奥本海姆《信号与系统(第二版)》习题参考答案
奥本海姆《信号与系统(第二版)》习题参考答案
Charpt 11.21—(a),(b),(c)一连续时间信号x(t)如图original所示，请画出下列信号并给予标注：a）x(t-1)b）x(2-t)c）x(2t+1)d）x(4-t/2)e）[x(t)=x(-t)]u(t)f）x(t)[δ(t+3/2)-δ(t-3/2)](d),(e),(f)1.22一离散时间信号x[n]如图original所示，请画出下列信号并给予标注。
a)x[n-4]b)x[3-n]c)x[3n]e)x[n]u[3-n]f)x[n-2]δ[n-2]1.23确定并画出图original信号的奇部和偶部，并给予标注。
1.25判定下列连续时间信号的周期性，若是周期的，确定它的基波周期。
a)x(t)=3cos(4t+π/3)T=2π/4=π/2;b)x(t)=e )1(t j T=2π/π=2;c)x(t)=[cos(2t-π/3)]2x(t)=1/2+cos[(cos(4t-2π/3))]/2, so T=2π/4=π/2;d)x(t)=E v {cos(4πt)u(t)}定义x(0)=1/2,则T=1/2; e)E v {sin(4πt)u(t)}非周期f ）x(t)=n n t e )2(假设其周期为T 则n n t e )2(=n T n t e )22(=n T n t e ))2(2(=n n t e )2(所以T=1/2(最小正周期)；1.26判定下列离散时间信号的周期性；若是周期的，确定他们的基波周期。
(a)x[n]=sin(6π/7+1) N=7(b)x[n]=cos(n/8-π) 不是周期信号（c ）x[n]=cos(πn 2/8)假设其周期为N ，则8/8/)(22n N n ＋k 2所以易得N=8（d ）x[n]=)4cos()2cos(n n N=8(e) x[n]=)62cos(2)8sin()4cos(2n n n N=16 1.31在本题中将要说明线性时不变性质的最重要的结果之一，即一旦知道了一个线性系统或线性时不变系统对某单一输入的响应或者对若干个输入的响应，就能直接计算出对许多其他输入信号的响应。
信号与系统奥本海姆第二版答案
信号与系统：
信号与系统是电子信息类本科生的专业课，学生应熟练地掌握本课程所讲述的基本概念、基本理论和基本分析方法，并利用这些经典理论分析、解释和计算信号、系统及其相互之间约束关系的问题。
奥本海姆（美国工程院院士）：
奥本海姆（Alan V. Oppenheim），教授，美国国家工程院（National Academy of Engineering）院士和IEEE会士，同时还担任古根海姆（Guggenheim）基金会和塞克勒尔（Sackler）基金会的客座研究员。
奥本海姆教授因其出色的研究和教学工作多次获奖，其中包括1988年IEEE教育勋章、IEEE成立百年杰出贡献奖、IEEE在声学、语音和信号处理领域的社会与技术成就奖等等。
著有《信号与系统》和《离散时间信号处理》，《信号与系统》一书是美国麻省理工学院经典教材之一，包括信号与系统分析的基本理论基本分析方法及其应用。
全书共分十一章：主要讲述了线性系统的基本理论、信号与系统的基本概念、线性时不变系统、连续采样、通信和反馈系统中的实例，并行讲座了连续系统、离散系统、时域系统和频域系统的分析方法，以使读者能透彻地理解各种信号系统的分析方法并比较其异同。
而《离散时间信号处理》也是离散（数字）信号处理的开山之作，这两本书是电子、通信等学科的权威之作，其权威性由国内相关教材对其参考、引用程度可见一斑。
奥本海姆信号与系统(第二版)复习题参考答案
第一章作业解答1.9解：（b ）jt t t j e e e t x --+-==)1(2)(由于)()(2)1()1())(1(2t x e e e T t x T j t j T t j ≠==++-+-++-，故不是周期信号；（或者：由于该函数的包络随t 增长衰减的指数信号，故其不是周期信号；） （c ）n j e n x π73][= 则πω70= 7220=ωπ是有理数，故其周期为N=2； 1.12解：]4[1][1)1(]1[1][43--=--==+---=∑∑∞=∞=n u m n mk k n n x m k δδ-3 –2 –1 0 1 2 3 4 5 6 n1…减去：-3 –2 –1 0 1 2 3 4 5 6 nu[n-4]等于：-3 –2 –1 0 1 23 4 5 6 n…故：]3[+-n u 即：M=-1，n 0=-3。
1.14解：x(t)的一个周期如图(a)所示，x(t)如图(b)所示：而：g(t)如图(c)所示……dtt dx )(如图（d ）所示：……故：)1(3)(3)(--=t g t g dtt dx 则：1t ,0t 3,32121==-==；A A 1.15解：该系统如下图所示： 2[n]（1）]4[2]3[5]2[2]}4[4]3[2{21]}3[4]2[2{]3[21]2[][][1111111222-+-+-=-+-+-+-=-+-==n x n x n x n x n x n x n x n x n x n y n y即：]4[2]3[5]2[2][-+-+-=n x n x n x n y（2）若系统级联顺序改变，该系统不会改变，因为该系统是线性时不变系统。
（也可以通过改变顺序求取输入、输出关系，与前面做对比）。
1.17解：（a ）因果性：)(sin )(t x t y =举一反例：当)0()y(,0int s x t =-=-=ππ则时输出与以后的输入有关，不是因果的；（b ）线性：按照线性的证明过程（这里略），该系统是线性的。
信号与系统奥本海姆第二版答案
《国外电子与通信教材系列:信号与系统(第2版)》是美国麻省理工学院（MIT）的经典教材之一，讨论了信号与系统分析的基本理论、基本分析方法及其应用。
全书共分11章，主要讲述了线性系统的基本理论、信号与系统的基本概念、线性时不变系统、连续与离散信号的傅里叶表示、傅里叶变换以及时域和频域系统的分析方法等内容。
作者使用了大量在滤波、采样、通信和反馈系统中的实例，并行讨论了连续系统、离散系统、时域系统和频域系统的分析方法，使读者能透彻地理解各种信号系统的分析方法并比较其异同。
上册：
下册。
信号与系统_第二版_奥本海默 _课后答案[1-10章]
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奥本海姆《信号与系统(第二版)》习题参考答案
第一章作业解答解：（b ）jt t t j e e e t x --+-==)1(2)(由于)()(2)1()1())(1(2t x e e e T t x T j t j T t j ≠==++-+-++-，故不是周期信号；（或者：由于该函数的包络随t 增长衰减的指数信号，故其不是周期信号；） （c ）n j e n x π73][= 则πω70= 7220=ωπ是有理数，故其周期为N=2；解：]4[1][1)1(]1[1][43--=--==+---=∑∑∞=∞=n u m n mk k n n x m k δδ-3 –2 –1 0 1 2 3 4 5 6 n1…减去：-3 –2 –1 0 1 2 3 4 5 6 nu[n-4]等于：-3 –2 –1 0 1 2 34 5 6 n…故：]3[+-n u 即：M=-1，n 0=-3。
解：x(t)的一个周期如图(a)所示，x(t)如图(b)所示：而：g(t)如图(c)所示……dtt dx )(如图（d ）所示：……故：)1(3)(3)(--=t g t g dtt dx 则：1t ,0t 3,32121==-==；A A 1.15解：该系统如下图所示： 2[n]（1）]4[2]3[5]2[2]}4[4]3[2{21]}3[4]2[2{]3[21]2[][][1111111222-+-+-=-+-+-+-=-+-==n x n x n x n x n x n x n x n x n x n y n y即：]4[2]3[5]2[2][-+-+-=n x n x n x n y（2）若系统级联顺序改变，该系统不会改变，因为该系统是线性时不变系统。
（也可以通过改变顺序求取输入、输出关系，与前面做对比）。
解：（a ）因果性：)(sin )(t x t y =举一反例：当)0()y(,0int s x t =-=-=ππ则时输出与以后的输入有关，不是因果的；（b ）线性：按照线性的证明过程（这里略），该系统是线性的。
信号与系统 奥本海姆 第二版 习题详解
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信号与系统奥本海姆第二版课后答案
信号与系统 奥本海姆第二版 习题解答Department of Computer Engineering2005.12ContentsChapter 1 (2)Chapter 2 (17)Chapter 3 (35)Chapter 4 (62)Chapter 5 (83)Chapter 6 (109)Chapter 7 (119)Chapter 8 (132)Chapter 9 (140)Chapter 10 (160)Chapter 1 Answers1.1 Converting from polar to Cartesian coordinates:111cos 222j eππ==- 111c o s ()222j e ππ-=-=- 2cos()sin()22jj j eπππ=+=2c o s ()s i n ()22jjj eπππ-=-=- 522j jj eeππ==4c o s ()s i n ())144jjj πππ+=+9441j jj ππ=-9441j j j ππ--==-41jj π-=-1.2 055j=, 22j e π-=,233jj e π--=212je π--=, 41j j π+=, ()2221jj eπ-=-4(1)j je π-=, 411j je π+=-12e π-1.3. (a) E ∞=4014tdt e∞-=⎰, P ∞=0, because E ∞<∞ (b) (2)42()j t t x eπ+=, 2()1t x =.Therefore, E ∞=22()dt t x +∞-∞⎰=dt +∞-∞⎰=∞,P ∞=211limlim222()TTTTT T dt dt TTt x --→∞→∞==⎰⎰lim11T →∞=(c) 2()t x =cos(t). Therefore, E ∞=23()dt t x +∞-∞⎰=2cos()dt t +∞-∞⎰=∞, P ∞=2111(2)1lim lim 2222cos()TTTTT T COS t dt dt T Tt --→∞→∞+==⎰⎰(d)1[][]12nn u n x =⎛⎫ ⎪⎝⎭,2[]11[]4nu n n x =⎛⎫ ⎪⎝⎭. Therefore, E ∞=24131[]4nn n x +∞∞-∞===⎛⎫∑∑ ⎪⎝⎭P ∞=0，because E ∞<∞.(e) 2[]n x =()28n j e ππ-+,22[]n x =1. therefore, E ∞=22[]n x +∞-∞∑=∞,P ∞=211limlim1122121[]NNN N n Nn NN N n x →∞→∞=-=-==++∑∑.(f) 3[]n x =cos 4nπ⎛⎫ ⎪⎝⎭. Therefore, E ∞=23[]n x +∞-∞∑=2cos()4n π+∞-∞∑=2cos()4n π+∞-∞∑,P ∞=1limcos 214nNN n NN π→∞=-=+⎛⎫∑ ⎪⎝⎭1cos()112lim ()2122NN n Nn N π→∞=-+=+∑ 1.4. (a) The signal x[n] is shifted by 3 to the right. The shifted signal will be zero for n<1, And n>7. (b) The signal x[n] is shifted by 4 to the left. The shifted signal will be zero for n<-6. And n>0. (c) The signal x[n] is flipped signal will be zero for n<-1 and n>2.(d) The signal x[n] is flipped and the flipped signal is shifted by 2 to the right. The new Signal will be zero for n<-2 and n>4.(e) The signal x[n] is flipped and the flipped and the flipped signal is shifted by 2 to the left. This new signal will be zero for n<-6 and n>0.1.5. (a) x(1-t) is obtained by flipping x(t) and shifting the flipped signal by 1 to the right. Therefore, x (1-t) will be zero for t>-2. (b) From (a), we know that x(1-t) is zero for t>-2. Similarly, x(2-t) is zero for t>-1, Therefore, x (1-t) +x(2-t) will be zero for t>-2. (c) x(3t) is obtained by linearly compression x(t) by a factor of3. Therefore, x(3t) will be zero for t<1.(d) x(t/3) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will bezero for t<9.1.6(a) x1(t) is not periodic because it is zero for t<0.(b) x2[n]=1 for all n. Therefore, it is periodic with a fundamental period of 1.(c) x3[n1.7. (a)()1[]vnxε={}1111[][]([][4][][4])22n n u n u n u n u nx x+-=--+----Therefore, ()1[]vnxεis zero for1[]nx>3.(b) Since x1(t) is an odd signal, ()2[]vnxεis zero for all values of t.(c)(){}11311[][][][3][3]221122vn nn n n u n u nx x xε-⎡⎤⎢⎥=+-=----⎢⎥⎢⎥⎣⎦⎛⎫⎛⎫⎪ ⎪⎝⎭⎝⎭Therefore, ()3[]vnxεis zero when n<3 and when n→∞.(d) ()1554411()(()())(2)(2)22vt tt t t u t u tx x x e eε-⎡⎤=+-=---+⎣⎦Therefore, ()4()vtxεis zero only when t→∞.1.8. (a) ()01{()}22cos(0)tt tx eπℜ=-=+(b) ()02{()}cos()cos(32)cos(3)cos(30)4tt t t tx eππℜ=+==+(c) ()3{()}sin(3)sin(3)2t tt t tx e eππ--ℜ=+=+(d) ()224{()}sin(100)sin(100)cos(100)2t t tt t t tx e e eππ---ℜ=-=+=+1.9. (a)1()tx is a periodic complex exponential.101021()j t j tt jx e eπ⎛⎫+⎪⎝⎭==(b)2()tx is a complex exponential multiplied by a decaying exponential. Therefore,2()tx is not periodic.