信号与系统奥本海姆中文规范标准答案chapter3
奥本海姆《信号与系统》(第2版)笔记和课后习题(含考研真题)详解(上册)-第3章 周期信号的傅里叶级
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（1）弼 x(t)为实函数时，由亍 x(t)＝x*(t)，傅里叶级数系数一定是共轭对称的，即
（2）若 x(t)为实偶函数，那么它的傅里叶级数系数也为实偶函数。 （3）若 x(t)为实奇函数，那么它的傅里叶级数系数为纯虚奇函数。 7．连续时间周期信号的帕斯瓦尔定理 （1）连续时间周期信号的帕斯瓦尔定理：
8．连续时间傅里叶级数性质列表 表 3-1 连续时间傅里叶级数性质
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圣才电子书 十万种考研考证电子书、题库规频学习平台
1．成谐波关系的复指数信号的线性组合 一般的周期序列的线性组合就有如下：
序列φk[n]只在 k 的 N 个相继值的匙间上是丌同的，因此上式的求和仅仅需要包括 N 项。 因此将求和限表示成 k＝（N），即离散时间傅里叶级数为
三、傅里叶级数的收敛 连续时间信号的傅里叶级数收敛的条件——狄里赫利条件： 1．条件 1 在仸何周期内，x(t)必须绝对可积，即
这一条件保证了每一系数 ak 都是有限值。 2．条件 2 在仸意有限匙间内，x(t)具有有限个起伏发化；也就是说，在仸何单个周期内，x(t)的
最大值和最小值的数目有限。 3．条件 3 在 x(t)的仸何有限匙间内，只有有限个丌连续点，而丏在这些丌连续点上，函数是有限
则
（1）施加亍连续时间信号上的时间反转会导致其对应的傅里叶级数系数序列的时间反 转。
（2）若 x(t)为偶函数，则其傅里叶级数系数也为偶，若 x(t)为奇函数，则其傅里叶级 数系数也为奇。
4．时域尺度发换 时间尺度运算是直接加在 x(t)的每一次谐波分量上的，傅里叶系数仍是相同的。 x（αt）的傅里叶级数表示：
信号与系统习题答案第三章
第三章习题基础题3.1 证明cos t , cos(2)t , …, cos()nt (n 为正整数)，在区间(0,2)π的正交集。
它是否是完备集?解：(积分???)此含数集在(0,2)π为正交集。
又有sin()nt 不属于此含数集02sin()cos()0nt mt dt π=⎰，对于所有的m 和n 。
由完备正交函数定义所以此函数集不完备。
3.2 上题的含数集在(0,)π是否为正交集？解：由此可知此含数集在区间(0,)π内是正交的。
3.3实周期信号()f t 在区间(,)22T T-内的能量定义为222()TT E f t dt -=⎰。
如有和信号12()()f t f t +(1)若1()f t 与2()f t 在区间(,)22T T-内相互正交，证明和信号的总能量等于各信号的能量之和;(2)若1()f t 与2()f t 不是相互正交的，求和信号的总能量。
解：(1)和信号f(t)的能量为[]222222222221212222()12()()()()()()T T T T T T T T T T E f t dt dtf t dt f t dt f t f t dtf t f t -----===+++⎰⎰⎰⎰⎰（少乘以2）由1()f t 与2()f t 在区间内正交可得2122()()0T T f t f t dt -=⎰则有 22221222()()T T T T E f t dt f t dt --=+⎰⎰即此时和信号的总能量等于各信号的能量之和。
和信号的能量为(2)[]222222222221212222()12()()()()()()T T T T T T T T T T E f t dt dtf t dt f t dt f t f t dtf t f t -----===+++⎰⎰⎰⎰⎰（少乘以2吧？）由1()f t 与2()f t 在区间(,)22T T-内不正交可得2122()()0T T f t f t dt K -=≠⎰则有2222222212122222()()()()T T T T T T T T E f t dt f t dt K f t dt f t dt ----=++≠+⎰⎰⎰⎰即此时和信号的总能量不等于各信号的能量之和。
《信号与系统》奥本海姆第三章
周期性方波序列的频谱
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成谐波关系的复指数信号集:
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Wang Zhengyong
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信号与系统奥本海姆课件第3章.
