Math 56 Homework 5 Michael Downs

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1 1. (a) Since f(x) = cos(6x) = ei6x 2 + e i6x 2, due to the orthogonality of each e inx, n Z, the only nonzero (complex) fourier coefficients are ˆf 6 and ˆf 6 and they re both 1 2 (which is also seen from Euler s identity). (b) The function f(x) = e 6ix lies outside the nyquist frequency N = 5 so aliasing will 2 occur. The DFT coefficients will approximate f m for integers N m < N each with 2 2 error + ˆf m 2N + ˆf m N + ˆf m+n ˆfm+2N + (these are the exact fourier coefficients). Since the only nonzero fourier coefficient is ˆf 6 = 1, the only nonzero approximation in the DFT will be f 4 since its error hits ˆf 6, the only nonzero fourier coefficient. The interpolated function will be f(x) = e 4ix. 2. (a) Code: 1 % s p e c t r a l d i f f e r e n t i a t i o n 3 N = 4 0 ; % number o f sample p o i n t s f x ) exp ( s i n ( x ) ) ; % f u n c t i o n we re sampling 5 df x ) exp ( s i n ( x ) ). cos ( x ) ; % i t s d e r i v a t i v e 11 7 gr = 0 : 1 e 3:2 pi ; % the f i n e s t g r i d 9 maxerrors = z e r o s ( 1, N/2) ; n = 2 : 2 :N; f o r j = 2 : 2 :N 13 x = l i n s p a c e ( 0, ( j 1)/ j 2 pi, j ) ; % sample x v a l u e s y = f ( x ). / j ; % samples 15 f t = f f t ( y ) ; % approximated c o e f f i c i e n t s 17 % get d e r i v a t i v e c o e f f i c i e n t s by m u l t i p l y i n g each approximated % c o e f f i c i e n t by i m where m i s i t s frequency / c o e f f i c i e n t number 19 f o r k = 1 : j /2 f t ( k ) = f t ( k ) i ( k 1) ; 21 f t ( j + 1 k ) = f t ( j + 1 k ) i k ; end 23 i n t e r x ) r e a l (sum( f t ( ( j /2 + 1) : j ). exp( 1 i x ( j /2: 1:1) ) ) + sum( f t ( 1 : j /2). exp ( i x ( 0 : ( j /2 1) ) ) ) ) ; 25 e r r o r = max( abs (df( gr ) arrayfun ( i n t e r, gr ) ) ) ; maxerrors ( j /2) = e r r o r ; 27 end 29 f i g u r e ; semilogy (n, maxerrors ) ; 31 x l a b e l ( N ) ; y l a b e l ( max e r r o r ) ; 33 t i t l e ( Error in truncated N term approximation using s p e c t r a l d i f f e r e n t i a t i o n ) 1

2 Plot: 10 2 Error in truncated N term approximation using spectral differentiation max error N (b) Replacing f and df with: 1 f x ) abs ( s i n ( x ) ). ˆ 3 ; % f u n c t i o n we re sampling df x ) 3 s i n ( x ). abs ( s i n ( x ) ). cos ( x ) ; % i t s d e r i v a t i v e and plotting log(y) vs x: 2

3 10 1 Error in truncated N term approximation using spectral differentiation max error N This is clearly not exponential convergence. I plot the errors again using log(y) vs log(x): 10 1 Error in truncated N term approximation using spectral differentiation max error N It appears to be algebraic of order 2 judging from the graph. The function is twice differentiable so by the theorem from HW4 1(e) it should have second order algebraic. It is not infinitely differentiable and therefore does not have super algebraic convergence. 3

