f k k = k , f k f ( ) k = k - PowerPoint PPT Presentation

Lesson 4 D ISCRETE F OURIER T RANSFORM

• We saw last lecture that Fourier series can be thought of as a projection onto the complex exponentials, using the continuous � 2 inner product: ∞ ∞ f k � � k θ = ˆ � � � � k θ , f � � k θ � � f ( θ ) ∼ k = −∞ k = −∞ • We will see now that the approximate Fourier series can also be thought of as a projection onto the complex exponentials, using the discretized inner product: β β k � � k θ = ˆ � � f m � � k θ , f m � � k θ � � f α , β ( θ ) = k = α k = α � Recall m = β − α + 1 so that the number of coefficients match the number of quadrature points • We will further view the map from the values to the approxi- mate coefficients as a linear operator (i.e., matrix), which is called the discrete Fourier transform • This will be used to see that the approximate Fourier series interpolates at the quadrature points

• We saw last lecture that Fourier series can be thought of as a projection onto the complex exponentials, using the continuous � 2 inner product: ∞ ∞ f k � � k θ = ˆ � � � � k θ , f � � k θ � � f ( θ ) ∼ k = −∞ k = −∞ • We will see now that the approximate Fourier series can also be thought of as a projection onto the complex exponentials, using the discretized inner product: β β k � � k θ = ˆ � � f m � � k θ , f m � � k θ � � f α , β ( θ ) = k = α k = α � Recall m = β − α + 1 so that the number of coefficients match the number of quadrature points • We will further view the map from the values f ( θ 1 ) , . . . , f ( θ m ) to the approxi- mate coefficients f m β as a linear operator (i.e., matrix), which is called the α , . . . , f m discrete Fourier transform • This will be used to see that the approximate Fourier series interpolates f at the quadrature points

• Recall the discrete orthogonal properties of complex exponentials: � � k θ , � � j θ � m = ( − 1) j − k for � j − k = . . . , − 2 m, − m, 0 , m, 2 m, . . . � � k θ , � � j θ � otherwise � m = 0 • When we restrict our attention to � � αθ , . . . , � � βθ this implies that the complex exponentials are orthogonal (assuming that m ≥ β − α + 1 ) • Thus we get that the approximate Fourier series is a projection � Because we are in the finite dimensional setting, we immediately see that β � � k θ = f α , β ( θ ) � � � k θ , f α , β � � k = α

� � � � � � � � � � ��� 5 θ = � 5 � θ + � − 5 � θ E XAMPLE 2 • For f ( θ ) = ��� 5 θ , we have m = 5 1.0 2 5 � � k θ = − 1 � � � k θ , f � � 0.5 Re k = − 2 q - 3 - 2 - 1 1 2 3 • Increasing to we still haven't resolved - 0.5 - 1.0 True function m = 5 1.0 0.5 • Finally, for we resolve the function Approximate Fourier series Im q - 3 - 2 - 1 1 2 3 - 0.5 - 1.0

� � � � ��� 5 θ = � 5 � θ + � − 5 � θ E XAMPLE 2 • For f ( θ ) = ��� 5 θ , we have m = 10 1.0 2 5 � � k θ = − 1 � � � k θ , f � � 0.5 Re k = − 2 q - 3 - 2 - 1 1 2 3 • Increasing m to 10 we still haven't resolved - 0.5 - 1.0 4 10 � � k θ = � − 5 � θ � � � k θ , f � � m = 10 1.0 k = − 5 0.5 • Finally, for we resolve the function Im q - 3 - 2 - 1 1 2 3 - 0.5 - 1.0

��� 5 θ = � 5 � θ + � − 5 � θ E XAMPLE 2 • For f ( θ ) = ��� 5 θ , we have m = 100 1.0 2 5 � � k θ = − 1 � � � k θ , f � � 0.5 Re k = − 2 q - 3 - 2 - 1 1 2 3 • Increasing m to 10 we still haven't resolved - 0.5 - 1.0 4 10 � � k θ = � − 5 � θ � � � k θ , f � � m = 100 1.0 k = − 5 0.5 • Finally, for m > 10 we resolve the function Im q - 3 - 2 - 1 1 2 3 49 100 � � k θ = f ( θ ) � � � k θ , f � � - 0.5 k = − 50 - 1.0

D ISCRETE F OURIER T RANSFORM

� � � � � � � � • We want to interpret the map from (exact) values � � f ( θ 1 ) . . f m := � � . � � f ( θ m ) to approximate Fourier coefficients ˆ f m � � α . ˆ . f α , β := � � . � � ˆ f m β as a linear operator • We can rewrite the approximate coefficients in terms of :

