# Phase response of a continuous transfer function

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

$$G(s)=\frac{1-\frac{s}{\omega_0}}{1+\frac{s}{\omega_0}}$$

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?

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}

• @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! Dec 12, 2018 at 9:44
• $|G(j\omega)| = 1$ 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) Dec 12, 2018 at 9:57
• 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. Dec 12, 2018 at 10:14
• 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. Dec 12, 2018 at 10:29
• What do you mean by computers take account with real position of functions? Thank you very much for your help! Dec 12, 2018 at 11:11

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.
$$arctan(*)$$ 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°.