信号与系统奥本海姆答案
奥本海姆版信号与系统课后答案
第七章7.6 解：见 8.17.8 解： (a) )]()([)21()(50πωδπωδπωk k j j X n k +--=∑= 信号截止频率 πω5=m采样频率 m s T ωπππω2102.022====对于正弦信号，会发生混叠 (b) ππω5==T c所以输出信号 )sin()21()(40t k t y k k π∑== 所以j e e t g tjk t jk k k 2)21()(40ππ-=-=∑ ∑-==44k t jk k e a π其中，⎪⎪⎩⎪⎪⎨⎧≤≤-=≤≤-=+-+14)21(0041)21(11k j k k j a k k k 7.10 解：(a) 错 信号时域为矩形波，频域为sinc 函数，无论怎么样都会混叠 (b) 符合采样定理，对(c) 符合采样定理，对7.15 解：要求 76N 2,76273ππππω>=⨯>即s 237max =<∴N N 取 7.16 解： 易见ππn n 2sin2满足性质1, 3对性质2，考虑时域乘积得频域卷积，易见2))2/sin((4][n n n x ππ=7.19 解：设x[n]经零值插入后得输出为z[n] (a) 531πω≤时， ⎪⎩⎪⎨⎧><=1101)(ωωωωωj e X ⎪⎪⎩⎪⎪⎨⎧>≤<=30531)(11ωωπωωωj e Z所以 ⎪⎪⎩⎪⎪⎨⎧><=3031)(11ωωωωωj e W因此可得，n n n w πω/)3(sin ][1=又由 ]5[][n w n y =可得 )5/()35(sin][1n n n y πω= (b) 531πω>时 ⎪⎪⎩⎪⎪⎨⎧>><=53031)(11πωωωωωj e Z)/()5(sin ][n nn w ππ=∴][51)5/()(sin ][n n n n y δππ== 7.21 解： 采样频率m s Tωππω2200002>== 即πω10000<m 时，可以恢复 (a) 可以(b) 不可以(c) 不能确定(d) 可以 (e) 不可以 (f) 可以 (g) 可以7.22 解：)(*)()(21t x t x t y = 则有πωωωω10000)()()(21>==j X j X j Y πω1000=∴m 因而 πωω20002=>m s故 s T s 3102-=<ωπ 7.23 解：见 8.27.24 解：见 8.37.29 解：见 8.107.31 解：见 8.157.35 解：见 8.247.38 解：见 8.97.41 解：见 8.197.45 解： 见 8.17。
奥本海姆 信号与系统 习题参考答案
.第三章作业解答3.1解：420ππω==T ， j a a 4*33-==- 则：t j t j t j t j k tjk ke a e a e a e a ea t x 00000333311)(ωωωωω----∞-∞=+++==∑-)243cos(84cos 443sin 84cos 4)](21[8)(2144422434344434344πππππππππππππ++=-=--⨯++⨯=-++=------t t tt e e je e jejeeet j t j t j t j t jt jt j t j3.3解：)35sin(4)32cos(2)(t t t x ππ++= 则3)32cos(1=→T t π 56)35s i n (2=→T t π故：6],[21==T T lcm T 320ππω==T )(214)(21235353232t j t j t j t j e e je e ππππ---⨯+++=则：20=a 2122==-a a 25j a -= 25j a =- 3.9x[n]波形如下图所示：0 1 4 5 n…- 4 -3则：N=4，220ππω==N ]84[41]}1[8][4{41][41][122302300πππωδδjk n jk n n jk n n jk N n k e e n n e n x e n x N a --=-=->=<+=-+===∑∑∑即：2112133210j a a j a a +=-=-==3.15解：6π=T ，1220==Tπω )(ωj H 如下图所示：则：⎩⎨⎧>≤=9||08||1)(0k k jk H ωtjk k kea t x 0)(ω∑∞-∞==tjk k ktjk k k ea ea jk H t y 00880)()(ωωω∑∑-=∞-∞===而：)()(t y t x =，即：t jk k k tjk k k e a t y ea t x 0088)()(ωω∑∑-=∞-∞====故：当9||≥k 时，0=k a3.22解：（a ）2=T ，ππω==T20 ]|[12121)(11111110dt e te jk dt te dt e t x T a tjk t jk t jk T t jk k ⎰⎰⎰---------===πππωπkjk t jk t jk k j k j k k je k j e jk te k j )1(k ]02[21]|1|[211111-=⎪⎪⎩⎪⎪⎨⎧-=--=---=-----πππππππππ为奇数为偶数021110==⎰-dt t a(注意：与性质验证，由于x(t)是实奇函数，则a k 为纯虚的奇函数，满足： *k k k a a a -=-=- 且：00=a ） (d) 2=T ，ππω==T20 ])1(21[21]21[21)]1(2)([21)(1200k jk t jk T tjk k e dt e t t dt e t x T a --=-=--==---⎰⎰--ππωδδ21)]1(2)([21200-=--=⎰--dt t t a δδ3.28（b ）解：)(21)(21)2cos()32sin(][223232nj n j n jnje e eejn n n x ππππππ--++== )(416/76/6/6/7n j n j n j n j e e e e j ππππ----+=12/2.712/2.12/2.12/2..7(41ππππn j jn jn n j e e e e j----+=⎪⎪⎪⎩⎪⎪⎪⎨⎧++=-++==othersrN rN k j rN rN k j a k 05,11417,141 则：⎪⎩⎪⎨⎧++++==othersrN rN rN rN k a k 05,11,7,141||⎪⎪⎪⎩⎪⎪⎪⎨⎧++=++=-=∠othersrN rN k rN rN k a k 05,1127,12ππ 3.34解：(b)∑∞-∞=--=n nn t t x )()1()(δ其波形如下图所示：其周期T=2，基波频率为：ππω==T20 ⎩⎨⎧=--=-=--==---⎰⎰--是偶数是奇数k 