（c）3[]nx is a periodic signal. 3[]n x=7j neπ=j neπ.3[]nx is a complex exponential with a fundamental period of 22ππ=.(d)4[]nx is a periodic signal. The fundamental period is given by N=m(23/5ππ)=10().3mBy choosing m=3. We obtain the fundamental period to be 10.(e)5[]nx is not periodic. 5[]nx is a complex exponential with 0w=3/5. We cannot find any integer m such that m(2wπ) is also an integer. Therefore,5[]nxis not periodic.1.10. x(t)=2cos(10t＋1)-sin(4t-1)Period of first term in the RHS =2105ππ=.Period of first term in the RHS =242ππ=.Therefore, the overall signal is periodic with a period which the least commonmultiple of the periods of the first and second terms. This is equal toπ.1.11. x[n] = 1+74j n e π−25j n e πPeriod of first term in the RHS =1. Period of second term in the RHS =⎪⎭⎫ ⎝⎛7/42π=7 （when m=2）Period of second term in the RHS =⎪⎭⎫ ⎝⎛5/22ππ=5 (when m=1)Therefore, the overall signal x[n] is periodic with a period which is the least common Multiple of the periods of the three terms inn x[n].This is equal to 35.1.12. The signal x[n] is as shown in figure S1.12. x[n] can be obtained by flipping u[n] and thenShifting the flipped signal by 3 to the right. Therefore, x[n]=u[-n+3]. This implies that M=-1 and no=-3.1.13y (t)=⎰∞-tdt x )(τ =dt t))2()2((--+⎰∞-τδτδ=⎪⎩⎪⎨⎧>≤≤--<2,022,12,0,t t tTherefore ⎰-==∞224d t E∑∑∞-∞=∞-∞=----=k k k t k t t g 12(3)2(3)(δδ)This implies that A 1=3, t 1=0, A 2=-3, and t 2=1.1.15 (a) The signal x 2[n], which is the input to S 2, is the same as y 1[n].Therefore ,y 2[n]= x 2[n-2]+21x 2[n-3] = y 1[n-2]+ 21y 1[n-3]=2x 1[n-2] +4x 1[n-3] +21( 2x 1[n-3]+ 4x 1[n-4]) =2x 1[n-2]+ 5x 1[n-3] + 2x 1[n-4] The input-output relationship for S isy[n]=2x[n-2]+ 5x [n-3] + 2x [n-4](b) The input-output relationship does not change if the order in which S 1and S 2 are connected series reversed. . We can easily prove this assuming that S 1 follows S 2. In this case , the signal x 1[n], which is the input to S 1 is the same as y 2[n].Therefore y 1[n] =2x 1[n]+ 4x 1[n-1]= 2y 2[n]+4 y 2[n-1]=2( x 2[n-2]+21 x 2[n-3] )+4(x 2[n-3]+21x 2[n-4]) =2 x 2[n-2]+5x 2[n-3]+ 2 x 2[n-4]The input-output relationship for S is once againy[n]=2x[n-2]+ 5x [n-3] + 2x [n-4]1.16 (a)The system is not memory less because y[n] depends on past values of x[n].(b)The output of the system will be y[n]= ]2[][-n n δδ=0(c)From the result of part (b), we may conclude that the system output is always zero for inputs of the form ][k n -δ, k ∈ ґ. Therefore , the system is not invertible .1.17 (a) The system is not causal because the output y(t) at some time may depend on future values of x(t). For instance , y(-π)=x(0).(b) Consider two arbitrary inputs x 1(t)and x 2(t).x 1(t) →y 1(t)= x 1(sin(t)) x 2(t) → y 2(t)= x 2(sin(t))Let x 3(t) be a linear combination of x 1(t) and x 2(t).That is , x 3(t)=a x 1(t)+b x 2(t)Where a and b are arbitrary scalars .If x 3(t) is the input to the given system ,then the corresponding output y 3(t) is y 3(t)= x 3( sin(t))=a x 1(sin(t))+ x 2(sin(t)) =a y 1(t)+ by 2(t)Therefore , the system is linear.1.18.(a) Consider two arbitrary inputs x 1[n]and x 2[n].x 1[n] → y 1[n] =][01k x n n n n k ∑+-=x 2[n ] → y 2[n] =][02k x n n n n k ∑+-=Let x 3[n] be a linear combination of x 1[n] and x 2[n]. That is :x 3[n]= ax 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 outputy 3[n] is y 3[n]=][03k x n n n n k ∑+-==])[][(2100k bx k ax n n n n k +∑+-==a ][001k x n n n n k ∑+-=+b ][02k x n n n n k ∑+-== ay 1[n]+b y 2[n]Therefore the system is linear.(b) Consider an arbitrary input x 1[n].Lety 1[n] =][01k x n n n n k ∑+-=be the corresponding output .Consider a second input x 2[n] obtained by shifting x 1[n] in time:x 2[n]= x 1[n-n 1]The output corresponding to this input isy 2[n]=][02k x n n n n k ∑+-== ]n [1100-∑+-=k x n n n n k = ][01011k x n n n n n n k ∑+---=Also note that y 1[n- n 1]=][01011k x n n n n n n k ∑+---=.Therefore , y 2[n]= y 1[n- n 1] This implies that the system is time-invariant.(c) If ][n x <B, then y[n]≤(2 n 0+1)B. Therefore ,C ≤(2 n 0+1)B.1.19 (a) (i) Consider two arbitrary inputs x 1(t) and x 2(t). x 1(t) → y 1(t)= t 2x 1(t-1)x 2(t) → y 2(t)= t 2x 2(t-1)Let x 3(t) be a linear combination of x 1(t) and x 2(t).That is x 3(t)=a x 1(t)+b x 2(t)where a and b are arbitrary scalars. If x 3(t) is the input to the given system, then the corresponding output y 3(t) is y 3(t)= t 2x 3 (t-1)= t 2(ax 1(t-1)+b x 2(t-1))= ay 1(t)+b y 2(t)Therefore , the system is linear.(ii) Consider an arbitrary inputs x 1(t).Let y 1(t)= t 2x 1(t-1)be the corresponding output .Consider a second input x 2(t) obtained by shifting x 1(t) in time:x 2(t)= x 1(t-t 0)The output corresponding to this input is y 2(t)= t 2x 2(t-1)= t 2x 1(t- 1- t 0)Also note that y 1(t-t 0)= (t-t 0)2x 1(t- 1- t 0)≠ y 2(t) Therefore the system is not time-invariant.(b) (i) Consider two arbitrary inputs x 1[n]and x 2[n]. x 1[n] → y 1[n] = x 12[n-2]x 2[n ] → y 2[n] = x 22[n-2].Let x 3(t) be a linear combination of x 1[n]and x 2[n].That is x 3[n]= ax 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 y 3[n] = x 32[n-2]=(a x 1[n-2] +b x 2[n-2])2=a 2x 12[n-2]+b 2x 22[n-2]+2ab x 1[n-2] x 2[n-2]≠ ay 1[n]+b y 2[n]Therefore the system is not linear.(ii) Consider an arbitrary input x 1[n]. Let y 1[n] = x 12[n-2]be the corresponding output .Consider a second input x 2[n] obtained by shifting x 1[n] in time:x 2[n]= x 1[n- n 0]The output corresponding to this input isy 2[n] = x 22[n-2].= x 12[n-2- n 0]Also note that y 1[n- n 0]= x 12[n-2- n 0] Therefore , y 2[n]= y 1[n- n 0] This implies that the system is time-invariant.(c) (i) Consider two arbitrary inputs x 1[n]and x 2[n].x 1[n] →y 1[n] = x 1[n+1]- x 1[n-1] x 2[n ]→y 2[n] = x 2[n+1 ]- x 2[n -1]Let x 3[n] be a linear combination of x 1[n] and x 2[n]. That is :x 3[n]= ax 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 y 3[n]= x 3[n+1]- x 3[n-1]=a x 1[n+1]+b x 2[n +1]-a x 1[n-1]-b x 2[n -1]=a(x 1[n+1]- x 1[n-1])+b(x 2[n +1]- x 2[n -1])= ay 1[n]+b y 2[n]Therefore the system is linear.(ii) Consider an arbitrary input x 1[n].Let y 1[n]= x 1[n+1]- x 1[n-1]be the corresponding output .Consider a second input x 2[n] obtained by shifting x 1[n] in time: x 2[n]= x 1[n-n 0]The output corresponding to this input isy 2[n]= x 2[n +1]- x 2[n -1]= x 1[n+1- n 0]- x 1[n-1- n 0] Also note that y 1[n-n 0]= x 1[n+1- n 0]- x 1[n-1- n 0] Therefore , y 2[n]= y 1[n-n 0] This implies that the system is time-invariant.(d) (i) Consider two arbitrary inputs x 1(t) and x 2(t).x 1(t) → y 1(t)= d O {}(t) x 1 x 2(t) → y 2(t)= {}(t) x 2d OLet x 3(t) be a linear combination of x 1(t) and x 2(t).That is x 3(t)=a x 1(t)+b x 2(t)where a and b are arbitrary scalars. If x 3(t) is the input to the given system, then the corresponding output y 3(t) is y 3(t)= d O {}(t) x 3={}(t) x b +(t) ax 21d O=a d O {}(t) x 1+b {}(t) x 2d O = ay 1(t)+b y 2(t)Therefore the system is linear.