# Bode diagram of a single complex pole and under-damped system

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?

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: $$H(s)=\dfrac{1}{s+(a-jb)}\:, \:a>0$$
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: $\small \dfrac{1}{s+(a-jb)}.\dfrac{1}{s+(a+jb)}=\dfrac{1}{s^2+2as+(a^2+b^2)}$