01])1(1[21]1[21)]1()([21)(1200k e dt e t t dt e t x T a k jk t jk T tjk k ππωδδ而：⎪⎩⎪⎨⎧<>==--00)(44||4t et e et h t tt则：240401684141)()(s s s dte e dt e e dt e t h s H st t st t st -=++-=+==--∞-∞--∞∞-⎰⎰⎰故：2)(168)(ππjk jk H -=故：⎪⎩⎪⎨⎧-==∑∞-∞=为偶数为奇数（k k e jk ea jk H t y tjk tjk k k 0)168)()(200πωπω3.357π=T ，1420==Tπω 解：)(ωj H 如下图所示：则：⎩⎨⎧<>=17||017||1)(0k k jk H ωtjk k kea t x 0)(ω∑∞-∞==tjk k k tjk k k ea ea jk H t y 0018||0)()(ωωω∑∑∞=∞-∞===而：)()(t y t x =，即：tjk k ktjk k kea t y ea t x 0018||)()(ωω∑∑∞=∞-∞====故：当18||<k 时，0=k a3.44解：（1）*k k a a =- （2）6=T ，320ππω==T （3）⎩⎨⎧===其他，不为02||1||0k k a k（4）k jk k k a e b t x a t x π--=→--→)3()(k jk k a ea π--= 则：当为偶数k a k 0=结合（3）则：⎩⎨⎧==其他不为01||0k a k（5）帕斯瓦尔关系式：21||21||||12121=⇒=+-a a a （6）211=a 211=-a 则t e e ea e a t x t j t j t j tj 3cos )(21)(333131πππππ=+=+=--- 故：03,1===C B A π。
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。
奥本海姆信号与系统第一章部分习题答案
(e)
x[n], n 1
y[n] 0,
n0
x[n 1], n 1
(e)
Байду номын сангаас[n], n 1
y[n] 0,
n0
x[n 1], n 1
(g )
y[n] x[4n 1]
+++
1.31 在本题中将要说明线性时不变性质的重要结果之一，即一旦知道了一个线性
∴ 1 = 3，1 = 0，2 = −3，2 =1（或-1）
1.19判定下列输入-输出关系的系统是否具有线性性质、时不变性质，或两者俱有。
线
性： 3 = 1 + 2
时不变性： 2 = 1 ( − 0 )
(a) = 2 ( − 1)
∴ 是线性的
∴ 不是时不变的
基波周期0 ： 使[] = + 成立的最小正整数。
离散时间复指数信号的周期： 0 ， 0 = 是有理数，则是周期的，
2
2 0
且和无公因子时，基波周期为，角频率为 =
常数通常不讨论它的周期性，但可以认为周期为1。
1
4
2
=
= ，1 = 7
2 7 × 2 7
1
3
1
[] = [cos + cos( )]
2
4
4
N1
2
* m 8, m 3
3 / 4
N1
∴ 是周期的，基波周期为 =8
2
* m 8, m 1
/4
+ + + 1.27 这一章介绍了系统的几个一般性质，这就是一个系统可能是或不是：
信号与系统_奥本海姆_中文答案_chapter
第九章 9.6 解：(a) 若是有限持续期信号Roc 为整个s 平面，故存在极点不可能，故不可能为有限持续期。
(b) 可能是左边的。
(c) 不可能是右边的，若是右边信号，它并不是绝对可积的。
(d) x(t)可能为双边的。
9.8 解：因为te t x t g 2)()(=的傅氏变换，)(ωj G 收敛 所以)(t x 绝对可积若)(t x 为左边或者右边信号，则)(t x 不绝对可积 故)(t x 为双边信号 9.10 解：(a) 低通 (b) 带通 (c) 高通 9.14 解：dt e t x s X st⎰∞∞--=)()(， 由)(t x 是偶函数可得)()()(t d e t x s X st--=⎰-∞∞dt et x ts ⎰∞∞----=)()(dt e t x t s ⎰∞∞---=)()( )(s X -=421πj e s =为极点，故421πj e s -=也为极点，由)(t x 是实信号可知其极点成对出现，故421πj e s -=与421πje s --=也为极点。
)21)(21)(21)(21()(4444ππππjjjjes es es es Ms X --++--=由⎰∞∞-=4)(dt t x 得 4)0(=x所以，M ＝1/4 即，42}Re{42<<-s 9.21 解：(a) 3121)(+++=s s s X 2}Re{->s(b) 25)5(541)(2++++=s s s X 4}R e {->s (c) 3121)(----=s s s X 2}R e {<s (d) 22)2(1)2(1)(--+=s s s X2}R e {2<<-s (e) 22)2(1)2(1)(-++-=s s s X 2}R e {2<<-s (f) 2)2(1)(-=s s X 2}R e {<s (g) )1(1)(s e ss X --=0}R e {>s (h) 22)1()(s e s X s -=-0}R e {>s如对您有帮助，欢迎下载支持，谢谢！(i) ss X 11)(+= 0}R e {>s (j) ss X 131)(+=0}R e {>s9.23 解：1． Roc 包括 Re{s}=3 2． Roc 包括 Re{s}=03． Roc 在最左边极点的左边 4． Roc 在最右边极点的右边图1：1，2}Re{>s2，2}Re{2<<-s 3，2}Re{-<s 4，2}Re{>s图2： 1，2}Re{->s 2，2}Re{->s 3，2}Re{-<s 4，2}Re{->s 图3： 1，2}Re{>s 2，2}Re{<s 3，2}Re{<s 4，2}Re{>s 图4： 1，S 为整个平面 2，S 为整个平面 3，S 为整个平面 4，S 为整个平面 9.25 解：图略 9.27 解：)(t x 为实信号，)(s X 有一个极点为j s +-=1 )(s X ∴另一个极点为j s --=1 )1)(1()(j s j s Ms X ++-+=∴又 8)0(=X16=∴M则，)1(8)1(8)(j s jj s j s X -+-++=1}Re{->s 或者1}Re{-<s 之一使其成立又 )(2t x e t不是绝对可积的∴对任一个s ，右移2，不一定在Roc 中因此，1}Re{-<s 9.35 解：(a) )(1)(*)(s X st u t x L−→− 那么方框图表示的方程为)(*)(*)(6)(*)()()(*)(*)()(*)(2)(t u t u t y t u t y t y t u t u t x t u t x t x --=++即 ⎰⎰⎰⎰⎰⎰∞-∞-∞-∞-∞-∞---=++t ttt ttdt d y d y t y dt d x d x t x ττττττττ)(6)()()()(2)(对两边求导可得)(6)()()()()(2222t x dt t dx dt t x d t y dt t dy dt t y d --=++ (b) 126)(22++--=s s s s s H121-==s s 是)(s H 的二重极点，由于系统是因果的所以 1}Re{->sRoc 包含虚轴，所以系统是稳定的。