4 3. Algorithm: % cooley tukey r e c u r s i v e algorithm f o r the f a s t f o u r i e r transform 2 % takes a row v e c t o r o f weighted samples o f l ength 2ˆn % and outputs the transformed v e c t o r 4 f u n c t i o n r e s = c t f f t ( f ) 6 N = l e n g t h ( f ) ; % base case in r e c u r s i o n when s i z e i s 2 8 i f N == 2 r e s = ( dftmtx ( 2 ) f. ). ; 10 r eturn ; end 12 % s h u f f l e i n t o even and odd v e c t o r s. KEEPING IN MIND THE ZERO INDEXING 14 % SO ACTUALLY IT S REVERSED f e = f ( 1 : 2 :N) ; 16 f o = f ( 2 : 2 :N) ; 18 % r e c u r s e to get t h e i r f f t s e = c t f f t ( f e ) ; 20 o = c t f f t ( f o ) ; 22 % multiply o by twiddle f a c t o r f o r combining l a t e r twid = exp( 1 1 i 2 pi 1/N ( 0 : (N/2 1) ) ) ; o = twid. o ; % combine them 28 r e s = z e r o s ( 1, N) ; r e s ( 1 :N/2) = e ( 1 :N/2) + o ( 1 :N/2) ; 30 r e s ( ( 1 :N/2) + N/2) = e ( 1 :N/2) o ( 1 :N/2) ; end driver: 1 % d r i v e r f o r my c t f f t code. computes norm o f e r r o r between matlab f f t and % my f f t. a l s o compares run times. 3 N = 2ˆ16; % number o f sample p o i n t s 5 f x ) exp ( s i n ( x ) ) ; % f u n c t i o n we re sampling 7 x = l i n s p a c e ( 0, ()/N 2 pi, N) ; % sample x v a l u e s y = f ( x ). /N; % samples 9 t i c ; 11 mlc = f f t ( y ) ; % matlab code l i b = toc ; 13 4

5 t i c ; 15 myc = c t f f t ( y ) ; % my code mine = toc ; 17 f p r i n t f ( norm o f e r r o r : \n ) 19 norm ( mlc myc) 21 f p r i n t f ( the l i b r a r y code i s %g times f a s t e r \n, mine/ l i b ) ; output: >> driver norm of error: ans = e-16 the library code is times faster 4. (a) Code: 1 [ f, f s, n b i t s ] = wavread ( s i g n o i s e. wav ) ; 3 x = l i n s p a c e ( 0, 2 pi ( length ( f ) 1)/ length ( f ), l ength ( f ) ) ; 5 p l o t ( x, f ) ; x l a b e l ( time ) 7 y l a b e l ( amplitude ) t i t l e ( s i g n a l f which c o n t a i n s one frequency component on top o f n o i s e ) 5

6 1 signal f which contains one frequency component on top of noise amplitude time There appears to be a subtle sinusoidal shape and some periodicity but it s difficult to make out. Turning to the fft to determine the dominant frequency: % use the f f t to r e c o n s t r u c t the o r i g i n a l s i g n a l by determining which 2 % f o u r i e r c o e f f i c i e n t s are dominant f t = f f t ( f ) ; 4 p l o t ( 1 : l e n g t h ( f t ), abs ( f t ) ) ; t i t l e ( magnitude o f approximated f o u r i e r c o e f f i c i e n t s ) ; 6 x l a b e l ( m ) ; y l a b e l ( abs ( f m ) ) ; A graph of the magntudes of the fourier coeffiients shows one coefficient pair to be much higher than the others: 6

7 450 magnitude of approximated fourier coefficients abs(f m ) m x 10 4 magnitude of approximated fourier coefficients abs(f m ) m zooming reveals that the dominant coefficient is at 655 in matlab so it s at m = 654 which means that the other coefficient in the pair should be at = (and it is). The indices are related because the negative coefficients get wrapped into the upper half of the DFT output vector. The 654th fourier coefficient is in the first half but the -654th is at position in the output vector. The coefficients are related in that they are eachother s complex conjugate. Confirming: >> ft(655) 7

8 ans = e e+02i >> ft(64883) ans = e e+02i They both correspond to the dominant frequency for the input signal F which is, presumably, some combination of sines and cosines. (b) The true frequency for m = 654, f s = 44100, and N = should be: BONUS I reconstruct the original signal: 1 % remove n o i s e f i x e d = z e r o s ( l e n g t h ( f t ), 1) ; 3 f i x e d (655) = f t (655) ; f i x e d (64883) = f t (64883) ; 5 f = mf s N % r e v e r s e f f t 7 s i g = i f f t ( f i x e d ) ; wavwrite ( s i g, , f i x e d. wav ) ; 440Hz I cannot hear the tone above the noise but since the fourier coefficients corresponding to the undistorted signal are larger in amplitude than the noise, they should be able to be heard over the noise. 5. (a) Given f = [1, 1, 1, 0, 0, 0] T and g = [1, 1, 1, 1, 0, 0] T, f g = [1, 2, 3, 3, 2, 1] T. (b) Given f = [1, 1, 1, 0, 0] T and g = [1, 1, 1, 1, 0] T, f g = [2, 2, 3, 3, 2] T (c) (a) has the same elements as in the non periodic convolution. Both vectors must be padded to length N 1 + N 2 1. (d) ft and ift denote the fourier transform and inverse fourier transform. f g = ift(ft(f g))) = ift(ft(f) ft(g)) = ift(ft(g) ft(f)) = ift(ft(g f)) = g f 8

