2信号与系统_奥本海默_华科_电信系_英文_试卷
英文版《信号与系统》第二章习题解答
Problems Solution
dx(t ) ∵ → 3 y (t ) + e 2t u (t ) dt
2e 3 h(t 1) = e 2t u (t )
the impulse 1 3 2(t +1) u (t + 1) response h(t ) = 2 e e
18
Chapter 2
∫
t
∞
h (τ )d τ =
0
t<0,t>2
sin (πτ )d τ =
1
∫
t
1
0
π
cos πτ | 0t
0< t<2
=
π
(1 cos π t )[u (t ) u (t 2 )]
12
Chapter 2
y (t ) = 2
Problems Solution
π
2
(1 cos π (t 1 ))[u (t 1 ) u (t 3 )] (1 cos π (t 3 ))[u (t 3 ) u (t 5 )]
∞
+∞
2
t
y (t ) = ∫ e (t τ 2 )u (t τ 2 )x(τ )dτ
∞
+∞
h(t ) = e
(t 2 )
u (t 2)
h(t τ )
(b ) y(t ) = x(t ) h(t ) = x′(t ) h (1) (t )
= h (1) (t + 1) h (1) (t 2)
0
e 3τ dτ
1 e 3(t 3 ) = 3 e 3(t 5 ) e 3(t 3 ) = 3
x(t τ )
t 3
2021《信号与系统》考研奥本海姆2021考研真题库
2021《信号与系统》考研奥本海姆2021考研真题库一、考研真题解析下列关于冲激函数性质的表达式不正确的是()。
[西安电子科技大学2012研] A.f(t)δ′(t)=f(0)δ′(t)B.f(t)δ(t)=f(0)δ(t)C.D.【答案】A查看答案【解析】A项,正确结果应该为f(t)δ′(t)=f(0)δ′(t)-f′(0)δ(t)。
2x(t)=asint-bsin(3t)的周期是()。
[西南交通大学研]A.π/2B.πC.2πD.∞【答案】C查看答案【解析】因为asint的周期为T1=2π/1=2π,bsin(3t)的周期为T2=2π/3,因为T1/T2=3/1为有理数,因此x(t)是周期信号,且x(t)=asint-bsin (3t)的周期是3T2=T1=2π。
3序列f(k)=e j2πk/3+e j4πk/3是()。
[西安电子科技大学2012研]A.非周期序列B.周期N=3C.周期N=6D.周期N=24【答案】B查看答案【解析】f1(k)=e j2πk/3的周期N1=2π/(2π/3)=3,f2(k)=e j4πk/3的周期N2=2π/(4π/3)=3/2,由于N1/N2=2为有理数,因此f(k)为周期序列,周期为2N2=N1=3。
4积分[西安电子科技大学2011研]A.2B.1C.0D.4【答案】A查看答案【解析】一电路系统H(s)=(10s+2)/(s3+3s2+4s+K),试确定系统稳定时系数K 的取值范围()。
[山东大学2019研]A.K>0B.0<K<12C.K>-2D.-2<K<2【答案】B查看答案【解析】H(s)=(10s+2)/(s3+3s2+4s+K)=B(s)/A(s),其中A(s)=s3+3s2+4s+K,系统稳定需要满足K>0,3×4>K,因此0<K<12。
7信号f(t)=6cos[π(t-1)/3]ε(t+1)的双边拉普拉斯变换F(s)=()。
[西安电子科技大学2012研]A.B.C.D.【答案】C查看答案【解析】信号f(t)变形为利用时移性质得到其拉式变换为8系统函数为H(s)=s/(s2+s+1),则系统的滤波特性为()。
奥本海姆《信号与系统》(第2版)(下册)名校考研真题-通信系统(圣才出品)
【答案】C
【 解 析 】 线 性 相 位 FIR 滤 波 器 必 满 足 某 种 对 称 性 , 即 h(n) = h( N −1− n) 或 者 h(n) = −h( N −1− n) 。答案中 C 为偶对称,且 N=8,为Ⅰ型 FIR 滤波器。
【答案】 h(n) = 0,n 0 h(t) = 0,t 0 【解析】①对于稳定的又是因果的离散系统,其系统函数 H (z) 的极点都在 z 平面的单 位圆内;②对于稳定的又是因果的连续系统,其系统函数 H (s) 的极点都在 s 平面的左半开 平面。
2.离散系统的模拟可由
【解析】LTI 连续时间系统总可被分解为全通网络和最小相移网络的级联的形式。
三、简答题
1.FIR 数字滤波器必为稳定系统,试说明。[清华大学 2006 研] 解:FIR 数字滤波器的冲击响应是有限长的,因而当有限输入时,必有有限输出,必为 稳定的。
2.已知
LTI
系统的输入
x[n]和输出
y[n]满足如下关系
圣才电子书
十万种考研考证电子书、题库视频学习平台
第 8 章 通信系统
一、选择题
1.下面给出了几个 FIR 滤波器的单位函数响应。其中满足线性相位特性的 FIR 滤波器 是( )。[东南大学 2007 研]
A.h(n)={1,2,3,4,5,6,7,8} B.h(n)={1,2,3,4,1,2,3,4} C.h(n)={1,2,3,4,4,3,2,1}
k +100
i=k −100
n) e(i
= +
k +n+100
e(i)
i=k +n−100
《信号与系统》考研奥本海姆版2021考研名校考研真题
《信号与系统》考研奥本海姆版2021考研名校考研真题第一部分考研真题精选一、选择题1已知信号f(t)的频带宽度为Δω,则信号y(t)=f2(t)的不失真采样间隔(奈奎斯特间隔)T等于()。
[西南交通大学研]A.π/(Δω)B.π/(2Δω)C.2π/(Δω)D.4π/(Δω)【答案】B查看答案【解析】根据卷积定理可知,y(t)=f2(t)→[1/(2π)]F(jω)*F(j ω)。
若信号f(t)的频带宽度为Δω,则y(t)的频带宽度为2Δω。
则奈奎斯特采样频率为4Δω,所以不失真采样间隔(奈奎斯特间隔)T等于2π/(4Δω)=π/(2Δω)。
2已知f(t)↔F(jω),f(t)的频带宽度为ωm,则信号y(t)=f(t/2-7)的奈奎斯特采样间隔等于()。
[西南交通大学研]A.2π/ωmB.2π/(2ωm-7)C.4π/ωmD.π/ωm【答案】A查看答案【解析】根据时域和频域之间关系,可知若时域扩展,则频域压缩。
所以若f(t)的频带宽度为ωm,则信号y(t)=f(t/2-7)的频带宽度为ωm/2。
所以,其奈奎斯特采样频率为(ωm/2)×2=ωm,即奈奎斯特采样间隔等于2π/ωm。
3有限长序列x(n)的长度为4,欲使x(n)与x(n)的圆卷积和线卷积相同,则长度L的最小值为()。
[中国科学院研究生院2012研]A.5B.6C.7D.8【答案】C查看答案【解析】x(n)的长度为4,则其线卷积的长度为4+4-1=7。
当x(n)与x(n)的圆卷积L≥7时,x(n)与x(n)的圆卷积和线卷积相同,可知L的最小值为7。
4下面给出了几个FIR滤波器的单位函数响应。
其中满足线性相位特性的FIR滤波器是()。
[东南大学研]A.h(n)={1,2,3,4,5,6,7,8}B.h(n)={1,2,3,4,1,2,3,4}C.h(n)={1,2,3,4,4,3,2,1}D.h(n)={1,2,3,4,-1,-2,-3,-4}【答案】C查看答案【解析】线性相位FIR滤波器必满足某种对称性,即h(n)=h(N-1-n)或者h(n)=-h(N-1-n)。
信号与系统奥本海姆习题答案
Chapter 1 Answers1.6 (a).NoBecause when t<0, )(1t x =0.(b).NoBecause only if n=0, ][2n x has valuable.(c).Yes Because ∑∞-∞=--+--+=+k k m n k m n m n x ]}414[]44[{]4[δδ ∑∞-∞=------=k m k n m k n )]}(41[)](4[{δδ ∑∞-∞=----=k k n k n ]}41[]4[{δδ N=4.1.9 (a). T=π/5Because 0w =10, T=2π/10=π/5.(b). Not periodic.Because jt t e e t x --=)(2, while t e -is not periodic, )(2t x is not periodic.(c). N=2Because 0w =7π, N=(2π/0w )*m, and m=7.(d). N=10Because n j j e e n x )5/3(10/343)(ππ=, that is 0w =3π/5, N=(2π/0w )*m, and m=3.(e). Not periodic. Because 0w =3/5, N=(2π/0w )*m=10πm/3 , it ’s not a rational number.1.14 A1=3, t1=0, A2=-3, t2=1 or -1dtt dx )( isSolution: x(t) isBecause ∑∞-∞=-=k k t t g )2()(δ, dt t dx )(=3g(t)-3g(t-1) or dtt dx )(=3g(t)-3g(t+1) 1.15. (a). y[n]=2x[n-2]+5x[n-3]+2x[n-4]Solution:]3[21]2[][222-+-=n x n x n y ]3[21]2[11-+-=n y n y ]}4[4]3[2{21]}3[4]2[2{1111-+-+-+-=n x n x n x n x ]4[2]3[5]2[2111-+-+-=n x n x n xThen, ]4[2]3[5]2[2][-+-+-=n x n x n x n y(b).No. For it ’s linearity.the relationship between ][1n y and ][2n x is the same in-out relationship with (a). you can have a try.1.16. (a). No.For example, when n=0, y[0]=x[0]x[-2]. So the system is memory. (b). y[n]=0.When the input is ][n A δ,then, ]2[][][2-=n n A n y δδ, so y[n]=0. (c). No.For example, when x[n]=0, y[n]=0; when x[n]=][n A δ, y[n]=0. So the system is not invertible.1.17. (a). No.For example, )0()(x y =-π. So it ’s not causal.(b). Yes.Because : ))(sin()(11t x t y = , ))(sin()(22t x t y =))(sin())(sin()()(2121t bx t ax t by t ay +=+1.21. Solution:We have known:(a).(b).(c).(d).1.22. Solution:We have known:(a).(b).(e).(g)1.23. Solution:For )]()([21)}({t x t x t x E v -+= )]()([21)}({t x t x t x O d --= then,(a).