Fourier Series Code. James K. Peterson. April 9, Department of Biological Sciences and Department of Mathematical Sciences Clemson University

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1 Fourier Series Code James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University April 9, 2018

2 Outline 1

3 We will need to approximate Fourier series expansions for arbitrary functions f when we solve the cable equation. All of these approximations require that we find inner products of the form < f, u n > for some function u n which is a sin or cos term. We must make sure our sin and cosine families are orthogonal so the Fourier coefficients can be accurate. If we have a familiy of functions which should be orthogonal, {g i } a standard test to see if our numerical inner product calculations are sufficiently accurate is to compute the N N matrix d = (< g i, g j >) which should be the essentially the N N identify matrix as the off diagonal entries should all be zero. However, if the inner product computations are inaccurate, off diagonal values need not be zero and our Fourier coefficients will be wrong.

4 1 f u n c t i o n c = i n n e r p r o d u c t ( f, g, a, b,n) h t ) f ( t ) g ( t ) ; 6 d e l x = ( b a ) /N; x = l i n s p a c e ( a, b,n+1) ; c = 0 ; f o r i =1:N c = c+ h ( x ( i ) ) d e l x ; 11

5 We can check how accurate we are with these inner product calculation by using them to do a Graham-Schmidt orthogonalization on the functions 1, t, t 2,..., t N on the interval [a, b]. The code to do this is in the function GrahamSchmidtTwo. This function returns the new orthogonal of length one functions g i and also prints out the matrix with terms < g i, g j > which should be an identity. The input NIP is the number of terms to use in the Riemann sum approximations to the inner product and N is size of the number of functions we perform GSO on.

6 f u n c t i o n g = GrahamSchmidtTwo ( a, b, N, NIP ) Perform Graham Schmidt O r t h o g o n a l i z a t i o n on a s e t o f f u n c t i o n s t ˆ 0,..., t ˆN 5 Setup f u n c t i o n h a n d l e s f = S e t U p F u n c t i o n s (N) ; g = c e l l (N+1,1) ; 10 n f = s q r t ( i n n e r p r o d u c t ( f {1}, f {1}, a, b, NIP ) ) ; g{1} x ) f {1}( x ) / n f ; d = z e r o s (N+1,N+1) ; f o r k =2:N+1 compute n e x t o r t h o g o n a l p i e c e 15 p h i x ) 0 ; f o r j = 1 : k 1 c = i n n e r p r o d u c t ( f {k }, g{ j }, a, b, NIP ) ; p h i x ) ( p h i ( x )+c g{ j }( x ) ) ; 20 p s i x ) ( f {k }( x ) p h i ( x ) ) ; n f = s q r t ( i n n e r p r o d u c t ( p s i, p s i, a, b, NIP ) ) ; g{k} x ) ( p s i ( x ) / n f ) ; 25 f o r i =1:N+1 f o r j =1:N+1 d ( i, j ) = i n n e r p r o d u c t ( g{ i }, g{ j }, a, b, NIP ) ; 30 d

7 Here is a typical run. g = GrahamSchmidtTwo ( 0, 2, 2, 5 0 ) ; This computes the GSO of the functions 1, t, t 2 on the interval [0, 2] using Riemann sum approximations with 50 points. The matrix < g i, g j > which is calculated is e e e e e e e e e+00 This is the 3 3 identify that we expect. Next, we do GSO on the functions 1, t,..., t 6. g = GrahamSchmidtTwo ( 0, 2, 6, 5 0 ) ;

8 This also generates the identity matrix we expect for < g i, g j > e e e e e e e e e e e e e e e e e e e e e e e e e+00

