习题Ch1

Exercise Ch. 11.1 Express each of the following complex numbers in Cartesian form (x+jy ):(3) e j π/2 = cos(π/2)+j sin(π/2)=j (6)2e j π/4 =2[cos(π/4)+j sin(π/4)]=1+j1.2 Express each of the following complex numbers in polar form (r e j θ, with -π< θ ≤ π):(6) (1-j)2 = 1-2j + j 2 = 1 + 2 e -j π/2- 1 = 2 e -j π/2(7) j(1-j) = j +1 = e j π/2 + 1 = e j π/4 ( e j π/4 + e -j π/4 ) = 2e j π/41.4 let x [n ] be a signal with x [n ] = 0 for n < -2, and n > 4. For each signal given below, determine thevalues of n for which it is guaranteed to be zero. (d) x [-n +2] Solution:(d) Since –n +2 < -2 ♑ n > 4; –n +2 > 4 ♑ n < -2, so, x [-n+2] will be zero for n < -2 and n > 4.1.11 Determine the fundamental period of the signal x[n]=1+e7/4n j π-e 5/2n j π.Solution: The period of e7/4n j π isN 1=7. The period of e5/2n j π isN 2=5. So N = N 1 N 2=351.12 Consider the discrete —time signal1[1][3kn n x k ---=∑∞=δDetermine the values of the integers M and 0n so that x[n] may be expressed as ].[][0n Mn u n x -=Solution: ∑∞=-=],[][k k n n u δ While ]4[]1[3-=--∑∞=n u k n k δSo, ]3[]4[]1[][]4[1][+-=---+-=--=n u n u n u n u n u n x we can find that easily: M = -1, 0n = -3.1.13 Consider the continuous-time signal )2()2()(--+=t t t x δδ.Calculate the value of ∞E for the signal ⎰∞-=td x t y .)()(ττSolution:Because )2()2()(--+=t t t x δδ)2()2()]2()2([)(-++=-++=⎰∞-t u t u d t y tττδτδ)2()2()2(2)2()(2-+-+-+=t u t u t u t u t y)2()2(2)2(-+--+=t u t u t u --+=t u t u ()2(2)So ⎰⎰+∞∞--∞===2224)(dt dt t y E1.14 Consider a periodic signal21,210,1{)(<<-≤≤=t t t xWith period T=2. The derivative of this signal is related to the ―impulse train ‖∑∞-∞=-=k k t t g )2()(δWith period T=2. It can be shown that)()()(2211t t g A t t g A dtt dx -+-=Determine the values of A 1, t 1, A 2, and t 2.Solution:We get the figure from the function,Then, in the interval of 20≤≤t ,)1(3)(3)(--=t t dtt dx δδ We getA 1＝3，t 1＝0，A 2＝－3，t 2＝1.)1.19 For each of the following input-output relationships, determine whether the corresponding system islinear, time invariant or both.(b) [][]22-=n x n y(i) Consider two arbitrary inputs x 1[n]and x 2[n] x 1[n] →y 1[n] = xı²[n - 2] x 2[n] → y 2[n] = x 2² [n - 2]Let x 3[n] be a linear combination of x 1[n] and x 2[n] That is,x 3[n] = a x 1[n] + b x 2[n]where a and b are arbitrary scalars. If x 3[n] is the input to the given system, then the corresponding output y 3[n] is[][][][]()[][][][][][]n by n ay n x n abx n xb n x a n bx n ax n x n y 212122221222123322222222+≠--+-+-=-+-=-=Therefore, the system is not linear.(ii)Consider an arbitrary input x 1[n]. Let [][]2211-=n x n yis th e corresponding output. Consider a second input x 2[n] obtained by shifting x 1[n] in time: [][]012n n x n x -=The output corresponding to this input is[][][][][][][]01202101021222222n n y n y n n x n n y n n x n x n y -=--=---=-=Also note that[][]021012n n xn n y --=-Therefore, [][]012n n y n y -=This implies that the system is time-invariant.1.31 In this problem we illustrate one of the most important consequences of the properties of linearity andtime invariance. Specifically, once we know the response of a linear system or a linear time-invariant (LTI) system to a single input or the responses to several inputs, we can directly compute the responses to many other input signals. Much of the remainder of this book deals with a thorough exploitation of this fact in order to develop results and techniques for analyzing and synthesizing LTI systems.(a) Consider an LTI system whose response to the signal x 1(t) in Figure P1.31 (a) is the signal y 1(t)illustrated in Figure P1.31(b). Determine and sketch carefully the response of the system to the input x 2(t) depicted in Figure 1.31(c).(b) Determine and sketch the response of the system considered in part (a) to the input x 3(t) shown inFigure P1.31 (d).x 2(t) x 3(t)Solution: ),2()()(112--=t x t x t x so, )2()()(112--=t y t y t y)()1()(113t x t x t x ++=, so, )()1()(113t y t y t y ++=。