( ) = ( ) ( ) ( ) ( ) τ τ. This is a more complete version of the solutions for assignment 2 courtesy of the course TA

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Christina Lang
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1 This is a mor complt vrsion o th solutions or assignmnt courtsy o th cours TA ) Find th Fourir transorms o th signals shown in Figur (a-b). -a) -b) From igur -a, w hav: g t 4 0 t = t 0 othrwis = j G g t dt j j = 4 dt + dt 0 4 = + j j j G g t dt = j 0 4 = + j j j4 ( ) ( ) 4 4+ = j4 4 = j j4 From igur -a, w hav: t τ τ t 0 g( t) = + t τ 0 t + τ 0 othrwis

2 j = G g t dt 0 t t = + τ τ τ + τ dt dt 0 By us o intgration by par, udv = uv vdu, w can ind: t j j j j j t. dt =. dt t = + j + = ( ) W can ind th Fourir intgration valu as ollow: = G g t dt 0 t = τ τ + = τ + t τ + τ j dt dt τ ( t dt t dt τ 0 ) 0 j ( ) j + j j + = + τ ( ) τ ( ) τ τ + + j τ j τ + τ + = + + τ ( ) ( ) ( ) ( ) + τ + = τ τ τ ( j τ j τ ) ( j τ j τ ) + 4 τ = + + τ τ τ j ( ) ( ) ( ) = + + τ ( ) ( cos( τ) τsin ( τ) ) + τ + j τ τ + j τ τ 0

4 ) Find th invrs Fourir transorms o th spctra shown in Figur (a-b). -a) By us o Eulr's ormula, w can xpand th cosin unction into two sparatd xponntial unction and tak th intgration sparatly, as ollow: = + j g t G d = + cos ( ) + j j j j = d + j t+ + j t = d d + = + j( t + ) j( t ) + + j t+ j t j ( t+ ) j ( t+ ) j ( t ) j ( t ) j( t + ) j( t ) = + ( t ) ( t ) ( t ) d ( + ) ( + ) ( ) ( ) j t j t j t j t = + t + j t j ( t ) ( t ) = sin ( t + ) + sin + ( t ) ( t ) = sin ( t ) + sin + ( t ) = sin ( t ) = cos 4 4 ( t)

5 -b) From igur -b, w hav: B B t 0 G( ) = + B 0 t + B 0 othrwis + j = g t G dt 0 + B + j + j = d + dt B B 0 B By us o intgration by part, w hav: = + j g t G dt ( ) ( t) ( t) B B jt jt + j + j + j d =. d + = + = B j t ( t) B ( t) + j + j + j 0 + B + j + j = d + dt B B 0 B 0 + B + + j + + j = + B ( t) ( t) B 0 = ( + ( jbt + ) ) + j Bt + B t B t ( ) ( cos( Bt) 4Bt sin ( Bt) ) (( ) ) 0 j Bt + j Bt 0 = + + B = + cos Bt + Bt sin Bt B ( )

6 3) Th Fourir transorm o th triangular puls g(t) in Figur 3 (a) is givn as, G =!!!!"!!!" Us this inormation, and th tim-shiing and tim-scaling proprtis, to ind th Fourir transorms o signal shown in Figur 3 (b), (c) and (d). 3-b) rom igur w can ind out: 3 = ( ) + ( ) = g( t ) + g ( t + ) g t g t g t G = j ( j j ) G = j ( ) 3 ( ( )) ( ) j j j( ) j( ) ( ) ( j ( ) ) + ( ) = j + + j ( ) j ( ) = ( j ) = ( ) ( cos( ))

7 3-c) rom igur w can ind out: g t = g t + g t 4 = g( t ) + g( ( t+ ) ) G = j 4 ( j j ) = + ( ) ( ) ( ( )) ( ) j ( ) G G G = j j j j j( ) j( ) j( ) ( j ( ) ) + j j j = j + + j ( ) j j = + ( j j j j ) j j 4 = j = ( ) sin ( )

8 3-c) rom igur w can ind out: g t =.5 g t 4 G = j ( j j ) 3 G4 ( ) = G = 3G ( ( )) j4 j j( ) j( ) j4 ( j ( ) ) 3 = 3 = ( j4 j4 ) 4

9 4) Th signals in Figur 4 ar modulatd signals with carrir cos 0t. Find th Fourir transorms o ths signals using th appropriat proprtis o th Fourir transorm. Sktch th amplitud and phas spctra or th Figur 4 (a) and (b). Not that cos! t = cos 0t, thror! = 5, 4-a) From Figur 4-a, w hav: t t F τ τ x ( t) =Δ cos0t Δ sin c τ F = g( t) cos0t cos 0.5 δ + + δ t g t G c =Δ = sin ( ) = cos0 F = *0.5 δ( + ) + δ( ) x t g t t X G ο X ( ) = sin c ( ( ο) ) sin c ( ( )) + + ο = sin c ( ω 0) sin c ( ω 0) + + ο ο ο ο 4-b) From Figur 4-b, w s that s only a shid vrsion o signal in igur 4-a in tim domain by sc, so w can ind th Fourir transorm by us o tim shiing proprty: = ( ) x t x t = X X j4 = sin c ( ( ο) ) sin c ( ( )) + + ο = sin c ( ω 0) sin c ( ω 0) + + j4 ω F ( ± ) ± j ο g t tο G

10 From abov igur it s obvious that th only chang in rquncy domain is th signal angl, and thir amplitud spctrum is xactly th sam. Not that th Y-axis o abov igur is scald or signal s phass.

11 4-c) From Figur 4-c, w hav: t x3 t =Π cos0 t ( ) ( g( t) cos0 t) * δ( t ) = t g t G c =Π = sin ( τ) t F Π τsin c τ ( τ) F ( ) δ( + ) + δ( ) cos 0.5 ο ο ο t F x3( t) = Π cos0 t * δ( ) 3 = *0.5 δ( + ) + δ( ). t X G ( ) ( ο ο ) ( ) ( sin *0.5 ο ο ). = τ ( τ) δ( + ) + δ( ) X3 c 5 5 = sin c + + sin. c j4 = sin 0 sin 0 c + + c. j4 = sin c ω+ 0 + sin c ω 0. ω

12 5) Th procss o rcovring a signal g t rom th modulatd signal g t cos! t is calld dmodulation. Show that th signal g t cos! t can b dmodulatd by multiplying it with cos! t and passing th product through a low-pass iltr o bandwidth B Hz [th bandwidth o g t ]. Assum B <!. Hint: cos!! t = + cos 4! t. Rcogniz that th spctrum o g cos 4! t is cntrd at! and will b supprssd by a low-pass iltr o bandwidth B Hz. Th modulatd signal is g t cos! t. Multiplying by cos! t yild, ( ) = + ( ) = + ( ) gtcos ο gt cos 4 ο gt gtcos 4 ο Obsrv that th rsulting signal contains th original signal g t and a modulatd copy o th signal movd to a cntr rquncy o!. I th bandwidth o th signal is B <!, thn th modulatd copy will not xtnd urthr than! rom its cntr rquncy and a low pass iltr rom! to! will pass only g(t) and iltr out th modulatd copy.

Introduction to the Fourier transform. Computer Vision & Digital Image Processing. The Fourier transform (continued) The Fourier transform (continued)

Introduction to the Fourier transform. Computer Vision & Digital Image Processing. The Fourier transform (continued) The Fourier transform (continued) Introduction to th Fourir transform Computr Vision & Digital Imag Procssing Fourir Transform Lt f(x) b a continuous function of a ral variabl x Th Fourir transform of f(x), dnotd by I {f(x)} is givn by: