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Is this transfer function unstable?

  1. Apr 12, 2012 #1
    Is this transfer function unstabile?
    Spoiler: This is a feedback control system related issue

    I am trying to sketch the bode plot of the following transfer function but I can't find a solution.
    I should use 20log( unknown / 3 * 10) to find where the magnitude starts.

    G(s) = (s^2+4s+25) / ((s+3)*(s+10))

    It is very appealing to solve the second grade equation, which give complex solutions.

    http://www.wolframalpha.com/input/?i=bode+plot+(s^2+4s+25)/((s+3)(s+10)

    Question 1) Is the system unstable because the poles/zeros are on the right side?
    Question 2) What should I do with the second grade equation to find the values I need to sketch the bode plot?
    Question 3) Does it have 3 break frequencies?

    What I'm looking for is something that looks like, i.e. ((s+923) / ((s+3)*(s+20))),
    so that I have 3 break frequencies and initial magnitude of 20log(923/(3*20)
     
  2. jcsd
  3. Apr 12, 2012 #2
    Stable. Both poles are on the LHS (left hand side) of the s plane, aka LHP (left half plane). With 2nd order functions, stability can be determined by inspection. THe 2nd order denominator has LHP poles due to all signs being positive. The function (s+3)*(s+10) has poles at s=-3, & s=-10 rad/sec. These are in the LHP.

    With 3rd order and higher, even if all signs are positive, i.e. s^3 + a*s^2 + b*s + c, where a, b, & c are all positive. There may be poles in the RHP (right half plane), which is UNstable. There are techniques to determine this. Check out the Routh-Horowitz criterion. It provides a quick way to evaluate high order functions to determine pole location.

    Did I help?

    Claude
     
  4. Apr 12, 2012 #3
    Yes for poles. To get a feeling why, you can transform the system back into the time domain, where all these "s" symbols will become exponentials. Complex poles will become trigonometric functions.

    Anyway, if the sign of the real part is negative, then the exponential has decreases with time. On the other hand, if the sign is positive, the exponential will blow up over time.

    No for zeros. The problem with RHP zeros is that the take away valuable phase margin. Though when you design a control system, you would want all RPH zeros below your crossover frequency.
     
  5. Apr 12, 2012 #4
    Does this make sense? I used an on-line program http://wims.unice.fr/wims/en_tool~analysis~fourierlaplace.en.html [Broken]
     

    Attached Files:

    Last edited by a moderator: May 5, 2017
  6. Apr 13, 2012 #5

    psparky

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    Gold Member

    Cognitive.....

    Don't forget that your transfer function is a mathematical equation that will give a gain and phase angle at any given frequency from 0 to infinity.

    G(s) = (s^2+4s+25) / ((s+3)*(s+10))

    Simply replace "s" with "Jω".

    G(s) = ((Jω)^2+4(Jω)+25) / ((Jω+3)*(Jω+10))

    So when ω=0 for example.......you get 25/30.....or a gain of .83 (-1.61 dB) at your lowest possible frequency.....also your phase angle will obviously be zero since all the "J's" went away. And yes, this corresponds exactly with the graphs you show in your first post.

    Take ω to equal 10 for example. Now you will get a magnitude with it's phase angle once you simplify all the J's. Don't forget that J simply equals 1<90.

    Plug in any value for ω and you will always get the exact magnitude and phase. It's a real help when you get "stuck" at certain points in your bode plot. Your program may not always be available....like during an exam.
     
    Last edited: Apr 13, 2012
  7. Apr 14, 2012 #6
    This was very helpful. I didn't realize I could do this to a second grade equation with complex solutions, very smart.

    This is wrong. What determines where the phase starts, depends on the type of system.
    A type zero system has no s in the denominator, aka s^0 = 1. If the denominator was s(s+3)(s+10), we'd have a type 1 system and the phase would start at -90 degrees.

    This is wrong and ridiculous, because J or often i is the exact definition of the square root of (-1).

    But perhaps you would be kind enough to tell me, how many break frequencies you see?
    I still don't know whether the nominator which has complex solutions, amount to any break frequencies. The denominator, obviously has break frequencies of 3 and 10.
     
  8. Apr 14, 2012 #7

    psparky

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    Gold Member

    I stand behind everything I just wrote 100%.

    You believed me that it worked when ω=0. Why do you not believe it for other values?

    The proof is in the pudding. Do the vector math for when ω=10. It will match the graph of your computer program exactly.

    All I'm doing is following the transfer function you gave. Your computer program is following the same transfer function.....neither one of us can help our results.

    To quote Miagi....not everything is as it seem.....

    What's 1<90 X 1<90?

    Let me know when the bells start ringing.

    If you do the vector math for ω=10....

    You will get (40j-75)/(130j-70)

    Change to polar vectors you get (85<152)/(147.6<118.27)

    The gain is .5758

    20 log .5758 = -4.79 dB

    Phase angle is 33.73 degrees. Again, we agree with your graph.

    I know you are trying to find the break frequencies......it is harder to see when the zeros are complex at -2 + or - 9.27j.

    But still your poles are -3, -10......and you have two complex zeros at -2 +9.27j and -2 -9.27j

    Follow your transfer function and the rest is elementary.
     
    Last edited: Apr 14, 2012
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