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Unformatted text preview: EE261 Raj Bhatnagar Summer 2010-2011 EE 261 The Fourier Transform and its Applications Problem Set 3 Due Wednesday 13 July 1. (15 points) Convolution and cross-correlation The cross-correlation (sometimes just called correlation) of two real-valued signals f ( t ) and g ( t ) is de ned by ( f ? g )( x ) = Z ∞-∞ f ( y ) g ( x + y ) dy . ( f ? g )( x ) is often described as a measure of how well the values of g , when shifted by x , correlate with the values of f . It depends on x ; some shifts of g may correlate better with f than other shifts. To get an intuitive sense of this, think about the following three scenarios, some of which are illustrated below: Positive correlation: ( f ? g ) is large and positive: 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 ( y ) and g ( x + y ) (as y varies) the larger the integral and the larger the cross-correlation. Negative correlation: ( f ?g ) is large and negative: If on the other hand, f ( y ) and g ( x + y ) maintain opposite signs as y varies, then the integral will be negative and ( f ?g )( x ) < . The more the negatively they are correlated, the more negative ( f ? g )( x ) . Uncorrelated: ( f ?g ) is near zero: Finally, it might be that the values of f ( y ) 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 says that f and g are uncorrelated if ( f ? g )( x ) ≈ for all x . Figure 1: Positive correlation Figure 2: Uncorrelated In this problem, we will look at how the cross-correlation of two functions and the convolution of two functions are related. (a) Is f ? g = g ? f , i.e. , is cross-correlation commutative? (b) Show that f ? g = f- * g = ( f * g- )- (c) Show that f ? ( τ b g ) = τ b ( f ? g ) . Recall our notation: τ b g = g ( t- b ) . 1 Solution: (a) We make a substitution in the de ntion of f ? g : ( f ? g )( t ) = Z ∞-∞ f ( u ) g ( u + t ) du = Z ∞-∞ f ( p- t ) g ( p ) dp (by substituting p = u + t ) = Z ∞-∞ g ( p ) f ( p + (- t )) dp = ( g ? f )(- t ) Hence, cross-correlation is not commutative unlike convolution. (b) Rewriting the equations from part (a), ( f ? g )( t ) = Z ∞-∞ g ( p ) f (- t + p ) dp = Z ∞-∞ g ( p ) f [- ( t- p )] dp The last integral is exactly the de nition of ( f- * g ) ( t ) . To prove the second portion, we again write the integral and see how we can coax the required result out of it. ( f ? g )( t ) = Z ∞-∞ f ( u ) g ( u + t ) du = Z ∞-∞ f (- p- t ) g (- p ) dp (by substituting u + t =- p ) = Z ∞-∞ g (- p ) f (- t- p ) dp = ( f * g- )(- t ) ....
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This document was uploaded on 07/28/2011.
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