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Oscillations in transient response

  1. Aug 18, 2012 #1
    Why oscillations occur in transient response of a system ?
    Suppose transfer function of a system is 1/(s+1). Then its impulse response will be c(t)=e[itex]^{}-t[/itex] . So why just we don't show only an exponential curve to graphically represent the impulse response, why we also show the oscillations ?
  2. jcsd
  3. Aug 18, 2012 #2
    I am really not getting what your point is. Can you elaborate it using pictures of waveforms etc?
  4. Aug 18, 2012 #3
    please see the attachment
    in all the transient responses , the dashed curves show the actual time dependency of the output signal and solid curves show the oscillatory behavior . My question is this in studying control system when we have to judge the stability of the system we use the oscillatory behavior , but why is there an oscillatory behavior ?
    We solved for the output of the system mathematically and got the output of the system in time domain as in example below , then how the oscillations come ?

    Suppose transfer function of a system is 1/(s+1).
    Then its impulse response will be c(t)=e^(-t) . So why just we don't show only an exponential curve to graphically represent the impulse response, why we also show the oscillations ?

    Attached Files:

    Last edited: Aug 18, 2012
  5. Aug 18, 2012 #4


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    The simple (and kind of obvious) answer is that not all systems have transfer functions that are as simple as your single pole example.
  6. Aug 19, 2012 #5


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    If it's a second-order or higher order system, then it can overshoot & oscillate, depending on damping.
  7. Aug 20, 2012 #6
    Yes, I think this is the answer.

    To elaborate a little bit, if a system is truly described by the transfer function, c(t)=e−t , it should not oscillate. This system does not oscillate.

    I think think what you are describing can be modeled by a decaying exponential with a delayed decaying sine wave. My control theory is a little rough, but I think the function you are thinking of would look something like c(t) = e-t *sin(t+T) (where T is a time delay.

    I think your "problem" is just a simple matter of modeling. Sometimes. the oscillations aren't significant enough for an engineer to spend the time factoring it in. It is much easier to run a regression on a curve to find one variable in an exponential function than multiple variables in a decaying sine function.
  8. Aug 23, 2012 #7
    sorry i posted wrong. c(t) is [itex]e^{-t}[/itex] . and no sin term

    the graph of c(t) is simply decaying exponentially , then why do we show oscillations. I mean mathematically i showed that c(t) is just decaying exponential , and there is no sin term. So tell me where i am missing ?
  9. Aug 23, 2012 #8
    The other curves you are showing are not described by the transferfunction

    [itex]X\left(s\right) = \frac{1}{s+1}[/itex]
    When you convert that particular transferfunction back to the time-domain, yes, you have an exponential.

    But if you are dealing with a 2nd Order System then your system could be described by the differential equations:

    [itex]y(t) + A\cdot y''(t) + B\cdot y'(t) = C\cdot x(t)[/itex]

    When you do the LaPlace transform you end up with

    [itex]Y(s) + As^{2}Y(s) + BsY(s) = CX(s)[/itex]

    You can then go ahead and describe the transferfunction as

    [itex]H(s) = \frac{Y(s)}{X(s)} = \frac{C/A}{s^{2}+ B/As + 1/A}[/itex]

    Now we are ready to solve this and get a clear description of the output.

    The BOTTOM term of the fraction is very important, since it is a basic second order polynomial it is straight forward to solve it.

    It will have two roots s1 and s2. When you know the roots it is possible to factorize it into

    [itex]H(s) = \frac{C/A}{(s-s1)(s-s2)}[/itex]

    Now these roots can be in three different formats:

    1. Two distinct real roots s1 and s2
    2. A double real root s1
    3. Two complex-conjugate roots s1 and s1*

    IF it happens that the roots are COMPLEX, when you go back to the time domain by doing the partial fraction expansion, you will find that you will get sines and cosines.-

    And that is where the oscilations come from.

    Now when are the roots complex? - It depends on the coefficients A, B and C.

    You will find that these coefficients are written in other ways also, and one very common way is to write the transferfunction in terms of a damping ratio [itex]\varsigma[/itex].

    It takes a bit of basic algebra to get to the point where the damping ratio will tell you all you need to know, namely:

    1. If the damping ratio is less than 1, you will have oscillations
    2. If the damping ratio is 1 or more, you will not have oscillations.

    The graph of a damped system (damping ratio more than 1) will LOOK like an exponetial, but is in fact NOT an exponential - which can be seen if one does the math and calculates the time domain behavior of y(t) in each of the three cases (real roots, 1 double real root, complex roots).
  10. Aug 23, 2012 #9
    thanks for the reply
    in books they took a first order system , calculated its response in time domain by using the inverse laplace of system function and got an decaying exponetial curve. Then they plot the exponential curve and marked the oscillations in that curve and said that " as the oscillations are decreasing , the system is stable". OK i know that in that particular impulse response if there are oscillations then they will die after sometime. But why there are oscillations ?
  11. Aug 23, 2012 #10
    If the book is talking about oscillations in a first order system the book is wrong.

    Oscillations only happen in higher order systems. (2nd and above) due to the facts I stated in my last reply.

    Mathematically the oscillations in higher order systems happen because of complex poles of the transfer function (roots of the polynomial in the bottom of the fraction). These complex poles are present because of the way the coefficients are chosen.

    I don't know what to say other than that. That IS the reason why there is oscillations.

    In an electrical circuit it comes because you dont have a big enough resistor in your LCR circuit.
  12. Aug 23, 2012 #11
    So you mean there is no transient response for 1st order system ?
    because transient response are associated with oscillations.
  13. Aug 24, 2012 #12


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    No, it doesn't need to oscillate to be a transient response. The transient response of a first order system is an exponential.
  14. Aug 24, 2012 #13
    Yes, as uart says. The steady-state response of a 1st order system is a DC. The transient will be an exponential.

    A 2nd order system can have oscillatory transients, and its steady-state response can also be oscillatory (if there is no damping at all).
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