Ch_05 - Fourier Series - Examples

# Ch_05 - Fourier Series - Examples - Examples of Fourier...

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Examples of Fourier Series, Integrals and Transforms

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The Fourier Series and Integrals, Transforms Fourier Series Fourier Series (p.498) Example 1: Complex Fourier Series Find the complex Fourier series of the following periodic functions () x f xe = for || x π < , (2 ) ( ) f xf x + = Solution:
The Fourier Series and Integrals, Transforms Fourier Series EXAMPLE: Complex Fourier Series Expansion Find the complex form of the Fourier series of the following periodic functions 1 for 0 1 () 0 for 1 2 x fx x < < = < < Solution: Method I: 2 1 ( ) 2 nx i aL L n a Cf x e d x L π + = and 1 L = 1 12 01 0 1 1 0 22 11 ( 1 ) ( 1 ( 1 ) ) 2 ( 2 1 ) in x in x in x n in n n e Ce d x e d x in ei i ni n n ππ −− ⎡⎤ ⎧⎫ =⋅ + = ⎢⎥ ⎨⎬ ⎩⎭ ⎣⎦ == = ∫∫ since /2 cos sin cos sin i i i i =+ = =− = and 2 cos sin ( 1) cos 2 sin 2 1 n en i n i n ± ± = ±= = Therefore, (2 1) in x i L n nn ie n ∞∞ =−∞ ∑∑ Method II: Use the relationship of 2 n ai b C = and 2 n b C + = Then, since 1 L = 1 1 0 0 1 1 0 0 1s i n 1 c o s 0 1 1c o s 1 o s 1(1 ) 2 1) n n n an x d x n bn x d x n n n = = = = Then, ( 2 1 ) n n b i b i C n =

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The Fourier Series and Integrals, Transforms Fourier Series (p.480) Example 1: Periodic Rectangular Wave Find the Fourier coefficients of the following periodic function () f x if 0 () if 0 kx fx π −− < < = < < and (2 ) ( ) f xf x + = Solution:
The Fourier Series and Integrals, Transforms Fourier Series (p.488) Example 1: Periodic Rectangular Wave Find the Fourier series of the following function 0 if 2 1 ( ) if 1 1 0 if 2 2 x fx k x x −<< =− < < < < 24 pL = = , 2 L = Solution:

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The Fourier Series and Integrals, Transforms Fourier Series (p.488) Example 2: Periodic Rectangular Wave Find the Fourier series of the following function if 2 0 () if 0 2 kx fx −− < < = < < 24 pL = = , 2 L = Solution:
The Fourier Series and Integrals, Transforms Fourier Series (p.489) Example 3: Half-wave Rectifier Fourier series expansion of half-wave rectifier. 0(0 ) () sin ( 0 ) Lx fx Ex x L ω −<< = < < L π = Solution: The period is 2/ πω , [] // 0 /0 00 ( ) s i n 22 sin cos sin(1 ) sin(1 ) 2 n E af x d x E x d x E aE x n x d x n x n x d x ωω ππ == = + + ∫∫ 2 1c o s ( 1 ) o s ( 21 1 2 212 1 2 (2 1) n En n a nn EE n −+ −− ⎛⎞ =+ ⎜⎟ +− ⎝⎠ = With the same manner, 1 2 E b = and 0 n b = for 2, 3, n = " The function () f x can be represented as 1 2c o s 2 s i n 2( 2 1 ) ( 2 1 ) n E n x x = + For 1 =

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The Fourier Series and Integrals, Transforms Fourier Series EXAMPLE: Fourier Series Expansion Establish the Fourier series which reproduces the periodic function for 0 () for 0 x fx xx ππ π −<< = < < Hence show that 2 2 0 1 8 (2 1) n n = = + Solution: Find the Fourier coefficients 0 22 , 4 1c o s 1 2 (2 12 c o s (1 ) n n n a n a nn n b =− =⋅ −− == 2 1 2cos(2 1 ) s in ( 1 2 )) 4 n n nx n x n n = =− − + − − ⎩⎭ At 0 x = 2 1 21 (0) 24 ( 2 1 ) n f n = or 2 0 4( 2 1 ) n n = = + 2 2 0 1 8 n n = = + NOTE: cos( ) ( 1) n n , cos( 1) n n + =−− cos( 1/ 2) 0 n + = sin( ( 1) n n += sin( ) 0 n = 1(1 2 n ⎢⎥ = ⎣⎦ if nm 0 n −= 2 = 2 n +− = 2
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## This note was uploaded on 04/27/2011 for the course CHEM 101 taught by Professor Suh during the Spring '11 term at University of Toronto.

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Ch_05 - Fourier Series - Examples - Examples of Fourier...

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