Eigenfunctions of LTI systems

Discussion in 'Math' started by jrv9090, Nov 29, 2014.

  1. jrv9090

    Thread Starter New Member

    Nov 26, 2014
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    Hello all,

    I have been studying about LTI system and Eigen functions. Here is a doubt I like to clarify.

    Say I have a LTI system defined by the system equation [y''(t) + 5y'(t) + 6y(t) = x(t) ]. It is evident that for the above system the transient portion will be of the form Aexp(-2t) + Bexp(-3t) [Same as Eigen functions of this system].

    My understanding is as follows: that the transients are independent of x(t). Hence no matter what signal I send into the signal via x(t), the transient will persist. Also, by definition of Eigen functions, if I send an Eigen function into the system, I will get the same as output with different gain.

    Say, I decide to send the signal exp(-2t) as input via x(t). Then the output has to be, by definition of Eigen functions, C *exp(-2t). I would like to know what has happened to the transient exp(-3t). Has it disappeared? If so how?

    My guess is that it has something to do with initial conditions. If so, I don't see how. Kindly let me know.
     
  2. studiot

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  3. jrv9090

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    Nov 26, 2014
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  4. t_n_k

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    Can you show how (mathematically) the other term might have disappeared.
     
  5. jrv9090

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    Nov 26, 2014
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    I don't know if the other term disappears or not.

    1.By definition of Eigen functions, if I send exp(-2t) [which by the way is the Eigen function for the above system] into the system, the output should only contain exp(-2t) and not exp(-3t) or any other signal for that matter.
    2.But since transient are independent of input signals, I believe transient response to contain both exp(-2t) and exp(-3t).

    Statements 1 and 2 above contradict each other based on my understanding and not by any mathematical derivation. Hence I require clarification.
     
  6. t_n_k

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    I'm afraid I don't follow your logic.
     
  7. jrv9090

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    Nov 26, 2014
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    Can you please let me know which part is ambiguous???
     
  8. t_n_k

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    I would probably firstly answer the question as to what the solution would be if the input is of the form of one of the eigenfunctions.

    I believe for

    x(t)=exp{-2t}

    The response is

    y(t)=exp{-3t}-\[1-t\] exp{-2t}

    So as you proposed in point 2 of your last post (#5) the response does indeed include both forms of the eigenvalues.

    I don't know what your point 1 is based upon. Perhaps you could cite a reference.

    If the system is not driven by a steady state input, one would expect there to be only the natural transient terms in the response at t=infinity. Since the natural transient terms are decaying exponentials, then at t=infinity these transients would have vanished and the system is "at rest". This would be the case if (say) there was simply some initial energy storage in the system such as a voltage on a capacitor or perhaps an impulse Dirac type input. Or indeed a decaying exponential.
    One could also have a driven system where x(t) is a known function - such as a steady state sinusoid or a DC step & so on - which persists indefinitely. One would anticipate the resultant response would therefore include the superposition of terms resulting from both the steady state input and the natural [decaying transient] responses.
     
    Last edited: Nov 29, 2014
  9. studiot

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    But these are not eigenfunctions.

    if y=exp(-2t) is a solution of the homogeneous equation this means that

    L[y] = 0

    Whereas the definition of an eigen function requires L[y] = hy where h is a number.

    Another way of saying the output has to be C *exp(-2t) is to substitute this into the LTI and seeing if this is possible.

    Now the A and B in your expression are purely arbitrary until determined by the boundary conditions and Aexp(-2t) + Bexp(-3t) = 0 is true for all values of A and B so taking A = B = 1 means no loss of generality

    So by my reckoning substituting y=C *exp(-2t) into [y''(t) + 5y'(t) + 6y(t) = x(t) ]. yields

    +4y -10y +6y = Cy

    Thus there is no C for which this is possible

    Note this will always happen if you try to substitute the solutions of the homogenous expression as the x(t) term in the inhomogenous equation so you cannot use these as trial functions to obtain a particular integral.

    There are standard proceedures for dealing with this, tnk has showed that for real exponential solutions you use the product of t and the exponenetial.

    A useful theorem connecting the solutions of the homogeneous and nonhomogenous equations is

    If L[y] = 0 is the homogeneous equation with solutions y1 and y2

    and L[y] = f(t) with solutions t1 and t2

    Then
    1)The difference of solutions of the homogeneous equation is a solution of the nonhomogeneous equation

    that is y1-y2 = t1 or y1-y2 = t2

    and

    2)

    The sum of one solution of the homogeneous equations and one solution of the non homogeneous equations is a solution of the homogeneous equation.
     
    Last edited: Nov 30, 2014
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