(b).(c).1.24.For: ])[][(21]}[{n x n x n x E v -+= ])[][(21]}[{n x n x n x O d --=then,(a).(b).1.25. (a). Periodic. T=π/2.Solution: T=2π/4=π/2.(b). Periodic. T=2.Solution: T=2π/π=2.(d). Periodic. T=0.5. Solution: )}()4{cos()(t u t E t x v π=)}())(4cos()()4{cos(21t u t t u t --+=ππ )}()(){4cos(21t u t u t -+=π )4cos(21t π= So, T=2π/4π=0.51.26. (a). Periodic. N=7Solution: N=m *7/62ππ=7, m=3.(b). Aperriodic.Solution: N=ππm m 16*8/12=, it ’s not rational number.(e). Periodic. N=16 Solution as follow:)62cos(2)8sin()4cos(2][ππππ+-+=n n n n x in this equation,)4cos(2n π, it ’s period is N=2π*m/(π/4)=8, m=1.)8sin(n π, it ’s period is N=2π*m/(π/8)=16, m=1.)62cos(2ππ+-n , it ’s period is N=2π*m/(π/2)=4, m=1. So, the fundamental period of ][n x is N=(8,16,4)=16.1.31. SolutionBecause )()1()(),2()()(113112t x t x t x t x t x t x ++=--=. According to LTI property ,)()1()(),2()()(113112t y t y t y t y t y t y ++=--=Extra problems:Sketch ⎰∞-=t dt t x t y )()(. 1. SupposeSolution:2. SupposeSketch:(1). )]1(2)1()3()[(--+++t t t t g δδδ(2). ∑∞-∞=-k k t t g )2()(δ(2).Chapter 22.1 Solution:Because x[n]=(1 2 0 –1)0, h[n]=(2 0 2)1-, then(a).So, ]4[2]2[2]1[2][4]1[2][1---+-+++=n n n n n n y δδδδδ (b). according to the property of convolutioin:]2[][12+=n y n y(c). ]2[][13+=n y n y][*][][n h n x n y =][][k n h k x k -=∑∞-∞= ∑∞-∞=-+--=k k k n u k u ]2[]2[)21(2 ][211)21()21(][)21(12)2(0222n u n u n n k k --==+-++=-∑ ][])21(1[21n u n +-= the figure of the y[n] is:2.5 Solution:We have known: ⎩⎨⎧≤≤=elsewhere n n x ....090....1][,,, ⎩⎨⎧≤≤=elsewhere N n n h ....00....1][,,,(9≤N ) Then, ]10[][][--=n u n u n x , ]1[][][---=N n u n u n h∑∞-∞=-==k k n u k h n h n x n y ][][][*][][ ∑∞-∞=-------=k k n u k n u N k u k u ])10[][])(1[][(So, y[4] ∑∞-∞=-------=k k u k u N k u k u ])6[]4[])(1[][( ⎪⎪⎩⎪⎪⎨⎧≥≤=∑∑==4,...14, (140)0N N k Nk =5, then 4≥N And y[14] ∑∞-∞=------=k k u k u N k u k u ])4[]14[])(1[][(⎪⎪⎩⎪⎪⎨⎧≥≤=∑∑==14,...114, (1145)5N N k Nk =0, then 5<N ∴4=N2.7 Solution:[][][2]k y n x k g n k ∞=-∞=-∑(a )[][1]x n n δ=-,[][][2][1][2][2]k k y n x k g n k k g n k g n δ∞∞=-∞=-∞=-=--=-∑∑(b) [][2]x n n δ=-,[][][2][2][2][4]k k y n x k g n k k g n k g n δ∞∞=-∞=-∞=-=--=-∑∑ (c) S is not LTI system..(d) [][]x n u n =,0[][][2][][2][2]k k k y n x k g n k u k g n k g n k ∞∞∞=-∞=-∞==-=-=-∑∑∑2.8 Solution: )]1(2)2([*)()(*)()(+++==t t t x t h t x t y δδ )1(2)2(+++=t x t xThen,That is, ⎪⎪⎪⎩⎪⎪⎪⎨⎧≤<-≤<-+-=-<<-+=others t t t t t t t t y ,........010,....2201,.....41..,.........412,.....3)(2.10 Solution:(a). We know:Then,)()()(αδδ--='t t t h)]()([*)()(*)()(αδδ--='='t t t x t h t x t y )()(α--=t x t xthat is,So, ⎪⎪⎩⎪⎪⎨⎧+≤≤-+≤≤≤≤=others t t t t t t y ,.....011,.....11,....0,.....)(ααααα(b). From the figure of )(t y ', only if 1=α, )(t y ' would contain merely therediscontinuities.2.11 Solution:(a). )(*)]5()3([)(*)()(3t u et u t u t h t x t y t----==⎰⎰∞∞---∞∞--------=ττττττττd t u e u d t u eu t t )()5()()3()(3)(3⎰⎰-------=tt t t d e t u d et u 5)(33)(3)5()3(ττττ⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧≥+-=-<≤-=<=---------⎰⎰⎰5,.......353,.....313.........,.........0315395)(33)(3393)(3t e e d e d e t e d e t tt t t t t t t t ττττττ(b). )(*)]5()3([)(*)/)(()(3t u e t t t h dt t dx t g t ----==δδ)5()3()5(3)3(3---=----t u e t u e t t(c). It ’s obvious that dt t dy t g /)()(=.2.12 Solution∑∑∞-∞=-∞-∞=--=-=k tk tk t t u ek t t u e t y )]3(*)([)3(*)()(δδ∑∞-∞=---=k k t k t u e)3()3(Considering for 30<≤t ,we can obtain33311])3([)(---∞=-∞-∞=--==-=∑∑ee e ek t u e e t y tk k tk kt. (Because k must be negetive ,1)3(=-k t u for 30<≤t ).2.19 Solution:(a). We have known:][]1[21][n x n w n w +-=(1) ][]1[][n w n y n y βα+-=(2)from (1), 21)(1-=E EE Hfrom (2), αβ-=E EE H )(2then, 212212)21(1)21)(()()()(--++-=--==E E E E E E H E H E H ααβαβ∴][]2[2]1[)21(][n x n y n y n y βαα=-+-+-but, ][]1[43]2[81][n x n y n y n y +-+--=∴⎪⎩⎪⎨⎧=⎪⎭⎫ ⎝⎛=+=143)21(:....812βααor ∴⎪⎩⎪⎨⎧==141βα(b). from (a), we know )21)(41()()()(221--==E E E E H E H E H21241-+--=E EE E ∴][)41()21(2][n u n h n n ⎥⎦⎤⎢⎣⎡-=2.20 (a). 1⎰⎰∞∞-∞∞-===1)0cos()cos()()cos()(0dt t t dt t t u δ(b). 0dt t t )3()2sin(5+⎰δπ has value only on 3-=t , but ]5,0[3∉-∴dt t t )3()2sin(5+⎰δπ=0(c). 0⎰⎰---=-641551)2cos()()2cos()1(dt t t u d u πτπττ⎰-'-=64)2cos()(dt t t πδ0|)2(s co ='=t t π 0|)2sin(20=-==t t ππ∑∞-∞=-==k t h kT t t h t x t y )(*)()(*)()(δ∑∞-∞=-=k kT t h )(∴2.27Solution()y A y t dt ∞-∞=⎰,()xA x t dt ∞-∞=⎰,()hA h t dt ∞-∞=⎰.