(ii) Consider an arbitrary inputs x 1(t).Lety 1(t)= d O {}(t) x 1=2)(x -(t) x 11t -be the corresponding output .Consider a second input x 2(t) obtained by shifting x 1(t) in time:x 2(t)= x 1(t-t 0)The output corresponding to this input isy 2(t)= {}(t) x 2d O =2)(x -(t) x 22t -=2)(x -)t -(t x 0101t t --Also note that y 1(t-t 0)= 2)(x -)t -(t x 0101t t --≠ y 2(t)Therefore the system is not time-invariant.1.20 (a) Givenx )(t =jt e 2 y(t)=t j e 3x )(t =jt e 2- y(t)=t j e 3- Since the system liner+=tj e t x 21(2/1)(jt e 2-))(1t y =1/2(tj e 3+tj e 3-)Thereforex 1(t)=cos(2t))(1t y =cos(3t)(b) we know thatx 2(t)=cos(2(t-1/2))= (j e -jte 2+je jt e 2-)/2Using the linearity property, we may once again writex 1(t)=21( j e -jt e 2+j e jte 2-))(1t y =(j e -jt e 3+je jte 3-)= cos(3t-1)Therefore,x 1(t)=cos(2(t-1/2)))(1t y =cos(3t-1)1.21.The signals are sketched in figure S1.21.1.24 The even and odd parts are sketched in Figure S1.24 1.25 (a) periodic period=2π/(4)= π/2 (b) periodic period=2π/(4)= 2(c) x(t)=[1+cos(4t-2π/3)]/2. periodic period=2π/(4)= π/2 (d) x(t)=cos(4πt)/2. periodic period=2π/(4)= 1/2 (e) x(t)=[sin(4πt)u(t)-sin(4πt)u(-t)]/2. Not period. (f) Not period.1.26 (a) periodic, period=7.(b) Not period.(c) periodic, period=8.(d) x[n]=(1/2)[cos(3πn/4+cos(πn/4)). periodic, period=8. (e) periodic, period=16. 1.27 (a) Linear, stable(b) Not period. (c) Linear(d) Linear, causal, stable(e) Time invariant, linear, causal, stable (f) Linear, stable(g) Time invariant, linear, causal 1.28 (a) Linear, stable(b) Time invariant, linear, causal, stable (c)Memoryless, linear, causal (d) Linear, stable (e) Linear, stable(f) Memoryless, linear, causal, stable (g) Linear, stable1.29 (a) Consider two inputs to the system such that[][][]{}111.S e x n y n x n −−→=ℜand [][][]{}221.Se x n y n x n −−→=ℜNow consider a third inputx3[n]=x2[n]+x 1[n]. The corresponding system outputWill be [][]{}[][]{}[]{}[]{}[][]33121212e e e e y n x n x n x n x n x n y n y n ==+=+=+ℜℜℜℜtherefore, we may conclude that the system is additive Let us now assume that inputs to the system such that [][][]{}/4111.Sj e x n y n e x n π−−→=ℜand[][][]{}/4222.Sj e x n y n e x n π−−→=ℜNow consider a third input x 3 [n]= x 2 [n]+ x 1 [n]. The corresponding system outputWill be[][]{}()[]{}()[]{}()[]{}()[]{}()[]{}()[]{}[]{}[]{}[][]/433331122/4/41212cos /4sin /4cos /4sin /4cos /4sin /4j e m e m e m e j j e e y n e x n n x n n x n n x n n x n n x n n x n e x n e x n y n y n πππππππππ==-+-+-=+=+ℜℜI ℜI ℜI ℜℜ therefore, we may conclude that the system is additive (b) (i) Consider two inputs to the system such that()()()()211111Sdx t x t y t x t dt ⎡⎤−−→=⎢⎥⎣⎦and ()()()()222211S dx t x t y t x t dt ⎡⎤−−→=⎢⎥⎣⎦ Now consider a third input x3[t]=x2[t]+x 1[t]. The corresponding system outputWill be()()()()()()()()()2333211111211dx t y t x t dt d x t x t x t x t dt y t y t ⎡⎤=⎢⎥⎣⎦⎡⎤+⎡⎤⎣⎦=⎢⎥+⎢⎥⎣⎦≠+ therefore, we may conclude that the system is not additiveNow consider a third input x 4 [t]= a x 1 [t]. The corresponding system output Will be()()()()()()()()2444211211111dx t y t x t dt d ax t ax t dt dx t a x t dt ay t ⎡⎤=⎢⎥⎣⎦⎡⎤⎡⎤⎣⎦=⎢⎥⎢⎥⎣⎦⎡⎤=⎢⎥⎣⎦=Therefore, the system is homogeneous.(ii) This system is not additive. Consider the fowling example .Let δ[n]=2δ[n+2]+2δ[n+1]+2δ[n] andx2[n]=δ[n+1]+ 2δ[n+1]+ 3δ[n]. The corresponding outputs evaluated at n=0 are [][]120203/2y andy ==Now consider a third input x 3 [n]= x 2 [n]+ x 1 [n].= 3δ[n+2]+4δ[n+1]+5δ[n]The corresponding outputs evaluated at n=0 is y 3[0]=15/4. Gnarly, y 3[0]≠ ]0[][21y y n +.This[][][][][]444442,1010,x n x n x n y n x n otherwise ⎧--≠⎪=-⎨⎪⎩[][][][][][]4445442,1010,x n x n ax n y n ay n x n otherwise ⎧--≠⎪==-⎨⎪⎩Therefore, the system is homogenous.1.30 (a) Invertible. Inverse system y(t)=x(t+4)(b)Non invertible. The signals x(t) and x 1(t)=x(t)+2πgive the same output (c) δ[n] and 2δ[n] give the same output d) Invertible. Inverse system; y(t)=dx(t)/dt(e) Invertible. Inverse system y(n)=x(n+1) for n ≥0 and y[n]=x[n] for n<0 (f) Non invertible. x (n) and –x(n) give the same result (g)Invertible. Inverse system y(n)=x(1-n) (h) Invertible. Inverse system y(t)=dx(t)/dt(i) Invertible. Inverse system y(n) = x(n)-(1/2)x[n-1] (j) Non invertible. If x(t) is any constant, then y(t)=0 (k) δ[n] and 2δ[n] result in y[n]=0 (l) Invertible. Inverse system: y(t)=x(t/2)(m) Non invertible x 1 [n]= δ[n]+ δ[n-1]and x 2 [n]= δ[n] give y[n]= δ[n] (n) Invertible. Inverse system: y[n]=x[2n]1.31 (a) Note that x 2[t]= x 1 [t]- x 1 [t-2]. Therefore, using linearity we get y 2 (t)= y 1 (t)- y 1 (t-2).this is shown in Figure S1.31(b)Note that x3 (t)= x1 [t]+ x1 [t+1]. .Therefore, using linearity we get Y3 (t)= y1 (t)+ y1 (t+2). this is2(4) y 2(t) periodic, period T; x(t) periodic, period T/2;1.33(1) True x[n]=x[n+N ]; y 1 (n)= y 1 (n+ N 0)i.e. periodic with N 0=n/2if N is even and with period N 0=n if N is odd.(2)False. y 1 [n] periodic does no imply x[n] is periodic i.e. Let x[n] = g[n]+h[n] where0,1,[][]0,(1/2),nn even n even g n and h n n odd n odd⎧⎧==⎨⎨⎩⎩ Then y 1 [n] = x [2n] is periodic but x[n] is clearly not periodic. (3)True. x [n+N] =x[n]; y 2 [n+N 0] =y 2 [n] where N 0=2N (4) True. y 2 [n+N] =y 2 [n]; y 2 [n+N 0 ]=y 2 [n] where N 0=N/2 1.34. (a) ConsiderIf x[n] is odd, x[n] +x [-n] =0. Therefore, the given summation evaluates to zero. (b) Let y[n] =x 1[n]x 2[n] .Theny [-n] =x 1[-n] x 2[-n] =-x 1[n]x 2[n] =-y[n]. This implies that y[n] is odd.(c)ConsiderUsing the result of part (b), we know that x e [n]x o [n] is an odd signal .Therefore, using the result of part (a) we may conclude thatTherefore,(d)ConsiderAgain, since x e (t) x o (t) is odd,Therefore,1.35. We want to find the smallest N 0 such that m(2π /N) N 0 =2πk or N 0 =kN/m,{}1[][0][][]n n x n x x n x n ∞∞=-∞==++-∑∑22[][]e o n n n n x x ∞∞=-∞=-∞=+∑∑222[][][]e on n n n n n x x x∞∞∞=-∞=-∞=-∞==+∑∑∑2[][]0eon n n x x ∞=-∞=∑222[][][].e on n n n n n xx x ∞∞∞=-∞=-∞=-∞==+∑∑∑2220()()()2()().eoet dt t dt t dt t t dt x x x x x ∞∞∞∞-∞-∞-∞-∞=++⎰⎰⎰⎰0()()0.et t dt x x ∞-∞=⎰222()()().e ot dt t dt t dt xx x ∞∞∞-∞-∞-∞=+⎰⎰⎰()()()()()().xy yx t x t y d y t x d t φττττττφ∞-∞∞-∞=+=-+=-⎰⎰where k is an integer, then N must be a multiple of m/k and m/k must be an integer .this implies that m/k is a divisor of both m and N .Also, if we want the smallest possible N 0, then m/k should be the GCD of m and N. Therefore, N 0=N/gcd(m,N). 1.36．(a)If x[n] is periodic0(),0..2/j n N T o e where T ωωπ+= This implies that022o T kNT k T T Nππ=⇒==a rational number . (b)T/T 0 =p/q then x[n] =2(/)j n p q e π,The fundamental period is q/gcd(p,q) and the fundmental frequencyis(c) p/gcd(p,q) periods of x(t) are needed .1.37.(a) From the definition of ().xy t φWe havepart(a) that()().xx xx t t φφ=-This implies that()xy t φis(b) Note from even .Therefore,the odd part of().xx t φis zero.(c) Here, ()().xy xx t t T φφ=-and ()().yy xx t t φφ= 1.38.