I would like to draw Bode plot for a complex pole as below. Let's assume that for now regardless that the single complex pole doesn't exist. This is the function:

$$H = \frac{1}{s - (a + jb)}$$ where \$a\$ and \$b\$ are real numbers.

Now I would like to draw Bode plot of this transfer function. First, substitute \$s\$ by \$j\omega\$ and we have:

$$H(j\omega) = \frac{1}{-a + j(\omega -b)}$$

From this I can calculate the magnitude and phase and draw Bode diagram.

I want to calculate break frequency (the frequency where the gain is 0.707 value of DC gain).

My questions are:

  1. Is that method correct to draw Bode plot of a single complex pole?

  2. Is my method of break frequency calculation correct?

  3. Assume that I have a second order transfer function as under-damped system. The transfer function has two complex poles. Does this mean it has two break frequencies?

Can I draw Bode plot of the under-damped system by plot Bode diagram of each single complex pole separately and then sum or subtract them?


1 Answer 1


Given that your first equation has the correct signs, \$ a\$ must be negative otherwise it's an unstable pole and \$ s\rightarrow j\omega\$ is not a valid operation, i.e. there is no steady state, therefore a frequency response does not exist.

Hence it would be better (less confusing) to let the complex pole be: $$ H(s)=\dfrac{1}{s+(a-jb)}\:, \:a>0$$

Now, determining the amplitude Bode plot for this isolated complex pole is mathematically simple, but in practice it's meaningless. Rather, you need to consider the frequency response of the complex conjugate pair.

To illustrate the problem, the DC gain of the single pole (obtained by taking \$\small s=0)\$ would be \$\small \dfrac{1}{(a-jb)}\$. A complex gain is not something that is physically realisable.

However, including the complex conjugate pole in the formulation would give a real DC gain of \$\small \dfrac{1}{(a^2+b^2)}\$, which complies with the 2nd order TF: \$\small \dfrac{1}{s+(a-jb)}.\dfrac{1}{s+(a+jb)}=\dfrac{1}{s^2+2as+(a^2+b^2)}\$

  • \$\begingroup\$ Thanks. For under-damped system, why there is only a single break frequency when the transfer function has two complex roots and each root creates a break frequency? \$\endgroup\$
    – emnha
    Nov 8, 2016 at 14:12
  • \$\begingroup\$ See update for more detail \$\endgroup\$
    – Chu
    Nov 8, 2016 at 14:39

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