()()*()()()y t x t h t x x t d τττ∞-∞==-⎰()()()()()()()()()(){()}y x hA y t dt x x t d dtx x t dtd x x t dtd x x d d x d x d A A ττττττττττξξτττξξ∞∞∞-∞-∞-∞∞∞∞∞-∞-∞-∞-∞∞∞∞∞-∞-∞-∞-∞==-=-=-===⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰(a) ()()(2)tt y t e x d τττ---∞=-⎰,Let ()()x t t δ=,then ()()y t h t =. So , 2()(2)(2)()(2)()(2)t t t t t h t ed e d e u t τξδττδξξ---------∞-∞=-==-⎰⎰(b) (2)()()*()[(1)(2)]*(2)t y t x t h t u t u t e u t --==+---(2)(2)(1)(2)(2)(2)t t u eu t d u e u t d ττττττττ∞∞-------∞-∞=+------⎰⎰22(2)(2)12(1)(4)t t t t u t e d u t e d ττττ---------=---⎰⎰(2)2(2)212(1)[]|(4)[]|t t t t u t e e u t ee ττ-------=--- (1)(4)[1](1)[1](4)t t e u t e u t ----=-----2.46 SolutionBecause)]1([2)1(]2[)(33-+-=--t u dtde t u e dt d t x dt d t t )1(2)(3)1(2)(333-+-=-+-=--t e t x t e t x t δδ.From LTI property ,we know)1(2)(3)(3-+-→-t h e t y t x dtdwhere )(t h is the impulse response of the system. So ,following equation can be derived.)()1(223t u e t h e t --=-Finally, )1(21)()1(23+=+-t u e e t h t 2.47 SoliutionAccording to the property of the linear time-invariant system: (a). )(2)(*)(2)(*)()(000t y t h t x t h t x t y ===(b). )(*)]2()([)(*)()(00t h t x t x t h t x t y --==)(*)2()(*)(0000t h t x t h t x --=012y(t)t4)2()(00--=t y t y(c). )1()1(*)(*)2()1(*)2()(*)()(00000-=+-=+-==t y t t h t x t h t x t h t x t y δ(d). The condition is not enough.(e). )(*)()(*)()(00t h t x t h t x t y --==τττd t h x )()(00+--=⎰∞∞-)()()(000t y dm m t h m x -=--=⎰∞∞-(f). )()]([)](*)([)(*)()(*)()(000000t y t y t h t x t h t x t h t x t y "=''='--'=-'-'==Extra problems:1. Solute h(t), h[n](1). )()(6)(5)(22t x t y t y dt dt y dtd =++ (2). ]1[][2]1[2]2[+=++++n x n y n y n y Solution:(1). Because 3121)3)(2(1651)(2+-++=++=++=P P P P P P P Hso )()()()3121()(32t u e e t P P t h t t ---=+-++=δ (2). Because )1)(1(1)1(22)(22i E i E EE E E E E E H -+++=++=++=iE Eii E E i -+-+++=1212 so []][)1()1(2][1212][n u i i i k i E E i i E E i n h n n +----=⎪⎪⎪⎪⎭⎫⎝⎛-+-+++=δChapter 33.1 Solution:Fundamental period 8T =.02/8/4ωππ==00000000033113333()224434cos()8sin()44j kt j t j t j t j tk k j t j t j t j tx t a e a e a e a e a e e e je je t t ωωωωωωωωωππ∞----=-∞--==+++=++-=-∑3.2 Solution:for, 10=a , 4/2πj ea --= , 4/2πj ea = , 3/42πj ea --=, 3/42πj ea =n N jk k N k e a n x )/2(][π∑>=<=n j n j n j n j e a e a e a e a a )5/8(4)5/8(4)5/4(2)5/4(20ππππ----++++=n j j n j j n j j n j j e e e e e e e e )5/8(3/)5/8(3/)5/4(4/)5/4(4/221ππππππππ----++++= )358cos(4)454cos(21ππππ++++=n n)6558sin(4)4354sin(21ππππ++++=n n3.3 Solution: for the period of )32cos(t πis 3=T , the period of )35sin(t πis 6=Tso the period of )(t x is 6 , i.e. 3/6/20ππ==w)35sin(4)32cos(2)(t t t x ππ++= )5sin(4)2cos(21200t w t w ++=)(2)(21200005522t w j t w j t w j t w j e e j e e ----++=then, 20=a , 2122==-a a , j a 25=-, j a 25-=3.5 Solution:(1). Because )1()1()(112-+-=t x t x t x , then )(2t x has the same period as )(1t x ,that is 21T T T ==, 12w w =(2). 212111()((1)(1))jkw t jkw tk T T b x t e dt x t x t e dt T--==-+-⎰⎰111111(1)(1)jkw tjkw t T Tx t e dt x t e dt T T --=-+-⎰⎰ 111)(jkw k k jkw k jkw k e a a e a e a -----+=+=3.8 Solution:kt jw k k e a t x 0)(∑∞-∞==while:)(t x is real and odd, then 00=a , k k a a --=2=T , then ππ==2/20wand0=k a for 1>kso kt jw k k e a t x 0)(∑∞-∞==t jw t jw e a e a a 00110++=--)sin(2)(11t a e e a t j t j πππ=-=-for12)(2121212120220==++=-⎰a a a a dt t x∴2/21±=a ∴)sin(2)(t t x π±=3.13 Solution:Fundamental period 8T =.02/8/4ωππ==kt jw k k e a t x 0)(∑∞-∞==∴t jkw k k e jkw H a t y 0)()(0∑∞-∞==0004, 0sin(4)()0, 0k k H jk k k ωωω=⎧==⎨≠⎩ ∴000()()4jkw t k k y t a H jkw e a ∞=-∞==∑Because 48004111()1(1)088T a x t dt dt dt T ==+-=⎰⎰⎰So ()0y t =.kt jw k k e a t x 0)(∑∞-∞==∴t jkw k k e jkw H a t y 0)()(0∑∞-∞== ∴dt e jkw H t y Ta t jkw Tk 0)()(10-⎰=for⎪⎩⎪⎨⎧>≤=100, (0100),.......1)(w w jw H ∴if 0=k a , it needs 1000>kwthat is 12100,........1006/2>>k kππand k is integer, so 8>K3.22 Solution:021)(1110===⎰⎰-tdt dt t x Ta Tdt te dt te dt e t x T a t jk t jk t jkw T k ππ-----⎰⎰⎰===1122112121)(10t jk tde jk ππ--⎰-=1121⎥⎥⎦⎤⎢⎢⎣⎡---=----111121ππππjk e te jk t jk tjk ⎥⎦⎤⎢⎣⎡---+-=--ππππππjk e e e e jk jk jk jk jk )()(21⎥⎦⎤⎢⎣⎡-+-=ππππjk k k jk )sin(2)cos(221[]πππππk jk k j k jk k)1()cos()cos(221-==-=0............≠k404402()()1184416tj tj t t j tt j t H j h t edt ee dte edt e e dtj j ωωωωωωωω∞∞----∞-∞∞----∞===+=+=-++⎰⎰⎰⎰A periodic continous-signal has Fourier Series:. 0()j kt k k x t a e ω∞=-∞=∑T is the fundamental period of ()x t .02/T ωπ=The output of LTI system with inputed ()x t is 00()()jk t k k y t a H jk e ωω∞=-∞=∑Its coefficients of Fourier Series: 0()k k b a H jk ω= (a)()()n x t t n δ∞=-∞=-∑.T=1, 02ωπ=11k a T==. 