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Why s is substitute with jw in transfer function?

  1. Jun 20, 2015 #1
    Can anybody explain to me briefly on why in a transfer function like G(s), we substitute s=jw such that it become G(jw) in order to find its gain value?

    Thanks
     
  2. jcsd
  3. Jun 21, 2015 #2
    You can explain it in a variety of ways, but I prefer this one, since it's quite general:

    The only difference between the bilateral Laplace transform:
    $$
    G(s) = \mathcal{L}\{f(t)\} = \int_{-\infty}^\infty e^{-st} f(t)dt\\
    $$
    and the Fourier transform:
    $$
    H(\omega) = \mathcal{F}\{f(t)\} = \int_{-\infty}^\infty e^{-j\omega t} f(t) dt
    $$
    is the variable substitution ##s = j\omega##, i.e. ##H(\omega) = G(s)\left.\right|_{s = j\omega} = G(j\omega)##.

    For some system with transfer function ##G(s)##, ##G(s)## is also its impulse response, which makes ##G(j\omega)## the Fourier transform of its impulse response.

    More intuitively:
    To find the gain of a linear system at some frequency ##\omega##, you could apply a sinusoidal input with frequency ##\omega## and unity amplitude, and observe the amplitude of the output signal, which directly gives you the gain of the system at ##\omega##.

    An ideal impulse has components with unity amplitude at all frequencies, and thus ##G(j\omega)## "picks out" the gain and phase of the system at ##\omega## from its impulse response.

    Makes sense?
     
  4. Jun 22, 2015 #3
    Wow, I like your answer. I am still a beginner in this kind of things and I can understand most of your answer. But one question. What is exactly a linear system?
    Thanks in advance
     
  5. Jun 24, 2015 #4

    donpacino

    User Avatar
    Gold Member

    a big thing in looking at systems like this is knowing if they are linear and if they are time invarient.

    you should know these

    Linear:: more or less to be a linear system the sum of two inputs have to equal the sum of the two outputs
    x1(n)=y1(n)
    x2(n)=y2(n)
    so in order to be linear
    x1(n)+x2(n)=y1(n)+Y2(n)

    then continue that out to inf and -inf

    for a system that is time invariant, delaying the input by a constant delays the output by the same amount.

    given the system x(t)=y(t)

    x(t-d)=y(t-d) for all values of t and d
     
  6. Jun 24, 2015 #5

    donpacino

    User Avatar
    Gold Member

     
  7. Jun 24, 2015 #6
    Generally,

    ## s = \sigma + j \omega##

    When ##\sigma = 0##, ##s = j \omega##

    ## \sigma ## is the Neper frequency. It tells us the rate at which the function decays (when it is negative).

    ## \omega ## is the radial frequency. It tells us the rate at which the function oscillates.
     
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