anonymous
  • anonymous
laplace cosat and sinat
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
L(cos at)=\[s/(s ^{2}+a ^{2})\] and L(sin at)=\[a/(s ^{2}+a ^{2})\]
anonymous
  • anonymous
if u are interested i can tell u the derivation for these
anonymous
  • anonymous
thanx friend...and ya i want to expain me in detail..

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anonymous
  • anonymous
ok ...
anonymous
  • anonymous
thnx
anonymous
  • anonymous
\[e ^{iat}=\cos(at)+isin(at)\] 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::
anonymous
  • anonymous
U GET : \[L(\cos at)=s/(s ^{2}+a ^{2})\] and \[L(\sin at)=a/(s ^{2}+a ^{2})\]
anonymous
  • anonymous
it is enough for me..
anonymous
  • anonymous
thanx dear..
UnkleRhaukus
  • UnkleRhaukus
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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