9 To find the GSO of arbitary functions, use the GrahamScmidtThree function. f u n c t i o n g = GrahamSchmidtThree ( f, a, b, NIP ) Perform Graham Schmidt O r t h o g o n a l i z a t i o n on a s e t o f f u n c t i o n s f Setup f u n c t i o n h a n d l e s 5 N = l e n g t h ( f ) ; g = c e l l (N, 1 ) ; n f = s q r t ( i n n e r p r o d u c t ( f {1}, f {1}, a, b, NIP ) ) ; g{1} x ) f {1}( x ) / n f ; 10 d = z e r o s (N,N) ; f o r k =2:N compute n e x t o r t h o g o n a l p i e c e p h i x ) 0 ; f o r j = 1 : k 1 15 c = i n n e r p r o d u c t ( f {k }, g{ j }, a, b, NIP ) ; p h i x ) ( p h i ( x )+c g{ j }( x ) ) ; p s i x ) ( f {k }( x ) p h i ( x ) ) ; n f = s q r t ( i n n e r p r o d u c t ( p s i, p s i, a, b, NIP ) ) ; 20 g{k} x ) ( p s i ( x ) / n f ) ; check o r t h o g o n a l i t y f o r i =1:N f o r j =1:N 25 d ( i, j ) = i n n e r p r o d u c t ( g{ i }, g{ j }, a, b, NIP ) ; d

10 To use this, we have to setup a cell structure to hold our list of functions. For example, if f (t) = t 2, g(t) = sin(4t + 3) and h(t) = 1/(1 + t 2 ), we would do the GSO as follows: f t ) t. ˆ 2 ; g t ) s i n (4 t +3) ; h t ) 1./(1+ t. ˆ 2 ) ; F = c e l l ( 3, 1 ) ; F{1} = f ; F{2} = g ; F{3} = h ; G = GrahamSchmidtThree (F, 1,1,300) ; d =

11 We can then see the graphs of the original functions using the basic code T = l i n s p a c e ( 1,1,51) ; p l o t (T, F{1}(T),T, F{2}(T),T, F{3}(T) ) ; x l a b e l ( Time ) ; y l a b e l ( O r i g i n a l F u n c t i o n s ) ; t i t l e ( O r i g i n a l F u n c t i o n s on [ 1,1] ) ;

12 We can then plot the new orthonormal basis using the code T = l i n s p a c e ( 1,1,51) ; p l o t (T, G{1}(T),T, G{2}(T),T, G{3}(T) ) ; x l a b e l ( Time ) ; y l a b e l ( B a s i s F u n c t i o n s ) ; t i t l e ( B a s i s F u n c t i o n s on [ 1,1] ) ;

13 To compute the first N terms of the Fourier Cosine series of a function f on ( the interval [0, L], we first need a way to encode all the functions nπ cos L ). x We do this in the function SetUpCosines. f u n c t i o n f = S e t U p C o s i n e s ( L,N) Setup f u n c t i o n h a n d l e s 5 f = c e l l (N+1,1) ; f o r i =1:N+1 f { i } x ) c o s ( ( i 1) p i x /L ) ;

14 ( π This generates handles to the functions 1, cos L ), x cos ( forth ing with cos Nπ L x ). A similar function encodes the ( 2π L x ) and so corresponding sin functions we need for the first N terms of a Fourier Sine series. The function is called SetUpSines with code f u n c t i o n f = S e t U p S i n e s ( L,N) Setup f u n c t i o n h a n d l e s 5 f = c e l l (N, 1 ) ; f o r i =1:N f { i } x ) s i n ( i p i x /L ) ; ( π This generates handles to the functions sin L ), x sin ( Nπ forth ing with sin L ). x ( 2π L x ) and so

15 Let s check how accurate our innerproduct calculations are on the sin and cos terms. First, we modify the functions which set up the sin and cos functions to return functions of length one on the interval [0, L]. This is done in the functions SetUpOrthogCos and SetUpOrthogSin. f u n c t i o n f = SetUpOrthogCos ( L,N) Setup f u n c t i o n h a n d l e s 5 f = c e l l (N+1,1) ; f {1} x ) s q r t (1/ L ) ; f o r i =2:N+1 f { i } x ) s q r t (2/ L ) c o s ( ( i 1) p i x /L ) ; and 1 f u n c t i o n f = SetUpOrthogSin ( L,N) Setup f u n c t i o n h a n d l e s f = c e l l (N, 1 ) ; 6 f o r i =1:N f { i } x ) s q r t (2/ L ) s i n ( ( i ) p i x /L ) ;