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3.0 引言 Introduction
• 时域分析方法的基础 ： 1)信号在时域的分解。 2)LTI系统满足线性、时不变性。
• 从分解信号的角度出发，基本信号单元必须满 足两个要求：
1.本身简单，且LTI系统对它的响应能简便得到。 2.具有普遍性，能够用以构成相当广泛的信号。
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Ch3 Fourier Series Representation of Periodic Signals
第3章 周期信号的 傅里叶级数表示
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Contents：
• Representation of Periodic Signals(周期信号描述 • Fourier Series（傅里叶级数） • Response of LTI System to Periodic Signals
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信号与系统课后习题与解答第三章
3-1 求图3-1所示对称周期矩形信号的傅利叶级数（三角形式和指数形式）。
图3-1解 由图3-1可知，)(t f 为奇函数，因而00==a a n2112011201)cos(2)sin(242,)sin()(4T T T n t n T n Edt t n E T T dt t n t f T b ωωωπωω-====⎰⎰所以，三角形式的傅利叶级数（FS ）为T t t t E t f πωωωωπ2,)5sin(51)3sin(31)sin(2)(1111=⎥⎦⎤⎢⎣⎡+++=指数形式的傅利叶级数（FS ）的系数为⎪⎩⎪⎨⎧±±=-±±==-= ,3,1,0,,4,2,0,021n n jE n jb F n n π所以，指数形式的傅利叶级数为T e jE e jE e jE e jE t f t j t j t j t j πωππππωωωω2,33)(11111=++-+-=--3-2 周期矩形信号如图3-2所示。
若：图3-22τT-2τ-重复频率kHz f 5= 脉宽 s μτ20=幅度 V E 10=求直流分量大小以及基波、二次和三次谐波的有效值。
解 对于图3-2所示的周期矩形信号，其指数形式的傅利叶级数（FS ）的系数⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛====⎰⎰--22sin 12,)(1112212211τωττωππωττωωn Sa T E n n E dt Ee T T dt e t f T F tjn TT t jn n则的指数形式的傅利叶级数（FS ）为∑∑∞-∞=∞-∞=⎪⎭⎫ ⎝⎛==n tjn n tjn n e n Sa TE eF t f 112)(1ωωτωτ其直流分量为TE n Sa T EF n ττωτ=⎪⎭⎫ ⎝⎛=→2lim100 基波分量的幅度为⎪⎭⎫ ⎝⎛⋅=+-2sin 2111τωπEF F 二次谐波分量的幅度为⎪⎭⎫ ⎝⎛⋅=+-22sin 122τωπEF F 三次谐波分量的幅度为⎪⎭⎫ ⎝⎛⋅=+-23sin 32133τωπE F F 由所给参数kHz f 5=可得s T s rad 441102,/10-⨯==πω将各参数的值代入，可得直流分量大小为V 110210201046=⨯⨯⨯--基波的有效值为())(39.118sin 210101010sin 210264V ≈=⨯⨯⨯- πππ二次谐波分量的有效值为())(32.136sin 251010102sin 21064V ≈=⨯⨯⨯- πππ三次谐波分量的有效值为())(21.1524sin 32101010103sin 2310264V ≈=⨯⨯⨯⨯- πππ3-3 若周期矩形信号)(1t f 和)(2t f 的波形如图3-2所示，)(1t f 的参数为s μτ5.0=，s T μ1= ，V E 1=； )(2t f 的参数为s μτ5.1=，s T μ3= ，V E 3=，分别求：（1）)(1t f 的谱线间隔和带宽（第一零点位置），频率单位以kHz 表示； （2）)(2t f 的谱线间隔和带宽； （3）)(1t f 与)(2t f 的基波幅度之比； （4）)(1t f 基波与)(2t f 三次谐波幅度之比。
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。
信号与系统奥本海姆中文规范标准答案chapter3
