This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: EE261 Raj Bhatnagar Summer 20102011 EE 261 The Fourier Transform and its Applications Problem Set 3 Due Wednesday 13 July 1. (15 points) Convolution and crosscorrelation The crosscorrelation (sometimes just called correlation) of two realvalued 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 crosscorrelation. 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 crosscorrelation of two functions and the convolution of two functions are related. (a) Is f ? g = g ? f , i.e. , is crosscorrelation 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, crosscorrelation 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 ) ....
View
Full
Document
This document was uploaded on 07/28/2011.
 Summer '09

Click to edit the document details