PS-9-2007-Solutions

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

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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

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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
(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

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

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