第三章 3.5 解：由于)(2t x 只是对)(1t x 做了平移变换 所以，21ωω=而由傅立叶级数的性质有，1121ωωjk k jk k k k k e a ea b b b ---+=+=)(1k k jk a a e +=--ω3.8 解：由1，k k k k a a a a ∴-==-,*是虚的奇函数由2，ππω===TT 2,2 由3，)(t x 至多有三个非零傅立叶级数系数，110,,-a a a 又⎰==Tdt t x T a 0)(10，11a a -=- )()(1t j tj e ea t x ππ--=∴由4，利用parseval 定理，21,1112121===+--a a a a 即 j a j a 22,2211μ=±=∴- )sin(2)(t t x π±=∴3.11 解：由1，k a 是实偶函数由2，3可知，55,101111==⇒==-a a a N由4，∑∑∑==-==⇒==909552225050][101n k k kka a n x又511==-a a ⎩⎨⎧±==∴取其它值k N k a k ,01,5综上，)5cos(10][552)(2n ea ea n x k nNj k N k nNj k πππ===∑∑-==故有，0,5,10===C B A π3.22 解：(a). (a) 2T =，()x t Q 是实的奇函数，00a ∴=11111111111(1)||,(0)22(1)(1)()kjk t jk t jk t k k k jk t jk tk k j a te te e k jk jk k j j x t e e k k ππππππππππ-------∞=-∞=-⎡⎤-==-+=≠⎢⎥⎣⎦--∴=+⎰∑∑b). 6T =，012a ∴=00,()(1)jkw t k k k k k even a x t a e j k odd k π∞=-∞⎧⎪=∴=⎨-⎪⎩∑c). T=3, 01a ∴=02/3/3223sin(2/3)2sin(/3),(0)2()jk jk k jk tkk ja e k e k k k x t a eππωπππ∞=-∞⎡⎤=+≠⎣⎦∴=∑3.28 解：a). a) N=7, 472675sin()117[]77sin()7jk j knk n ek a x n ek ππππ--===∑ b). N=6, ∑∑=--=----•===503345033116161][61n k j k j n kn j kn j k e e e e n x a ππππ =k kek j 6sin 32sin612πππ- 51≤≤k ; 320=ac). ∑=-=----+++-==223233323]212[61][61n n k j k j k j k j kn j k e e e e e n x a πππππ=k k 32cos 313cos 3261ππ-+, 50≤≤k c). (c) )(2114sin 1][44nj n j e e jnn x πππ---=-=, (30≤≤n )∑∑∑=+-=--=-+-=80)21(280)21(2802818141n k n j n k n j n kn j k ej e j e a πππ=)21(2)21(2)21(2)21(222118111811141+-+---------•+--•---•k j k j k j k j k j kj e e j e e j e e ππππππ =22cos 22221114122----•--k e e k j k j πππ 即： ,423)21(4110-=+-=a ),2cos 21()1(411πk a k k +-=+ 3,2,1=k (d) ∑∑∑=+-=--=-+-=110)23(6110)23(61106241241121n k n j n k n j n kn j k e j e j e a πππ =)23(6)23(2)23(6)23(262112411124111121+-+---------•+--•---•k j k j k j k j k j kj e ej e e j e e ππππππ = 26cos 222611112162-•---•--k e e k j k j πππ 即： 1221122226110-=-•-=a ,3cos 16cos212126cos 22121kk k a k πππ+-=--= 111≤≤k 3.30 N=6, a). 0111,1/2,a a a -=== b). */411/2,j b b eπ--==c). 22k l k ll c a b-=-=∑,可求得：*/4*/401122cos(/4)/2,/2,/2j j c c c ec c e πππ----===== 3.34解：设0(),jk tkk y t b eω∞=-∞=∑则0();k k b a H k ω=其中k k a b 、分别是()x t 和()y t 的傅里叶级数系数。