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 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。
《信号与系统》奥本海姆英文版课后答案
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2 (f) x [n] = . Therefore, E = = = , 3 cos x cos( n) cos( n) 3[n]
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2 )1 lim 1 ( 2 2 N 2 N 1 n N 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 of 3. Therefore, x(3t) will be zero for t<1.
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。
(完整版)信号与系统奥本海姆_习题答案
∑ {δ [n + 4m - 4k ] - δ [n + 4m - 1 - 4k ]}∑ {δ [n - 4(k - m )] - δ [n - 1 - 4(k - m )]}∑ {δ [n - 4k ] - δ [n - 1 - 4k ]}s Because g (t ) =∑ δ (t - 2k ) ,Chapter 1 Answers1.6 (a).NoBecause when t<0, x (t ) =0. 1(b).NoBecause only if n=0, x [n ] has valuable.2(c).Y esBecause x[n + 4m ] ===∞ k =-∞ ∞ k =-∞ ∞ k =-∞N=4.1.9 (a). T=π /5Because w =10, T=2π /10= π /5.(b). Not periodic.Because x (t ) = e -t e - jt , while e -t is not periodic, x (t ) is not periodic.2 2(c). N=2Because w =7 π , N=(2 π / w )*m, and m=7.0 0(d). N =10Because x (n) = 3e j 3π / 10 e j (3π / 5)n , that is w =3 π /5, N=(2 π / w )*m, and m=3.4 0(e). Not periodic.Because w =3/5, N=(2 π / w )*m=10π m/3 , it ’not a rational number .1.14 A1=3, t1=0, A2=-3, t2=1 or -1Solution: x(t) isdx(t )dtis∞ k =-∞1.15. (a). y[n]=2x[n-2]+5x[n-3]+2x[n-4]dx(t ) dx(t )=3g(t)-3g(t -1) or =3g(t)-3g(t+1)d t dt2 22 12Solution:y [n ] = x [n - 2] + 1x [n - 3] 2 2 1= y [n - 2] + y [n - 3]1 1= {2 x [n - 2] + 4 x [n - 3]} + {2 x [n - 3] + 4 x [n - 4]}1 1 1 1 =2 x [n - 2] + 5x [n - 3] + 2 x [n - 4]1 11Then, y[n ] = 2 x [n - 2] + 5x[n - 3] + 2 x [n - 4](b).No. For it ’s linearity .the relationship be tw e en y [n ] and x [n ] is the same in-out relationship with (a).1 2you 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 A δ [n ] ,then, y[n] = A 2δ [n]δ [n - 2] , so y[n]=0.(c). No.For example, when x[n]=0, y[n]=0; when x[n]= A δ [n ] , y[n]=0.So the system is not invertible.1.17. (a). No.For example, y(-π ) = x(0) . So it ’s not causal.(b). Y es.Because : y (t ) = x (sin(t )) ,y (t ) = x (sin(t ))1 122ay (t ) + by (t ) = ax (sin(t )) + bx (sin(t ))1 2121.21. Solution:W e(a).have known:(b).(c).(d).1.22.Solution:W e have known:(a).(b).(e).22 E {x(t )} =(g)1.23. Solution:For1[ x (t ) + x(-t )] v 1O {x(t )} = [ x (t ) - x(-t )] dthen, (a).(b).(c).1.24.2Solution:For:E {x[n ]} = v 1 2( x [n ] + x[-n ])1O {x[n]} = ( x [n ] - x[-n ]) dthen,(a).(b).Solution: x(t ) = E {cos(4π t )u(t )}s(c).