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DSP - Differential Equation for linear system H

  1. Dec 30, 2011 #1
    Hello.

    I need to find a differential equation for a linear system, H, which is characterised by the impulse response h(t) = e-2t, where t denotes time.

    But I am not sure how to go about this. Would anyone be able to point me in the correct direction?

    Thanks in advance.

    Seán
     
  2. jcsd
  3. Jan 3, 2012 #2

    rude man

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    1. Transform your impulse equation into the frequency (Laplace) domain: h(t) --> H(s)

    2. Inverse-transform the resultant H(s) transfer function back to the time domain for your diff. eq.

    Hint: H(s) = Y(s)/X(s) where X is transform of time input x(t) and Y is transform of output y(t).
     
  4. Jan 3, 2012 #3
    Hello,

    Thanks for your reply.

    I will have a go at this in the morning, and I will let you know how I get on.

    Thanks a lot. I just wasn't sure how to tackle it.

    Yours
    Seán
     
  5. Jan 4, 2012 #4
    Hello again.

    Can I just make sure I have this right?

    I got the Laplace transfer of h(t), which is H(s) = 1/(s+a).

    So, now I need to get the transform of the input, which is

    0.5 cos(ω2 - ω1)t) + 0.5 cos(ω2 + ω1)t)

    But I am really not sure about the Laplace transform of this. I was thinking it was ...0.5 s/(s2 - ω2) ...but I don't think that is right because in the time domain it is ω2 - ω1, and I am not too sure how to get this in the frequency domain.

    Seán
     
  6. Jan 4, 2012 #5
    You're supposed to describe system behavior for any x(t) input. There should be no specific expression for "transform of the input" for this problem besides "X(s)".

    The goal is to have and equation for the system relating the output y and its derivatives to the input x and its derivatives.
     
  7. Jan 4, 2012 #6
    Hello,

    And thanks fot the reply.

    I am afriad I am getting lost on this one.

    So, if I can ask, once I have the transform of the impulse, where do I go from there?

    I thought I needed to get the transform of the input to the system, and from that, get an equation for the output by relating the input to the system H.

    Seán
     
  8. Jan 4, 2012 #7
    Do you know what is the transform of the derivative of a function in terms of that function's transform?
     
  9. Jan 4, 2012 #8
    I'm sorry, I don't.

    Seán
     
  10. Jan 4, 2012 #9
    equation 11 is the general rule for the nth derivative. For the first derivative, this would be as follows.

    [itex] x(t) \stackrel{\mathcal{L}}{\leftrightarrow} X(s) [/itex]

    [itex] \frac{dx(t)}{dt} \stackrel{\mathcal{L}}{\leftrightarrow} sX(s) - x(0) [/itex]
     
  11. Jan 4, 2012 #10
    Hey,

    Okay, yes ...I get that ....but (and I am very sorry if this sounds rude, I don't mean it to be) I don't understand how this helps me.

    I am more used to the hands on stuff ...building different things ....so, the maths behind it all is very new to me.

    But from what you said earlier, that I need to relate the input to the output ...I don't actually know what the output of the system is ...or, at least I don't know if I know it.

    If I knew specific steps to take, I might be able to figure it out. Kind of what like Rude Man suggested, except it seem to me like the inverse transform from step one would just take me back to where I was at the start.

    Again, sorry for not getting this ...maybe I have just been trying to figure it out for a bit, I just have a bit of a mental block.

    Seán
     
  12. Jan 4, 2012 #11
    Oh, I think I may have got a bit further on ...I will try it and get back either way :)

    Seán
     
  13. Jan 4, 2012 #12

    rude man

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    The input in your case is an impulse = δ(t). That's what's meant by 'impulse response'.
    y(t) is your output = exp(-2t), and Y(s) = 1/(s+2) as you have correctly ascertained.

    So the question now is, I know what my response is if the input is δ(t), so what is the response to a general input x(t)? If the input is x(t), what is the output y(t)?

    Well, you got the transform of the output right, but you blew it big-time on the input.
    What does your Laplace table say for X(s) if x(t) = δ(t)?

    Once you have the transform of δ(t) right = X(s), you can then say Y(s)/X(s) = transfer function of your system = H(s).

    Do that next.
     
  14. Jan 4, 2012 #13
    Hello,

    Well, for the table δ(t) = 1. So X(s) = 1.

    If 1/(s+2) = Y(s), then H(s) = (1/(s+2))/1 ...?
     
  15. Jan 4, 2012 #14

    rude man

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    δ(t) is not 1, δ(t) = δ(t). But L{δ(t)} = 1, that's right. And H(s) is exactly what you said. So write an equation with X(s), Y(s) and H(s)?
     
  16. Jan 4, 2012 #15
    Do you mean something like Y(s) = H(s)X(s)?
     
  17. Jan 4, 2012 #16

    rude man

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    Yes I do. You know what H(s) is now, so substitute in your equation.
     
  18. Jan 4, 2012 #17
    Okay, well, if

    X(s) = 1
    Y(s) = 1/(s+2)
    H(s) = (1/(s+2))/1

    I get

    [itex]\frac{1}{s+2}[/itex] = [itex]\frac{1/s+2}{1}[/itex][itex]\times[/itex]1

    Which is the same as

    [itex]e^{-2t}[/itex] = [itex]\frac{e^{-2t}}{1}[/itex][itex]\times[/itex]1

    I think.
     
  19. Jan 4, 2012 #18
    No, X(s) and Y(s) represent arbitrary functions, constrained only by the transfer function Y(s)/X(s). You don't get to decide this X(s) is equal to 1!

    Think of Y(s)/X(s), as describing a relationship between the output in the input in general. The impulse response h(t) and corresponding Y(s)/X(s) characterize the response of a linear system to any input.

    When X(s) = 1 (a transformed impulse), Y(s) becomes the transformed impulse response, of course. But the solution you seek is a differential equation for the system with any input, and so it is not proper to have X(s) equal to 1.
     
    Last edited: Jan 4, 2012
  20. Jan 5, 2012 #19

    rude man

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    You have determined what H(s) is. You did that by knowing that if x(t) = δ(t), so that
    X(s) = 1, so since H(s) = Y(s)/X(s), and Y(s) you have determined to be 1/(s+2) if X(s) = 1, now you know Y(s)/X(s) for any x(t) which, when transformed is X(s). So, stay with your equation: Y(s) = H(s)X(s). That is the key to coming up with your sought-after differential equation.

    The next step is transforming that algebraic equation in the frequency domain (never mind for now what I mean by that) into a differential equation in the time domain.

    Clue: use cross-multiplication in your equation!
     
  21. Jan 5, 2012 #20
    Hello.

    I get what you mean about needing to find a solution for any input. It is finally sinking in.

    So, since I need to find a silution for any x(t), which is X(s), so would a better form of the equation to work with be

    X(s) = Y(s)/H(s)?

    I thought to go from the frequency domain (Laplace) to the time domain, I just needed to use the Laplace tables in reverse basically. I take it that is not the case?

    And if I stick with Y(s) = H(s)X(s) ....how do I cross multiply this? Anytime I have cross multiplied something it has been in the form of (a/b) x (c/d).

    Seán
     
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