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b) A Fourier series can be written as 10 f(t) = 2/2 + (ancos (17) + bn n=1 Given that Fourier coefficient an i) Find a.. 25 (2(nn)³ 2²7) sin nπ, + 4nn. 5

b) A Fourier series can be written as 10 f(t) = 2/2 + (ancos (17) + bn n=1 Given that Fourier coefficient an i) Find a.. 25 (2(nn)³ 2²7) sin nπ, + 4nn. 5

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.4: Values Of The Trigonometric Functions
Problem 22E
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b) A Fourier series can be written as Given that Fourier coefficient i) i) Find ao. Show that bn = f(t) = 2/2 + Σ (ancos (17²) + b₂ sin ( n=1 25 2(nn)³ +--/-) nπ 25 2(nn)³ + -2) 4nn cos(5 4nn. sin (+7) + 5 (nt)). 25 2(nn)³ 15 (nπ)²*
Transcribed Image Text:b) A Fourier series can be written as Given that Fourier coefficient i) i) Find ao. Show that bn = f(t) = 2/2 + Σ (ancos (17²) + b₂ sin ( n=1 25 2(nn)³ +--/-) nπ 25 2(nn)³ + -2) 4nn cos(5 4nn. sin (+7) + 5 (nt)). 25 2(nn)³ 15 (nπ)²*
Function a (constant) f t" sin at cos at sinh at cosh at eat f(t) f(t-a)u(t-a) t" f(t) f(n) (t) f(u) du f(u)g(t-u) du Trigonometric Identities: APPENDIX TABLE OF LAPLACE TRANSFORMS Laplace Transforms a S 1 s-a n! 50+1 a s² + a² S + a 2 a s²-a² S 2 s² - a² 2 F(s-a) e-as F(s) (-1)"F(")(s) s"F(s)-sn-1f(0)-sn-2f (¹) (0) -- F(s) S F(s) G(s) 2sinAcosB = sin (A + B) + sin (A - B) 2cosAcosB = cos (A - B) + cos (A + B) 2sinAsinB = cos (A-B)- cos (A + B) -f(n-1) (0) sin 2A = 2 sin A cos A cos 2A = cos²A - sin³A = 2 cos²A-1 = 1-2 sin²A sin (A + B) = sin A cos B ± cos A sin B sin A sin B cos (A + B) = cos A cos B cosh?A – sinh?A = 1 sinh (A + B) = sinh A cosh B + cosh A sinh B cosh (A + B) = cosh A cosh B+ sinh A sinh B
Transcribed Image Text:Function a (constant) f t" sin at cos at sinh at cosh at eat f(t) f(t-a)u(t-a) t" f(t) f(n) (t) f(u) du f(u)g(t-u) du Trigonometric Identities: APPENDIX TABLE OF LAPLACE TRANSFORMS Laplace Transforms a S 1 s-a n! 50+1 a s² + a² S + a 2 a s²-a² S 2 s² - a² 2 F(s-a) e-as F(s) (-1)"F(")(s) s"F(s)-sn-1f(0)-sn-2f (¹) (0) -- F(s) S F(s) G(s) 2sinAcosB = sin (A + B) + sin (A - B) 2cosAcosB = cos (A - B) + cos (A + B) 2sinAsinB = cos (A-B)- cos (A + B) -f(n-1) (0) sin 2A = 2 sin A cos A cos 2A = cos²A - sin³A = 2 cos²A-1 = 1-2 sin²A sin (A + B) = sin A cos B ± cos A sin B sin A sin B cos (A + B) = cos A cos B cosh?A – sinh?A = 1 sinh (A + B) = sinh A cosh B + cosh A sinh B cosh (A + B) = cosh A cosh B+ sinh A sinh B
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