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1 PROBLEM ANSWER X Y x, x, rect, () X Y, otherwise D Fourier trasform is defied as ad i D case it ca be defied as We ca write give fuctio from Eq. () as It follows usig Eq. (3) it ( ) ( ) F f t e dt () i( ux+ v) F u v f x e dxd (, ) (, ) (3) (, ) ( ) ( ) f x g x h (4) i( ux+ v) iux iv F u v f x e dxd g x h e e dxd (, ) (, ) iv g x e dx h e d iux ( ) ( ) ( ) ( ) X/ X/ iux iux iux g ( x) e dx + g ( x) e dx + g ( x) e dx X/ X/ Y/ Y/ iv iv iv h( ) e d + h( ) e d + h( ) e d Y/ Y/ (5) I regio to X /ad / Similarl for h( ). We have X to, fuctio g( x ) ad first ad third itegral will be zero.

2 X / Y / X / Y / iux iv iux iv (, ) ( ) ( ) F u v g x e dx h e d e dx e d X / Y / X / Y / X/ Y/ iux iv e e iu iv X/ Y/ X X Y Y iu iu iv iv e e e e iu iv Y Y X X X X iv iv cosu i si u cos u si iu i u e e iv X Y si u si v u v (6) We have obtaied D Fourier trasform of rect fuctio give i Eq. (). It ca be also writte i terms of sic fuctio which is defied as We have sic ( ax) si ( ax) (7) ax X Y si u si v X Y F ( u, v) si u si v XY u v X Y u v X Y XYsicu sicv (8)

3 PROBLEM ANSWER a. Samplig the sigal f ( t ) ivolves multiplig that sigal with the impulse trai (also kows as comb or shah) fuctio which is defied as Sampled sigal will be thus Ш t ( t) ( t t ) (9) ( ) ( ) ( ) f f t f t Ш t t ( ) ( ) f t t t ( ) ( ) f t t t () Because comb fuctio is periodic, we ca fid its Fourier series represetatio Coefficiets c are foud as Ш t( t) ce e t i t i t t t ()

4 t/ t/ i t i t t t t( ) ( ) t t/ t/ c Ш t e dt t t e dt t t/ t/ i t i t t t ( t t) e dt ( t ) e dt t t t/ t/ () where we used the fact that fuctio ( t t) t t is zero i iterval, i case whe. Kowig that i t t Fe, it follows that Fourier trasform of comb fuctio is t i t t F Шt ( t) F e t t t (3) Multiplicatio i time domai is equal to covolutio i frequec domai. That meas whe takig Fourier trasform of sampled fuctio we will have covolutio of them. It follows ( ) ( ) ( ) ( ) f F f t Ш t F f t F Ш t t t ( ) ( ) F f t F t t F( ) F( ) t t t t F t t (4) where shiftig propert of delta fuctio was used whe fidig covolutio of two fuctios. The fial expressio Eq. (4) shows that Fourier trasform of the sampled fuctio is a periodic fuctio cosistig of the repeated copies of the trasform of the origial cotiuous-time sigal. b. We have that discrete Fourier trasform (DFT) is f i t fte t (5) ad iverse DFT

5 F N i t f e (6) t I our case we have that. Now, if we add M zero poits at the ed of our sigal, we get f f e f e + M i t i t + M + M t t t t (7) The sum does t chage because extra zeroes do t cotribute to it. Now, we have + M spectral samples with the same Nquist frequec but with differet lie spacig. Takig the iverse DFT, we get g F f e f e e N i t N i t i t + M t t N t t i + M ft e t (8) The term N e t t i + M c. ca be solved aalticall which leads to sic fuctio (aliased sic fuctio, Zero paddig meas added extra zeros oto the ed of f t before performig the DFT. Because the zeropadded sigal is loger the resultig DFT provides better frequec resolutio. Zero-paddig i the timedomai results i iterpolatio i the frequec-domai. PROBLEM 3

6 ANSWER possible sstem trasfer fuctio: Output sigals: Ax Bcost Bxt Fourier trasform of output sigals ( ) AX ( ) Y Y i ( ) Be X ( ) Trasfer fuctio

7 G ( ) G G ( ) ( ) ( ) ( ) Y G ( ) A X ( ) B ( ) j Y G ( ) X G ( ) G G ( ) ( ) A i Be Impulse respose fuctio as iverse Fourier trasform of trasfer fuctio g ( t) G( ) ( t) A B t This is a a liear time-ivariat sstem due to for uit sample sigal we have uit sample respose. Let see that this is a liear time-ivariat sstem for various iputs ad correspods outputs: For the first iput ad output: Iput x ( t) cos( t) si( t / ) - iput is a siusoidal sigal Output ( t) Acos( t) Asi( t / ) ( t) Bsi( t) - is a siusoidal sigal - is a siusoidal sigal For the secod pair Iput

8 x ( t) si( t) - iput is a siusoidal sigal Output ( t) Asi( t) - is a siusoidal sigal ( t) Bcos( t) Bsi( t / ) - is a siusoidal sigal