Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

One question about the stability and pole location

  1. May 7, 2009 #1
    I have a question related to that is explained on the textbook Sedra & Smith.

    for a pole at s=a+bj,say, 1/(s-a-bj) why can we have the time domain of the signal to
    be exp(at+jbt)? It is strange because s=jw, so Re(s)=0 <a ,which leads to the solution to
    be -exp(-at-bjt)u(-t), so that the laplace transform will be convergent.

    Recall that inverse LT of 1/(s-a) is exp(-at)u(t) for Re(s)>a ,-exp(at)u(-t) for Re(s)<a

    Could you explain it to me ? Thank you.

    Best wishes,
    WANG Lu, Mike
     
  2. jcsd
  3. May 7, 2009 #2

    Defennder

    User Avatar
    Homework Helper

    That's something I haven't seen before. In particular, complex poles always occurs in pairs, so you'll get another pole at s=a-jb. Taking the Laplace inverse transform of both poles would give you 1/b exp(at) sin (bt).
     
  4. May 8, 2009 #3
    but for an expressiong 1/(s-a-bj) + 1/(s-a+bj) where s is only for physical frequencies and is equal to jw. if a>0 , we have Re(s) < a .

    The laplace transform is different for Re(s) <a ,which gives a result of -exp(-at)u(-t).
     
  5. May 9, 2009 #4

    Defennder

    User Avatar
    Homework Helper

    Hi mike, since you specifically cited Sedra and Smith, perhaps you could point me to the relevant section which you have problems with. I own a copy of that book, 5th edn.
     
  6. May 9, 2009 #5
    The question is at section 8.9.1

    How come for different theta's , the inverse laplace transforms are the same for three kinds of theta. you know s=jw and Re(s)=0, so Re(s)<theta , when theta >0
     
  7. May 9, 2009 #6
    The question is at section 8.9.1

    How come for different theta's , the inverse laplace transforms are the same for three kinds of theta. you know s=jw and Re(s)=0, so Re(s)<theta , when theta >0
     
  8. May 9, 2009 #7

    Defennder

    User Avatar
    Homework Helper

    I really don't see how your question is related to that section. The section explains how stability of a system may be deduced from pole locations and it also explains how the envelope of the transient response is determined from Re(s).

    In particular, there's no theta in that section. Which edition are you using?
     
  9. May 9, 2009 #8
    I don't know how to spell that greek character. I use * to replace it.
    But it doesn't explain how the transient response is determined from Re(s). It gives the result immediately, just an expression of v(t). But with a pole pair at s=(*)+/-jw. The expression is like A/(s-*-wj)+B(s-*+wj). We must compare Re(s) with *, isn't it? When *>0 , because s=jw for physical frequencies, Re(s)=0 < *. The result of inverse laplace transform is changed.

    So the result gained is not right from inverse Laplace transform.
     
  10. May 9, 2009 #9

    Defennder

    User Avatar
    Homework Helper

    The denominator of your transfer function should be written as a quadratic expression. Then use completing the square and s-shifting (I think it's called that) to inverse laplace the transfer function. You should get an expression with an exponential envelope and a sinusoidal function.
     
  11. May 9, 2009 #10
    It's the same,just to have different forms. The requirement for the inverse laplace transform is still Re(s) > *
     
  12. May 10, 2009 #11

    Defennder

    User Avatar
    Homework Helper

    ?? What's the same? The real part of the pole tells you whether the system is stable. Any pole in the right-hand side of the diagram tells you that it'll be unstable. More specifically the magnitude of the real part of the pole gives you the exponential envelope which accompanies the sinusoidal response.

    Why is there a need to express the transfer function in terms of complex roots? If you want to do so then you're unnecessarily complicating the problem. But I'll give it a shot. First express the TF in terms of http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/PartialFraction/PartialFraction.html" [Broken].
    [tex]\frac{1}{(s-a-jb)(s-a+jb)} = \frac{-j\frac{1}{2b}}{s-a-jb} + \frac{j\frac{1}{2b}}{s-a+jb}[/tex].

    Taking the inverse laplace transform of both sides, we apply the formal definition of it, resulting in a http://en.wikipedia.org/wiki/Inverse_Laplace_transform" [Broken].