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.v1= {cos(4πt )u (t ) + cos(4π (-t ))u (-t )}2 1= cos(4π t ){u (t ) + u(-t )}2 1= cos(4π t )2So, T=2π /4 π =0.51.26. (a). Periodic. N=7Solution: N= 2π* m =7, m=3.6π / 7(b). Aperriodic.Solution: N= 2π 1/ 8* m = 16m π , it ’not rational number .(e). Periodic. N =16Solution as follow:2 cos( n ) , it ’s period is N=2π *m/( π /4)=8, m=1.sin( n ) , it ’s period is N=2π *m/( π /8)=16, m=1.(2). g (t ) ∑δ (t - 2k )π π π πx[n ] = 2 cos( n ) + sin( n ) - 2 cos( n + 4 8 2 6)in this equation,π4 π8π π- 2 cos( n + 2 6) , it ’s period is N=2π *m/( π /2)=4, m=1.So, the fundamental period of x[n ] is N=(8,16,4)=16.1.31. SolutionBecausex (t ) = x (t ) - x (t - 2), x (t ) = x (t + 1) + x (t ) .2 11311According to LTI property ,y (t ) = y (t ) - y (t - 2), y (t ) = y (t + 1) + y (t )2 11311Extra problems:1. SupposeSketch y(t ) = ⎰t-∞x(t )dt .Solution:2. SupposeSketch:(1). g (t )[δ (t + 3) + δ (t + 1) - 2δ (t - 1)]∞k =-∞Because x[n]=(1 2 0 –1) , h[n]=(2 0 2) , the nSolution: (1).(2).Chapter 22.1 Solution:-1(a).So,y [n ] = 2δ [n + 1] + 4δ [n ] + 2δ [n - 1] + 2δ [n - 2] - 2δ [n - 4]1(b). according to the property of convolutioin:y [n ] = y [n + 2]2 1(c). y [n] = y [n + 2]31=∑ x[k ]h [n - k ]( ) 0 - ( ) (n +2)-2+1= ∑ ( ) k -2 u[n] = 2 u[n]2 ⎩0， elsewhere W e have known: x[n] = ⎨ ⎩0，elsewhere , h[n] = ⎨ ,( N ≤ 9 )， ， ∑ h[k ]u[n - k ]∑ (u[k ] - u[k - N - 1])(u[n - k ] - u[n - k - 10])∑ (u[k ] - u[k - N - 1])(u[4 - k ] - u[-k - 6])⎧∑ 1,...N ≤ 4⎪∑1,...N ≥ 4 ⎪⎩∑ (u[k ] - u[k - N - 1])(u[14 - k ] - u[4 - k ])2.3 Solution:y[n ] = x[n ]* h [n ]∞ k =-∞ ∞1= ∑ ( ) k -2 u [k - 2]u [n - k + 2]2k =-∞1 1 n +2 121 k =2 1 -21= 2[1 - ( ) n +1 ]u [n ]2the figure of the y[n] is:2.5 Solution:⎧1 ....0 ≤ n ≤ 9 ....⎧1 0≤ n ≤ N .... Then,x[n] = u[n] - u[n - 10] , h[n] = u[n] - u[n - N - 1]y[n] = x[n]* h[n] =∞k =-∞=∞ k =-∞So, y[4] =∞ k =-∞N⎪ ⎪ = ⎨k =04k =0=5, the n N ≥ 4And y[14] =∞ k =-∞⎧∑ 1,...N ≤ 14⎪∑1,...N ≥ 14 ⎪⎩ ∑ x[k ]g [n - 2k ]∑ x[k ]g [n - 2k ] = ∑ δ [k - 1]g [n - 2k ] = g [n - 2]∑ x[k ]g [n - 2k ] = ∑ δ [k - 2]g [n - 2k ] = g [n - 4]∑ x[k ]g [n - 2k ] = ∑ u[k ]g [n - 2k ] = ∑ g [n - 2k ]N⎪ ⎪= ⎨ k =514k =5∴N = 4=0, the n N < 52.7 Solution:y[n] =∞k =-∞（a ） x[n] = δ [n - 1] , y[n] =∞∞k =-∞ k =-∞ (b)x[n] = δ [n - 2] , y[n] =∞∞k =-∞k =-∞(c) S is not LTI system..(d) x[n] = u[n] , y[n] =∞ ∞∞k =-∞k =-∞ k =02.8 Solution:y(t ) = x(t ) * h (t ) = x(t ) *[δ (t + 2) + 2δ (t + 1)]= x(t + 2) + 2 x (t + 1)Then,⎩ = ⎰ u(τ - 3)e -3(t -τ )u(t - τ )d τ - ⎰ u(τ - 5)e -3(t -τ )u(t - τ )d τ⎩= u(t - 3)⎰ e -3(t -τ ) d τ - u(t - 5)⎰ e -3(t -τ ) d τ⎧t + 3,..... - 2 < t < -1 ⎪4,.......... t = -1 ⎪⎪That is, y(t ) = ⎨t + 4,..... - 1 < t ≤ 0⎪2 - 2t,....0 < t ≤ 1 ⎪ ⎪0,....... others2.10 Solution:(a). W e know:Then,h '(t ) = δ (t ) - δ (t - α )y '(t ) = x(t ) * h '(t ) = x(t ) *[δ (t ) - δ (t - α )]= x(t ) - x(t - α )that is,⎧t,.....0 ≤ t ≤ α ⎪α ,....