# Bode Phase Plot for negative transfer function

My question is relatively simple. Supposing that we have a transfer function of the type $$\\frac{-Kj\omega}{\omega_0}\$$. Then for the phase Bode plot, I should start from $$\\pi\$$ and then add $$\\frac{\pi}{2}\$$ giving me a constant phase of $$\\frac{\pi}{2}\$$?

Thank you in advance(I don't know why latex doesn't work).

• You have to use \$ instead of $ for inline math, and  for their own paragraph. Jan 25, 2022 at 11:08

Your transfer function $$\H(j\omega) = \frac{-Kj\omega}{\omega_0} \$$ has a zero at $$\\omega=0\$$ and no poles.

In other words, your system has a pure differentiator and no integrators. Therefore, you should have constant phase $$\\angle H(j\omega)=90^\circ\$$ but due to the minus sign in your tf you obtain $$\\angle H(j\omega) = -90^\circ\$$.

By the way, your transfer function describes a non-causal system which is unrealizable in the real world.

• Yeah I know. My actual transfer function is much more complicated but the only point that I was unsure of was the minus sign. Thanks a lot Jan 25, 2022 at 12:07
• @GeorgiosDemeteiou You are welcome. If you feel you have received what you sought from one of the answers on this post then you could (and should for the record) select an answer by clicking on the checkmark.
– Carl
Jan 25, 2022 at 12:25

You could start from the definition of the phase:

$$\phi(\omega)=\arctan{\left[\dfrac{\Im\left(H(j\omega)\right)}{\Re\left(H(j\omega)\right)}\right]}=\arctan{\left(\dfrac{-K/\omega_0}{0}\right)}$$

Since that's absurd, the limit needs to be considered and that gives $$\-\pi/2\$$.

The argument of the constant $$\-\frac{K}{\omega_0}\$$ is $$\-\pi\$$ (just imagine where a negative real number lies on the Gaussian plane) whilst the zero in the origin, $$\j\omega\$$, gives you a constant phase of $$\+\frac{\pi}{2}\$$.

The total phase is $$\arg[-\frac{Kj\omega}{\omega_0}]=arg[-\frac{K}{\omega_0}]+arg[j\omega]=-\pi+\frac{\pi}{2}=-\frac{\pi}{2}\$$