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UnkleRhaukus

  • 2 years ago

\[\mathcal L\big\{f(x-a)h(x-a)\big\}\]

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  1. UnkleRhaukus
    • 2 years ago
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    \[\begin{align*} \mathcal L\big\{f(x-a)h(x-a)\big\}&=\int\limits_0^\infty f(x-a)h(x-a)e^{-px}\text dx\\ \text{let } t=x-a\\ \text dt=\text dx\\ x=0\rightarrow t=-a\\ x=\infty\rightarrow t=\infty\\ &=\int\limits_{-a}^\infty f(t)h(t)e^{-p(t+a)}\text dt\\ &= \\ \end{align*}\]

  2. UnkleRhaukus
    • 2 years ago
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    \[\begin{align*} &=\int\limits_{-a}^\infty f(t)h(t)e^{-p(t+a)}\text dt\\ &=e^{-pa}\int\limits_{0}^\infty f(t)e^{-pt}\text dt\\ &=e^{-pa}F(p)\ \\ \end{align*}\] is this right?

  3. UnkleRhaukus
    • 2 years ago
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    i dont get it

  4. ChmE
    • 2 years ago
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    Thats the same thing I show in my book. I just learned this last week. I don't get it either

  5. ChmE
    • 2 years ago
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    The h(x-a) or in my book u(t-a) as I understand it is just an on off switch for the discontinuous function

  6. UnkleRhaukus
    • 2 years ago
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    yeah , h, u , or \(\theta\) are used for the heaviside unit step function

  7. UnkleRhaukus
    • 2 years ago
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    \[\mathcal L\big\{f(x-a)h(x-a)\big\}=e^{-pa}F(p)\] \[f(x-a)h(x-a)=\mathcal L^{-1}\big\{e^{-pa}F(p)\big\}\]

  8. UnkleRhaukus
    • 2 years ago
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    @gohangoku58

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