enter image description here

The above graph is pulled from the book "Automatic Control Systems" by Kuo.

The subject loop transfer function is $$L(s)=\frac{K}{s(1+T_1s)}$$ or $$L(j\omega)=\frac{-jK(1-jT_1\omega)}{\omega(1+T_1^2\omega^2)}$$ or $$L(j\omega)=\frac{K(-j-T_1\omega)}{\omega(1+T_1^2\omega^2)}$$

Hence the phase equation should be $$tan(\theta)=\frac{1}{T_1\omega}$$

Now at frequency infinity, which corresponds to origin in the above plot, the value of the phase function is $$tan(\theta)= \lim_{\omega \to \infty} \frac{1}{T_1\omega}=0$$ Also $$tan(\theta)= \lim_{\omega \to 0} \frac{1}{T_1\omega}=\infty$$ At infinite frequency the phase angle can assume value of either zero degree or 180 degree. In this plot it is taken as 180 degree, I can't understand why, when zero degree is also a perfect candidate.

Also at zero frequency, the phase angle should be 90 degree but here it is taken as -90 degree. I can't understand this phase thing.


1 Answer 1



Here the coordinate comes in the third quadrant so you need to take


so at \$w\to0\$,\$\theta=180+90=270=-90\$

and at \$w\to\infty\$,\$\theta=180+0=180\$ enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.