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Laplace transform physical meaning

  1. Sep 26, 2010 #1
    When comparing time and frequency domains, it is easy to imagine the meaning of the Fourier transform.

    In time domain, our function takes time as a parameter, and returns the value (result) of our process.

    When we make a Fourier transform of the same function, we take it to the frequency domain, meaning we get a different function which accepts frequency as a parameter, and returns the complex value representing the amplitude and phase of the spectral (sinusoidal) component of our initial function.

    Now, Laplace transform should be a generalization of Fourier if I got it right. Function in Laplace domain is a function of s, where s=σ+jw, where σ is attenuation, and w corresponds to the same frequency as in FT. So, for σ=0, it returns the same result as FT.

    So, what I don't understand it:

    1. What does attenuation exactly mean? This should be an exponential function multiplying the input sinusoidal function of frequency w?

    2. What does the result of LT correspond to, "physically"? Is it amplitude+phase again? The problem is, in FT, we passed a one-dimensional variable and got a two-dimensional variable. Here, we have a two-dimensional input, and again get two-dimensional output. In this case, what happens to attenuation information at output? Surely the output of LT should not represent a periodical function if the input function is attenuated. But obviously, it represents a periodical function for σ=0.
  2. jcsd
  3. Sep 27, 2010 #2
    Ok, so in other words, speed of the wheel (RPM) when observed from the stationary frame, should increase as the bike moves faster, and be proportional to the speed of bike.

    Meanwhile, RPM will be even larger in the moving frame, because road will be contracted and less time will pass between full turns.
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