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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?



    2.
    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]




    3.
    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
    [tex]H(s)=\frac{10}{s^2+11s+10}[/tex]
    As you probably know, bode plots deal with frequency response, so s = jw
    [tex]H(j\omega)=\frac{10}{(j\omega)^2+11j\omega+10}[/tex]
    [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


    http://ocw.mit.edu/courses/mechanic...ntrol-i-spring-2005/exams/secondorderbode.pdf

    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
     
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