Let $\omega_0 \in \mathbb{R}$ and $\omega_0 > 0$. Be $G(s)$ a transfer function defined as:


We're interested into evaluating its phase response in a continuous LTI system. First, we split $G(j\omega)$ into its real and imaginary parts (assuming $\tau=\frac{\omega}{\omega_0}$):

$$G(j\omega) = \frac{1}{1+\tau^2}\left[1-\tau^2+j\left(-2\tau\right)\right]$$

Then using the definition of phase of a transfer function we have:

$$\angle G(j\omega)=\arctan\left(\frac{\mathbb{Im}\{G(j\omega)\}} {\mathbb{Re}\{G(j\omega)\}}\right)=\arctan\left(\frac{-2\tau}{1-\tau^2}\right)$$

So, for any $\tau \gg 1$ (which implies that $\omega \gg 1$) then the following should hold for Taylor's series:

$$\angle G(j\omega) \simeq \arctan\left(\frac{2}{\tau}\right) \simeq \frac{2}{\tau}$$ which approaches $0$ as $\omega$ gets bigger.

Here comes the discrepancy. Let $\omega_0 = 1$ (for plotting reasons), why does Wolfram Alpha give me the following Bode plot?

Wolfram Alpha Bode plot

Also, I expected one discontinuity point at $\omega = \omega_0 = 1$ (and another one for $\omega = -\omega_0$ not shown in the plot), but I guess I messed up something since the beginning.

2018-12-12 (in response to Sam F.) Here's how I've separated the real from the imaginary part of $G(j\omega)$.

$$ \begin{align} G(j\omega) &=\frac{1-j\frac{\omega}{\omega_0}}{1+j\frac{\omega}{\omega_0}} \\ &=\frac{1-j\tau}{1+j\tau} \\ &=\frac{1-j\tau}{1+j\tau}\cdot\frac{1-j\tau}{1-j\tau} \\ &=\frac{(1-j\tau)^2}{1^2-(j\tau)^2} \\ &=\frac{1^2+(j\tau)^2-2\cdot 1 \cdot j\tau }{1+\tau^2} \\ &=\frac{1- \tau^2+j(-2\tau)}{1+\tau^2} \\ \end{align} $$

  • $\begingroup$ @SamFarjamirad I've just added how I separated the real and the imaginary part of $G(j\omega)$. I'm stuck, where's the mistake? Thank you very much! $\endgroup$ Dec 12, 2018 at 9:44
  • $\begingroup$ $|G(j\omega)| = 1$ everywhere using $\mathbb{Re}\{G(j\omega)\}=1-\tau^2$ as it should be. I'm afraid I still don't get what you're trying to explain to me, my bad! (I expected two discontinuity points in the phase response, not the magnitude, forgot to say this before) $\endgroup$ Dec 12, 2018 at 9:57
  • $\begingroup$ Actually given my $\angle G(j\omega)$ there should be only one discontinuity point (a jump from $-\pi/2$ to $\pi/2$) in the phase response at $\omega = \omega_0 = 1$ because of the positive $\omega$ as you already noted. $\endgroup$ Dec 12, 2018 at 10:14
  • $\begingroup$ And I agree with you, just I don't get if the $\angle G(j\omega)$ I got was right and I am misreading it or if $\angle G(j\omega)$ is just wrong. $\endgroup$ Dec 12, 2018 at 10:29
  • $\begingroup$ What do you mean by computers take account with real position of functions? Thank you very much for your help! $\endgroup$ Dec 12, 2018 at 11:11

1 Answer 1


The plots are correct.

This is how we do it on our heads in about a minute after years of practice.

The magnitude:

Consider the numerator and denominator separately. Numerator rise with a slope of $20$ $dB/dec$, but at the same time denominator falls with the exact same slope, now it's clear that these two tend to cancel each other out, the results is a constant number. As you know the $20log1 = 0$ so it explains the magnitude plot. Notice if you look at the magnitude of the transfer function you'll see no real pole.

The phase

$arctan(*)$ is a continue function, at the break frequency it is equal to $-\frac{\pi}{2}$, as we plot the phase in logarithmic scale we have to add the phases. Both denominator and numerator drop $-\frac{\pi}{2}$ over the whole frequency band. So $-\pi$ or 180°.


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