信号与系统课后答案第三章作业答案
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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》配套题库【课后习题】(周期信号的傅里叶级数表示)
第3章周期信号的傅里叶级数表示基本题3.1 有一实值连续时间周期信号x（t），其基波周期了T＝8，x（t）的非零傅里叶级数系数为a1＝a－1＝2，a3＝a－3＝4j。
试将x（t）表示成：解：3.2 有一实值离散时间周期信号x[n]，其基波周期N＝5，x[n]的非零傅里叶级数系数为，试将x[n]表示成：解：3.3 对下面连续时间周期信号求基波频率ω0和傅里叶级数系数a k，以表示成解：即非零的傅里叶级数系数为3.4 利用傅里叶级数分析式计算下连续时间周期信号（基波频率ω0＝π）的系数a k：解：因ω0＝π，故3.5 设x1（t）是一连续时间周期信号，其基波频率为叫ω1，傅里叶系数为a k，已知x2（t）＝x1（1－t）十x1（t－1），问x2（t）的基波频率ω2与ω1是什么关系？求x2（t）的傅里叶级数系数b k与系数a k之间的关系。
解：x1（1－t）和x1（t－1）的基波频率都是ω1，则它们的基波周期都是T1＝2π/π。
因为x2（t）是x1（1－t）和x1（t－1）的线性组合，所以x2（t）的基波周期，即ω2＝ω1。
又故即3.6 有三个连续时间周期信号，其傅里叶级数表示如下：利用傅里叶级数性质回答下列问题：（a）三个信号中哪些是实值的？（b）哪些又是偶函数？解：（a）与式对照可知，对于x1（t），有由共轭对称性可知，若x1（t）为实信号，则有显然故x1（t）不是实信号。
同理，对于x2（t），对于x3（t），由于故可知x2（t）和x3（t）都是实信号。
（b）由于偶函数的傅里叶级数是偶函数，由上可知，只有x2（t）的a k是偶函数，故只有x2（t）是偶信号。
3.7 假定周期信号x（t）有基波周期为T，傅里叶系数为，的傅里叶级数系数为b k。
已知，试利用傅里叶级数的性质求a k用b k和T表达的表达式。
解：当k＝0时，故3.8 现对一信号给出如下信息：（1）x（t）是实的且为奇函数；（2）x（t）是周期的，周期T＝2，傅里叶级数为a k；（3）对|k|>1，a k＝0；（4）试确定两个不同的信号都满足这些条件。
Chapter3-1信号与系统ppt(所有系列)(奥本海姆+中文)
3.4 连续时间傅里叶级数的收敛
Convergence of the Fourier series 这一节来研究用傅氏级数表示周期信号的普遍性 问题，即满足什么条件的周期信号可以表示为傅里 叶级数。
系统对该信号加权的常数称为系统与特征函数相对 应的特征值。
结论： 复指数函数
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四.连续时间傅里叶级数系数的确定 如果周期信号 x(t ) 可以表示为傅里叶级数 2 jk0t 0 x (t ) ak e , T0 则有
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,。
离散时间信号处理奥本海姆第二版课后答案第三章
第三章连续时间信号的采样3.1 序列[]⎪⎭⎫⎝⎛=n n x 4cos π， ∞<<∞-n ， 用采样模拟信号()()t t x c 0cos Ω=， ∞<<∞-t 。
而得到，采样率为1000样本/每秒，问有哪两种可能的0Ω值以同样的采样率能得到该序列[]n x ？解：对模拟信号 ()()()t f t t x c 002cos cos π=Ω=以采样率s f 进行采样产生离散时间序列[]()()n f f nT x n x ss c 02cos π==，又对任意整数k ，⎪⎪⎭⎫⎝⎛+±=⎪⎪⎭⎫ ⎝⎛±n f kf f n f f s ss 002cos 2cos ππ ∴ 当以采样频率为s kf f f +±=0的正弦波都会产生相同的序列，对于[]⎪⎭⎫⎝⎛=n n x 4cos π∴ 420ππ=s f f ∴ 125810==s f f （样本/秒），π2500=Ω或π2250rad/s 均可。
所以0Ω取π250或π2250都能以同样的采样率得到该序列。
3.2 令()t h c 记作某一线性时不变连续时间滤波器的冲击响应，()n h d 为某一线性时不变离散时间滤波器的冲击响应。
()a 若()⎩⎨⎧<≥=-00t t e t h atc 求该连续时间滤波器的频率响应，并画出它的幅度特性。