16 Then, we can check the accuracy of the inner product calculations by computing the matrix d = (< f i, f j >). We do this with the new function CheckOrtho We input the function f and interval [a, b] points a and b and the number of terms to use in the Riemann sum approximation of the inner product, NIP. f u n c t i o n CheckOrtho ( f, a, b, NIP ) Perform Graham Schmidt O r t h o g o n a l i z a t i o n on a s e t o f f u n c t i o n s f 5 Setup f u n c t i o n h a n d l e s N = l e n g t h ( f ) ; f o r i =1:N 10 f o r j =1:N d ( i, j ) = i n n e r p r o d u c t ( f { i }, f { j }, a, b, NIP ) ; d 15 Let s try it with the cos functions. We use a small number of terms for the Riemann sums, NIP = 50 and compute < g i, g j > for the first 7 functions.

17 f = SetUpOrthogCos ( 5, 7 ) ; CheckOrtho ( f, 0, 5, 5 0 ) ; We do not get the identity matrix as we expect some of the off diagonal values are too large e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e+00 Resetting NIP = 200, we find a better result.

18 1. 0 e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e+00 We can do even better by increasing NIP, of course. We encourage you to do these experiments yourself. We did not show the results for the sin terms, but they will be similar. We can then calculate the first N + 1 terms of a Fourier series or the first N terms of a Fourier Sine series using the functions FourierCosineApprox and FourierSineApprox, respectively.

19 Let s look at the Fourier Cosine approximation first. This function returns a handle to the function p which is the approximation p(x) = N n=0 < f, g i > g i (x) ( iπ where g i (x) = cos L ). x It also returns the Fourier cosine coefficients A 1 = 1 L < f, g 0 > through A N+1 = 2 L < f, g N >. We must choose how many points to use in our Riemann sum inner product estimates this is the input variable N. The other inputs are the function handle f and the length L. We also plot our approximation and the original function together so we can see how we did.

20 f u n c t i o n [ A, p ] = F o u r i e r C o s i n e A p p r o x ( f, L,M,N) p i s t h e Nth F o u r i e r A p p r o x i m a t i o n f i s t h e o r i g i n a l f u n c t i o n 5 M i s the number of Riemann sum terms i n the i n n e r product N i s the number of terms i n the approximation L i s t h e i n t e r v a l l e n g t h g e t t h e f i r s t N+1 F o u r i e r c o s a p p r o x i m a t i o n s 10 g = SetUpCosines ( L,N) ; g e t F o u r i e r C o s i n e C o e f f i c i e n t s A = z e r o s (N+1,1) ; A( 1 ) = i n n e r p r o d u c t ( f, g {1},0, L,M) /L ; f o r i =2:N+1 15 A( i ) = 2 i n n e r p r o d u c t ( f, g{ i },0, L,M) /L ; g e t Nth F o u r i e r C o s i n e A p p r o x i m a t i o n p x ) 0 ; f o r i =1:N+1 20 p x ) ( p ( x ) + A( i ) g{ i }( x ) ) ; x = l i n s p a c e ( 0, L, ) ; f o r i =1:101 y ( i ) = f ( x ( i ) ) ; 25 yp = p ( x ) ; f i g u r e s = [ Fourier Cosine Approximation with, i n t 2 s t r (N+1), term ( s ) ] ; p l o t ( x, y, x, yp ) ; 30 x l a b e l ( x axis ) ; y l a b e l ( y axis ) ; t i t l e ( s ) ;

