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Linear system, input/output signal

  1. Nov 10, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I'm stuck on the following problem:
    A linear system has an output of [itex]G(\omega ) e^{-i \omega t}[/itex] to an input signal of [itex]e^{-i\omega t}[/itex] where omega is arbitrary.
    If the input signal has the form [itex]f(t) = \begin{cases} 0 \\ e^{-\lambda t} \end{cases}[/itex] where lambda is a constant, then the output signal is [itex]F(t)= \begin{cases} 0 \\ (1-e^{-\alpha t })e^{-\lambda t} \end{cases}[/itex] where alpha is another constant.
    1)Calculate [itex]G(\omega )[/itex].
    2)Calculate the output signal of the input signal [itex]f(t)=A \delta (t)[/itex].

    2. Relevant equations
    Not really sure.


    3. The attempt at a solution
    Let [itex]\lambda t =i\omega t \Rightarrow \lambda = i \omega[/itex]. Since alpha and lambda are constants, I can write [itex]\frac{\alpha}{\lambda}=c[/itex]. Therefore [itex]G(\omega ) =1-e^{-c\lambda t}=1-e^{-\frac{i\alpha \omega t}{\lambda}}[/itex].
    This would be my answer for part 1). Does this look correct?
    2)I've no idea how to solve this. I guess I must use my expression for [itex]G (\omega )[/itex] but I really don't see how this help.
    Any tip is welcome.
     
  2. jcsd
  3. Nov 10, 2012 #2

    I like Serena

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    Hi fluidistic,

    I suspect λ is supposed to be real.
    In that case your answer to (1) is not correct.

    If you have a linear system, the typical way to deal with it, is to make fourier transforms.
    If the system gets an input i(t) and has an output o(t), then the response function H(ω) is given by:
    $$H(ω) = \frac {\mathscr{F}_t[o(t)](ω)} {\mathscr{F}_t[i(t)](ω)}$$
    Do you know how to calculate that in your respective cases?
     
  4. Nov 10, 2012 #3

    fluidistic

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    Hi and thanks for the lesson, I had no idea about this!
    But I'm totally confused. Is your [itex]H (\omega )[/itex] what they call [itex]G( \omega )[/itex]?
     
  5. Nov 11, 2012 #4

    I like Serena

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    Yes.
    TBH, I expected that you were studying exactly this when you came up with this problem.
    So I'm a bit surprised that you aren't.

    I avoided the use of your G(ω) since its introduction in your problem is unusual to me.
    However, if you do the math you'll see that G(ω) and H(ω) are the same.

    Btw, as an alternative to the fourier transform you could also use the laplace transform.
    The theory is the same. Both transforms are identical with just a different choice of the arguments.
    Seeing that you have another thread on the laplace transform, did you perhaps study that?
     
    Last edited: Nov 11, 2012
  6. Nov 11, 2012 #5

    fluidistic

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    Ok thanks. Basically I'm self studying a mathematical course (in view of taking a final exam in December); I'm solving all the problems I can from a list of problems on a sheet given by a professor. That's the only exercise talking about "signals", I had never seen this before.
    Ok so yeah I've showed that H(\omega ) = G (\omega ). I in fact reached [itex]H(\omega ) =G (\omega ) \frac{\int _{-\infty }^\infty e^{-2i \omega t dt }}{\int _{-\infty }^\infty e^{-2i \omega t dt }}=G(\omega )[/itex].
    For part 1), I reached that [itex]G(\omega ) = \frac{\alpha}{\alpha + \lambda + i \omega}[/itex]. (I hope it's OK, if not I'll post my work).
    I'll try part 2) now, thanks so far.

    Edit: I notice I forgot to mention in my OP that f(t) and F(t) are worth 0 for t<0 and are different from 0 for t>0!

    Edit 2: For part 2) I get that o(t) is worth [itex]\frac{A\alpha }{2\pi}\int _{-\infty}^{\infty } \frac{e^{i\omega t }}{\alpha + \lambda + i \omega }d\omega[/itex]. This does not make any sense, I've made some errors somewhere.
     
    Last edited: Nov 11, 2012
  7. Nov 11, 2012 #6

    I like Serena

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    Last edited: Nov 11, 2012
  8. Nov 11, 2012 #7

    fluidistic

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    Ok thanks!
    Well no, I have not looked into any table for part 1). So I looked at the wikipedia's article for part 2) and the answer is ... [itex]o(t)=A\alpha \Omega (t)e^{-(\lambda + \alpha ) t}[/itex] where the omega function is the Heaviside step "function". I hope this result is right.
    Unfortunately we cannot look at any integral/DE/integral transforms tables during the final exam (worth 100% of the grade). So I guess I'll have to solve more problems than what they give us.
    Thank you very much for all your help I like Serena!
     
  9. Nov 11, 2012 #8

    I like Serena

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    I'm kind of curious.
    In this problem a number of the symbols deviate from the conventions I'm used to.

    Just now, you used ##\Omega(t)## for the Heaviside step function.
    But I only know ##H(t)##, ##u(t)##, or ##\Theta(t)## to be used for this function.
    Where does the ##\Omega(t)## come from?

    As for solving the integral, it helps when you realize that in part 1 you already did the same transform in the opposite direction.
     
  10. Nov 11, 2012 #9

    fluidistic

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    Sorry for the "omega" rather than the "theta" convention for the Heaviside step function, I made a mistake in calling it omega while it was a capital theta, my fault.
    And yes you are right, I did not notice I did the same transform but in the opposite direction. So my result looks good (maybe I made a sign mistake in the last result though. I'll have to recheck this up).
     
  11. Nov 11, 2012 #10

    I like Serena

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    Ah, okay.
    Btw, your result for part 2 is also correct.
     
  12. Nov 11, 2012 #11

    fluidistic

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    Ok thanks for all. :biggrin:
     
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