# Bode plot of the transfer function with unstable pole

Let's say I have a LTI system described by following transfer function $$G(s) = \frac{(s+1)}{s(\frac{s}{10} - 1)}.$$ As soon as I run following Matlab code

s = tf('s');
G = (s + 1)/(s*(s/10 - 1));
bode(G);


The magnitude part is in accordance with my expectation (rough sketch created by hand). As far as the phase part there is a discrepancy between what Matlab gave me and what I expected. Namely I expected that until the first break point $$\omega = 1\,rad\cdot s^{-1}$$ the phase should be $$-90^{\circ}$$ due to the pole at the origin. It seems to me that there a phase shift of $$-180^{\circ}$$ but I don't understand why.

If you pop in a very small s you get

$$G(s) = \frac{(s+1)}{s(\frac{s}{10} - 1)} \approx\frac{1}{-s} = \frac{j}{\omega}.$$

That has indeed a phase of +90 or -270 degrees.

That has nothing to do with the pole location but it's a simple sign flip from:

$$G(s) = \frac{(s+1)}{s(1-\frac{s}{10})}$$

The transfer function phase is the numerator phase minus the numerator phase

The phase of the numerator can be expressed as $$atan(\omega)$$

For the denominator, it is a bit more complex. You need to expand the denominator to $$\frac{s^2}{10} - s$$

Replacing $$s = j\omega$$ in the denominator and you get $$-\frac{\omega^2}{10} - j\omega$$. For a positive value of $$\omega$$ this yields a complex number with negative real value and negative imaginary value there you cannot simply use Atan as the function is limited to $$±\frac{\pi}{2}$$. However, you can use atan and simply add $$\pi$$ to the denominator phase.

Therefore the denominator phase can be expressed as $$\pi + atan(\frac{10\omega}{\omega^2})$$ or $$\pi + atan(\frac{10}{\omega})$$

Finally the transfer function is the numerator phase minus the numerator phase which gives you

$$\theta(\omega) = Atan(\omega) - (\pi + (Atan(\frac{10}{\omega}))$$

For small values of $$\omega$$ , $$atan(\omega) \approx 0$$ and $$atan(\frac{10}{\omega}) \approx \frac{\pi}{2}$$.

Therefore, for small values of $$\omega$$ , $$\theta(\omega) \approx -\frac{3\pi}{2}$$

Edit : Thanks Matt. L for pointing out my original mistake.

• Note that for $\omega\to 0$ the result is not $-\pi/2$ but $+\pi/2$ (or, equivalently, $-3\pi/2$), as shown in the OP's plot. Dec 31, 2021 at 9:50
• Fixed it, I made a rookie mistake. We had this question on my final exam 20 years ago (which I got right). 20 year-old me is dissapointed!
– Ben
Dec 31, 2021 at 14:26