(a) We know that /22(2)().t t δδ=ThereforeThis implies that1(2)().2t t δδ=(b)The plot are as shown in Figure s3.18. 1.39 We havelim ()()lim (0)()0.u t t u t δδ→→==Also,0022gcd(,)gcd(,)gcd(,)gcd(,).T pp q p q p q p q q p q p pωωππ===/21lim (2)lim ().2t t δδ→∞→∞=01lim ()()().2u t t t δδ→=u Δ'（t ） 1 1/2Δ/2-Δ/2t 0tu Δ'（t ）12Δ t 0tu Δ'（t ） 1 1/2Δ-Δttu Δ'（t ）1 1/2Δ-Δt 0t⎰⎰∞∞∞--=-=0)()()()()(ττδτττδτd t u d t u t gTherefore,0,0()1,00t g t t undefined for t >⎧⎪=<⎨⎪=⎩()0()()()t u t t δττδτδτ-=-=-1.40.(a) If a system is additive ,then also, if a system is homogeneous,then(b) y(t)=x 2(t) is such a systerm . (c) No.For example,consider y(t) ()()ty t x d ττ-∞=⎰with ()()(1).x t u t u t =--Then x(t)=0for t>1,but y(t)=1 for t>1.1.41. (a) y[n]=2x[n].Therefore, the system is time invariant.(b) y[n]=(2n-1)x[n].This is not time-invariant because y[n- N 0]≠(2n-1)2x [n- N 0]. (c) y[n]=x[n]{1+(-1)n +1+(-1)n-1}=2x[n].Therefore, the system is time invariant .1.42.(a) Consider two system S 1 and S 2 connected in series .Assume that if x 1(t) and x 2(t) arethe inputs to S 1..then y 1(t) and y 2(t) are the outputs.respectively .Also,assume thatif y 1(t) and y 2(t) are the input to S 2 ,then z 1(t) and z 2(t) are the outputs, respectively . Since S 1 is linear ,we may write()()()()11212,s ax t bx t ay t by t +→+where a and b are constants. Since S 2 is also linear ,we may write()()()()21212,s ay t by t az t bz t +→+We may therefore conclude that)()()()(212121t b t a t b t a z z x x s s +−→−+Therefore ,the series combination of S 1 and S 2 is linear. Since S 1 is time invariant, we may write()()11010s x t T y t T -→-and()()21010s y t T z t T -→-Therefore,()()121010s s x t T z t T -→-Therefore, the series combination of S 1 and S 2 is time invariant.(b) False, Let y(t)=x(t)+1 and z(t)=y(t)-1.These corresponds to two nonlinear systems. If these systems are connected in series ,then z(t)=x(t) which is a linear system.00.()().00x t y t =→=0()()()()0x t x t y t y t =-→-=(c) Let us name the output of system 1 as w[n] and the output of system 2 as z[n] .Then11[][2][2][21][22]24y n z n w n w n w n ==+-+-[][][]241121-+-+=n x n x n xThe overall system is linear and time-invariant.1.43. (a) We have())(t y t x s−→−Since S is time-invariant.())(T t y T t x s-−→−-Now if x (t) is periodic with period T. x{t}=x(t-T). Therefore, we may conclude that y(t)=y(t-T).This impliesthat y(t) is also periodic with T .A similar argument may be made in discrete time . (b)1.44 (a) Assumption : If x(t)=0 for t<t 0 ,then y(t)=0 for t< t 0.To prove That : The system is causal.Let us consider an arbitrary signal x 1(t) .Let us consider another signal x 2(t) which is the same as x 1(t)fort< t 0. But for t> t 0 , x 2(t) ≠x 1(t),Since the system is linear,()()()()1212,x t x t y t y t -→-Since ()()120x t x t -=for t< t 0 ,by our assumption =()()120y t y t -=for t< t 0 .This implies that()()12y t y t =for t< t 0 . In other words, t he output is not affected by input values for 0t t ≥. Therefore, thesystem is causal .Assumption: the system is causal . To prove that :If x(t)=0 for t< t 0 .then y(t)=0 for t< t 0 .Let us assume that the signal x(t)=0 for t< t 0 .Then we may express x(t) as ()()12()x t x t x t =-, Where()()12x t x t = for t< t 0 . the system is linear .the output to x(t) will be()()12()y t y t y t =-.Now ,since the system is causal . ()()12y t y t = for t< t 0 .implies that()()12y t y t = for t< t 0 .Therefore y(t)=0 for t< t 0 .(b) Consider y(t)=x(t)x(t+1) .Now , x(t)=0 for t< t 0 implies that y(t)=0 for t< t 0 .Note that the system is nonlinear and non-causal .(c) Consider y(t)=x(t)+1. the system is nonlinear and causal .This does not satisfy the condition of part(a). (d) Assumption: the system is invertible. To prove that :y[n]=0 for all n only if x[n]=0 for all n . Consider[]0[]x n y n =→. Since the system is linear :2[]02[]x n y n =→.Since the input has not changed in the two above equations ,we require that y[n]= 2y[n].This implies that y[n]=0. Since we have assumed that the system is invertible , only one input could have led to this particular output .That input must be x[n]=0 .Assumption: y[n]=0 for all n if x[n]=0 for all n . To prove that : The system is invertible . Suppose that11[][]x n y n → and21[][]x n y n →Since the system is linear ,1212[][][][]0x n x n y n y n -=→-=By the original assumption ,we must conclude that 12[][]x n x n =.That is ,any particular y 1[n] can be produced that by only one distinct input x 1[n] .Therefore , the system is invertible. (e) y[n]=x 2[n]. 1.45. (a) Consider ,()111()()shx x t y t t φ→= and()222()()shx x t y t t φ→=.Now, consider ()()()312x t ax t bx t =+. The corresponding system output will be()()12331212()()()()()()()()()hx hx y t x h t d a x h t d b x t h t d a t b t ay t by t ττττττττφφ∞-∞∞∞-∞-∞=+=+++=+=+⎰⎰⎰Therefore, S is linear .Now ,consider x 4(t)=x 1(t-T).The corresponding system output will be()14411()()()()()()()hx y t x h t d x T h t d x h t T d t T τττττττττφ∞-∞∞-∞∞-∞=+=-+=++=+⎰⎰⎰Clearly, y 4(t)≠ y 1(t-T).Therefore ,the system is not time-invariant.The system is definitely not causal because the output at any time depends on future values of the input signal x(t).(b) The system will then be linear ,time invariant and non-causal. 1.46. The plots are in Figure S1.46.1.47.(a) The overall response of the system of Figure P1.47.(a)=(the response of the system to x[n]+x 1[n])-the response of the system to x 1[n]=(Response of a linear system L to x[n]+x 1[n]+zero input response of S)- (Response of a linear system L to x 1[n]+zero input response of S)=( (Response of a linear system L to x[n]).Chapter 2 answers2.1 (a) We have know that 1[]*[][][]k y x n h n h k x n k ∞=-∞==-∑1[][1][1][1][1]y n h x n h x n =-++-2[1]2[1]x n x n =++-This gives1[]2[1]4[]2[1]2[2]2[4]y n n n n n n δδδδδ=+++-+--- (b)We know that2[][2]*[][][2]k y n x n h n h k x n k ∞=-∞=+=+-∑Comparing with eq.(S2.1-1),we see that21[][2]y n y n =+(c) We may rewrite eq.(S2.1-1) as1[][]*[][][]k y n x n h n x k h n k ∞=-∞==-∑Similarly, we may write3[][]*[2][][2]k y n x n h n x k h n k ∞=-∞=+=+-∑Comparing this with eq.(S2.1),we see that31[][2]y n y n =+2.2 Using given definition for the signal h[n], we may write{}11[][3][10]2k h k u k u k -⎛⎫=+-- ⎪⎝⎭The signal h[k] is non zero only in the rang 1[][2]h n h n =+. From this we know that the signal h[-k] is non zero only in the rage 93k -≤≤.If we now shift the signal h[-k] by n to the right, then the resultant signal h[n-k] will be zero in the range (9)(3)n k n -≤≤+. Therefore ,9,A n =- 3B n =+ 2.3 Let us define the signals11[][]2nx n u n ⎛⎫= ⎪⎝⎭and1[][]h n u n =. We note that1[][2]x n x n =- and 1[][2]h n h n =+ Now,。
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。
信号与系统第二版课后习题解答(6-7-9)奥本海姆
