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S_Student

  • 3 years ago

laplace cosat and sinat

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  1. nitz
    • 3 years ago
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    L(cos at)=\[s/(s ^{2}+a ^{2})\] and L(sin at)=\[a/(s ^{2}+a ^{2})\]

  2. nitz
    • 3 years ago
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    if u are interested i can tell u the derivation for these

  3. S_Student
    • 3 years ago
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    thanx friend...and ya i want to expain me in detail..

  4. nitz
    • 3 years ago
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    ok ...

  5. S_Student
    • 3 years ago
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    thnx

  6. nitz
    • 3 years ago
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    \[e ^{iat}=\cos(at)+isin(at)\] taking laplace on both sides:: \[L(e ^{iat})=L(\cos (at))+iL(\sin at)\]...............(1) we know that :\[L(e ^{iat})=1/(s-ia)\] now rationalizing(ie multiplying numerator and denominator by (s+ia): \[L(e ^{iat})=(s+ia)/((s+ia)(s-ia))\] \[L(e ^{iat})=(s+ia)/(s ^{2}+a ^{2})\] \[L(e ^{iat})=(s/(s ^{2}+a ^{2}))+i (a/(s ^{2}+a ^{2}))\]................(2) compare 1 and 2::

  7. nitz
    • 3 years ago
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    U GET : \[L(\cos at)=s/(s ^{2}+a ^{2})\] and \[L(\sin at)=a/(s ^{2}+a ^{2})\]

  8. S_Student
    • 3 years ago
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    it is enough for me..

  9. S_Student
    • 3 years ago
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    thanx dear..

  10. UnkleRhaukus
    • 3 years ago
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    Slight variation ____________ \[\begin{align*} \mathcal L\left\{ \sin({n}t)\right\}(p)&=\int\limits_0^\infty \sin({n}t)e^{-{p}t}\text dt\\ \\&=\int\limits_0^\infty {\frak I}\left(e^{i{n}t}\right)e^{-{p}t}\text dt\\ \\&={\frak I}\left(\int\limits_0^\infty e^{i{n}t}e^{-{p}t}\text dt\right)\\ \\&={\frak I}\left(\int\limits_0^\infty e^{-({p}-i{n})t}\text dt\right)\\ \\&={\frak I}\left(\left.\frac{e^{-({p}-i{n})t}}{-({p}-i{n})}\right|_0^\infty\right)\\ \\&={\frak I}\left(\frac{1}{{p}-i{n}}\right)\\ \\&={\frak I}\left(\frac{1}{{p}-i{n}}\times\frac{{p}+i{n}}{{p}+i{n}}\right)\\ \\&={\frak I}\left(\frac{{p}+i{n}}{{p}^2+{n}^2}\right)\\ \\&= \frac{n}{{p}^2+{n}^2}\\ \end{align*}\] ____________ \[\begin{align*} \mathcal L\left\{\cos ({n}t)\right\}(p)&=\int\limits_0^\infty \cos({n}t)e^{-{p}t}\text dt\\ \\&=\int\limits_0^\infty {\frak R}\left(e^{i{n}t}\right)e^{-{p}t}\text dt\\ \\&={\frak R}\left(\int\limits_0^\infty e^{i{n}t}e^{-{p}t}\text dt\right)\\ \\&={\frak R}\left(\int\limits_0^\infty e^{-({p}-i{n})t}\text dt\right)\\ \\&={\frak R}\left(\left.\frac{e^{-({p}-i{n})t}}{-({p}-i{n})}\right|_0^\infty\right)\\ \\&={\frak R}\left(\frac{1}{{p}-i{n}}\right)\\ \\&={\frak R}\left(\frac{1}{{p}-i{n}}\times\frac{{p}+i{n}}{{p}+i{n}}\right)\\ \\&={\frak R}\left(\frac{{p}+i{n}}{{p}^2+{n}^2}\right)\\ \\&= \frac{{p}}{{p}^2+{n}^2}\\ \end{align*}\]

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