right]$$  2. When we have more capacitors or inductors or a combination of these two (called resonance frequency): $$\Im\left[\text{H}_{\text{T}}\left(j\omega\right)\right]=0$$

Question: For which transfer functions does this hold, that you can calculate the poles and zeros of the Laplace domain transfer function, using the complex frequency domain as stated above? Because it does not work for all transfer function? And how can we prove that it only hold for some transfer functions?

Circuits where it work, for example a simple RC series circuit. Because:




They give the same result (poles and zeros):





When I have a series RRL circuit, my transfer function looks like:


Finding the poles and zeros of that function gives my different values for \$ \omega \$ then when I solve:




  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Dave Tweed
    Nov 16, 2016 at 12:04

2 Answers 2


Equating real with imaginery locates the turning frequencies of the system and this hints at the poles. In the example of an RL high pass filter but with an extra resistor in series with the inductor, there are two frequencies of interest.

The first (lower) is when the inductor's impedance equals that of R2 (the extra resistor in series with L) and this marks the point at which output starts to rise with frequency.

The 2nd frequency of interest is equivalent to the original -3 dB point where the |reactance| of L equals that of R1.

Now, when you do the math and equate real and imaginery, sure you can solve it - it's a bit of a complicated quadratic and, of course with a quadratic, it produces two values for \$\omega\$ and these are the turning frequencies.

So, what is the problem?

  • \$\begingroup\$ The problem is: I dont get the same values for omega When I find the poles and zeros and set real to imaginary \$\endgroup\$
    – kloepas
    Nov 18, 2016 at 17:34
  • \$\begingroup\$ Well, you ought to show your final maths \$\endgroup\$
    – Andy aka
    Nov 18, 2016 at 17:34
  • \$\begingroup\$ Try it for yourself, set values for R1 and R2 and L and compute the pole and zero and compute real equals imaginary and you find two different values \$\endgroup\$
    – kloepas
    Nov 18, 2016 at 17:44
  • \$\begingroup\$ Because there are two different values for omega! \$\endgroup\$
    – Andy aka
    Nov 18, 2016 at 17:46
  • \$\begingroup\$ Yes indeed, for the pole one the for the zero one but for the real equals imaginary we get also two different so we get 4 different values for omega \$\endgroup\$
    – kloepas
    Nov 18, 2016 at 17:47

Both terms "pole" and "zero" are defined in the s-domain only. In some simple cases (first-order functions) you can observe an identity between these terms and the 3dB frequencies in the jw-domain. Example: The pole frequency is found equalizing the real and imag. parts of the denominator.

Please note that you have equalized the real and imag. parts of the whole transfer function (instead of the denominator). This approach gives the 3dB frequency only if (1) the transfer function is of 1st order and (2) if the numerator is real. It is very easy to demonstrate these restrictions. In your example, the numerator is also complex and the real/imag. parts of the denominator and the whole function will be different.

For all classical 2nd-oder functions (filter circuits) we have different values for pole and 3dB frequencies (exception: Butterworth response).


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