21 We will test our approximations on two standard functions: the sawtooth curve and the square wave. To define these functions, we will use an auxiliary function splitfunc which defines a new function z on the interval [0, L] as follows { f (x) 0 x < L z(x) = 2, L g(x) 2 x L The arguments to splitfunc are the functions f and g, the value of x for we wish to find the output z and the value of L. f u n c t i o n z = s p l i t f u n c ( x, f, g, L ) i f x < L/2 z = f ( x ) ; e l s e z = g ( x ) ;

22 It is easy then to define a sawtooth and a square wave with the following code. The square wave, Sq has value H on [0, L 2 ) and value 0 on [ L 2, L]. In general, the sawtooth curve, Saw, is the straight line connecting the point (0, 0) to ( L 2, H) on the interval [0, L 2 ] and the line connecting ( L 2, H) to (L, 0) on the interval ( L 2, L). Thus, and Sq(x) = { H 0 x < L 0 L 2, 2 x L Saw(x) = { 2 L Hx 0 x < L 2, 2H 2 L Hx L 2 x L f u n c t i o n f = sawtooth ( L,H) f l e f t x ) 2 x H/L ; f r i g h t x ) 2 H 2 H x /L ; f x ) s p l i t f u n c ( x, f l e f t, f r i g h t, L ) ;

23 As an example, we build a square wave and sawtooth of height 10 on the interval [0, 10]. It is easy to plot these functions as well, although we won t show the plots yet. We will wait until we can compare the functions to their Fourier series approximations. However, the plotting code is listed below for your convenience. Note the plot must be set up in a for loop as the inequality checks in the function splitfunc do not handle a vector argument such as the x from a linspace command correctly. f x ) 1 0 ; g x ) 0 ; Saw = s awtooth ( 1 0, 1 0 ) ; Sq x ) s p l i t f u n c ( x, f, g, 1 0 ) ; x = l i n s p a c e ( 0, 1 0, ) ; f o r i =1:101 ysq ( i ) = Sq ( x ( i ) ) ; ysaw ( i ) = Saw ( x ( i ) ) ; p l o t ( x, ysq ) ; a x i s ( [. 1, ] ) ; p l o t ( x, YSaw) ;

24 We can then test the Fourier Cosine Approximation code on a saw function. f x ) 1 0 ; g x ) 0 ; Saw = s awtooth ( 1 0, 1 0 ) ; Sq x ) s p l i t f u n c ( x, f, g, 1 0 ) ; [ Acos, pcos ] = F o u r i e r C o s i n e A p p r o x ( Saw, 1 0, 1 0 0, 5 ) ;

25 We can then test the Fourier Cosine Approximation code on a square wave function. f x ) 1 0 ; g x ) 0 ; Saw = s awtooth ( 1 0, 1 0 ) ; Sq x ) s p l i t f u n c ( x, f, g, 1 0 ) ; [ Acos, pcos ] = F o u r i e r C o s i n e A p p r o x ( Sq, 1 0, 1 0 0, 5 ) ;

26 The code for FourierSineApprox is next. It is quite similar to the cosine approximation code and so we will say little about it. It returns the Fourier sine coefficients as A with ( A 1 = 2 L < f, g 1 > through A N = 2 L < f, g iπ N > where g i (x) = sin L ). x It also returns the handle to the Fourier sine approximation p(x) = N < f, g i ) > g i (x). n=1

