Math 56
Homework 5
i6x
Michael Downs
−i6x
1. (a) Since f (x) = cos(6x) = e 2 + e 2 , due to the orthogonality of each einx , n ∈ Z, the
only nonzero (complex) fourier coefficients are fˆ6 and fˆ−6 and they’re both 12 (which
is also seen from Euler’s identity).
(b) The function f (x) = e6ix lies outside the nyquist frequency N2 = 5 so aliasing will
occur. The DFT coefficients will approximate f˜m for integers −N
≤ m < N2 each with
2
error · · ·+ fˆm−2N + fˆm−N + fˆm+N fˆm+2N +· · · (these are the exact fourier coefficients).
Since the only nonzero fourier coefficient is fˆ6 = 1, the only nonzero approximation
in the DFT will be f˜−4 since its error hits fˆ6 , the only nonzero fourier coefficient.
The interpolated function will be f˜(x) = e−4ix .
2. (a) Code:
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% spectral differentiation
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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 ’ r e s a m p l i n g
dF = @( x ) exp ( s i n ( x ) ) . ∗ c o s ( x ) ; % i t s d e r i v a t i v e
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g r = 0 : 1 e −3:2∗ p i ; % t h e f i n e s t g r i d
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m a x e r r o r s = z e r o s ( 1 , N/ 2 ) ;
n = 2 : 2 :N;
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f o r j = 2 : 2 :N
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
f t = f f t ( y ) ; % approximated c o e f f i c i e n t s
% g e t 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 f r e q u e n c y / c o e f f i c i e n t number
f o r k = 1 : j /2
f t ( k ) = f t ( k ) ∗ i ∗ ( k−1) ;
f t ( j + 1 − k ) = f t ( j + 1 − k ) ∗ i ∗ −k ;
end
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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 ) ) ) ) ) ;
e r r o r = max( abs ( dF ( g r ) − a r r a y f u n ( i n t e r , g r ) ) ) ;
maxerrors ( j /2) = e r r o r ;
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end
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figure ;
semilogy (n , maxerrors ) ;
x l a b e l ( ’N ’ ) ;
y l a b e l ( ’max e r r o r ’ ) ;
t i t l e ( ’ E r r o r i n t r u n c a t e d N term a p p r o x i m a t i o n u s i n g s p e c t r a l
differentiation ’)
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Math 56
Homework 5
Michael Downs
Plot:
Error in truncated N term approximation using spectral differentiation
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0
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−2
10
−4
max error
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−6
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−8
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−10
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−12
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−14
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−16
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N
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(b) Replacing f and df with:
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f = @( x ) abs ( s i n ( x ) ) . ˆ 3 ; % f u n c t i o n we ’ r e s a m p l i n g
dF = @( x ) 3∗ s i n ( x ) . ∗ abs ( s i n ( x ) ) . ∗ c o s ( x ) ; % i t s d e r i v a t i v e
and plotting log(y) vs x:
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Math 56
Homework 5
Michael Downs
Error in truncated N term approximation using spectral differentiation
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0
10
−1
max error
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−2
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−3
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−4
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N
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This is clearly not exponential convergence. I plot the errors again using log(y) vs
log(x):
Error in truncated N term approximation using spectral differentiation
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0
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max error
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−2
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−3
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−4
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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.
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Math 56
Homework 5
Michael Downs
3. Algorithm:
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% c o o l e y −tukey r e c u r s i v e a l g o r i t h m f o r t h e f a s t f o u r i e r t r a n s f o r m
% t a k e s a row v e c t o r o f w e i g h t e d s a m p l e s o f l e n g t h 2ˆn
% and o u t p u t s t h e t r a n s f o r m e d v e c t o r
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function res = c t f f t ( f )
N = length ( f ) ;
% b a s e c a s e i n r e c u r s i o n when s i z e i s 2
i f N == 2
r e s = ( dftmtx ( 2 ) ∗ f . ’ ) . ’ ;
return ;
end
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% 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
% SO ACTUALLY IT ’ S REVERSED
f e = f ( 1 : 2 : N) ;
f o = f ( 2 : 2 : N) ;
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% r e c u r s e to get t h e i r f f t s
e = ctf ft ( fe ) ;
o = c t f f t ( fo ) ;
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% m u l t i p l y o by t w i d d l e f a c t o r f o r combining l a t e r
twid = exp(−1 ∗ 1 i ∗ 2 ∗ p i ∗ 1/N ∗ ( 0 : (N/2 − 1 ) ) ) ;
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o = twid . ∗ o ;
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% combine them
r e s = z e r o s ( 1 , N) ;
r e s ( 1 :N/ 2 ) = e ( 1 :N/ 2 ) + o ( 1 :N/ 2 ) ;
r e s ( ( 1 :N/ 2 ) + N/ 2 ) = e ( 1 :N/ 2 ) − o ( 1 :N/ 2 ) ;
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end
driver:
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% 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 t i m e s .
