I'm currently an undergraduate electrical engineering student who recently took an elective mathematical course on complex analysis, where we learned about all kinds of singularities for complex numbers, like poles.

This semester, in a circuit analysis course, we are finding the poles of transfer functions, but the concepts don't make sense in my head.

Using an easy example (from Wikipedia), say we have a transfer function of

\$H(s) = \frac{1}{1 + RCs}\$, where they say that the pole is at \$s = -\frac{1}{RC}\$, which, sure, works out at the moment. However, \$s = j\omega\$, so equating the two,

\$-\frac{1}{RC} = j\omega\$

\$\frac{j}{RC} = \omega\$

But now this means that our angular frequency, \$\omega\$, is imaginary, which I thought was not possible due to \$\omega\$ being a real number, as the frequency of a circuit would only be real (please let me know if my assumption is wrong).

So basically, I don't understand how there can possibly be a pole for that transfer function, as we have a constant non-zero real part on the denominator, accompanied by a variable imaginary term, \$j\omega\$. Given that I am assuming that \$\omega\$ can only ever be real, then the denominator can never approach 0, thus not making it a pole.

Have I made an incorrect assumption somewhere, is my understanding flawed, or is there a mismatch between mathematics and electrical engineering in what a "pole" is?


3 Answers 3


The problem is that you equate \$s\$ with \$j\omega\$, which - in the context of poles and zeros of a transfer function - does not make sense. In general, \$s=\sigma+j\omega\$ is a complex variable, and for the example that you gave the pole \$s_{\infty}=-1/RC\$ is purely real. For a system to be causal and stable (i.e. at least in theory realizable), all poles of the corresponding transfer function must lie in the left half plane of the complex \$s\$-plane.

If you have a transfer function \$H(s)\$ of a stable system, then you can evaluate its frequency response by setting \$s=j\omega\$ to obtain \$H(j\omega)\$. But then you're only talking about the frequency response of the system, and not anymore of poles and zeros of \$H(s)\$. There can be no poles on the imaginary \$j\omega\$ axis if the system is stable. Of course there can be zeros on the imaginary axis. These are the frequencies that are completely suppressed by the system.

  • \$\begingroup\$ I understand OP question, but literally not a single line in your answer. \$\endgroup\$
    – KMC
    Oct 29, 2020 at 18:22

I agree with everything Matt L. has written. Nevertheless, I will try a different kind of explanation.

1.) If we would be interested in the frequency response of such a circuit only there is no necessity to use complex frequencies. In this case, \$s\$ is simply an abbreviation for \$j\omega\$ - nothing else.

2.) It is to be noted that the complex frequency \$s = \sigma +j\omega\$ is based on a pure mathematical definition - such a frequency does NOT exist in reality. It is, however, very convenient to work with this complex quantity.

3.) This has several reasons - one of which is to describe frequency-dependent circuits using pole-zero locations in the complex frequency plane. This leads - for example - to the situation that the magnitude of the system's transfer function in the complex plane approaches infinity at the (fictive and complex) pole frequency. Of course, such a situation is not possible in reality. However, in reality setting \$s = j\omega\$ we have at this point something like a corner frequency (for a first order system it is the 3dB frequency).

4.) Summary: It is very convenient to work with complex frequencies - in particular for describing/specifying the properties of frequency-dependent circuits. In this context, it is to be noted that the solution of the differential equation in the time domain for a frequency-dependent system gives a "characteristic polynomial" with the complex variable \$s\$, which is identical to the denominator of a circuits transfer function. Hence, this complex variable \$s\$ occurs "automatically" during the process of solving the differential equation.


The concept of negative frequency is an interesting thing to explore. If you are interested you can refer to B.P.Lathi book on communication systems.


The frequency response of modulated signals are considered to be having both positive and negative contents. It is because the exponential notation of the sinusoids are a handy tools in signal analysis.


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