,-.). '.)*' / 0 f ( x ) 0 1 - PowerPoint PPT Presentation

�� ! "#� $ "#%& '()�* ��+ �,-.)*. '*.)*' /�� 0 f ( x ) 0 ��1�� [ − L 2 , L 2 ] ! ( ) f x − L L 2 2 �� 2 2�� 3�� f ( x ) �� [ − L 2 , L 2 ] 2�� 0 ˜ f ( x ) # � �� 4�� 4�� 0 { φ n ( x ) = e 2 πinx/L , n ∈ � � } 0 �� ∞ ˜ c n e 2 πinx/L f ( x ) = n = −∞ 2�� c n �� 5��#

�� ! "#� $ "#%& '()�* 6��+ �� 4�� 2� �� # �� & 7*.),�)/ 2�� L 0 �� φ n ( x + L ) = φ n ( x ) 0 �� & �6& ,.�8,�,9�( �� [ − L 2 , L 2 ] 0 �� 0 , �� n � = k � L 2 e 2 πinx/L e − 2 πikx/L dx = L, �� n = k − L ��& 2 7�� & �� ˜ f ( x + L ) = ˜ f ( x ) # �� ˜ f ( x ) �� 3�� f ( x ) 0 4�� f ( x ) �� [ − L 2 , L 2 ] 0 �� p ( x ) ! ( ) p x x − 3 3 L − L L L 2 2 2 2

�� ! "#� $ "#%& '()�* +��+ 7�� 6& ��4�� 5��0 c n 0 �� ˜ f ( x ) ! � � L ∞ L � 2 2 f ( x ) e − 2 πikx/L dx = e 2 πinx/L e − 2 πikx/L dx c n − L − L n = −∞ 2 2 = 0 + 0 + . . . + c k × L + . . . + 0 + 0 � L c n = 1 2 f ( x ) e − 2 πinx/L dx ��### L − L 2 *3�� ! �� ˜ f ( x ) 2�� f ( x ) = | x | �� [ − 1 , 1] � 1 � 1 c n = 1 f ( x ) e − πinx dx = 1 f ( x ) [cos( nπx ) − i sin( nπx )] dx 2 2 − 1 − 1 � 1 � 1 � x � 1 = 2 0 − 1 x cos( nπx ) dx = nπ sin( nπx ) sin( nπx ) dx 2 nπ 0 0 � 0 , � � 1 = ( − 1) n − 1 �� n �� ( n � = 0) = − 1 − 1 nπ cos( nπx ) = − 2 n 2 π 2 n 2 π 2 , �� n �� nπ 0

�� ! "#� $ "#%& '()�* "��+ c 0 = 1 2 ��& ∞ � e inπx �� f ( x ) = 1 2 − 2 ˜ π 2 n 2 � � −∞ � � �� ∞ = 1 2 − 4 cos( nπx ) π 2 n 2 n =1 , 3 , 5 ,... ��& '�� 8�2 �� 5�� | c n | �� 2�� n → ∞ : �� *3�� 2� ��! 2 | c n | = n 2 π 2 , n �� | c 0 | = 1 2 | c n | = 0 , n �� ( n � = 0) �� 5�� | c n | �� 1 n 2

�� ! "#� $ "#%& '()�* ;��+ abs ( ) c n 05 . 0203 . 0023 . n − 2 − 1 1 2 Coefficients of LOW frequency modes Coefficients of HIGH frequency modes *3�� 6 �� ˜ f ( x ) 2�� f ( x ) = x �� [ − 1 , 1] c n = i ( − 1) n , c 0 = 0 nπ | c n | = 1 nπ, | c 0 | = 0 �� 5�� | c n | �� 1 n �� ∞ � ∞ ( − 1) n sin( nπx ) ( − 1) n e inπx f ( x ) = i = − 2 ˜ ��& π n π n n = −∞ n =1

�� ! "#� $ "#%& '()�* %��+ abs ( ) c n 031 . 016 . 011 . n − 2 − 1 1 2 Coefficients of LOW frequency modes Coefficients of HIGH frequency modes )� �� 2� �� 2�� ### �� p ( x ) �� f ( x ) �� [ − L 2 , L 2 ] 0 �� 7�� /�� 1 p ( x ) �� # n /�� 1 p ( x ) �� p ′ ( x ) �� n 2 p ( x ) �� /�� 1 p ′ ( x ) �� n 3 p ′′ ( x ) ��

�� ! "#� $ "#%& '()�* <��+ ��0 �� f ( x ) ��0 �� 2�� 2�� (��*.& *3�� + �� ˜ f ( x ) 2�� f ( x ) = x, x ∈ [0 , 1] 1 Option (1) − 1 1 1 1 1 Option (2) − 1 1 �� & 4��:

�� ! "#� $ "#%& '()�* =��+ �� >�� *3$ �� 6 �� 2�� >�� n ### Example 1 till n = 0 till n = 1 till n = 3 till n = 5 till n = 7

�� ! "#� $ "#%& '()�* ?��+ Example 2 1 till n = 0 0.5 0 −0.5 −1 −3 −2 −1 0 1 2 3 1 till n = 1 0.5 0 −0.5 −1 −3 −2 −1 0 1 2 3 1 till n = 3 0.5 0 −0.5 −1 −3 −2 −1 0 1 2 3 2 till n = 5 1 0 −1 −2 −3 −2 −1 0 1 2 3 2 till n = 7 1 0 −1 −2 −3 −2 −1 0 1 2 3

,-.). '.)*' / 0 f ( x ) 0 1 - PowerPoint PPT Presentation

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,-.)*. '*.)*' / 0 f ( x ) 0 1 - PowerPoint PPT Presentation

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Outline Evolution of neurocomputing Artificial neural networks Feed forward

Joint Feature Selection in Distributed Stochastic Learning for Large-Scale Discriminative SMT

Feature Selection CE-725: Statistical Pattern Recognition Sharif University of Technology

We are familiar with the Fourier Series in which a function x of period T can be represented as a

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Topic 4: Continuous-Time Fourier Transform (CTFT) o Introduction to Fourier Transform o Fourier

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The Fourier Transform Saravanan Vijayakumaran sarva@ee.iitb.ac.in Department of Electrical

GCT535- Sound Technology for Multimedia Fourier Representations of Audio Graduate School of

Signals and Systems Fall 2003 Lecture #6 23 September 2003 1. CT Fourier series reprise,

Classification and feat u re engineering MAC H IN E L E AR N IN G FOR TIME SE R IE S DATA IN

Chapter 14: Fourier Transforms and Boundary Value Problems in an Unbounded Region Department of

Fourier Transform Saravanan Vijayakumaran sarva@ee.iitb.ac.in Department of Electrical

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,-.). '.)*' / 0 f ( x ) 0 1 - PowerPoint PPT Presentation