• We want to interpret the map from (exact) values � � f ( θ 1 ) . . f m := � � . � � f ( θ m ) to approximate Fourier coefficients ˆ f m � � α . ˆ . f α , β := � � . � � ˆ f m β as a linear operator • We can rewrite the approximate coefficients in terms of f m : m m = 1 ˆ � � � k θ , f f ( θ j ) � − � k θ j f m � � = k m j =1 = 1 � − � k θ 1 , . . . , � − � k θ m � � f m m

• We want to interpret the map from (exact) values • It follows that ˆ f α , β = F α , β f m � � f ( θ 1 ) . for the discrete Fourier transform matrix . f m := � � . � � f ( θ m ) � − � αθ 1 � − � αθ m   · · · F α , β := 1 . . ... to approximate Fourier coefficients   . . . .   m � − � βθ 1 � − � βθ m · · · ˆ f m � � α . ˆ . f α , β := � � . � � ˆ f m β as a linear operator • We can rewrite the approximate coefficients in terms of f m : m m = 1 ˆ � � � k θ , f f ( θ j ) � − � k θ j f m � � = k m j =1 = 1 � − � k θ 1 , . . . , � − � k θ m � � f m m

I NTERPOLATION

� � • We say that g interpolates f at the points θ = ( θ 1 , . . . , θ m ) � if f ( θ ) = g ( θ ) • In other words, f ( θ i ) = g ( θ i ) for i = 1 , . . . , m • Problem: choose coefficients so that interpolates at

• We say that g interpolates f at the points θ = ( θ 1 , . . . , θ m ) � if f ( θ ) = g ( θ ) • In other words, f ( θ i ) = g ( θ i ) for i = 1 , . . . , m • Problem: choose coefficients c k so that β � c k � � k θ g ( θ ) = k = α interpolates f at θ

� � • Represent the coefficients as a length m vector: � � c α . . c = � � . � � c β • The Vandermonde matrix is the map from coefficients to values at θ : � � αθ 1 � � βθ 1 � � · · · . . ... � � � � α θ | · · · | � � β θ . . V = = � � . . � � � � αθ m � � βθ m · · · • Therefore, • What coefficients interpolate at ? Precisely

• Represent the coefficients as a length m vector: � � c α . . c = � � . � � c β • The Vandermonde matrix is the map from coefficients to values at θ : � � αθ 1 � � βθ 1 � � · · · . . ... � � � � α θ | · · · | � � β θ . . V = = � � . . � � � � αθ m � � βθ m · · · • Therefore, β c k � � k θ = V c � g ( θ ) = k = α • What coefficients interpolate f at θ ? Precisely c = V − 1 f ( θ )

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � • We now show that F = V − 1 • In other words, the discrete Fourier transform interpolates f at the points θ • This follows immediately: � � � � − � � θ . F V = 1 � � � � � θ | · · · | � � � θ . � � . � � m � � � � − � � θ �� � � � � . . ... . . . . . . ... . . . .

� � � � � � � � � � � � � � � � • We now show that F = V − 1 • In other words, the discrete Fourier transform interpolates f at the points θ • This follows immediately: � � � � − � � θ . F V = 1 � � � � � θ | · · · | � � � θ . � � . � � m � � � � − � � θ � � � θ � � � � � θ � � � � � � � � � θ � � � θ · · · . . = 1 ... � � . . � � . . m � � � � � θ � � � � � θ � � � � � � � � � θ � � � θ · · · �� � � � � . . ... . . . .

• We now show that F = V − 1 • In other words, the discrete Fourier transform interpolates f at the points θ • This follows immediately: � � � � − � � θ . F V = 1 � � � � � θ | · · · | � � � θ . � � . � � m � � � � − � � θ � � � θ � � � � � θ � � � � � � � � � θ � � � θ · · · . . = 1 ... � � . . � � . . m � � � � � θ � � � � � θ � � � � � � � � � θ � � � θ · · · � � � �� , � � �� � � � � �� , � � �� � � � · · · m m . . ... . . = � � . . � � � � � �� , � � �� � � � � �� , � � �� � · · · m m = I

A LIASING F ORMULA

��� � • Define the following � p norms: � � � f � � 1 = | f k | , k = �� Absolutely converging � � � | f k | 2 , � � � f � � 2 = Fourier series � k = �� � f � � ∞ = | f k | �� <k< � � � � C � for f = � . . . , f � 1 , f 0 , f 1 , . . . � More generally, we can define • The norm defines the spaces: if