form yofxthee with f g the x thus the fourier

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 x dx = u f).! ) form yofx,the}e= with f= )g (the .x Thus the Fourier Transform find (ˆ((!exists.. 5.31 ( however the respectyto Fourier= g any value(of)y , x exists. g )y,)ySince of xe and in class). ˆ( lim exists, 1 i! second integral diverges (this wasZshown of f x Z Z s i! Z Z 01 Assuming that g is differentiable, show that ZF {uy (x, y )} = uyx ! , y ). (1 i )x 0 ˆ( x 6=(x) +i g (y ))e i x dx x= iZ x dx(x thei left diverges dx = i value (b) g (y )(f (x)ethisx means theeintegralfon )e lim dx0+(1 (iy )x for anylim . eof y and = f 0, e dx Z (c) dx x = e = i x eg ) s s s 1 i( 2. (a) The Fourier y )e i on the right y ) does )e since the Fourier Transform of f!x)s f (x g ( Tranform of u(x, f (x not dx. the first)integral dx = g (y ) converges exist. s Zi s 1 ee 1 Since the Fourier Transform of f (xthe i Fourier was shown in class). Since the a lim = a integral =)eexistsusing integral on the right conThus 3. (a) exists, however the second)e i. x dx diverges, (this Transform of f (xproperty. F{ (x a)} = (x the substitution ) exists. s 1 the integralintegral on the left divergesany value of y , and we imeans the on the left converges for for any value of y and ! 1 i! verges, thus g (y ) 6= 0, this Z Z ˆˆ ˆ (b) Let F{f (x)Tranform(of).Using theˆy (! , y )xof Zpart(a) ). Further, have u(! , y )}== (f (! ). u Clearly u result = g (y )f (! and the Convolution ˆ g y )f ! x, i 2. (a) the Fourier y )e Z x dx = g ((y ) y )i does )not exist. f ( x) g ( f (x e i dx. x Z ˆ Theorem ywe = F{uy (x, )} have f (x)g (y )i x dx = i a y ) e g( f (x)e i x dx = g (y )f (! ). Sincexthe)} = Fourier TransformZ ofdx(=)eexistsusing integral on the right confx the Z the Z 3. (a) F{ ( a ( x a) e , substitution property. Z 1 ˆ x {F{f i of y F 1 { f (x) +i the)) verges, thus t y integral on (the f (x (e i tx }} = g ( any ez ) xz t) and f ( (b)Let u(f (! )e= g (}x=eFig (ydxwithxg (yleft )converges fory )f (xvalue(show, dz = wex t). = )}F{ x ) dx + dx. ˆ( ˆ b) Let F{f (x)}f = )f+! ). ) Using the 6= 0. Use...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online