0
\$\begingroup\$

I know how to construct the Bode diagrams of first order and seconde order(Complexe conjugate) transfer functions. But here in the two examples below i confront some difficulties. I hope someone can explain it to me how can i overcome the problem.

As a first example consider the transfer function: $$T_1(s) = \frac{1}{900+s^2\tau^2}$$

hier is: $$s=j\omega$$

with j the imaginary unit. The Bode diagram of phase is: enter image description here

I begin to construct the phase diagram with low frequencies, so the transfer function is a real number nl.: $$\frac{1}{900}$$ so it lies on the right hand side of complex coordinate system so the phase should be zero. My interpretation is consistent with the transfer function here. As the frequencies grows the transfer function gets negative and more negative so it's still a real number but it lies in the left hand side of the complexe coordinate system. so i guess its phase should be: $$\pi$$ My interpretation is not consistent with the diagram anymore.

As the second example consider the transfer function: $$T_2(s) = 1600 + s^2\tau^2$$ with Bode diagram of phase: enter image description here

I follow exactly the same principal, so with low frequencies i get only a real positive real number so its phase is zero again, it is consistent with my interpretation and as the frequency increases it gets negative so it's is still a real number but negative so its phase should be 180°. You see my interpretation is totally consistent with with de diagram.

I'm pretty much sure that i make a mistake somewhere but i don't see it. I hope someone can explain why i'm wrong or where i make a naive mistake.

\$\endgroup\$
4
  • \$\begingroup\$ phase is defined as arctg( Im(F(jω))/Re(F(jω)) ), so as Im(F(jω)) = 0 for all values of jω the phase is zero all the time, regardless of the sign. Both the pictures and your reasoning seem wrong. \$\endgroup\$
    – user117884
    Dec 1, 2017 at 20:53
  • \$\begingroup\$ @user117884 Actually those pictures are MATLAB generated. I also suspected but i tried to reconstruct 'em with wolfram math online. The results are exactly the same as above. \$\endgroup\$
    – user140508
    Dec 1, 2017 at 20:57
  • \$\begingroup\$ @user117884 And another things is it could be pi or zero and as far as i know the phase is defined with arctan2(x) so it would't be surprise if you get some pi or -pi here. \$\endgroup\$
    – user140508
    Dec 1, 2017 at 21:00
  • \$\begingroup\$ imaginary part of the transfer function is zero for all jω. If you divide zero by some number ( the real part ) it is still zero. Sometimes a bit of common sense is necessary. \$\endgroup\$
    – user117884
    Dec 2, 2017 at 16:43

0

Your Answer

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