    Since we know the roots (or singularities) are at [tex]a\pm jb[/tex], we can evaluate the line integral by applying Cauchy's residue theorem ie. for example inverse Laplace transform the first term:

    [tex]L^{-1} \left( \frac{-j\frac{1}{2b}}{s-a-jb} \right) = -j\frac{1}{2b} \cdot \frac{1}{j2\pi} \oint_c e^{st} \frac{1}{s-a-jb} ds = -j\frac{1}{2b} \cdot \frac{1}{j2\pi} \cdot j2\pi Res(e^{st}F(s),a+jb)[/tex]

    [tex]\text{Res}(F(s)e^{st},a+jb) = e^{(a+jb)t}[/tex].

    Using this result, we plug this back into the expression above:

    [tex]L^{-1} \left( \frac{-j\frac{1}{2b}}{s-a-jb} \right) = -\frac{1}{2b} e^{t(a+jb)}[/tex].

    Applying the same procedure for the other partial fraction term gives the following result:

    [tex]-j\frac{1}{2b}e^{t(a+jb)} + j\frac{1}{2b}e^{t(a-jb)} = -\frac{1}{b}e^{at} \frac{1}{j2} (e^{jbt}-e^{-jbt}) = -\frac{1}{b}e^{at}\sin(bt)[/tex]

    All the above was done using the fundamental definition of the Laplace inverse transform.

    Now you should be able to see from the above that the time-domain form of the response is stable only if a<0, since otherwise the e^(at) term grows exponentially. If the poles are strictly imaginary ie. s=jw, then e^(at)=1 where a=0 and you have a marginally stable where the system responses oscillates, which is consistent with what the textbook says.

    Hope this helps.
     
    Last edited by a moderator: May 4, 2017
  13. May 10, 2009 #12
    At first, thank you for you detailed mathematical expressions. But I still have this question.

    why such expression don't need to compare Re(s) with a? The inverse laplace transform I leant needs to compare Re(s) with a and gets different expressions.

    [tex]L^{-1} \left( \frac{-j\frac{1}{2b}}{s-a-jb} \right) = -j\frac{1}{2b} \cdot \frac{1}{j2\pi} \oint_c e^{st} \frac{1}{s-a-jb} ds = -j\frac{1}{2b} \cdot \frac{1}{j2\pi} \cdot j2\pi Res(e^{st}F(s),a+jb)[/tex]

    [tex]\text{Res}(F(s)e^{st},a+jb) = e^{(a+jb)t}[/tex]
     
    Last edited by a moderator: May 4, 2017
  14. May 10, 2009 #13

    Defennder

    User Avatar
    Homework Helper

    I'm not too sure if you've learnt the formal mathematical definition of the inverse laplace transform. But all that is required for the inverse laplace transform to exist is the following conditions to be satisfield:

    [tex]\lim_{s\rightarrow \infty} F(s) = 0[/tex]
    [tex]\lim_{s\rightarrow \infty} s\cdot F(s) = c[/tex] where c is any constant. In other words, the latter limit must converge.

    I'm not too sure by what you mean by why we are not required to compare Re(s) with a or something, but I guess you must be referring to the Laplace transform, not the inverse transform. Once you have the expression in s-domain, we only care about the conditions for which the inverse Laplace transform exists, and not about how the expression was obtained. That's just my 2 cents.
     
  15. May 10, 2009 #14
    I kind of start to know where I am wrong. Because the laplace transform we are studying is very primary, the requirement on the existence of the inverse laplace transform is much higher.

    For example for 1/(s-a) we must make Re(s) >a to get the form as you wrote. If Re(s)<a, the expression becomes to only have value on t<0 and with different expression.

    I think it's maybe because of the Cauchy's residue theorem. Thank you for letting me know this.
     
  16. May 11, 2009 #15

    Defennder

    User Avatar
    Homework Helper

    The fundamental definition of the inverse Laplace transform which I invoked above is rarely used. I only did so to show that one need not check the condition for Laplace transformed-function F(s) Re(s)>a before one could use that.

    I suppose one could easily simply take the inverse Laplace of [tex]\frac{1}{s-a-jb}[/tex] directly without checking that condition. Let a+jb=z then do the laplace transform. That way you avoid the whole business of doing a closed-path line integral over the complex plane.
     
  17. May 11, 2009 #16
    Yes, you are right. I think I get the point. Thank you very much.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: One question about the stability and pole location
Loading...