9 (e) The sum of the elements in f g in (a) is 12 which is 3 4 while 3 and 4 are the sums of the elements in f and g respectively. Proposition: j=0 (f g) j = i = 0 f i j=0 (f g) j = j=0 = = j=0 i=0 i=0 i=0 f i j=0 f i j=0 f i g (j i)modn g (j i)modn g j j=0 g (j i)modn = j=0 g j for all i. It s the same summation in a different order. 6. % s c r i p t to use the f f t and i f f t to convolve an audio f i l e with an impulse 2 % response 4 [ f1, fs1, n b i t s 1 ] = wavread ( invoker. wav ) ; [ f2, fs2, n b i t s 2 ] = wavread ( impulseresponse. wav ) ; 6 f 1 = f1 ; f 2 = f2 ; 8 % pad with z e r o s so that the c o n v o l u t i o n i s c o r r e c t 10 p a d s i z e = l e n g t h ( f 1 ) + l e n g t h ( f 2 ) 1 ; padf1 = z e r o s ( 1, p a d s i z e l e ngth ( f 1 ) ) ; 12 padf2 = z e r o s ( 1, p a d s i z e l e ngth ( f 2 ) ) ; f 1 = [ f 1 padf1 ] ; 14 f 2 = [ f 2 padf2 ] ; 16 % convert to f o u r i e r space. f 1 = f f t ( f 1 ) ; 18 f 2 = f f t ( f 2 ) ; 20 % perform component wise m u l t i p l i c a t i o n r e s = f 1. f 2 ; 22 % s h i f t back to r e g u l a r space 24 r e s = i f f t ( r e s ) ; 26 r e s = res ; 28 r e s = r e s /max( r e s ) ; 9

30 wavwrite ( res, 4 4 1 0 0, r e s u l t. wav ) ; 7. (a) blur = reshape ( t e x t r e a d ( b l u r r y. txt ),512,512) ; 2 aper = reshape ( t e x t r e a d ( aperture.

10 30 wavwrite ( res, , r e s u l t. wav ) ; 7. (a) blur = reshape ( t e x t r e a d ( b l u r r y. txt ),512,512) ; 2 aper = reshape ( t e x t r e a d ( aperture. txt ),512,512) ; 4 a x i s equal ; colormap ( gray (256) ) 6 imagesc ( blur ) ; (b) % convert to f o u r i e r space 2 f b l u r = f f t 2 ( blur ) ; f a p e r = f f t 2 ( aper ) ; 4 % deconvolve and convert back 6 c l e a r = f b l u r. / f a p e r ; c l e a r = i f f t 2 ( c l e a r ) ; 8 imagesc ( c l e a r ) ; 10

50 100 150 200 250 300 350 400 450 500 50 100 150 200 250 300 350 400 450 500 (c) 1 % add a small amount o f e r r o r n = l e n g t h ( blur ) ; 3 e r b l u r = blur + randn ( n ) 3 10ˆ 5 5 %

11 (c) 1 % add a small amount o f e r r o r n = l e n g t h ( blur ) ; 3 e r b l u r = blur + randn ( n ) 3 10ˆ 5 5 % convert to f o u r i e r space f b l u r = f f t 2 ( e r b l u r ) ; 7 f a p e r = f f t 2 ( aper ) ; 9 % deconvolve and convert back c l e a r 2 = f b l u r. / f a p e r ; 11 c l e a r 2 = i f f t 2 ( c l e a r 2 ) ; imagesc ( c l e a r 2 ) ; Why does the error get amplified? In the component wise division in fourier space, most of the entries in the aperture matrix are small, some on the order of Divison by such small numbers amplifies the 10 5 error. A visualization of the dft of the aperture function is below. The blue signifies relatively smaller values while the 11

12 red/green signifies relatively larger values. The majority of values in the dft of the aperture matrix are small and this particular deconvolution process amplifies errors 10 5 or larger

SEISMIC WAVE PROPAGATION. Lecture 2: Fourier Analysis

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