α ≤ t ≤ 1So, y(t ) = ⎨⎪1 + α - t,.....1 ≤ t ≤ 1 + α ⎪0,.....others(b). From the figure of y '(t ) , only if α = 1 , y '(t ) would contain merely therediscontinuities.2.11 Solution:(a).y(t ) = x(t ) * h(t ) = [u (t - 3) - u (t - 5)]* e -3t u (t )∞ ∞-∞-∞tt35= ⎨⎰ e -3(t -τ ) d τ = ,.....3 ≤ t < 5 ⎪ 3 ⎪⎰ e -3(t -τ ) d τ - ⎰ e -3(t -τ ) d τ = - e ⎪ t9-3t + e 15-3t ⎪⎩ s y(t ) = e -t u (t ) * ∑ δ (t - 3k ) = ∑ [e = ∑ e -(t -3k )u (t - 3k )y(t ) = e -t [ ∑ e 3k u (t - 3k )] = e -t∑ ew [n ] = 1w [n - 1] + x[n ]⎧⎪ ⎪0,................. t < 3⎪ t1 - e 9-3t3t353,...... t ≥ 5(b). g (t ) = (dx(t ) / dt ) * h(t ) = [δ (t - 3) - δ (t - 5)]* e -3t u (t )= e -3(t -3) u (t - 3) - e -3(t -5) u (t - 5)(c). It ’obvious that g (t ) = d y (t ) / dt .2.12 Solution∞∞k =-∞k =-∞∞k =-∞Considering for 0 ≤ t < 3 ,we can obtain-t u (t ) * δ (t - 3k )]∞k =-∞0 k =-∞3k= e -t 11 - e -3.(Because k mu st be negetive , u (t - 3k ) = 1 for 0 ≤ t < 3 ).2.19 Solution:(a). W e have known:2 (1)y[n ] = αy[n - 1] + βw [n ](2)then, H ( E ) = H ( E ) H ( E ) =βE 2= .... or : (α + ) = ∴⎨ 2 8 ⎝ 2 = - E ∴ h [n ] = ⎢2( ) n - ( ) n ⎥u [n ] ⎩Θ⎰⎰ sin(2πt )δ (t + 3)dt has value only on t = -3 , but - 3 ∉ [0,5]⎰ sin(2πt )δ (t + 3)dt =0Θ⎰-4from (1), H ( E ) =E1E -1 2from (2), H ( E ) =2 βEE - α121 ( E - α )(E - )2 = β1 α 1 - (α + ) E -1 + E -22 21 α∴ y[n ] - (α + ) y[n - 1] + y[n - 2] = βx[n ]2 21 3but, y[n ] = - y[n - 2] + y[n - 1] + x[n ]8 4⎧α 1 ⎛1 ⎪ 3 ⎫ ⎪4 ⎭ ⎧ 1 ⎪α = ∴⎨ 4⎪β = 1(b). from (a), we know H ( E ) = H ( E ) H ( E ) =1 22E +1 1 E - E -4 2⎡ 1 1 ⎤ ⎣ 24 ⎦2.20 (a). 1⎪⎩β = 1E 21 1 ( E - )(E - ) 4 2(b). 0∞-∞ u (t ) cos(t )dt =⎰∞ δ (t ) cos(t )dt = cos(0) = 1-∞Θ∴(c). 05 0 5 05-5 u (1 - τ ) cos(2πτ )d τ = -⎰6 u (t ) cos(2πt )dt1 1= -⎰6 δ '(t ) cos(2πt )dt-4= cos '(2π t ) |t =0= -2π sin(2πt ) |t =0= 0∑ δ (t - kT ) * h (t )∑ h (t - kT )⎰ y(t )d t , A = ⎰ x(t )dt ,A = ⎰ h(t )d t .⎰ x(τ ) x (t - τ )d τ⎰ y(t )dt = ⎰ ⎰ x(τ ) x (t - τ )d τd t= ⎰ ⎰ x(τ ) x (t - τ )dtd τ = ⎰ x(τ ) ⎰ x(t - τ )dtd τ⎰ x(τ ) ⎰ x(ξ )d ξ d τ = ⎰ x(τ )d τ{ ⎰ x(ξ )d ξ}2.23 Solution:Θ y(t ) = x(t ) * h (t ) =∞k =-∞=∞ k =-∞∴2.27 SolutionA = y∞ ∞ ∞ x h-∞ y(t ) = x(t )* h(t ) = -∞ -∞ ∞-∞A = y∞ ∞ ∞-∞ -∞ -∞∞ ∞∞∞-∞ -∞-∞ -∞= ∞ ∞ ∞ ∞-∞= A Ax h-∞ -∞ -∞⎰e ⎰ eδ (τ - 2)d τ = ⎰ e⎰ u(τ + 1)eu(t - 2 - τ )d τ - ⎰ u(τ - 2)e= u(t - 1) ⎰ ed τ - u(t - 4) ⎰ e-(t -2-τ )d τ2.40 Solution(a) y(t ) = t-(t -τ) x(τ - 2)d τ ,Let x(t ) = δ (t ) ,then y(t ) = h (t ) .-∞So , h(t ) = t t -2-(t -τ ) -∞-∞-(t -2-ξ )δ (ξ )d ξ = e -(t -2)u(t - 2)(b)y(t ) = x(t )* h(t ) = [u(t + 1) - u(t - 2)]* e -(t -2)u(t - 2)=∞ ∞ -(t -2-τ )-∞-∞-(t -2-τ )u(t - 2 - τ )d τt -2-1-(t -2-τ ) t -2 2= u(t - 1)[e -(t -2) e τ ]| t -2 -u(t - 4)[e -(t -2) e τ ]| t -2-1 2= [1- e -(t -1) ]u(t - 1) - [1- e -(t -4) ]u(t - 4)2.46 SolutionBecaused d dx(t ) = [ 2e -3t ]u (t - 1) + 2e -3t [ u (t - 1)] d t dt d t= -3x(t ) + 2e -3t δ (t - 1) = -3x(t ) + 2e -3δ (t - 1) .From LTI property ,we knowdd tx(t ) → -3 y (t ) + 2e -3 h (t - 1)whereh (t ) is the impulse response of the system.So ,following