tailieunhanh - Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 2

Tham khảo tài liệu 'proakis j. (2002) communication systems engineering - solutions manual (299s) episode 2', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Problem z i x t a t a To x - I 0 otherwise Thus Zoo i-a To xa t e-j2nftdt x t e-j2nftdt -o da Evaluating Xa f for f T we obtain Xa n T x t e -j7ĩTtdt Toxn To J a where xn are the coefficients in the Fourier series expansion of x t . Thus Xa T is independent of the choice of a. Problem o E x t - nTs n -o x t f ỗ t - nTs nx t f ej2nis . T s . n -o s n -o 1 I k r o I X f OL ổ f - n Ta L n -o s J 1 I k o Ễ T ể f - n n -o s s o Tsn o. X n j2n n t e T T s If we set t 0 in the previous relation we obtain Poisson s sum formula o o I o z 1 n E x -nTs E x mTs T E X ụn Problem 1 We know that e-a t -- a a2 4n2f2 Applying Poisson s sum formula with Ts 1 we obtain 5 e-a n - 2a Er a2 4n2n2 n -o n -o 2 Use the Fourier transform pair n t sinc f in the Poisson s sum formula with Ts K. Then CXD -i oo I n ỵ n nK K sinc K But n nK 1 for n 0 and n nK 0 for n 1 and K G 1 2 . . Thus the left side of the previous relation reduces to 1 and oo K Y. sinc KÌ 18 3 Use the Fourier transform pair A t - sinc2 f in the Poisson s sum formula with Ts K. Then o 1 0 . E A nK K sinc K n -o n -o Reasoning as before we see that 52n -o A nK 1 since for K G 1 2 . 0 otherwise Thus K E0 -o sinc2 K Problem Let H f be the Fourier transform of h t . Then H f F e-atu-1 t F ỗ t H f a j2nf 1 H f a j2nf The response of the system to e-at cos pt u-1 t is y t F-1 H f F e-at cos pt u-1 t But -rl -at O-U _ ra1 - at .í- - Fdt I 1-at a jM F e cos pt u-1 t F e u-1 t eJ e u-1 t e 1 1 1 2 _a j2n f - a j2n f so that Y f F y t a 2 1 1 a j2n f - 2n a j2n f 2n Using the linearity property of the Fourier transform the Convolution theorem and the fact that Ổ t j2nf we obtain y t ae-atcos pt u-1 t e-atcos pt u-1 t ổz t e-at cos pt ỗ t pe-at sin pt u-1 t ỗ t pe-at sin pt u-1 t Problem 1 y t x t h t x t ỗ t ỗ t x t dtx t With x t e altl we obtain y t e 1 ae altlsgn t . 2 o y t h r x t T dr -o e-aTe-P t-T dT e-pt e- a-A TdT Jo Jo 19 If a p y t te atU-1 t a p y t e-fit - - u-1 t -1- e-at - u-1 t p a 0 p a L