01/221/21()()1jkw t jk tk T a x t e dt t e dt Tπδ---===⎰⎰ (Note :If ()()n x t t nT δ∞=-∞=-∑,1k a T=) So 2282(2)16(2)4()k k b a H jk k k πππ===++ (b)()(1)()n n x t t n δ∞=-∞=--∑ .T=2, 0ωπ=,11k a T== 01/23/21/21/2111()()(1)(1)221[1(1)]2jkw t jk tjk t k T k a x t e dt t e dt t e dtT ππδδ----==+--=--⎰⎰⎰So 24[1(1)]()16()k k k b a H jk k ππ--==+, (c) T=1,02ωπ=01/421/4sin()12()jk t jk tk T k a x t e dt e dt Tk ωπππ---===⎰⎰28sin()2()[16(2)]k k k b a H jk k k ππππ==+ 3.35 Solution: T= /7π,02/14T ωπ==.kt jw k k e a t x 0)(∑∞-∞==∴t jkw k k e jkw H a t y 0)()(0∑∞-∞==∴0()k k b a H jkw =for⎩⎨⎧≥=otherwise w jw H ,.......0250,.......1)(,01,. (17)()0,.......k H jkw otherwise ⎧≥⎪=⎨⎪⎩that is 0250250, (14)k k ω<<, and k is integer, so 18....17k or k <≤. Let ()()y t x t =,k k b a =, it needs 0=k a ,for 18....17k or k <≤.3.37 Solution:11()[]()212()21312411511cos 224nj j nj n n n n j nn j nn n j j j H e h n ee ee e e e ωωωωωωωωω∞∞--=-∞=-∞-∞--=-∞=-===+=+=---∑∑∑∑A periodic sequence has Fourier Series:2()[]jk n Nk k N x n a eπ=<>=∑.N is the fundamental period of []x n .The output of LTI system with inputed []x n is 22()[]()jk jk n NNk k N y n a H eeππ=<>=∑.Its coefficients of Fourier Series: 2()jk Nk k b a H eπ=(a)[][4]k x n n k δ∞=-∞=-∑.N=4, 14k a =.So 2314()524cos()44j k Nk k b a H e k ππ==-3165cos()42k b k π=-3.40 Solution: According to the property of fourier series: (a). )2cos(2)cos(20000000t Tka t kw a e a ea a k k t jkw k t jkw k k π==+='- (b). Because 2)()()}({t x t x t x E v -+=}{2k v k k k a E a a a =+='-(c). Because 2)(*)()}({t x t x t x R e +=2*kk k a a a -+='(d). k k k a Tjka jkw a 220)2()(π=='(e). first, the period of )13(-t x is 3T T ='then 3)(1)13(131213120dme m x T dt e t x T a m T jk T t T jk T k +'--'-'-'⎰⎰'=-'='ππTjkk m T jk T T jk T jk m T jk T ea dm e m x T e dm e e m x T πππππ221122211)(1)(1---------=⎥⎦⎤⎢⎣⎡==⎰⎰3.43 (a) Proof:(i )Because ()x t is odd harmonic ,(2/)()jk T t k k x t a e π∞=-∞=∑,where 0k a = for everynon-zero even k.(2/)()2(2/)(2/)()2T jk T t k k jk jk T tk k jk T tk k T x t a ea e e a e ππππ∞+=-∞∞=-∞∞=-∞+===-∑∑∑It is noticed that k is odd integers or k=0.That means()()2Tx t x t =-+(ii )Because of ()()2Tx t x t =-+,we get the coefficients of Fourier Series222/200/222(/2)/2/20022/2/200111()()()11()(/2)11()()(1)jk t jk t jk t T T T T T T k T jk t jk t T T T T Tjk t jk t T T k TT a x t e dt x t e dt x t e dtT T T x t e dt x t T e dt T T x t e dt x t e dt T T πππππππ-----+--==+=++=--⎰⎰⎰⎰⎰⎰⎰ 2/21[1(1)]()jk t T kT x t e dt T π-=--⎰It is obvious that 0k a = for every non-zero even k. So ()x t is odd harmonic ,(b)Extra problems:∑∞-∞=-=k kT t t x )()(δ, π=T(1). Consider )(t y , when )(jw H ist(2). Consider )(t y , when )(jw H isSolution:∑∞-∞=-=k kT t t x )()(δ↔π11=T , 220==Tw π(1).kt j k k tjkw k k e k j H a ejkw H a t y 20)2(1)()(0∑∑∞-∞=∞-∞===ππ2=(for k can only has value 0)(2).kt j k k tjkw k k e k j H a e jkw H a t y 20)2(1)()(0∑∑∞-∞=∞-∞===πππte e t j t j 2cos 2)(122=+=- (for k can only has value –1 and 1)。
奥本海姆 信号与系统 习题参考答案
第四章作业解答4.1解:ωωωj e dt eet u eF j tj t t +==--∞------⎰2)}1({1)1(2)1(2 4.2解:ωωδδj j e e F -+=-++)}1t )1t {((4.3)}(21{}42{sin )42()42(ππππππ+-+-=+t j t j e e jF t F)2()2()2(221)2(221)}(21{444424)24πωδππωδππωπδπωπδππππππππ++--=+⨯--⨯=-=----jjj j t j j t j jej e j e j e j e e e e jF 4.4(b )解:⎩⎨⎧>≤=1||01||2)(ωωωj G 定义 则ttt g πsin 2)(=而:))1(())1(()(2--+=ωωωj G j G j X故由频移特性:tt j t t t j e e tt e e t g t x jt jtjt jt πππ22sin 4sin sin 4)(sin 2))(()(-=-=-=-=-- (也可以直接用反变换公式求解)解:由公式{ωωa 00)(||1)}(tj e aj X a t at x F -=-直接得到结果(见书后答案)4.21(a)解:)}()(21{)}(cos {000t u e e eF t tu e F t j tj atat ωωω---+= 而:ωj a t u e F at+=-1)}({则根据频移特性:)()(1)(121)}()(21{0000ωωωωωω+++-+=+--j a j a t u e e eF t j t j at4.22解:(a )因为⎩⎨⎧>≤=-3||03||1}3sin 2{1t t F ωω根据频移特性:⎩⎨⎧>≤=3||03||)(2t t e t x t j π (b))4(21)4(21}2121{)}4{cos(4411-++=+=---t t e e F F j j δδωωω则根据频移特性:t j e t t t x 3)]4()4([21)(πδδ--++=(d))(23)(1)]}2()2([3)]1()1([2{221tj t j jt jt e e e e F πππππωδπωδωδωδ----+-=+--++--解:(a )设)1()(1+=t x t x ,如下图所示,则)1()(1-=t x t x故:ωωωj e j X j X -=)()(1又因为:)(1t x 是实偶信号,则)(1ωj X 也为实偶,故:⎩⎨⎧<-≥-=-∠=∠0)(0)()()(111ωωπωωωωωj X j X j X j X(b) 因为:dt e t x j X t j ⎰∞∞--=ωω)()(则:dt t x j X ⎰∞∞-=)()0(即为x(t)的面积,故:7)()0(==⎰∞∞-dt t x j X(c) 因为:ωωπωd e j X t x t j ⎰∞∞-=)(21)(则:ωωπd j X x ⎰∞∞-=)(21)0( ππωω4)0(2)(==⇒⎰∞∞-x d j X(d) 令:ωωωsin 2)(=j G 则:)()()(ωωωj G j X j Y =则:)(*)()(t g t x t y = 其中g(t)如下图所示:则:)2(2)(sin 2)(22y d e j Y d e j X j j πωωωωωωωω==⎰⎰∞∞-∞∞-而:τττd t g x t g t x t y )()()(*)()(⎰∞∞--==τττππd g x y )2()(2)2(2⎰∞∞--=1 2 3 τ则:ππτττππ3232)2()(2)2(2=⨯=-=⎰∞∞-d g x y(e )根据帕斯瓦尔关系式:πππωω26132|)(|2|)(|22=⨯==⎰⎰∞∞-∞∞-dt t x d j X(f )2)()()()}({1t x t x t x j X F e -+==-ω(图略)4.28解:(a )以为p(t)是周期信号,其傅里叶级数为:tjn n n o e a t p ω∑∞-∞==)(则其傅里叶变换为:)(2)(on nn a j P ωωδπω-=∑∞-∞=由于:)()()(t p t x t y =则:])(2*)([21)(*)(21)(0∑∞-∞=-==n n n a j X j P j X j Y ωωδπωπωωπω ))((0∑∞-∞=-=n nn j X aωω(b)(1)22cos )(22t jtjeet t p -+== 则:πω4210==T⎪⎩⎪⎨⎧=±==00121n n a n则:))21((21))21((21))(()(0++-=-=∑∞-∞=ωωωωωj X j X n j X a j Y n n(6) 解:∑∞-∞=-=n n t t p )()(πδ周期π=T20=ω π11==T a n故:))2((1))(()(0∑∑∞-∞=∞-∞=-=-=n n nn j X n j X aj Y ωπωωω4.32解:⎩⎨⎧<=otherst tF 04||1}4sin {ωπ则根据时移特性:⎩⎨⎧<=--=otherse t t F j H j 04||})1()1(4sin {)(ωπωω(a) 因为][21)26cos()()26()26(1πππ+-++=+=t j t j e et t x 则:60=ω 根据特征函数特征值的概念:0])6([)6([21)()26()26(1=-+=+-+ππt j t j e H eH t y。
《信号与系统》考研奥本海姆版配套2021考研真题库
《信号与系统》考研奥本海姆版配套2021考研真题库第一部分考研真题精选一、选择题1下列关于冲激函数性质的表达式不正确的是()。
[西安电子科技大学2012研]A.f(t)δ′(t)=f(0)δ′(t)B.f(t)δ(t)=f(0)δ(t)C.D.【答案】A查看答案【解析】A项,正确结果应该为f(t)δ′(t)=f(0)δ′(t)-f′(0)δ(t)。
2x(t)=asint-bsin(3t)的周期是()。