Chap 66.1 Consider a continuous-time LTI system with frequency response()()|()|H j H j H j e ωωω=and real impulse response h(t). Suppose that we apply an input 00()cos()x t t ωφ=+ to this system .The resulting output can be shown to be of the form0()()y t Ax t t =-Where A is a nonnegative real number representing anamplitude-scaling factor and 0t is a time delay.(a)Express A in terms of |()|H j ω.(b)Express 0t in terms of0()H j ω Solution:(a) For 0()()y t Ax t t =-So 0()()jt Y j AX j e ωωω-=0()()()j t Y j H j Ae X j ωωωω-== So |()|A H j ω=(b) for 0()H j t ωω=- So 0()H j t ωω=-6.3 Consider the following frequency response for a causal and stable LTI system:1()1j H j j ωωω-=+ (a) Show that |()|H j A ω=,and determine the values of A. (b)Determine which of the following statements is true about ()τω,the group delay of the system.(Note()(())/d H j d τωωω=-,where ()H j ωis expressed in aform that does not contain any discontinuities.)1.()0 0for τωω=>2.()0 0for τωω>>3 ()0 0for τωω<>Solution:(a) for |()|1H j ω== So A=1(b) for )(2)()()1()1()(ωωωωωωarctg arctg arctg j j j H -=--=+∠--∠=∠ 212)()(ωωωωτ+=∠-=d j H d So ()0 0for τωω>>6.5 Consider a continuous-time ideal bandpass filter whose frequency response is⎩⎨⎧≤≤=elsewherej H c c,03||,1)(ωωωω (a) If h(t) is the impulse response of this filter, determine a functiong(t) such that)(sin )(t g tt t h c πω= (b) As c ω is increased, dose the impulse response of the filter get more concentrated or less concentrated about the origin?Solution(a) Method 1. Let1()()()()()()2h t x t g t H j X j G j ωωωπ=↔=* They are shown in the figures,where1,sin ()(){0,c c ctx t X j t ωωωωωωπ<=↔=> So we can get()2cos(2)()2[(2)(2)]c c c g t t G j ωωπδωωδωω=↔=-++Method 2. Using the inverse FT definition,it is obtained331(){}2c c c cj t j t h t e d e d ωωωωωωωωπ--=+⎰⎰ 11{sin 3sin }{sin }{2cos 2}c c c c t t t t t tωωωωππ=-= (b) more concentrated.Chap 77.1 A real-valued signal x(t) is know to be uniquely determined by its samples when the sampling frequency is 10,000s ωπ=.For what values of ω is ()X j ω guaranteed to be zero?Solution:According to the sampling theorem 2s M w w > That is 110000500022M s w w ππ<== So if 5000M w w π>=,0)(=jw X7.2 A continuous-time signal x(t) is obtained at the output of an ideal lowpass filter with cutoff frequency 1,000c ωπ=.If impulse-train sampling is performed on x(t), which of the following sampling periods would guarantee that x(t) can be recovered from its sampled version using an appropriate lowpass filter?(a) 30.510T -=⨯(b) 3210T -=⨯(c) 410T -= Solution: π1000==c M w wFrom the sampling theorem,∴π20002=>M s w w ,that is 3102000222-==<πππM s w T ∴the conditions (a) and (c) are satisfied with the sampling theorem,(b) is not satisfied.7.3 The frequency which, under the sampling theorem,must be exceeded by the sampling frequency is called the Nyquist rate. Determine the Nyquist rate corresponding to each of the following signals:(a)()1cos(2,000)sin(4,000)x t t t ππ=++ (b)sin(4,000)()t x t tππ=(c) 2sin(4,000)()()t x t t ππ= Solution: (a))4000sin()2000cos(1)(t t t x ππ++=max(0,2000,4000)4000M w πππ==∴ the Nyquist rate is 28000s M w w π>= (b) sin(4000)()t x t tππ= 4000M w π=∴ the Nyquist rate is 28000s M w w π>= (c) 2sin(4000)()t x t t ππ⎛⎫= ⎪⎝⎭ 2sin(4000)()t x tt ππ⎛⎫= ⎪⎝⎭221(1cos(8000))2t t ππ=- ∴8000M w π=∴the Nyquist rate is 216000s M w w π>=7.4 Let x(t) be a signal with Nyquist rate 0ω. Determine the Nyquist rate for each of the following signals:(a)()(1)x t x t +- (b)()dx t dt(c)2()x t(d)0()cos x t t ωSolution:(a) we let 1()()(1)y t x t x t =+-So 1()()()(1)()j j Y j X j e X j e X j ωωωωωω--=+=+So the Nyquist rate of signal (a) is 0ω.(b) we let 2()()dx t y t dt= So 2()()Y j j X j ωωω=So the Nyquist rate of signal (b) is 0ω.(c) we let 23()()y t x t = So 31()()*()2Y j X j X j ωωωπ= So the Nyquist rate of signal (c) is 20ω.(d) we let 40()()cos y t x t t ω=For 000cos [()()]FT t ωπδωωδωω→-++ So 4001()((()(())2Y j X j X j ωωωωω=-++ So the Nyquist rate of signal (d) is 03ω7.9 Consider the signal 2sin 50()()t x t tππ= Which we wish to sample with a sampling frequency of 150s ωπ= to obtain a signal g(t) with Fourier transform ()G j ω.Determine the maximum value of 0ω for which it is guaranteed that0()75() ||G j X j for ωωωω=≤Where ()X j ω is the Fourier transform of x(t).Solution: 2sin(50)()t x t t ππ⎛⎫= ⎪⎝⎭))100cos(1(2122t t ππ-= ∴100M w π=But π150=s wthe figure about before-sampling and after-sampling of )(jw H isWe can see that only when π500≤w , the before-sampling and after-sampling of )(jw H have the same figure.So if 0..)..(75)(w w for jw X jw G ≤=The maximum value of 0w is π50.Chap 99.2 Consider the signal 5()(1)tx t e u t -=- and denote its Laplace transform by X(s).(a)Using eq.(9.3),evaluate X(s) and specify its region of convergence. (b)Determine the values of the finite numbers A and 0t such that theLaplace transform G(s) of 50()()t g t Ae u t t -=-- has the same algebraic form as X(s).what is the region of convergencecorresponding to G(s)?Solution:(a). According to eq.(9.3), we will getdt e t x s X st -∞∞-⎰=)()(dt e t u e st t --∞∞--=⎰)1(5dt e t s )5(1+-∞⎰= )5()5()5()5()5(1)5(+=+--=+-=+-+-∞+-s e s e s e s s t s ROC: Re{s}>-5(b). )()(05t t u Aet g t --=-−→←LT 0)5(5)(t s e s A s G ++-=, Re{s}<-5 ∴If )()(s X s G =then it ’s obviously that A=-1, 10-=t , Re{s}<-5.9.5 For each of the following algebraic expressions for the Laplace transform of a signal, determine the number of zeros located in the finite s-plane and the number of zeros located at infinity: (a)1113s s +++ (b) 211s s +- (c) 3211s s s -++ Solution :(a).1, 1 )3)(1(423111+++=+++s s s s s ∴it has a zero in the finite s-plane, that is 2-=sAnd because the order of the denominator exceeds the order of the numerator by 1∴ X(s) has 1 zero at infinity.(b). 0, 1 11)1)(1(1112-=-++=-+s s s s s s ∴it has no zero in the finite s-plane.And because the order of the denominator exceeds the order of the numerator by 1∴ X(s) has 1 zero at infinity.(c). 1, 0 11)1)(1(112223-=++++-=++-s s s s s s s s s ∴it has a zero in the finite s-plane, that is 1=sAnd because the order of the denominator equals to the order of the numerator∴ X(s) has no zero at infinity.9.7 How many signals have a Laplace transform that may be expressed as 2(1)(2)(3)(1)s s s s s -++++ in its region of convergence?Solution:There are 4 poles in the expression, but only 3 of them have different real part.∴ The s-plane will be divided into 4 strips which parallel to the jw-axis and have no cut-across.∴ There are 4 signals having the same Laplace transform expression.9.8 Let x(t) be a signal that has a rational Laplace transform with exactly two poles located at s=-1 and s=-3. If2()() ()t g t e x t and G j ω=[ the Fourier transform of g(t)]converges, determine whether x(t) is left sided, right sided, or two sided.Solution:)()(2t x e t g t =∴)2()(-=s X s G ROC: R(x)+Re{2}And x(t) have three possible ROC strips:),1(),1,3(),3,(+∞-----∞∴g(t) have three possible ROC strips: ),1(),1,1(),1,(+∞---∞ IF jw s s G jw G ==|)()(Then the ROC of )(s G is (-1,1)∴)(t x is two sides. 