equation can be derived.2e -3h(t - 1) = e -2t u (t )Finally, h (t ) = 12e 3e -2(t +1)u (t + 1)2.47 SoliutionAccording to the property of the linear time-invariant system:(a). y(t ) = x(t ) * h(t ) = 2 x (t ) * h (t ) = 2 y (t )0 0(b). y(t ) = x(t ) * h(t ) = [ x (t ) - x (t - 2)]* h(t )1y(t)= x (t ) * h (t ) - x (t - 2) * h (t )0 2 4t= [ y (t )] = y (1). Because H ( P ) = 1so h (t ) = (1= 2 + E - E ⎪ [ ]⎪δ [k ] = i (-1 - i) n- (-1 + i) n u [n] so h [n ] = 2 2 i= y (t ) - y (t - 2)0 0(c). y(t ) = x(t ) * h(t ) = x (t - 2) * h (t + 1) = x (t - 2) * h (t ) * δ (t + 1) = y (t - 1)0 0(d). The condition is not enough.(e). y(t ) = x(t ) * h(t ) = x (-t ) * h (-t )0 0= ⎰∞ x (-τ )h (-t + τ )d τ-∞ = ⎰∞x (m )h (-t - m )dm = y (-t )-∞(f). y(t ) = x(t ) * h (t ) = x ' (-t ) * h ' (-t ) = [ x ' (-t ) * h (-t )] ' ' ' " (t )Extra problems:1. Solute h(t), h[n](1). d 2 dy(t ) + 5 y(t ) + 6 y(t ) = x(t )dt 2 dt(2). y[n + 2] + 2 y[n + 1] + 2 y[n ] = x[n + 1]Solution:1 1 - 1= = +P 2 + 5P + 6 ( P + 2)( P + 3) P + 2 P + 3- 1+)δ (t ) = (e -2t - e -3t )u (t )P + 2P + 3(2). Because H ( E ) = E E E= =E 2 + 2E + 2 ( E + 1) 2 + 1 ( E + 1 + i)( E + 1 - i)i i E - E2E + 1 + i E + 1 - i⎛ i ⎫+E + 1 + i E + 1 - i ⎪ 2 ⎪ ⎝ ⎭x(t ) = ∑ for the period of cos( 5πt ) is T = 63the period of sin( 22⎰ x 2 (t )e - jkw 2t d t = ⎰ ( x 1 (1- t ) + x 1 (t - 1))e - jkw 1t dtT T TChapter 33.1 Solution:Fundamental period T = 8 . ω = 2π / 8 = π / 4∞a e j ω0kt = a e j ω0t + a e - j ω0t + a e j 3ω0t + a e - j 3ω0tk 1 -1 3 -3k =-∞ = 2ej ω0t+ 2e - j ω0t + 4 je j 3ω0t - 4 je - j3ω0t π 3π= 4cos( t ) - 8sin( t )4 43.2 Solution:for , a = 1 , a0 -2= e - j π / 4 , a = e j π / 4 , a 2-4= 2e - j π / 3 , a = 2e j π / 34x[n] = ∑ a e jk (2π / N )nkk =< N >= a + a e j (4π / 5)n + a e - j (4π / 5)n + a e j (8π / 5)n + a e - j (8π / 5)n0 2-24-4= 1 + e j π / 4 e j (4π / 5)n + e - j π / 4 e - j (4π / 5)n + 2e j π / 3e j (8π / 5)n + 2e - j π / 3e - j (8π / 5)n4 π 8 π= 1 + 2 cos( πn + ) + 4 cos( πn + )5 4 5 3 4 3π 8 5π= 1 + 2sin( πn + ) + 4sin( πn + )5 4 5 63.3 Solution:2πt ) is T= 3 , 3so the period of x(t ) is 6 , i.e. w = 2π / 6 = π / 32π 5π x(t ) = 2 + cos(t ) + 4sin(t )331= 2 + cos(2w t ) + 4sin(5w t )0 0 1= 2 + (e j 2w 0t + e - j 2w 0t ) - 2 j(e j5w 0t - e - j5w 0t )2 then, a = 2 , a 0 -2 1= a = , a 2 -5 = 2 j , a = -2 j 53.5 Solution:(1). Because x (t ) = x (1 - t ) + x (t - 1) , the n x (t ) has the same period as x (t ) ,21121that is T = T = T ,w = w2121(2). b = 1 k⎰ x 1 (1- t )e - jkw 1t d t + 1 ⎰ x 1 (t - 1)e - jkw 1t dt ∑∑⎰ x(t ) 2 dt = a 0 2 + a -1 2 + a 1 2 = 2 a 1 2 = 1 Fundamental period T = 8 . ω = 2π / 8 = π / 4∑∑ a H ( jkw )ejkw 0tk ω ⎩0,......k ≠ 0⎧ ∑t Because a =⎰ x(t )d t = 1⎰4 1d t + 1 ⎰ 8(-1)d t = 0TT88 4= 1 T T T T= a e - jkw 1 + a e - jkw 1 = (a -k k3.8 Solution:-k+ a )e - jkw 1 kΘx(t ) =∞ k =-∞a e jw 0ktkwhile:andx(t ) is real and odd, the n a = 0 , a = -a 0 kT = 2 , the n w = 2π / 2 = πa = 0 for k > 1k-ksox(t ) =∞ a e jw 0kt = a + a e - jw 0t + a e jw 0tk 0 -1 1k =-∞= a (e j πt - e - j πt ) = 2a sin(π t )11for1 2 2 0∴∴a = ± 2 /21x(t ) = ± 2 sin(π t )3.13 Solution:Θx(t ) =∞ k =-∞a e jw 0ktk∴ y(t ) =∞k 0k =-∞H ( jk ω ) = sin(4k ω0 ) =⎨4,...... k = 00 0 ∴ y(t ) =∞a H ( jkw )e jkw 0= 4a k 00 k =-∞1Soy(t ) = 0 .