[西南交通大学研]A.π/2B.πC.2πD.∞【答案】C查看答案【解析】因为asint的周期为T1=2π/1=2π,bsin(3t)的周期为T2=2π/3,因为T1/T2=3/1为有理数,因此x(t)是周期信号,且x(t)=asint-bsin (3t)的周期是3T2=T1=2π。
3序列f(k)=e j2πk/3+e j4πk/3是()。
[西安电子科技大学2012研]A.非周期序列B.周期N=3C.周期N=6D.周期N=24【答案】B查看答案【解析】f1(k)=e j2πk/3的周期N1=2π/(2π/3)=3,f2(k)=e j4πk/3的周期N2=2π/(4π/3)=3/2,由于N1/N2=2为有理数,因此f(k)为周期序列,周期为2N2=N1=3。
4积分[西安电子科技大学2011研]A.2B.1C.0D.4【答案】A查看答案【解析】5序列乘积δ(k+1)δ(k-1)=()。
[西安电子科技大学研]A.0B.δ(k)C.δ(k+1)D.δ(k-1)【答案】A查看答案【解析】根据f(k)δ(k-k0)=f(k0)δ(k-k0),因此δ(k+1)δ(k-1)=δ(2)δ(k-1)=0。
6信号f1(t)=2,f2(t)的波形如图1-1-1所示,设y(t)=f1(t)*f2(t),则y(11)=()。
[西安电子科技大学2011研]图1-1-1A.1B.0C.2D.3【答案】B查看答案【解析】7已知一连续系统在输入f(t)作用下的零状态响应为y(t)=f(4t),则该系统为()。
奥本海姆《信号与系统》第二版信号与系统答案
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信号与系统 奥本海姆1-4答案.doc
Signals and SystemChap11.6 Determine whether or not each of the following signals is periodic:(a): (/4)1()2()j t x t e u t π+= (b): 2[][][]x n u n u n =+-(c): 3[]{[4][14]}k x n n k n k δδ∞=-∞=----∑Solution:(a).No 【周期信号无始无终,单边肯定不周期】Because 12cos()2sin(),0()440,0t j t t x t t ππ⎧+++>⎪=⎨⎪<⎩ when t<0, )(1t x =0. (b).No 【注意n =0】 Because 21,0[]2,01,0n n n n x >⎧⎪==⎨⎪<⎩(c).Y es 【画图、归纳】 Because∑∞-∞=--+--+=+k k m n k m n m n x ]}414[]44[{]4[3δδ∑∞-∞=------=k m k n m k n )]}(41[)](4[{δδ{[4][14]}k n k n k δδ∞=-∞=----∑N=4.1.9 Determine whether or not each of the following signals is periodic, if a signal is periodic, specify its fundamental period:(a): 101()j tx t je =(b): (1)2()j t x t e -+=(c): 73[]j n x n e π=(d): 3(1/2)/54[]3j n x n e π+= (e): 3/5(1/2)5[]3j n x n e += Solution: (a). T=π/5Because 0w =10, T=2π/10=π/5. (b). Aperiodic.Because jt t e e t x --=)(2, while t e -is not periodic, )(2t x is not periodic. (c). N=2Because 0w =7π, N=(2π/0w )*m, and m=7. (d). N=10Because n j j e e n x )5/3(10/343)(ππ=, that is 0w =3π/5,N=(2π/0w )*m, and m=3. (e). Aperiodic.Because 0w =3/5, N=(2π/0w )*m=10πm/3 , it ’s not a rational number.1.14 consider a periodic signal 1,01()2,12t x t t ≤≤⎧=⎨-<<⎩with periodT=2. The derivative of this signal is related to the “impulsetrain ”()(2)k g t t k δ∞=-∞=-∑, with period T=2. It can be shownthat1122()()()dx t A g t t A g t t dt=-+-. Determine the values of1A , 1t , 2A , 2t .Solution:A 1=3, t 1=0, A 2=-3, t 2=1 or -1 Because∑∞-∞=-=k k t t g )2()(δ,)1(3)(3)(--=t g t g dtt dx1.15. Consider a system S with input x[n] and output y[n].This system is obtained through a series interconnection of a system S 1 followed by a system S2. The input-output relationships for S 1 and S 2 areS 1: ],1[4][2][111-+=n x n x n y S 2: ]3[21]2[][222-+-=n x n x n yWhere ][1n x and ][2n x denote input signals.(a) Determine the input-output relationship for system S.(b)Does the input-output relationship of system S change if the order in which S 1 and S 2 are connected in series is reversed(ie., if S2 follows S 1)? Solution: (a)]3[21]2[][222-+-=n x n x n y]3[21]2[11-+-=n y n y]}4[4]3[2{21]}3[4]2[2{1111-+-+-+-=n x n x n x n x]4[2]3[5]2[2111-+-+-=n x n x n xThen, ]4[2]3[5]2[2][-+-+-=n x n x n x n y【可以考虑先求取单位脉冲响应,再做卷积】(b).No. because it ’s linear, S 1 and S 2 do not diverge.1.16. Consider a discrete-time system with input x[n] and output y[n].The input-output relationship for this system is]2[][][-=n x n x n y(a) Is the system memory less?(b) Determine the system output when the input is ][n A δ, where A is any real or complex number . (c) Is the system invertible? Solution: (a). No.For example, when n=0, y[0]=x[0]x[-2]. So the system is memory. (b). y[n]=0.When the input is ][n A δ,]2[][][2-=n n A n y δδ, so y[n]=0.(c). No.For example, when x[n]=0, y[n]=0; when x[n]=][n A δ, y[n]=0. So the system is not invertible.1.17.Consider a continuous-time system with input x(t) and output y(t) related by ))(sin()(t x t y =, (a) Is this system causal? (b) Is this system linear? Solution: (A). No.For example,)0()(x y =-π. So it ’s not causal.【得到什么启示?】 (b). Y es.Because : ))(sin()(11t x t y = , (sin()(22tx t y =)()())(sin())(sin()(21213t by t ay t bx t ax t y +=+=1.21. A continuous-time signal ()x t is shown in Figure P1.21. Sketch and label carefully each of the following signals:(a): (1)x t - (b): (2)x t - (c): (21)x t + (d): (4/2)x t - (e): [()()]()x t x t u t +-(f): ()[(3/2)(3/2)]x t t t δδ+--Solution: (a).(b).(c). (d).1.22. A discrete-time signal ][n x is shown in as the following. Sketch and label carefully each of the following signals: (a): [4]x n - (b): [3]x n - (c): [3]x n(d): [31]x n + (e): [][3]x n u n -(f): [2][2]x n n δ--(g): 11[](1)[]22nx n x n +-(h): 2[(1)]x n -Solution:(a).(b).(e).(f) ]2[-n δ(g)1.25. Determine whether or not each of the following continuous-time signals is periodic. If the signal is periodic, determine its fundamental period.(a): ()3cos(4)3x t t π=+ (b): (1)()j t x t e π-=(c): 2()[cos(2)]3x t t π=-(d): (){cos(4)()}x t t u t ενπ=(e): (){sin(4)()}x t t u t ενπ= (f): (2)()t n n x t e∞--=-∞=∑Solution:(a).Periodic. T=π/2. Solution: T=2π/4=π/2. (b). Periodic. T=2.Solution: T=2π/π=2.(c). Periodic. T=π/2.【括号内周期,平方后仍然周期,或者做三角变换】 (d). Periodic. T=0.5. Solution: )}()4{cos()(t u t E t x v π= )}())(4cos()()4{cos(21t u t t u t --+=ππ )}()(){4cos(21t u t u t -+=π)4cos(21t π=So, T=2π/4π=0.5【值得商榷】 (e)、(f)非周期信号。
