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PS-4-2007-Solutions

PS-4-2007-Solutions - EE 261 The Fourier Transform and its...

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EE 261 The Fourier Transform and its Applications Fall 2007 Solutions to Problem Set Four 1. (15 points) Cross Correlation The cross-correlation (sometimes just called correlation) of two real-valued signals f ( t ) and g ( t ) is defined by ( f g )( x ) = −∞ f ( y ) g ( x + y ) dy . (a) f g is often described as ‘a measure of how well g , shifted by x , matches f ’, thus how much the values of g are correlated with those of f after a shift. Explain this, e.g. what would it mean for f g to be large? small? positive? negative? You could illustrate this with some plots. (b) Cross-correlation is similar to convolution, with some important differences. Show that f g = f g = ( f g ) . Is it true that f g = g f ? (c) Cross-correlation and delays Show that f ( τ b g ) = τ b ( f g ) . What about ( τ b f ) g ? Solutions: (a) The cross-correlation is ( f g )( x ) = −∞ f ( x ) g ( x + y ) dy . To get a sense of this, think about when it’s positive (and large) or negative (and large) or zero (or near zero). If, for a given x , the values f ( y ) and g ( x + y ) are tracking each other – both positive or both negative – then the integral will be positive and so the value ( f g )( x ) will be positive. The closer the match between f ( x ) and g ( x + y ) (as y varies) the larger the integral and the larger the cross-correlation. In the other direction, if, for example, f ( y ) and g ( x + y ) maintain opposite signs as y varies (so are negatively correlated) them the integral will be negative and f g )( x ) < 0 The more the negatively they are correlated the more negative ( f g )( x ). Finally, it might be that the values of f ( x ) and g ( x + y ) jump around as y varies; sometimes positive and sometimes negative, and it may then be that the values cancel out in taking the integral, making ( f g )( x ) near zero. One might say – one does say – that f and g are uncorrelated if ( f g )( x ) = 0 for all x . Here are some plots. First take two Π’s, one shifted, Their cross-correlation looks like 1

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0 5 10 15 20 25 30 35 40 0 1 2 3 4 5 6 7 8 9 10 big cross correlation between 2 shifted rects Here’s a plot of the cross-correlation between a Π and a shifted, negative Π. Finally, here’s a plot of the cross-correlation between a rect and a random signal (noise). 2
(b) For the cross-correlation we have ( f g )( x ) = −∞ f ( y ) g ( x + y ) dy = −∞ f ( u x ) g ( u ) du (substituting u = x + y ) = −∞ f ( ( x u )) g ( u ) du = −∞ f ( x u ) g ( u ) du = ( f g )( x ) . Next, starting from the convolution ( f g ) we have ( f g ) ( x ) = ( f g )( x ) = −∞ f ( y ) g ( x y ) dy = −∞ f ( y ) g ( ( x y )) dy = −∞ f ( y ) g ( x + y ) dy = ( f g )( x ) . Finally, no, cross-correlation does not, in general, commute. Using the previous identities f g = f g while then g f = g f 3

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and while we have f g = g f we do not necessarily then have g f = g f .
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PS-4-2007-Solutions - EE 261 The Fourier Transform and its...

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