信号与系统

1、Another approach for computing the Fourier Series coefficientThe Fourier series coefficient ak can be computed by using the Fourier transform X0(j) of x0(t), where x0(t) is one period of x(t). i.e.,0)(10kkjXTExampleDetermine the Fourier series coefficient of the following periodic signal.ooThe full-wave rectifier(全波整流器)Lt)(4tx)(4t)10cs(tSolution:Let x0(t) = x(t) = cos(100t)u(t+0.005)-u(t-0.005) for -0.005 N + 9, there is no overlap between xk and hn-k, in this case, yn = 0. Associating this with

2、 the condition y14 = 0, we can conclude thatN + 9 5, as illustrated in Figure P2.11.c, in this case, 53)()()( dthdhxy )(13()553)( ttt ee2.12 Let kttuy)(*)()(Show that for 0 t 2, , i.e., .)2(te()(tutht(b) Determine the response. )(tx)(th41)(tx)(tht2t1 4t)(tx)(th210)(tx)(th214t2Figure P2.40Figure P2.40.a Figure P2.40.b Figure P2.40.cFor t 8, ak = 0.3.34 Consider a continuous-time LTI system with impulse response)(txt481)(jH10teh4)(Find the Fourier series representation of the output y(t) for each

3、of the following inputs:(a). nttx)()(b). nntt)(1)(c). x(t) is the periodic wave depicted in Figure P3.34.Solution:The frequency response of the system 168)( 20404 dtedtedtexjjtj(c). For the periodic signal depicted in Figure P3.34, the fundamental frequncy is 210TThe Fourier series coefficients ak kkdtetetxjkTjk 2)/sin()4/sin()( 04/1/2/ 00 Then the output signal is expressed asktjkeajHty0)()(tkjek22)/si(16)(8So the Fourier series representation of the output y(t) is given byk tkjety 22)/sin()()S

4、olutions for Problems in Chapter 4The Fourier transform is an important tool for signal and system analysis. It is important to have a good command of the following emphases1. DefinitionExistance condition: dtetxjXj)( dtx)(The inverse transform is defined bytejXtxj)(21)(2. Understand the physical meaning of the Fourier transform.3. Properties)(txt114LLFigure P3.34Specifically, Time-shifting, Differentiation in the time-domain, Duality, Convolution and Multiplication properties.4. Exercises(a) Pe

5、rform the following evaluations)(tx dtx)(t t12)(12Determine the Fourier transform X(j) of x(t)(jX1Determine the inverse Fourier transform x(t) of X(j)2(b) Problems 4.25 in Page 341 of our textbook.4.10 (a) Using Tables 4.1 and 4.2 to help determine the Fourier transform of the following signal:2)sin()(ttxSolution: x(t) can be alternately expressed as ttttx)sin()sin()(2Let , and then .ttx)sin(001t)()(1xFrom Table 4.2 we see that ,)(0jXThe Fourier transform of 1txdepicted in the figure.)(*)(2)(001

6、jjjFrom Table 4.1, we see that)()(1jXdj )2()2 uu4.11 Given the relationships)(0jX1djX)(12/j)(0jX12)(j/)(*)(thxtyand )3(*)(thxtgand given that x(t) has Fourier transform X(j) and h(t) has Fourier transform H(j), use Fourier transform properties to show that g(t) has the formg(t) = Ay(Bt).Determine the values of A and B.Solution:From time scaling property)3(1)(jXtxF)3(1)(jHthFThen from the convolution property, we have)()(3)(jHjjGand Ythus we have )3()()(jjXjComparing G(j) and Y(j) we conclude tha

7、t the relationship between G(j) and Y(j) is )3(91)(jYjGBy taking the inverse Fourier transform to we get)3(91)(jYjG31)(tytgThis implies that A = 1/3, B = 3.4.13 Let x(t) be a signal whose Fourier transform is )5()()( jXand let h(t) = u(t) u(t-2)(a). Is x(t) periodic?(b). Is x(t)*h(t) periodic?(c). Can the convolution of two aperiodic signals be periodic?Solution:(a). The inverse Fourier transform of X(j) is computed as dedejXtx tjtj )5()()21)(21)(5tjtjAlthough that and periodic signals respectiv

8、ely, they have no common period. So x(t) is not tjetjperiodic.(b). The Fourier transform of h(t) jtj edehjH)sin(2)(From the convolution property, )()(jjXjY je)sin(2)5()( jjj eee )si()(sin2sin2 50 in2)i()i( jjj 5)sin2jeThe inverse Fourier transform of Y(j) 5)sin(2(0)( jety dedejYty tjtj i)21)(21)( 55sin5tjtThe above expression of y(t) implies that y(t) is a periodic signal.(c). Part (a) and (b) show that the convolution of two aperiodic signals may be periodic.4.14 Consider a signal x(t) with Fou

9、rier transform X(j). Suppose we are given the following facts:1. x(t) is real and nonnegative.2. F-1(1+j) X(j) = Ae-2tu(t), where A is independent of t.3. .2)(djXDetermine a closed-form expression for x(t).Solution:From fact 2, F-1(1+j) X(j) = Ae-2tu(t), we have 2)(1(jAXj 21)2(1)( jAjjjAjXTaking the inverse transform to X(j) we arrive at)()(2tuetxtFrom fact 3, , and applying Parsevals Relation we have2djX, this implies dtxdjX22)()(1)(2dtxFrom fact 1, we have x(t) = |x(t)|.and 04320222)( teeAteAtx ttt1)431(22So

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