奥本海姆 信号与系统 习题参考答案 电子科技大学
Chapter 22.1 Solution:Because x[n]=(1 2 0 –1)0, h[n]=(2 0 2)1-, then (a).So, ]4[2]2[2]1[2][4]1[2][1---+-+++=n n n n n n y δδδδδ(b). according to the property of convolutioin:]2[][12+=n y n y(c). ]2[][13+=n y n y2.3 Solution:][*][][n h n x n y =][][k n h k x k -=∑∞-∞=∑∞-∞=-+--=k k k n u k u ]2[]2[)21(2][211)21()21(][)21(12)2(0222n u n u n n k k --==+-++=-∑][])21(1[21n u n +-=the figure of the y[n] is:2.5 Solution:We have known: ⎩⎨⎧≤≤=elsewhere n n x ....090....1][,,, ⎩⎨⎧≤≤=elsewhere N n n h ....00....1][,,,(9≤N)Then, ]10[][][--=n u n u n x , ]1[][][---=N n u n u n h∑∞-∞=-==k k n u k h n h n x n y ][][][*][][∑∞-∞=-------=k k n u k n u N k u k u ])10[][])(1[][(So, y[4] ∑∞-∞=-------=k k u k u N k u k u ])6[]4[])(1[][(⎪⎪⎩⎪⎪⎨⎧≥≤=∑∑==4,...14, (14)N N k Nk =5, then 4≥NAnd y[14] ∑∞-∞=------=k k u k u N k u k u ])4[]14[])(1[][(⎪⎪⎩⎪⎪⎨⎧≥≤=∑∑==14,...114, (114)55N N k Nk =0, then 5<N∴ 4=N2.7 Solution:[][][2]k y n x k g n k ∞=-∞=-∑(a ) [][1]x n n δ=-,[][][2][1][2][2]k k y n x k g n k k g n k g n δ∞∞=-∞=-∞=-=--=-∑∑(b) [][2]x n n δ=-,[][][2][2][2][4]k k y n x k g n k k g n k g n δ∞∞=-∞=-∞=-=--=-∑∑(c) S is not LTI system.. (d) [][]x n u n =,0[][][2][][2][2]k k k y n x k g n k u k g n k g n k ∞∞∞=-∞=-∞==-=-=-∑∑∑2.8 Solution:)]1(2)2([*)()(*)()(+++==t t t x t h t x t y δδ)1(2)2(+++=t x t xThen,That is, ⎪⎪⎪⎩⎪⎪⎪⎨⎧≤<-≤<-+-=-<<-+=others t t t t t t t t y ,........010,....2201,.....41..,.........412,.....3)(2.10 Solution:(a). We know: Then, )()()(αδδ--='t t t h)]()([*)()(*)()(αδδ--='='t t t x t h t x t y)()(α--=t x t xthat is,So, ⎪⎪⎩⎪⎪⎨⎧+≤≤-+≤≤≤≤=others t t t t t t y ,.....011,.....11,....0,.....)(ααααα(b). From the figure of )(t y ', only if 1=α, )(t y ' would contain merely therediscontinuities.2.11 Solution:(a). )(*)]5()3([)(*)()(3t u e t u t u t h t x t y t ----==⎰⎰∞∞---∞∞--------=ττττττττd t u e u d t u eu t t )()5()()3()(3)(3⎰⎰-------=tt t t d e t u d et u 5)(33)(3)5()3(ττττ⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧≥+-=-<≤-=<=---------⎰⎰⎰5,.......353,.....313.........,.........0315395)(33)(3393)(3t e e d e d e t e d e t tt t t t t t t t ττττττ(b). )(*)]5()3([)(*)/)(()(3t u e t t t h dt t dx t g t----==δδ)5()3()5(3)3(3---=----t u e t u e t t(c). It ’s obvious that dt t dy t g /)()(=.2.12 Solution∑∑∞-∞=-∞-∞=--=-=k tk tk t t u ek t t u e t y )]3(*)([)3(*)()(δδ∑∞-∞=---=k k t k t u e)3()3(Considering for 30<≤t ,we can obtain33311])3([)(---∞=-∞-∞=--==-=∑∑e e e ek t u e e t y tk k tk kt.(Because k must be negetive , 1)3(=-k t u for 30<≤t ).2.19 Solution:(a). We have known: ][]1[21][n x n w n w +-=(1)][]1[][n w n y n y βα+-=(2)from (1), 21)(1-=E EE Hfrom (2), αβ-=E EE H )(2then, 212212)21(1)21)(()()()(--++-=--==E E E E E E H E H E H ααβαβ∴ ][]2[2]1[)21(][n x n y n y n y βαα=-+-+-but, ][]1[43]2[81][n x n y n y n y +-+--=∴ ⎪⎩⎪⎨⎧=⎪⎭⎫ ⎝⎛=+=143)21(:....812βααor∴⎪⎩⎪⎨⎧==141βα (b). from (a), we know )21)(41()()()(221--==E E E E H E H E H21241-+--=E E E E∴ ][)41()21(2][n u n h n n ⎥⎦⎤⎢⎣⎡-=2.20 (a). 1⎰⎰∞∞-∞∞-===1)0cos()cos()()cos()(0dt t t dt t t u δ(b). 0 dt t t )3()2sin(5+⎰δπ has value only on 3-=t , but ]5,0[3∉-∴dt t t )3()2sin(5+⎰δπ=0(c). 0⎰⎰---=-641551)2cos()()2cos()1(dt t t u d u πτπττ⎰-'-=64)2cos()(dt t t πδ0|)2(s co ='=t t π0|)2sin(20=-==t t ππ2.23 Solution:∑∞-∞=-==k t h kT t t h t x t y )(*)()(*)()(δ∑∞-∞=-=k kT t h )(∴2.27 Solution()y A y t dt ∞-∞=⎰,()xA x t dt ∞-∞=⎰,()hA h t dt ∞-∞=⎰.()()*()()()y t x t h t x x t d τττ∞-∞==-⎰()()()()()()()()()(){()}y x hA y t dt x x t d dtx x t dtd x x t dtd x x d d x d x d A A ττττττττττξξτττξξ∞∞∞-∞-∞-∞∞∞∞∞-∞-∞-∞-∞∞∞∞∞-∞-∞-∞-∞==-=-=-===⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰2.40 Solution(a) ()()(2)tt y t ex d τττ---∞=-⎰,Let ()()x t t δ=,then ()()y t h t =.So , 2()(2)(2)()(2)()(2)tt t t t h t ed e d e u t τξδττδξξ---------∞-∞=-==-⎰⎰(b) (2)()()*()[(1)(2)]*(2)t y t x t h t u t u t eu t --==+---(2)(2)(1)(2)(2)(2)t t u eu t d u e u t d ττττττττ∞∞-------∞-∞=+------⎰⎰22(2)(2)12(1)(4)t t t t u t ed u te d ττττ---------=---⎰⎰(2)2(2)212(1)[]|(4)[]|t t t t u t e e u t ee ττ-------=--- (1)(4)[1](1)[1](4)t t e u t e u t ----=-----2.46 SolutionBecause)]1([2)1(]2[)(33-+-=--t u dtde t u e dt d t x dt d t t)1(2)(3)1(2)(333-+-=-+-=--t e t x t et x tδδ.From LTI property ,we know)1(2)(3)(3-+-→-t h e t y t x dtdwhere )(t h is the impulse response of the system. So ,following equation can be derived.)()1(223t u e t h e t --=-Finally, )1(21)()1(23+=+-t u e e t h t 2.47 SoliutionAccording to the property of the linear time-invariant system: (a). )(2)(*)(2)(*)()(000t y t h t x t h t x t y ===(b). )(*)]2()([)(*)()(00t h t x t x t h t x t y --==)(*)2()(*)(0000t h t x t h t x --=)2()(00--=t y t y(c). )1()1(*)(*)2()1(*)2()(*)()(00000-=+-=+-==t y t t h t x t h t x t h t x t y δ(d). The condition is not enough.(e). )(*)()(*)()(00t h t x t h t x t y --== τττd t h x )()(00+--=⎰∞∞-)()()(000t y dm m t h m x -=--=⎰∞∞-(f). )()]([)](*)([)(*)()(*)()(000000t y t y t h t x t h t x t h t x t y "=''='--'=-'-'==Extra problems:1. Solute h(t), h[n](1). )()(6)(5)(22t x t y t y dt dt y dtd =++ (2). ]1[][2]1[2]2[+=++++n x n y n y n y012y(t)t4Solution:(1). Because 3121)3)(2(1651)(2+-++=++=++=P P P P P P P H so )()()()3121()(32t u e e t P P t h t t ---=+-++=δ (2). Because )1)(1(1)1(22)(22i E i E EE E E E E E H -+++=++=++=iE Ei i E E i -+-+++=1212so []][)1()1(2][1212][n u i i i k i E E i i E E i n h n n +----=⎪⎪⎪⎪⎭⎫⎝⎛-+-+++=δ。