9.9 Given that1(),{}Re{}sat e u t Re s a s a -↔>-+ Determine the inverse Laplace transform of22(2)(),Re{}3712s X s s s s +=>-++ Solution: It is obtained from the partial-fractional expansion:22(2)2(2)42()712(4)(3)43s s X s s s s s s s ++-===+++++++,Re{}3s >-We can get the inverse Laplace transform from given formula and linear property.43()4()2()t t x t e u t e u t --=-9.10 Using geometric evaluation of the magnitude of the Fourier transform from the corresponding pole-zero plot ,determine, for each of the following Laplace transforms, whether the magnitude of the corresponding Fourier transform is approximately lowpass, highpass, or bandpass. (a):1}Re{,.........)3)(1(1)(1->++=s s s s H (b):221(),{}12s H s e s s s =ℜ>-++ (c):232(),{}121s H s e s s s =ℜ>-++ Solution:(a). 1}Re{,.........)3)(1(1)(1->++=s s s s H It ’s lowpass.(b).21}Re{,.........1)(22->++=s s s s s H It ’s bandpass.(c). 1}Re{., (1)2)(223->++=s s s s s H It ’shighpass.9.13 Let ()()()g t x t x t α=+- ,Where ()()t x t e u t β-=. And the Laplace transform of g(t) is 2(),1{}11s G s e s s =-<ℜ<-. Determine the values of the constants αand βSolution:()()()g t x t x t α=+-,and ()()t x t e u t β-=The Laplace transform : ()()()G s X s X s α=+- and ()1X s s β=+,Re{}1s >-From the scale property of Laplace transform,()1X s s β-=-+,Re{}1s < So 2(1)(1)()()()111s G s X s X s s s s βαββαβαα--+=+-=+=+-+-,1Re{}1s -<< From given 2()1s G s s =-,1Re{}1s -<< We can determine : 11,2αβ=-=。
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，
信号与系统奥本海姆第二版答案
第1章信号与系统1.1复习笔记1，连续时间和离散时间信号1个连续时间信号和离散时间信号（1）连续时间信号（图1-1（a））①定义连续时间信号是指自变量是连续变量的信号，并且该信号是在自变量的连续值上定义的。
②代表自变量由T表示，表示连续时间。
连续时间信号表示为X（T）。
（2）离散时间信号（图1-1（b））①定义离散时间信号的自变量仅在一组离散值中选择，并且仅在离散时间点定义信号。
②代表自变量由N表示，N表示离散时间。
离散时间信号表示为x [n]。
说明：hwocrtemp_ ROC60图1-1信号的图形表示（a）连续的时间表示；（b）离散时间信号2.信号能量和功率（1）有限间隔内信号的总能量和功率①描述中的连续时间信号x（T）：hwocrtemp_ roc120中的总能量说明：hwocrtemp_ ROC130哪里x |是X的模块（可能是复数）。
通过将上述公式除以长度t2-t1，可以获得平均功率。
②描述中的离散时间信号x [n]：hwocrtemp_ roc140中的总能量说明：hwocrtemp_ ROC150将其除以interval_中的点数即可。
Roc160获得该范围内的平均功率。
（2）无限间隔内信号的总能量和功率①无限时间连续时间信号的总能量x（T）说明：hwocrtemp_ ROC180无限时间连续时间信号x（T）的平均功率说明：hwocrtemp_ ROC220②无限时间中离散时间信号x [n]的总能量说明：hwocrtemp_ ROC190无限时间间隔内离散时间信号x [n]的平均功率说明：hwocrtemp_ ROC230（3）根据信号能量和功率的限制进行分类①该信号的总能量有限，即：hwocrtemp_ Roc240，该信号的平均功率为零。
②如果平均功率P∞是有限的，则其能量是无限的。
③具有无限大的P∞和E∞的信号。
2，自变量的变换基本转型（1）时移①X（t-t0）表示具有延迟|的X（T）。
信号与系统第二版课后答案_西安交大_奥本海姆(汉语)
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。
信号与系统第二版课后习题解答(6_7_9)奥本海姆
Chap 66.1 Consider a continuous-time LTI system with frequency response()()|()|H j H j H j e ωωω=S and real impulse response h(t). Suppose that we apply an input 00()cos()x t t ωφ=+ to this system .The resulting output can be shown to be of the form0()()y t Ax t t =-Where A is a nonnegative real number representing anamplitude-scaling factor and 0t is a time delay.(a)Express A in terms of |()|H j ω.(b)Express 0t in terms of 0()H j ωSSolution:(a) For 0()()y t Ax t t =-So 0()()jt Y j AX j eωωω-= 0()()()j t Y j H j Ae X j ωωωω-== So |()|A H j ω=(b) for 0()H j t ωω=-S So 0()H j t ωω=-S6.3 Consider the following frequency response for a causal and stable LTI system:1()1j H j j ωωω-=+ (a) Show that |()|H j A ω=,and determine the values of A. (b)Determine which of the following statements is true about ()τω,the group delay of the system.(Note()(())/d H j d τωωω=-S ,where ()H j ωS is expressed in aform that does not contain any discontinuities.)1.()0 0for τωω=>2.()0 0for τωω>>3 ()0 0for τωω<>Solution:(a) for |()|1H j ω== So A=1(b) for )(2)()()1()1()(ωωωωωωarctg arctg arctg j j j H -=--=+∠--∠=∠ 212)()(ωωωωτ+=∠-=d j H d So ()0 0for τωω>>6.5 Consider a continuous-time ideal bandpass filter whose frequency response is⎩⎨⎧≤≤=elsewherej H c c,03||,1)(ωωωω (a) If h(t) is the impulse response of this filter, determine a functiong(t) such that)(sin )(t g tt t h c πω= (b) As c ω is increased, dose the impulse response of the filter get more concentrated or less concentrated about the origin?Solution(a) Method 1. Let1()()()()()()2h t x t g t H j X j G j ωωωπ=↔=* They are shown in the figures,where1,sin ()(){0,c c ctx t X j t ωωωωωωπ<=↔=> So we can get()2cos(2)()2[(2)(2)]c c c g t t G j ωωπδωωδωω=↔=-++Method 2. Using the inverse FT definition,it is obtained331(){}2c c c cj t j t h t e d e d ωωωωωωωωπ--=+⎰⎰ 11{sin 3sin }{sin }{2cos 2}c c c c t t t t t tωωωωππ=-= (b) more concentrated.Chap 77.1 A real-valued signal x(t) is know to be uniquely determined by its samples when the sampling frequency is 10,000s ωπ=.For what values of ω is ()X j ω guaranteed to be zero?Solution:According to the sampling theorem 2s M w w > That is 110000500022M s w w ππ<== So if 5000M w w π>=,0)(=jw X7.2 A continuous-time signal x(t) is obtained at the output of an ideal lowpass filter with cutoff frequency 1,000c ωπ=.If impulse-train sampling is performed on x(t), which of the following sampling periods would guarantee that x(t) can be recovered from its sampled version using an appropriate lowpass filter?(a) 30.510T -=⨯(b) 3210T -=⨯(c) 410T -=Solution: Q π1000==c M w wFrom the sampling theorem,∴π20002=>M s w w ,that is 3102000222-==<πππM s w T ∴the conditions (a) and (c) are satisfied with the sampling theorem,(b) is not satisfied.7.3 The frequency which, under the sampling theorem, must be exceeded by the sampling frequency is called the Nyquist rate. Determine the Nyquist rate corresponding to each of the following signals:(a)()1cos(2,000)sin(4,000)x t t t ππ=++ (b)sin(4,000)()t x t tππ=(c) 2sin(4,000)()()t x t tππ= Solution: (a) Q )4000sin()2000cos(1)(t t t x ππ++=Θmax(0,2000,4000)4000M w πππ==∴ the Nyquist rate is 28000s M w w π>= (b) sin(4000)()t x t tππ= Q 4000M w π=∴ the Nyquist rate is 28000s M w w π>= (c) 2sin(4000)()t x t t ππ⎛⎫= ⎪⎝⎭Q 2sin(4000)()t x t t ππ⎛⎫= ⎪⎝⎭221(1cos(8000))2t t ππ=- ∴8000M w π=∴the Nyquist rate is 216000s M w w π>=7.4 Let x(t) be a signal with Nyquist rate 0ω. Determine the Nyquist rate for each of the following signals:(a)()(1)x t x t +- (b)()dx t dt(c)2()x t(d)0()cos x t t ωSolution:(a) we let 1()()(1)y t x t x t =+-So 1()()()(1)()j j Y j X j e X j e X j ωωωωωω--=+=+So the Nyquist rate of signal (a) is 0ω.(b) we let 2()()dx t y t dt= So 2()()Y j j X j ωωω=So the Nyquist rate of signal (b) is 0ω.(c) we let 23()()y t x t = So 31()()*()2Y j X j X j ωωωπ= So the Nyquist rate of signal (c) is 20ω.(d) we let 40()()cos y t x t t ω=For 000cos [()()]FT t ωπδωωδωω→-++ So 4001()((()(())2Y j X j X j ωωωωω=-++ So the Nyquist rate of signal (d) is 03ω7.9 Consider the signal 2sin 50()()t x t tππ= Which we wish to sample with a sampling frequency of 150s ωπ= to obtain a signal g(t) with Fourier transform ()G j ω.Determine the maximum value of 0ω for which it is guaranteed that0()75() ||G j X j for ωωωω=≤Where ()X j ω is the Fourier transform of x(t).Solution: Θ2sin(50)()t x t t ππ⎛⎫= ⎪⎝⎭))100cos(1(2122t t ππ-= ∴100M w π=But π150=s wthe figure about before-sampling and after-sampling of )(jw H isWe can see that only when π500≤w , the before-sampling and after-sampling of )(jw H have the same figure.So if 0..)