1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Control problem. Transfer function given. Sketch Bode plot

  1. Aug 10, 2010 #1
    1. For the control system described by the transfer function H(s)=Y(s)/U(s)=10/(s^2+11s+10) Sketch the bode plot: amplitude and phase diagram. What is the bandwidth of the system?

    General form of a second order system H(s)=wn^2/(s^2+2*zeta*wn*s+wn^2)
    Magnitude characteristic in logarithmic form is A(w)[dB]=-20log[(2*zeta*w/wn)^2+(1-w^2/wn^2)^2]^0.5
    The phase of second-order system is theta(w)=-tan^(-1) [2*zeta*w/wn]/[1-(w/wn)^2]

    I really want to understand the algorithm (step by step process) how to do this.

    1. Using the general form of a transfer function, I can find resonance frequency wn=10^0.5. I use it as my cut of frequency when plotting magnitude. Before wn magnitude is a straight line at 0 dB. After wn=10^0.5 it is a line going down.
    I can also find zeta, which is = 11/(2*10^0.5). I know that it influences the shape of the function near the corner frequency.
    No ideas how to do the phase plot though.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 10, 2010 #2
    As you probably know, bode plots deal with frequency response, so s = jw
    [tex]H(j\omega)=\frac{10}{-\omega^2 + 11\omega j + 10}[/tex]
    When w is extremely small, the equation becomes
    [tex]H(j\omega) = \frac{10}{10} = 1[/tex]
    which has a phase of 0.
    when omega is very big we have
    [tex]H(j\omega)= \frac{10}{-\omega^2}[/tex]
    which has a phase of -pi


    from google, here is an example. You can see the angles start at 0 degrees and approach -180 degrees or -pi radians, as we just explored. This is how all phases for second order underdamped systems will turn out. The larger zeta is, the faster the transition from 0 to -pi.
  4. Aug 10, 2010 #3
    Thank you for some useful insights.
    However, I can understand why [tex]H(j\omega) = \frac{10}{10} = 1[/tex] results in phase 0 and [tex]H(j\omega)= \frac{10}{-\omega^2}[/tex] in phase -pi ?
  5. Aug 10, 2010 #4
    omega is just a real-valued number. What is the phase of a negative, real-valued number? It is pi radians (aka on the negative real axis). However, we are dividing by this negative, real number. When you divide a complex numerator by a complex denominator, you do regular division of the numerator's and denominator's magnitudes and you subtract their angles.

    In that example, you do |10|/|-w| = 10/w to find the new magnitude. 10 has an angle of 0 and -w has an angle of pi (remember that we are dealing with positive omegas). So 0 - pi = -pi
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook