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:

$$\text{H}_{\text{T}}\left(\text{s}\right)=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}$$

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

$$\text{s}=j\omega$$

Where $j^2=-1$.

Now we substitute that into the Laplace domain transfer function:

$$\text{H}_{\text{T}}\left(j\omega\right)=\frac{\text{Y}\left(j\omega\right)}{\text{X}\left(j\omega\right)}$$

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:

$$\text{H}_{\text{T}}\left(\text{s}\right)=\frac{\frac{1}{\text{RC}}}{\text{s}+\frac{1}{\text{RC}}}$$

And:

$$\text{H}_{\text{T}}\left(j\omega\right)=\frac{\frac{1}{\text{RC}}}{j\omega+\frac{1}{\text{RC}}}$$

They give the same result (poles and zeros):

$$\left|\text{s}\right|=\frac{1}{\text{RC}}$$

And:

$$\Re\left[\text{H}_{\text{T}}\left(j\omega\right)\right]=\Im\left[\text{H}_{\text{T}}\left(j\omega\right)\right]\space\Longleftrightarrow\space\omega=\frac{1}{\text{RC}}$$

If $H(s)$ is a rational transfer function then the poles are the roots of the denominator, or equivalently the characteristic values of the system matrix. It doesn't change if you substitute $j\omega$ for $s$--you are just either (1) taking the Fourier transform, or (2) enforcing the fact that $s$ is assumed to have no real part when doing so. The idea of ''poles'' strictly only works for rational transfer functions (e.g. what are the poles of $e^{-sT}$?) but typically just refers to any roots of a polynomial denominator if present. For linear systems, poles contain all information about system stability and even a significant amount of information about the transient response.

  • Ok, but why does it not work for the function I stated in my question?? – kloepas Nov 18 '16 at 16:49

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