..(75)(w w for jw X jw G ≤=The maximum value of 0w is π50.Chap 99.2 Consider the signal 5()(1)tx t e u t -=- and denote its Laplace transform by X(s).(a)Using eq.(9.3),evaluate X(s) and specify its region of convergence. (b)Determine the values of the finite numbers A and 0t such that theLaplace transform G(s) of 50()()t g t Ae u t t -=-- has the same algebraic form as X(s).what is the region of convergencecorresponding to G(s)?Solution:(a). According to eq.(9.3), we will getdt e t x s X st -∞∞-⎰=)()(dt e t u e st t --∞∞--=⎰)1(5dt e t s )5(1+-∞⎰=)5()5()5()5()5(1)5(+=+--=+-=+-+-∞+-s e s e s e s s t s ROC:Re{s}>-5 (b). Θ )()(05t t u Ae t g t --=-−→←LT 0)5(5)(t s e s A s G ++-=, Re{s}<-5 ∴ If )()(s X s G =then it ’s obviously that A=-1, 10-=t , Re{s}<-5.9.5 For each of the following algebraic expressions for the Laplace transform of a signal, determine the number of zeros located in the finite s-plane and the number of zeros located at infinity: (a)1113s s +++ (b) 211s s +- (c) 3211s s s -++ Solution :(a).1, 1Θ)3)(1(423111+++=+++s s s s s ∴ it has a zero in the finite s-plane, that is 2-=sAnd Θ because the order of the denominator exceeds the order of the numerator by 1∴ X(s) has 1 zero at infinity.(b). 0, 1Θ11)1)(1(1112-=-++=-+s s s s s s ∴ it has no zero in the finite s-plane.And Θ because the order of the denominator exceeds the order of the numerator by 1∴ X(s) has 1 zero at infinity.(c). 1, 0Θ 11)1)(1(112223-=++++-=++-s s s s s s s s s ∴ it has a zero in the finite s-plane, that is 1=sAnd Θ because the order of the denominator equals to the order of the numerator∴ X(s) has no zero at infinity.9.7 How many signals have a Laplace transform that may be expressed as 2(1)(2)(3)(1)s s s s s -++++ in its region of convergence?Solution:Θ There are 4 poles in the expression, but only 3 of them have different real part.∴ The s-plane will be divided into 4 strips which parallel to the jw-axis and have no cut-across.∴ There are 4 signals having the same Laplace transform expression.9.8 Let x(t) be a signal that has a rational Laplace transform with exactly two poles located at s=-1 and s=-3. If2()() ()t g t e x t and G j ω=[ the Fourier transform of g(t)]converges, determine whether x(t) is left sided, right sided, or two sided.Solution:Θ)()(2t x e t g t =∴)2()(-=s X s G ROC: R(x)+Re{2}And x(t) have three possible ROC strips:),1(),1,3(),3,(+∞-----∞∴g(t) have three possible ROC strips: ),1(),1,1(),1,(+∞---∞ IF jw s s G jw G ==|)()(Then the ROC of )(s G is (-1,1)∴)(t x is two sides. 9.9 Given that1(),{}Re{}sat e u t Re s a s a -↔>-+ Determine the inverse Laplace transform of22(2)(),Re{}3712s X s s s s +=>-++ Solution: It is obtained from the partial-fractional expansion:22(2)2(2)42()712(4)(3)43s s X s s s s s s s ++-===+++++++,Re{}3s >-We can get the inverse Laplace transform from given formula and linear property.43()4()2()t t x t e u t e u t --=-9.10 Using geometric evaluation of the magnitude of the Fourier transform from the corresponding pole-zero plot ,determine, for each of the following Laplace transforms, whether the magnitude of the corresponding Fourier transform is approximately lowpass, highpass, or bandpass. (a): 1}Re{,.........)3)(1(1)(1->++=s s s s H (b): 221(),{}12s H s e s s s =ℜ>-++(c): 232(),{}121s H s e s s s =ℜ>-++ Solution:(a). 1}Re{,.........)3)(1(1)(1->++=s s s s H It ’s lowpass.(b).21}Re{,.........1)(22->++=s s s s s H It ’s bandpass.(c). 1}Re{., (1)2)(223->++=s s s s s H It ’s highpass.9.13 Let ()()()g t x t x t α=+- ,Where ()()t x t e u t β-=. And the Laplace transform of g(t) is 2(),1{}11s G s e s s =-<ℜ<-. Determine the values of the constants αand βSolution:Θ ()()()g t x t x t α=+-,and ()()tx t e u t β-= The Laplace transform : ()()()G s X s X s α=+- and()1X s s β=+,Re{}1s >- From the scale property of Laplace transform, ()1X s s β-=-+,Re{}1s < So 2(1)(1)()()()111s G s X s X s s s s βαββαβαα--+=+-=+=+-+-,1Re{}1s -<< From given 2()1s G s s =-,1Re{}1s -<< We can determine : 11,2αβ=-=。
信号与系统第二版课后习题解答(6-7-9)奥本海姆
Chap 66.1 Consider a continuous-time LTI system with frequency response()()|()|H j H j H j e ωωω=and real impulse response h(t). Suppose that we apply an input 00()cos()x t t ωφ=+ to this system .The resulting output can be shown to be of the form0()()y t Ax t t =-Where A is a nonnegative real number representing anamplitude-scaling factor and 0t is a time delay.(a)Express A in terms of |()|H j ω.(b)Express 0t in terms of0()H j ω Solution:(a) For 0()()y t Ax t t =-So 0()()jt Y j AX j eωωω-= 0()()()j t Y j H j Ae X j ωωωω-== So |()|A H j ω=(b) for 0()H j t ωω=- So 0()H j t ωω=-6.3 Consider the following frequency response for a causal and stable LTI system:1()1j H j j ωωω-=+ (a) Show that |()|H j A ω=,and determine the values of A. (b)Determine which of the following statements is true about ()τω,the group delay of the system.(Note()(())/d H j d τωωω=-,where ()H j ωis expressed in aform that does not contain any discontinuities.)1.()0 0for τωω=>2.()0 0for τωω>>3 ()0 0for τωω<>Solution:(a) for |()|1H j ω== So A=1(b) for )(2)()()1()1()(ωωωωωωarctg arctg arctg j j j H -=--=+∠--∠=∠ 212)()(ωωωωτ+=∠-=d j H d So ()0 0for τωω>>6.5 Consider a continuous-time ideal bandpass filter whose frequency response is⎩⎨⎧≤≤=elsewherej H c c,03||,1)(ωωωω (a) If h(t) is the impulse response of this filter, determine a functiong(t) such that)(sin )(t g t t t h c πω=(b) As c ω is increased, dose the impulse response of the filter get more concentrated or less concentrated about the origin?Solution(a) Method 1. Let1()()()()()()2h t x t g t H j X j G j ωωωπ=↔=* They are shown in the figures,where1,sin ()(){0,c c ctx t X j t ωωωωωωπ<=↔=> So we can get()2cos(2)()2[(2)(2)]c c c g t t G j ωωπδωωδωω=↔=-++Method 2. Using the inverse FT definition,it is obtained331(){}2c c c cj t j t h t e d e d ωωωωωωωωπ--=+⎰⎰ 11{sin 3sin }{sin }{2cos 2}c c c c t t t t t tωωωωππ=-= (b) more concentrated.Chap 77.1 A real-valued signal x(t) is know to be uniquely determined by its samples when the sampling frequency is10,000s ωπ=.For what values ofω is ()X j ω guaranteed to be zero? Solution:According to the sampling theorem 2s M w w > That is 110000500022M s w w ππ<== So if 5000M w w π>=,0)(=jw X7.2 A continuous-time signal x(t) is obtained at the output of an ideal lowpass filter with cutoff frequency 1,000c ωπ=.If impulse-train sampling is performed on x(t), which of the following sampling periods would guarantee that x(t) can be recovered from its sampled version using an appropriate lowpass filter?