信号与系统第二版课后答案西安交大奥本海姆
第一章1.3 解:(a). 2401lim(),04Tt T TE x t dt e dt P ∞-∞∞→∞-====⎰⎰(b) dt t x TP T TT ⎰-∞→∞=2)(21lim121lim ==⎰-∞→dt T TTT∞===⎰⎰∞∞--∞→∞dt t x dt t x E TTT 22)()(lim(c).222lim()cos (),111cos(2)1lim()lim2222TT TTTT T TTE x t dt t dt t P x t dt dt TT∞∞→∞--∞∞→∞→∞--===∞+===⎰⎰⎰⎰(d) 034121lim )21(121lim ][121lim 022=⋅+=+=+=∞→=∞→-=∞→∞∑∑N N n x N P N Nn n N N N n N 34)21()(lim202===∑∑-∞=∞→∞nNNn N n x E (e). 2()1,x n E ∞==∞211lim []lim 112121N NN N n N n NP x n N N ∞→∞→∞=-=-===++∑∑ (f) ∑-=∞→∞=+=NNn N n x N P 21)(121lim 2∑-=∞→∞∞===NNn N n x E 2)(lim1.9. a). 00210,105T ππω===; b) 非周期的; c) 00007,,22mN N ωωππ=== d). 010;N = e). 非周期的; 1.12 解:∑∞=--3)1(k k n δ对于4n ≥时,为1即4≥n 时,x(n)为0,其余n 值时,x(n)为1易有:)3()(+-=n u n x , 01,3;M n =-=- 1.15 解:(a)]3[21]2[][][222-+-==n x n x n y n y , 又2111()()2()4(1)x n y n x n x n ==+-, 1111()2[2]4[3][3]2[4]y n x n x n x n x n ∴=-+-+-+-,1()()x n x n = ()2[2]5[3]2[4]y n x n x n x n =-+-+- 其中][n x 为系统输入。
奥本海姆《信号与系统》(第2版)(下册)名校考研真题-采样(圣才出品)
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圣才电子书 十万种考研考证电子书、题库视频学习平台
C.1 + 2
D. 12
【答案】C
【解析】由题意知,信号
x1 (t )
的最高频率为
1 2
,
x2 (t)
的最高频率为
2 2
。由傅里叶变
换性质和卷积定理可知:
x(t)
=
x1(t
+1)x2 (t
+ 2
=
2
+
3 2
1
。其奈
奎斯特采样频率: s = 2m = 22 + 31 。
2.已知数字音乐的抽样频率是 44.1KHz,由此,我们可以推测,人的听力频率范围的 上限接近( )。[华南理工大学 2008 研]
A.10 KHz B.20 KHz C.40 KHz D.80 KHz 【答案】B
【解析】根据奈圭斯特抽样定理,抽样频率 fs 2 fm 可得。
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圣才电子书 十万种考研考证电子书、题库视频学习平台
3.若信号
f
(t)
的奈奎斯特采样频率为
fs
,则信号
g (t )
=
f
(t)
f
t 2
的奈奎斯特采样频
率为( )。[北京邮电大学 2009 研]
A.
1 2
fs
B. fs
C.1.5 fs
D. 2 fs
【答案】C
C. 4π m
D. π m
【答案】A
【解析】根据时域和频域之间关系,可知若时域扩展,则频域压缩。所以若 f (t) 的频带
宽度为 m ,则信号
y(t) =
英文版《信号与系统》第二章习题解答
c xt x0t 2 ht h0t 1 yt y0t 1
d xt x0 t ht h0t
yt
1
01
2
t
We have not enough information to determine the output
e xt x0 t ht h0 t
yt
x0 t h0 t
0
2
t
Information to determine the output yt
yt
a xt 2x0t ht h0t
2
yt 2 y0 t
0
2
t
b xt x0 t x0 t 2 ht h0 t yt y0 t y0 t 2
yt
1
0
2
4t
19Chapter 2来自Problems Solution
xt
1
(a) yt et ut x 2d
1 0
yt et 2u t 2x d
ht et2ut 2
ht
2t
b yt xt ht xt h1t
h1t 1 h1t 2 16
Chapter 2
Problems Solution
h1 t t e 2u 2 d t e 2d
discontinuities,what is the value of a?
Solution : xt
ht
1
1
0 1 t 0a t
yt
a
0 a 1 1+a t
5
Chapter 2
d yt dt
1
0 a 1 1+a t
-1
Problems Solution
信与系统奥本海默第二版
或
x[k] zX(z)
10.1 双边 Z 变换 The z-Transform
一.双边Z变换的定义:
X(z) x(n)z-n 其中 z r是ej一个复数。 n-
当 r 时 1, z 即 e为j离散时间傅立叶变换。
这表明:DTFT就是在单位圆上进行的Z变换。
X(rej) x(n)r-ne-jnF [x(n)r-n] n-
当ROC包括 z 1时,Z 变换在单位圆上的情 况就是 X (e j ) ,因此也可以利用零极点图对其 进行几何求值。
其方法与拉氏变换时完全类似:
考查动点在单位圆上移动一周时,各极点矢 量和零点矢量的长度与幅角变化的情况,即可 反映系统的频率特性。
例1. 一阶系统
h(n) anu(n)
y(n)-a y(n- 1 )x(n)
幂级数展开法适合用来求解非有理函数形式 的反X 变( z )换。
3. 留数法: 对有理函数的 X (由z ) 留数定理有:
i x (n )1 2j
cX (z )zn - 1 d zR e s [X (z )zn - 1 ,z i]
i z i 是C内的极点。
n 0 时,x(n) R es[X(z)zn-1,zi]
可见:对 x ( n ) 做 Z 变换就等于对 x(n)r -n做DTFT。
因此,Z 变换是对DTFT的推广。
二. Z变换的收敛域(ROC):
Z变换与DTFT一样存在着收敛的问题。
1. 并非任何信号的Z变换都存在。
2. 并非Z平面上的任何复数都能使 X ( z ) 收敛。 Z平面上那些能使 X ( z ) 收敛的点的集合,就构 成了X ( z ) 的收敛域(ROC)。 X(z)存在或级数收敛的充分条件是
奥本海姆《信号与系统》(第2版)(下册)章节题库-采样(圣才出品)
s = 2m = 21 。
2.对信号
进行均匀冲激抽样,为使抽样信号不产生混叠,应使抽
样频率
【答案】
【解析】
即
3.已知信号 f(t)的最高频率为
,信号
的最高频率是
。
【答案】
【解析】根据傅里叶变换的乘积特性可得
,若 F(jω)的最高
频率为
,则 F(jω)和 F(jω)卷积后的最高频率为
,信号 的
最高频率是
根据傅里叶变换的乘积特性,两
信号时域相乘,其频谱为该两信号频谱的卷积,故 f(t/4)f(t/Z)的最高角频率为
根据时域抽样定理,可得对信号 f(t/4)f(t/2)取样时,其频谱不混叠的最大取样间 隔为
5.抽样信号 Sa(100t) 的最低抽样率是
,奈奎斯特(Nyquist)间隔是
。
【答案】100 / π ; π / 100
圣才电子书
十万种考研考证电子书、题库视频学习平台
第三部分 章节题库
第 7 章 采样
一、选择题
1.下列说法中正确的是( )。
A.罗斯-霍维茨准则也能判断离散系统的稳定性
B.信号经调制后带宽一定增加
C.抽样频率必须是信号最高频率的 2 倍以上才不产生混叠D.积器是线性运算,不改变信号的带宽
图 7-7
6.图 7-8 为一“信号采样及恢复”的原理线路。x(t)、y(t)为模拟信号,F1、F2 为
滤波器,K 为理想冲激采样器。采样时间间隔为 1ms。今要在下面提供的 5 种滤波器中选
用两只,分别作为 F1 及 F2(每种滤波器只准用一次),使输出端尽量恢复原信号。该如何选
择?申述理由。
(1)高通滤波器 fc=2kHz,
信号与系统 (奥本海默) 总结 复习
第一章:Singnals and System(信号与系统)1-1:continuous-time and discrete-time signals(连续时间与离散时间信号)信号:信息的载体。
在信号与系统分析中,信号的表达式为函数(functions)P3:Signals are represented mathematically as functions of one or more independent variables (独立自变量)。
例如:关于某导线电流强度对应不同时间的函数I(t);等比数列的某一个数对应其序号的函数a[n]=b^n自变量的定义域为连续的时间段(有限或无限)的信号(函数)称为连续时间信号x(t)自变量的定义域为间断的时间点(一般地,归一为整数点…-1,0,1,2…)的信号称为离散时间信号x[n]又叫序列(sequences)。
两者有相似处,离散时间函数(又称为离散时间序列)可以看作连续时间函数对整数点时间进行抽样得到,但两者计算上有很大区别。
信号(函数)对应某一自变量值的信号函数值大小称为信号的幅度(phenomenon)。
例如x(t)=2t,在t=3时x(t)=x(3)=6就是此刻的幅度。
