PS-9-2007-Solutions

PS-9-2007-Solutions - EE 261 The Fourier Transform and its...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
EE 261 The Fourier Transform and its Applications Fall 2007 Problem Set Nine Solutions 1. (10 points) 2D Convolution Find and sketch the function defined by the following convo- lution: g ( x, y )=Π( x )Π( y ) Π( x )Π( y ) Solution: Let h ( x, y x )Π( y ). Then F h ( ξ,η )=s inc ξ sinc η Now by the convolution theorem F g ( x, y )= F h ( ) F h ( ) so F g ( x, y 2 ξ sinc 2 η In turn, this is separable, so g ( x, y )=( F 1 sinc 2 )( x )( F 1 sinc 2 )( y ) or g ( x, y )=Λ( x )Λ( y ) Here’s a plot – it looks like a pyramid with a square base: 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2. (10 points) 2D Radial Stretch Theorem If g ( r ) is a circularly symmetric function with Hankel transform G ( ρ ), show that the Hankel transform of g ( ar )is 1 | a | 2 G ( ρ a ) . Solution: Recall that G ( ρ )=2 π ± 0 g ( r ) J 0 (2 πρr ) rdr Let u = ar so that r = u a and dr = du a .S in c e g ( r )= g ( r ) (a property of radial functions) we can assume a> 0 without loss of generality. Then F{ g ( ar ) } =2 π ± 0 g ( ar ) rJ 0 (2 ) dr = 1 a 2 2 π ± 0 g ( u ) uJ 0 (2 πρ u a ) du = 1 a 2 G ² ρ a ³ = 1 | a | 2 G ² ρ a ³ 3. (20 points) 2D Fourier Transforms Find the 2D Fourier Transforms of: 2
Background image of page 2
(a) sin 2 πax 1 sin 2 πbx 2 Solution: Because the function is separable we have F (sin 2 1 sin 2 2 )= F sin 2 1 F sin 2 2 = 1 2 i ( δ ( ξ 1 a ) δ ( ξ 1 + a )) 1 2 i ( δ ( ξ 2 b ) δ ( ξ 2 + b )) = 1 4 ( δ ( ξ 1 a ) δ ( ξ 2 b ) δ ( ξ 1 a ) δ ( ξ 2 + b ) δ ( ξ 1 + a ) δ ( ξ 2 b )+ δ ( ξ (1 + a ) δ ( ξ 2 + b )) = 1 4 ( δ ( ξ 1 a, ξ 2 b ) δ ( ξ 1 a, ξ 2 + b ) δ ( ξ 1 + a, ξ 2 b δ ( ξ 1 + a ) δ ( ξ 2 + b ) The plot of this would be four spikes, two up, two down, at the four vertices of the rectangle ( a, b ), ( a, b ), ( a, b ), ( a, b ). (b) e ar 2 Solution: You can use the radial stretch theorem, above, on this, knowing the Fourier transform of a Gaussian. That gives, in polar coordinates, F f ( ρ π a e π 2 a ρ 2 You can also do it directly using that f ( x, y ) is separable, f ( x, y e ax 2 e ay 2 , and appealing to the formulas for the Fourier transform of a one-dimensional Gaussian. (c) e 2 πi ( ax + by ) cos(2 πcx ) Solution: Again, this is separable F ± e 2 ( ax + by ) cos(2 ) ² = F ± e 2 πiax cos(2 ) e 2 πiby ² = F ³ e 2 πiax cos(2 ) ´ F ± e 2 πiby ² = µ δ ( ξ + a ) 1 2 [ δ ( ξ + c δ ( ξ c )] δ ( η + b ) = 1 2 [ δ ( ξ + a + c δ ( ξ + a c )] δ ( η + b ) This can also be expressed using the multi-dimensional notation for the product of delta functions of different variables F f ( ξ,η 1 2 [ δ ( ξ + a + c, η + b δ ( ξ + a c, η + b )] 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
(d) cos(2 π ( ax + by )) Hint: Use the addition formula for the cosine. Solution: One way to do this is to write cos(2 π ( ax + by )) = cos 2 πax cos 2 πby sin 2 sin 2 πby, i.e. , as a separable function. Then (much like part (a)) F (cos(2 π ( ax + by ))) = F (cos 2 cos 2 ) −F (sin 2 sin 2 ) = F (cos 2 ) F (cos 2 ) (sin 2 ) F (sin 2 ) = 1 4 ( δ ( ξ a )+ δ ( ξ + a ))( δ ( η b δ ( η + b )) + 1 4 ( δ ( ξ a ) δ ( ξ + a ))( δ ( η b ) δ ( η + b )) = 1 4 ( δ ( ξ a ) δ ( η b δ ( ξ a ) δ ( η + b ) + δ ( ξ + a ) δ ( η b δ ( ξ + a ) δ ( η + b )) + 1 4 ( δ ( ξ a ) δ ( η b ) δ ( ξ a ) δ ( η + b ) δ ( ξ + a ) δ ( η b δ ( ξ + a ) δ ( η + b )) These are some combinations and cancelations here, leading to F (cos(2 π ( ax + by ))) = 1 2 δ ( xi a ) δ (
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 07/28/2011.

Page1 / 12

PS-9-2007-Solutions - EE 261 The Fourier Transform and its...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online