# 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! – Giulio Scattolin Dec 12 '18 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) – Giulio Scattolin Dec 12 '18 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. – Giulio Scattolin Dec 12 '18 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. – Giulio Scattolin Dec 12 '18 at 10:29
• What do you mean by computers take account with real position of functions? Thank you very much for your help! – Giulio Scattolin Dec 12 '18 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°.