Signal energy and power(信号的能量与功率)把信号看作电流,该电流在某一段时间内流过1欧姆的电阻产生的能量和平均功率(average power)便是信号在该段时间的能量与功率。
因此可得在t1~~t2内信号x(t)的能量为:E=∫(t1~t2)(|x(t)|^2)dt,而相应这段时间的功率则为P=E/(t2-t1)信号在整个定义域的能量E∞=(limT→∞)∫(-T~T)(|x(t)|^2)dt信号在整个定义域的平均功率P∞=(limT→∞)(1/2T)∫(-T~T)(|x(t)|^2)dt相应的,对于离散时间信号则有P6-7(1,7)(1,9)(这个东西要输入太困难了,呵呵)显然,对于一个信号在无穷区间的能量与平均功率有三种可能:平均功率无穷大,总能量无穷大(2)平均功率有限,总能量无穷大(3)总能量有限,平均功率无穷小(也是有限)1-2:Transformations of the independent variable(自变量的变换)自变量的变换就是对信号x(t)或x[n]的自变量t或n进行相应变换,由此会影响信号。
信号与系统奥本海默第二版
❖ 由于右边序列的展开式中应包含无数多个Z的 负幂项,所以要按降幂长除。
ROC的公共部分。若没有公共区域则表明 x ( n ) 的Z变换不存在。
5)当 X ( z ) 是有理函数时,其ROC的边界总是 由 X ( z ) 的极点所在的圆周界定的。
6)若 X ( z ) 的ROC包括单位圆,则有
X(ej)X(z)|zej
三. X (的z )几何表示——零极点图:
4
3
1 z 1
4
3
将X ( z ) 展开为部分分式有:
X(z) 1 2
1-1z-1 1-1z-1
4
3
ROC1:|z|1/4 ROC2:|z|1/3
ROC1 ROC2
x(n)(1)nu(n)-2(1)nu(-n-1)
4
3
2. 幂级数展开法:(长除法) 由X ( 的z ) 定义,将其展开为幂级数,有 X (z ) x (- n )z n x (- 1 )z
n-
n1
- a-1z 1 1-a-1z 1-az-1
ROC: z a
Z平面 I m 单位圆
Re
a1
例6.1和例6.3的结论是应该熟记的,在以后的学习将经常用到。
例4. x(n)(1)nu(n)-2nu(-n-1)
2
X (z) (1)n z-n - -1 2n z-n
X(z)
i
Ai 1-aiz-1
步骤 :1. 求出X ( z ) 的所有极点 a i ;
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TEST OF HUAZHONG UNIVERSITY OF SCIENCE & TECHNOLOGY (A)
Course: SIGNALS & SYSTEMS (Closed Book) (2008/05/24)
SPECIALTY_________CLASS_________NAME__________No.____________
1. (20 points)Consider the following problems, then fill in the blanks. (2 points for each blank)
(a)
()()=-⎰-dt t t 3sin 2π
ππδ__________________;
(b) The fundamental period of sequence ⎪⎭
⎫ ⎝⎛+=376cos ][n n x π is________________; (c) If a continuous-time system is defined by ()()t x e t y t -=1, then we can determine that it ’s a (linear / nonlinear) _____________, (time invariant / time variant) _________________, (causal / noncausal) _____________ system;
(d) Consider a discrete-time system with the input and output relationship being
[][][2]y n x n x n =-, if the input [][]n A n x δ=, here A is an arbitrary real or complex number,
the output []y n =___________;
(e) If an LTI system with impulse response ()t h 1 is invertible, and its inverse system has an impulse response ()t h 2, then we have 12()()h t h t *=______________;
(f) The constant component of the continuous-time periodic signal ()sin()x t t ω= is________; (g) A signal ()x t with Fourier transform ()ωj X undergoes impulse-train sampling. If
()0=ωj X for s rad /105>ω, then the Nyquist sampling period is___________ second ; (h) Consider a signal ()t x 1 with FT ()ωj X 1. If ()01=ωj X for m ωω>, then for
signal ()⎪⎭
⎫ ⎝⎛=2312t x t x with FT ()ωj X 2, there must be ()02=ωj X for >ω_________.
2.(10 points) ()t x is shown in Figure 1, sketch and label carefully the signals ()ττd x t
⎰∞
--2
and ⎥⎦
⎤⎢⎣⎡⎪⎭⎫ ⎝
⎛+12
t x dt d .
Figure 1
3. (8 points) Verify that if the input signal []x n to an LTI system is periodic with period N , then the output []y n is also a periodic signal with the same period.
4. (10 points)
(a) Evaluate the Laplace transform of signal 2()t
x t t e -= and specify its ROC.
(b) Evaluate the inverse z-transform of ()a z z a z X >⎪⎭
⎫ ⎝⎛+=1ln .
5. (10 points) Consider the discrete-time periodic sequence [][4]k x n n k δ∞
=-∞
=
-∑, determine its
Fourier series coefficients k a and its Fourier transform, and plot []x n and ()j X e ω.
6. (10 points) Consider an ideal low-pass filter with frequency response 1,()0,c
c
H j ωωωωω⎧<⎪=⎨>⎪⎩.
Suppose that the input signal sin()
()at x t t
π=
, please answer the following questions: (a) Determine the output ()y t in the case of c a ω<; (b) Determine the output ()y t in the case of c a ω>; (c) In which case, the input signal is not distorted? And evaluate 2
()y t dt ∞-∞
⎰in this
case.
7. (16 points) Suppose we are given the following information about a causal LTI system described by a linear constant-coefficient second-order differential equation with impulse response ()h t and system function ()H s : 1. If ()1x t =, then ()1y t =-; 2. ()H s has a pole at
1s =-and a zero at 1s =; 3. ()h t doesn’t contain any impulse s and ()
20=+h . Try to answer the following questions: (a) Determine the differential equation and draw a direct-form block diagram for the system; (b) Is the system stable? (c) Determine the output of the system when the input ()()x t u t =and the initial conditions (0)4,(0)3y y --'=-=.
8. (16 points) A discrete-time causal system with system function 1()H z depicted in Figure 2(a), (a) If the system 1()H z is cascaded with another causal system with system function 2()H z to construct a system with system function ()a H z , just as Figure 2(b) shown, to make
21
()10.25j a j H e e
ωω
-=
-, determine 2()H z ; (b) Determine the unit sample response []a h n of the system ()a H z ; (c) Let ()b H z be the system function of a causal system shown in Figure 2(c), to make ()()
ωωj a j b e H e H =, determine the values of 1b 、2b 、3b in Figure 2(c); (d) Write
]n
[]x n
[]x n
[]n a
[]x n []y n
(c)
Figure 2。