Bode Phase Plot for negative transfer function

Question

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).

Answer 1

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.

Answer 2

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\$.

Answer 3

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}\$