∑∑a H(jkw)e jkw0tT t H(jw)=⎨if a=0,it needs kw>100T ⎰T⎰t dt=0T ⎰x(t)e-jkw0t dt=⎰te-jk22t dt=1⎰1te-jkπt dt11⎰1tde-jkπt2jkπ⎢-1⎦⎢(e-jkπ+e jkπ)-⎥-jkπ2c os(kπ)+-jkπ⎥⎦[2cos(kπ)]=j cos(kπ)=j(-1)k............k≠03.15Solution:Θx(t)=∞k=-∞a e jw0kt k∴y(t)=∞k=-∞k0∴a=1k ⎰Ty(t)H(jkw)e-jkw0d tfor⎧⎪1,......w≤100⎪⎩0,......w>100∴k0that is k2π100 >100,.......k>π/612and k is integer,so K>8 3.22Solution:a=10x(t)dt=112-1a= k 1T2-12-1π=-1 2jkπ-1=-1⎡⎢te-jkπt⎣1-1-e-jkπt-jkπ1⎤⎥⎥=-=-12jkπ12jkπ⎡(e-jkπ-e jkπ)⎤⎣⎦⎡2sin(kπ)⎤⎢⎣=-12jkπkπkπ⎰ h (t )e - j ωt d t = ⎰ e -4 t e - j ωt d t= ⎰ e e d t + ⎰ e -4t e - j ωt d t∑0 ∑∑Ta = ⎰ x(t )e - jkw 0t d t = ⎰1/ 2 δ(t )e - jk 2πt d t = 1T T-1/ 2 ∑T∑ (-1) δ (t - n ) .T=2, ω = π , a = 1T a = ⎰ x(t )e - jkw 0t d t = ⎰ δ (t )e - jk πt d t + ⎰ 3/ 2 (-1)δ (t - 1)e - jk πt d tT 2 -1/ 2 2 1/ 2 T 16 + (k π )23.34 Solution:∞ ∞H ( j ω ) =-∞-∞0 ∞ 4t - j ωt-∞118=+=4 - j ω 4 + j ω 16 + ω 2A periodic continous-signal has Fourier Series:. x(t ) =T is the fundamental period of x(t ) . ω = 2π / T∞ k =-∞a e j ω ktkThe output of LTI system with inputed x(t ) is y(t ) =Its coefficients of Fourier Series: b = a H ( jk ω )k k 0∞ k =-∞a H ( jk ω )e jk ω tk 0(a) x(t ) =∞ n =-∞ δ (t - n ) .T=1, ω = 2π a = 1 = 1 .0 k1 k（N ot e ：If x(t ) =∞ n =-∞δ (t - nT ) , a =1 k）So b = a H ( jk 2π ) = k k 8 2=16 + (2k π )2 4 + (k π )2(b) x(t ) = ∞n =-∞n0 k= 11 1 1/2 1 k1= [1- (-1)k ] 24[1-(-1)k ]So b = a H ( jk π ) = ,k k(c) T=1, ω = 2π⎰ x(t )e - jk ω0t d t = ⎰1/ 4e - jk 2πt d t =∑∑ a H ( jkw )ejkw 0t⎪⎩0,......otherwise ⎩0,......otherwise H ( jw) = ⎨⎪, 14Let y(t ) = x(t ) , b = a , it needs a = 0 ,for k < 18..or .. k ≤ 17 .∑∑∑ 2n e - j ωn + ∑ ( )n e - j ωn1 =2 41 1 5∑a ejk ( N )n .a = k1 T T -1/ 4 k π sin(2 k π)b = a H ( jk π ) =k k k π8sin( )2 k π [16 + (2k π )2 ]3.35 Solution: T= π / 7 , ω = 2π / T = 14 .Θx(t ) =∞a e jw 0ktk∴y(t ) =k =-∞ ∞ k =-∞k 0∴b = a H ( jkw )k k 0for ⎧1,...... w ≥ 250 ⎧1,...... k ≥ 170 that is k ω 0 < 250,....... k < 250, and k is integer , so k < 18..or .. k ≤ 17 .kkk3.37 Solution:H (ej ω) = ∞n =-∞h [n ]e- j ωn=∞ n =-∞1 ( ) ne - j ωn 2-1∞1= 2n =-∞ n =0 1 3e j ω+ =1 - e j ω 1 - e - j ω - cos ω2 2 4A periodic sequen ce has Fourier Series: x [n ] =N is the fundamental period of x[n ] .k =< N >k2πThe output of LTI system with inputed x[n ] is y[n ] =∑ a H (ekj 2π k N)ejk ( 2π )n N .k =< N >∑4 .So b = a H (e j N k ) = 1 4 45 - cos( 2π k ) k =2 21 T ' 1 3T '-1 = ⎰ x(3t - 1)e T ' dt = ⎰ x(m )e = ⎰ x(m )e e⎡ 1T -1 T ⎢⎰∑a e jk (2π/T )t ,where a = 0 for every2π Its coefficients of Fourier Series: b = a H (ejN k )kk3(a) x[n ] =∞ k =-∞δ [n - 