27 f u n c t i o n [ A, p ] = F o u r i e r S i n e A p p r o x ( f, L,M,N) p i s t h e Nth F o u r i e r A p p r o x i m a t i o n f i s t h e o r i g i n a l f u n c t i o n M i s the number of Riemann sum terms i n the i n n e r product 5 N i s the number of terms i n the approximation L i s t h e i n t e r v a l l e n g t h g e t t h e f i r s t N F o u r i e r s i n e a p p r o x i m a t i o n s g = SetUpSines ( L,N) ; g e t F o u r i e r S i n e C o e f f i c i e n t s 10 A = z e r o s (N, 1 ) ; f o r i =1:N A( i ) = 2 i n n e r p r o d u c t ( f, g{ i },0, L,M) /L ; g e t Nth F o u r i e r S i n e A p p r o x i m a t i o n 15 p x ) 0 ; f o r i =1:N p x ) ( p ( x ) + A( i ) g{ i }( x ) ) ; x = l i n s p a c e ( 0, L, ) ; 20 f o r i =1:101 y ( i ) = f ( x ( i ) ) ; yp = p ( x ) ; f i g u r e 25 s = [ Fourier Sine Approximation with, i n t 2 s t r (N), term ( s ) ] ; p l o t ( x, y, x, yp ) ; x l a b e l ( x axis ) ; y l a b e l ( y axis ) ; t i t l e ( s ) ; 30

28 Let s test the approximation code on the square wave. using 22 terms. The sin approximations are not going to like the starting value at 10 as you can see in the figure below. The approximation is generated with the command f x ) 1 0 ; g x ) 0 ; Saw = s awtooth ( 1 0, 1 0 ) ; Sq x ) s p l i t f u n c ( x, f, g, 1 0 ) ; [ Asin, p s i n ] = F o u r i e r S i n e A p p r o x ( Sq, 1 0, 1 0 0, 2 2 ) ;

29 Here is another generic type of function to work with: a pulse. This is a function constant of an interval of the form [x 0 r, x 0 + r] and 0 off of that interval. We implement it in pulsefunc. f u n c t i o n z = p u l s e f u n c ( x, x0, r,h) i f x > x0 r && x < x0+r z = H; 5 e l s e z = 0 ; It is easy to use. Here is a pulse centered at 3 of radius 0.5 on the interval [0, 10]. P x ) p u l s e f u n c ( x, 3, 0. 5, 1 0 ) ; [ Asin, p s i n ] = F o u r i e r S i n e A p p r o x (P, 1 0, 1 0 0, 2 2 ) ;

31 We can combine the Fourier Sin and Fourier Cos Approximations by adding them and dividing by two. Here is the code with the plotting portion removed to make it shorter. f u n c t i o n [ A, B] = F o u r i e r A p p r o x ( f, L,M,N) 3 g = SetUpCosines ( L,N) ; B = z e r o s (N+1,1) ; B( 1 ) = i n n e r p r o d u c t ( f, g {1},0, L,M) /L ; f o r i =2:N+1 B( i ) = 2 i n n e r p r o d u c t ( f, g{ i },0, L,M) /L ; 8 p x ) 0 ; f o r i =1:N+1 p x ) ( p ( x ) + B( i ) g{ i }( x ) ) ; 13 h = SetUpSines ( L,N) ; A = z e r o s (N, 1 ) ; f o r i =1:N A( i ) = 2 i n n e r p r o d u c t ( f, h{ i },0, L,M) /L ; 18 q x ) 0 ; f o r i =1:N q x ) ( q ( x ) + A( i ) h{ i }( x ) ) ; 23 x = l i n s p a c e ( 0, L, ) ; f o r i =1:101 y ( i ) = f ( x ( i ) ) ; yp = 0. 5 ( p ( x )+q ( x ) ) ; 28

32 P x ) p u l s e f u n c ( x, 3, 0. 5, 1 0 ) ; [ A, B] = F o u r i e r A p p r o x (P, 1 0, 1 0 0, 2 2 ) ;

33 Homework Generate the Fourier Sine Approximation with varying number of terms and different values of NIP for the periodic square wave. Draw figures as needed Generate the Fourier Sine Approximation with varying number of terms and different values of NIP for pulses applied at various locations. Draw figures as needed Generate the Fourier Cosine Approximation with varying number of terms and different values of NIP for pulses applied at various locations. Draw figures as needed.

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