(a) 30.510T -=⨯(b) 3210T -=⨯(c) 410T -= Solution: π1000==c M w wFrom the sampling theorem,∴π20002=>M s w w ,that is 3102000222-==<πππM s w T ∴the conditions (a) and (c) are satisfied with the sampling theorem,(b) is not satisfied.7.3 The frequency which, under the sampling theorem, must be exceeded by the sampling frequency is called the Nyquist rate. Determine the Nyquist rate corresponding to each of the following signals:(a)()1cos(2,000)sin(4,000)x t t t ππ=++ (b)sin(4,000)()t x t tππ=(c) 2sin(4,000)()()t x t t ππ= Solution: (a) )4000sin()2000cos(1)(t t t x ππ++=max(0,2000,4000)4000M w πππ==∴ the Nyquist rate is 28000s M w w π>= (b) sin(4000)()t x t tππ= 4000M w π=∴ the Nyquist rate is 28000s M w w π>= (c) 2sin(4000)()t x t t ππ⎛⎫= ⎪⎝⎭ 2sin(4000)()t x tt ππ⎛⎫= ⎪⎝⎭221(1cos(8000))2t t ππ=- ∴8000M w π=∴the Nyquist rate is 216000s M w w π>=7.4 Let x(t) be a signal with Nyquist rate 0ω. Determine the Nyquist rate for each of the following signals:(a)()(1)x t x t +- (b)()dx t dt(c)2()x t(d)0()cos x t t ωSolution:(a) we let 1()()(1)y t x t x t =+-So 1()()()(1)()j j Y j X j e X j e X j ωωωωωω--=+=+ So the Nyquist rate of signal (a) is 0ω.(b) we let 2()()dx t y t dt= So 2()()Y j j X j ωωω=So the Nyquist rate of signal (b) is0ω. (c) we let 23()()y t x t = So 31()()*()2Y j X j X j ωωωπ= So the Nyquist rate of signal (c) is 20ω.(d) we let 40()()cos y t x t t ω=For 000cos [()()]FT t ωπδωωδωω→-++ So 4001()((()(())2Y j X j X j ωωωωω=-++ So the Nyquist rate of signal (d) is 03ω7.9 Consider the signal 2sin 50()()t x t tππ= Which we wish to sample with a sampling frequency of 150s ωπ= to obtain a signal g(t) with Fourier transform ()G j ω.Determine the maximum value of 0ω for which it is guaranteed that0()75() ||G j X j for ωωωω=≤Where ()X j ω is the Fourier transform of x(t).Solution: 2sin(50)()t x t t ππ⎛⎫= ⎪⎝⎭))100cos(1(2122t t ππ-= ∴100M w π=But π150=s wthe figure about before-sampling and after-sampling of )(jw H isWe can see that only when π500≤w , the before-sampling and after-sampling of )(jw H have the same figure.So if 0..)..(75)(w w for jw X jw G ≤=The maximum value of 0w is π50.Chap 99.2 Consider the signal 5()(1)t x t e u t -=- and denote its Laplace transform by X(s).(a)Using eq.(9.3),evaluate X(s) and specify its region of convergence. (b)Determine the values of the finite numbers A and 0t such that the Laplace transform G(s) of 50()()t g t Ae u t t -=-- has the same algebraic form as X(s).what is the region of convergencecorresponding to G(s)?Solution:(a). According to eq.(9.3), we will getdt e t x s X st -∞∞-⎰=)()(dt e t u e st t --∞∞--=⎰)1(5dt e t s )5(1+-∞⎰=)5()5()5()5()5(1)5(+=+--=+-=+-+-∞+-s e s e s e s s t s ROC:Re{s}>-5 (b). )()(05t t u Ae t g t --=-−→←LT 0)5(5)(t s e s A s G ++-=, Re{s}<-5 ∴ If )()(s X s G =then it ’s obviously that A=-1, 10-=t , Re{s}<-5.9.5 For each of the following algebraic expressions for the Laplace transform of a signal, determine the number of zeros located in the finite s-plane and the number of zeros located at infinity: (a)1113s s +++ (b) 211s s +- (c) 3211s s s -++ Solution :(a).1, 1)3)(1(423111+++=+++s s s s s ∴ it has a zero in the finite s-plane, that is 2-=sAnd because the order of the denominator exceeds the order of the numerator by 1∴ X(s) has 1 zero at infinity.(b). 0, 111)1)(1(1112-=-++=-+s s s s s s ∴ it has no zero in the finite s-plane.And because the order of the denominator exceeds the order of the numerator by 1∴ X(s) has 1 zero at infinity.(c). 1, 011)1)(1(112223-=++++-=++-s s s s s s s s s ∴ it has a zero in the finite s-plane, that is 1=sAnd because the order of the denominator equals to the order of the numerator∴ X(s) has no zero at infinity.9.7 How many signals have a Laplace transform that may be expressed as 2(1)(2)(3)(1)s s s s s -++++ in its region of convergence?Solution:There are 4 poles in the expression, but only 3 of them have different real part.∴ The s-plane will be divided into 4 strips which parallel to the jw-axis and have no cut-across.∴ There are 4 signals having the same Laplace transform expression.9.8 Let x(t) be a signal that has a rational Laplace transform with exactly two poles located at s=-1 and s=-3. If2()() ()t g t e x t and G j ω=[ the Fourier transform of g(t)]converges, determine whether x(t) is left sided, right sided, or two sided.Solution:)()(2t x e t g t =∴)2()(-=s X s G ROC: R(x)+Re{2}And x(t) have three possible ROC strips:),1(),1,3(),3,(+∞-----∞∴g(t) have three possible ROC strips: ),1(),1,1(),1,(+∞---∞ IF jw s s G jw G ==|)()(Then the ROC of )(s G is (-1,1)∴)(t x is two sides. 9.9 Given that1(),{}Re{}sat e u t Re s a s a -↔>-+ Determine the inverse Laplace transform of22(2)(),Re{}3712s X s s s s +=>-++ Solution: It is obtained from the partial-fractional expansion:22(2)2(2)42()712(4)(3)43s s X s s s s s s s ++-===+++++++,Re{}3s >-We can get the inverse Laplace transform from given formula and linear property.43()4()2()t t x t e u t e u t --=-9.10 Using geometric evaluation of the magnitude of the Fourier transform from the corresponding pole-zero plot ,determine, for each of the following Laplace transforms, whether the magnitude of the corresponding Fourier transform is approximately lowpass, highpass, or bandpass. (a): 1}Re{,.........)3)(1(1)(1->++=s s s s H (b): 221(),{}12s H s e s s s =ℜ>-++(c): 232(),{}121s H s e s s s =ℜ>-++ Solution:(a). 1}Re{,.........)3)(1(1)(1->++=s s s s H It ’s lowpass.(b).21}Re{,.........1)(22->++=s s s s s H It ’s bandpass.(c). 1}Re{., (1)2)(223->++=s s s s s H It ’s highpass.9.13 Let ()()()g t x t x t α=+- ,Where ()()t x t e u t β-=. Andthe Laplace transform of g(t) is 2(),1{}11s G s e s s =-<ℜ<-. Determine the values of the constantsαand βSolution: ()()()g t x t x t α=+-,and ()()t x t e u t β-=The Laplace transform : ()()()G s X s X s α=+- and()1X s s β=+,Re{}1s >- From the scale property of Laplace transform, ()1X s s β-=-+,Re{}1s < So 2(1)(1)()()()111s G s X s X s s s s βαββαβαα--+=+-=+=+-+-,1Re{}1s -<< From given 2()1s G s s =-,1Re{}1s -<< We can determine : 11,2αβ=-=。
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Charpt 1
1.21—(a),(b),(c)
一连续时间信号x(t)如图original所示，请画出下列信号并给予标注：a）x(t-1)
b）x(2-t)
c）x(2t+1)
d）x(4-t/2)
e）[x(t)=x(-t)]u(t)
f）x(t)[δ(t+3/2)-δ(t-3/2)]
(d),(e),(f)
1.22
一离散时间信号x[n]如图original所示，请画出下列信号并给予标注。
a)x[n-4]
b)x[3-n]
c)x[3n]
e)x[n]u[3-n]
f)x[n-2]δ[n-2]
1.23
确定并画出图original信号的奇部和偶部，并给予标注。
1.25
判定下列连续时间信号的周期性，若是周期的，确定它的基波周期。
a) x(t)=3cos(4t+π/3)
T=2π/4=π/2;
b) x(t)=e )1(-t j π
T=2π/π=2;
c) x(t)=[cos(2t-π/3)]2
x(t)=1/2+cos[(cos(4t-2π/3))]/2, so T=2π/4=π/2;
d) x(t)=E v {cos (4πt)u(t)}
定义x(0)=1/2,则T=1/2;
e) E v {sin(4πt)u(t)}
非周期
f ）x(t)=∑∞
-∞=--n n t e )2(
假设其周期为T 则
∑∞-∞=--n n t e )2(=∑∞-∞=+--n T n t e )22(=∑∞-∞=---n T n t e ))2(2(=∑∞
-∞=--n n t e )2( 所以T=1/2(最小正周期)；
1.26
判定下列离散时间信号的周期性；若是周期的，确定他们的基波周期。
(a) x[n]=sin(6π/7+1)
N=7
(b) x[n]=cos(n/8-π)
不是周期信号
（c ）x[n]=cos(πn 2
/8)
假设其周期为N ，则8/8/)(22n N n ππ=+＋πk 2 所以易得N=8
（d ）x[n]=)4
cos()2cos(n n π
π N=8 (e) x[n]=)62cos(2)8sin()4cos(2π
ππ
π+-+n n n N=16
1.31
在本题中将要说明线性时不变性质的最重要的结果之一，即一旦知道了一个线性系统或线性时不变系统对某单一输入的响应或者对若干个输入的响应，就能直接计算出对许多其他输入信号的响应。
（a ） 考虑一个LTI 系统它对（a ）的信号x1（t ）的响应y1（t ）示于（b ）,确定并画出
该系统对于图（c ）的信号x2（t ）的响应。
（b ） 确定并画出（a ）中的系统对于（d ）的信号x3（t ）的响应。