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N = 2 ˆ 1 6 ; % 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 ’ r e s a m p l i n g
x = l i n s p a c e ( 0 , (N−1)/N ∗ 2 ∗ pi , N) ; % sample x v a l u e s
y = f ( x ) . /N; % s a m p l e s
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tic ;
mlc = f f t ( y ) ; % matlab code
l i b = toc ;
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Math 56
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Homework 5
Michael Downs
tic ;
myc = c t f f t ( y ) ; % my code
mine = t o c ;
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f p r i n t f ( ’ norm o f e r r o r : \n ’ )
norm ( mlc − myc )
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f p r i n t f ( ’ t h e l i b r a r y code i s %g t i m e s f a s t e r \n ’ , mine / l i b ) ;
output:
>> driver
norm of error:
ans =
6.9905e-16
the library code is 4290.64 times faster
4. (a) Code:
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[ f , f s , n b i t s ] = wavread ( ’ s i g n o i s e . wav ’ ) ;
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x = l i n s p a c e ( 0 , 2∗ p i ∗ ( l e n g t h ( f ) −1)/ l e n g t h ( f ) , l e n g t h ( f ) ) ;
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plot (x , f ’ ) ;
x l a b e l ( ’ time ’ )
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 f r e q u e n c y component on top o f n o i s e ’
)
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Math 56
Homework 5
Michael Downs
signal f which contains one frequency component on top of noise
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0.8
0.6
0.4
amplitude
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−0.2
−0.4
−0.6
−0.8
−1
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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:
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% u s e t h e f f t t o r e c o n s t r u c t t h e o r i g i n a l s i g n a l by d e t e r m i n i n g which
% f o u r i e r c o e f f i c i e n t s a r e dominant
ft = fft ( f ) ;
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 ’ ) ;
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:
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Math 56
Homework 5
Michael Downs
magnitude of approximated fourier coefficients
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abs(fm)
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m
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x 10
magnitude of approximated fourier coefficients
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abs(fm)
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651
652
653
654
655
m
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658
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 65536−654+1 = 64883
(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 64883 in the output vector. The coefficients are
related in that they are eachother’s complex conjugate. Confirming:
>> ft(655)
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Math 56
Homework 5
Michael Downs
ans =
3.1442e+02 - 2.8471e+02i
>> ft(64883)
ans =
3.1442e+02 + 2.8471e+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, fs = 44100, and N = 65536 should be:
f=
mfs
≈ 440Hz
N
BONUS I reconstruct the original signal:
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% remove n o i s e
f i x e d = z e r o s ( length ( f t ) , 1) ;
fixed (655) = f t (655) ;
fixed (64883) = f t (64883) ;
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% reverse f f t
sig = i f f t ( fixed ) ;
wavwrite ( s i g , 4 4 1 0 0 , ’ f i x e d . wav ’ ) ;
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 N1 + N2 − 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
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Math 56
Homework 5
Michael Downs
(e) The sum of the elements in f ∗ g in (a) is 12 which is 3 ∗ 4 while
3 and 4 are
PN −1
the sums of the elements in f and g respectively. Proposition:
j=0 (f ∗ g)j =
P
PN −1
N −1
i=0
fi ∗ j=0
N
−1
X
(f ∗ g)j =
j=0
N
−1 N
−1
X
X
j=0 i=0
=
=
N
−1
X
j=0
6.
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4
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g(j−i)modN =
PN −1
j=0
fi
N
−1
X
i=0
j=0
N
−1
X
N
−1
X
i=0
PN −1
fi g(j−i)modN
fi
g(j−i)modN
gj
j=0
gj for all i. It’s the same summation in a different order.
% s c r i p t t o u s e t h e f f t and i f f t t o c o n v o l v e an a u d i o f i l e with an i m p u l s e
% response
[ f1 ,
[ f2 ,
f1 =
f2 =
f s 1 , n b i t s 1 ] = wavread ( ’ i n v o k e r . wav ’ ) ;
f s 2 , n b i t s 2 ] = wavread ( ’ i m p u l s e r e s p o n s e . wav ’ ) ;
f1 ’ ;
f2 ’ ;
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% pad with z e r o s s o t h a t t h e c o n v o l u t i o n i s c o r r e c t
padsize = length ( f1 ) + length ( f2 ) − 1;
padf1 = z e r o s ( 1 , p a d s i z e − l e n g t h ( f 1 ) ) ;
padf2 = z e r o s ( 1 , p a d s i z e − l e n g t h ( f 2 ) ) ;
f 1 = [ f 1 padf1 ] ;
f 2 = [ f 2 padf2 ] ;
% convert to f o u r i e r space .
f1 = f f t ( f1 ) ;
f2 = f f t ( f2 ) ;
% perform component w i s e m u l t i p l i c a t i o n
res = f1 .∗ f2 ;
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% s h i f t back t o r e g u l a r s p a c e
res = i f f t ( res ) ;
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res = res ’ ;
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r e s = r e s /max( r e s ) ;
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Math 56
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Homework 5
Michael Downs
wavwrite ( r e s , 4 4 1 0 0 , ’ r e s u l t . wav ’ ) ;
7. (a)
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blur = reshape ( textread ( ’ blurry . txt ’ ) ,512 ,512) ;
aper = reshape ( textread ( ’ aperture . txt ’ ) ,512 ,512) ;
axis equal ;
colormap ( gray ( 2 5 6 ) )
imagesc ( blur ) ;
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(b)
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% convert to f o u r i e r space
fblur = f f t 2 ( blur ) ;
f a p e r = f f t 2 ( aper ) ;
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% d e c o n v o l v e and c o n v e r t back
clear = fblur ./ faper ;
clear = ifft2 ( clear ) ;
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imagesc ( c l e a r ) ;
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Math 56
Homework 5
Michael Downs
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(c)
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% add a s m a l l amount o f e r r o r
n = length ( blur ) ;
e r b l u r = b l u r + randn ( n ) ∗3∗10ˆ −5
% convert to f o u r i e r space
fblur = fft2 ( erblur ) ;
f a p e r = f f t 2 ( aper ) ;
% d e c o n v o l v e and c o n v e r t back
clear2 = fblur ./ faper ;
clear2 = i f f t 2 ( clear2 ) ;
imagesc ( c l e a r 2 ) ;
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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 10−6 .
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
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Math 56
Homework 5
Michael Downs
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.
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