4k ] .N=4, a = 1 k k k 2π 4 4b =k3 165 π- cos( k ) 4 23.40 Solution:According to the property of fourier series:(a). a k '= a e - jkw 0t 0 + a e jkw 0t 0 = 2a cos(kw t ) = 2a cos(k k k k 0 0 k 2π t )T 0(b). Because E {x(t )} =v x(t ) + x(-t )2a ' a + a k 2-k= E {a }v k(c). Because R {x(t )} = x(t ) + x * (t )e'a + a *a = k-k k(d). a '= ( jkw ) 2 a = ( jk k 0 k 2πT) 2 ak(e). first, the period of x(3t - 1) is T ' =T3th e n ak ' 2π - jk t T ' 0 T ' -11 T -12π 2π - jkm - jk dmT TT -1- jk 2π m +1 dm T ' 3 3= e- jk 2π ⎣ T -1x(m )e2π- jk m T⎤dm ⎥⎦2π = a e- jk Tk3.43 (a) Proof:（ i ） Because x(t ) is odd harmonic , x(t ) =non-zer o even k.∞ k =-∞k kx(t + ) = ∑ a e jk (2π /T )(t + 2 )T 2∑= - ∑ a e jk (2π /T )t（ii ）Because of x(t ) = - x (t + ) ,we get the coefficients of Fourier Seriesa = ⎰ x(t )e - jk 2T π t d t = 1 ⎰ T / 2 x(t )e - jk 2T π t d t + 1 ⎰ T x(t )e - jk 2T π t d tT 0 T 0 T T /2 1 T /2 1 T /2 = ⎰ T dt + ⎰ x(t + T / 2)e x(t )e 1 T /2 1 T /2 = ⎰ x(t )eT dt - ⎰ x(t )(-1)k e T dt 1T /2It is obvious that a = 0 for every non-zer o even k. So x(t ) is odd harmonic ,-11x(t ) = ∑ δ (t - kT ) , T = π∞ T k k =-∞= ∞a e jk π e jk (2π /T )tkk =-∞∞kk =-∞It is noticed that k is odd integers or k=0.That meansTx(t ) = - x (t + )2T21 T k2π - jk t T 0 T 0 2π- jk (t +T / 2) Tdt2π 2π- jk t - jk t T 0 T 0= [1- (-1)k ] ⎰T 02π x(t )e- jk Tt d tk(b) x(t )1......-2-12 tExtra problems:∞ k =-∞(1). Consider y(t ) , when H ( jw) isx(t ) = ∑ δ (t - kT ) ↔T π T∑ a H ( jkw )ejkw 0t=1k =-∞ π∑∑π∑1(2). Consider y(t ) , when H ( jw) isSolution:∞k =-∞ 1 1 2π= , w = = 2 0(1).y(t ) =∞k 0∞k =-∞a H ( j 2k )e j 2ktk=2π (for k can only has value 0)(2).y(t ) =∞ k =-∞a H ( jkw )e jkw 0t =1k 0∞k =-∞a H ( j 2k )e j 2ktk=1π (e - j 2t + e j 2t ) =2 cos 2tπ(for k can only has value – and 1)。
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信号与系统奥本海姆答案
“信号与系统”是与通信、信息及自动控制等专业有关的一门基础学科。
它的主要任务是：（1）在“时间域”及“频率域”下研究时间函数x(t)及离散序列x（n）的各种表示方式，（2）在“时间域”及“频率域”下研究系统特性的各种描述方式，（3）在“时间域”及“频率域”下研究激励信号通过系统时所获得的响应。
信号与系统是通信和电子信息类专业的核心基础课，其中的概念和分析方法广泛应用于通信、自动控制、信号与信息处理、电路与系统等领域。
时间函数x(t)及自变量代表时间的离散序列x(n)是信号的时域的基本表示方式。
此外，无论是连续信号还是离散序列，都可以在频域中用它的傅里叶变换（即频谱函数）表示，也可以在“复频域”中用相应的拉普拉斯变换或z变换表示。
与此对应，连续或离散系统的特性可在时域中用微分方程或差分方程描述，也可分别用冲激响应h(t)或h（n）描述；或频域中用频率响应H(jω)、H（ejω）或传递函数H(s)、H(z)描述。
研究在给定的激励下通过系统的响应，可以在时域下用解微分方程或差分方程的方法，或通过激励信号与冲激响应的卷积进行，也可以在频域下，分别将连续信号或离散序列的傅里叶变换、拉普拉斯变换或z变换与系统的频率响应或传递函数相乘，得出响应的傅里叶变换、拉普拉斯变换或z变换，再进行反变换。
就得到了以时间函数或离散序列表示的响应。
信号与系统是通信和电子信息类专业的核心基础课，其中的概念和分析方法广泛应用于通信、自动控制、信号与信息处理、电路与系统等领域。
本课程从概念上可以区分为信号分解和系统分析两部分，但二者又是密切相关的，根据连续信号分解为不同的基本信号，对应推导出线性系统的分析方法分别为：时域分析、频域分析和复频域分析；离散信号分解和系统分析也是类似的过程。
本课程采用先连续后离散的布局安排知识，可先集中精力学好连续信号与系统分析的内容，再通过类比理解离散信号与系统分析的概念。
状态分析方法也结合两大块给出，从而建立完整的信号与系统的概念。
信号与系统课程研究信号与系统理论的基本概念和基本分析方法。
初步认识如何建立信号与系统的数学模型，经适当的数学分析求解，对所得结果给以物理解释、赋予物理意义。
课程范围限定于确定性信号（非随机信号）经线性、时不变系统传输与处理的基本理论。
本课程涉及的数学内容包括微分方程、差分方程、级数、复变函数、线性代数等。
本课程与先修课程"电路分析基础"联系密切，电路分析基础课程是从电路分析的角度研究问题，本课程则从系统的观点进行分析。