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));

I receive following Bode plot

enter image description here

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.


2 Answers 2


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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – Matt L.
    Dec 31, 2021 at 9:50
  • 1
    $\begingroup$ 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! $\endgroup$
    – Ben
    Dec 31, 2021 at 14:26

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