Laplace Transform cos(ax)

Laplace Transform cos(ax) dated 25-06-2024 shahbaz ahmed June 2024 Introduction to Laplace Transform Let f(x) be defined for 0 ≤ x < ∞ and let s denote an arbitrary real variable. The Laplace transform of f (x), designated by either L or F (s), is R∞ L {f (x)}=F (s)= 0 e−sx f (x)dx for all values of s for which the improper integral converges. Convergence occur when the limit RR limR−→∞ 0 e−sx f (x)dx exists. Find L {cos ax} Solution Since R∞ L {f (x)}=F (s)= 0 e−sx f (x)dx Putting f (x) = cos ax R∞ L {cos ax} = F (s) = 0 e−sx cos ax dx 1 L {cos ax} = F (s) = limR−→∞ RR 0 e−sx cos ax dx d(sin ax) = cos ax d(ax) = a cos ax dx 1 d(sin ax) a = cos ax dx L {cos ax} = limR−→∞ RR 0 e−sx a1 d(sin ax) RR limR−→∞ 0 e−sx d(sin ax) RR aL {cos ax} = limR−→∞ 0 e−sx d(sin ax) RR aL {cos ax} = limR−→∞ [ sinesxax − 0 sin ax d(e−sx )] RR aL {cos ax} = limR−→∞ [ sinesxax + s 0 e−sx sin ax dx] L {cos ax} = 1 a d(cos ax) = − sin ax d(ax) = −a sin ax dx d(cos ax) = −a sin ax dx − a1 d(cos ax) = sin ax dx aL {cos ax} = limR−→∞ [ sinesxax + s RR 0 e−sx (− a1 )d(cos ax)] RR aL {cos ax} = limR−→∞ [ sinesxax − as 0 e−sx d(cos ax)] i h RR sin ax s cos ax −sx aL {cos ax} = limR−→∞ [ esx − a [ esx − 0 cos ax d(e )] h i RR sin ax s cos ax −sx aL {cos ax} = limR−→∞ [ esx − a [ esx + s 0 cos ax e dx] aL {cos ax} = limR−→∞ [ sinesxax − aL {cos ax} + (a + s2 )L {cos ax} a 2 +s2 (a s2 L {cos ax} a a s cos ax a esx − s2 L {cos ax}] a = limR−→∞ [ sinesxax − = limR−→∞ [ sinesxax − s cos ax R ] a esx 0 s cos ax R ] a esx 0 1 1 )L {cos ax} = limR−→∞ [sin aR( esR ) − as cos aR( esR ) + as ] Since sin aR and cos aR are finite numbers for all values of R .Also limR−→∞ 2 +s2 (a a 1 esR = 0.So )L {cos ax} = L {cos ax} = s a L {cos ax} = s a2 +s2 × s a a a2 +s2 for (s > 0) 2 Or F (s) = s a2 +s2 for (s > 0) Applications of Laplace Transform Find area under the curve y = e−7x cos 8x Solution Here s = 7, a = 8 Putting in the formula Area= F (s) = s a2 +s2 Area= F (s) = 7 82 +72 Area= F (s) = 7 113 3