Question

laplace cosat and sinat

Answer 1

L(cos at)=\[s/(s ^{2}+a ^{2})\]
and L(sin at)=\[a/(s ^{2}+a ^{2})\]

Answer 2

if u are interested i can tell u the derivation for these

Answer 3

thanx friend...and ya i want to expain me in detail..

Answer 4

ok ...

Answer 5

thnx

Answer 6

\[e ^{iat}=\cos(at)+isin(at)\]
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::

Answer 7

U GET :
\[L(\cos at)=s/(s ^{2}+a ^{2})\]
and \[L(\sin at)=a/(s ^{2}+a ^{2})\]

Answer 8

it is enough for me..

Answer 9

thanx dear..

Answer 10

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