()b 若()()nT Th n h c d =，()t h c 如()a 所给，求该离散时间滤波器的频率响应，并画出它的幅度特性。
()c 若给定a 的值，作为T 的函数，求离散时间滤波器频率响应的最小幅度值。
解：（a ）由连续时间信号的傅氏变换得：()ωωj a j H c +=1()221ωω+=a j H c(b) []()()()∑∞-∞=-==n c c d nT t t Th nT Th n h δ()()∑∞-∞=⎪⎭⎫ ⎝⎛-Ω*⋅=k c j d T jkj Tj H T eH πδπωπω2221 =∑∞-∞=⎪⎭⎫ ⎝⎛-k cT k j T j H πω2 =πωω<⎪⎭⎫ ⎝⎛T jH c=πωω<+Tja 1（c ）若a 为定值，当πω=时，幅度最小为：()22min1Ta e H j d πω+=（它是T 的函数）3.3 图P3.3-1表示一种多径信道的简单模型。
奥本海姆信号与系统中文版课后习题答案
1．对一个LTI 系统，我们已知如下信息：输入信号2()4()tx t e u t =-；输出响应22()()()t t y t e u t e u t -=-+(a) 确定系统的系统函数H(s)及收敛域。
(b) 求系统的单位冲激响应h(t)(c) 如果输入信号x(t)为(),tx t e t -=-∞<<+∞ 求输出y(t)。
解：(a)4114(),Re{}2,(),2Re{}2222(2)(2)X s s Y s s s s s s s ---=<=+=<-<--+-+1(),Re{}22H s s s =>-+(b)2()()t h t e u t -= (c)()2()()t ty t e e u d e τ+∞---τ--∞=ττ=⎰; ()(1)t ty t H e e --=-=.2. 已知因果全通系统的系统函数1()1s H s s -=+，输出信号2()()ty t e u t -=(a) 求产生此输出的输入信号x(t). (b) 若已知dt ∞∞<∞⎰+－|x(t)|，求输出信号x(t).(c) 已知一稳定系统当输入为2()te u t -时，输出为上述x(t)中的一个，确定是哪个？求出系统的单位冲激响应h(t).解：(a)1()2Y s s =+。
Re{}2s >-，()1()()(1)(2)Y s s X s H s s s +==-+ 由于()H s 的ROC 为Re{}1s >-，()X s ∴的ROC 为2Re{}1s -<<或Re{}1s >若 1ROC 为-2<Re{s}<1，则2112()()()33t tx t e u t e u t -=--若2ROC 为Re{s}>1，221()(2)()3t t x t e e u t -=+(b) 若 dt ∞∞<∞⎰+－|x(t)|，则只能是1()()x t x t =即：212()()()33t t x t e u t e u t -=--(c)212()()()()33t ty t x t e u t e u t-==--;1(),2Re{}1(1)(2)sY s ss s+=-<<-+()1()()1Y s sH sX s s+∴==-, 这就是(a)中系统的逆系统。
_奥本海姆信号与系统二版中文版答案
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 TTTT∞===∫∫∞∞−−∞→∞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 N n n N NNn 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,,22m N 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。
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第三章 3.5 解：由于)(2t x 只是对)(1t x 做了平移变换 所以，21ωω=而由傅立叶级数的性质有，1121ωωjk k jk k k k k e a ea b b b ---+=+=)(1k k jk a a e +=--ω3.8 解：由1，k k k k a a a a ∴-==-,*是虚的奇函数由2，ππω===TT 2,2 由3，)(t x 至多有三个非零傅立叶级数系数，110,,-a a a 又⎰==Tdt t x T a 0)(10，11a a -=- )()(1t j tj e ea t x ππ--=∴由4，利用parseval 定理，21,1112121===+--a a a a 即 j a j a 22,2211μ=±=∴- )sin(2)(t t x π±=∴3.11 解：由1，k a 是实偶函数由2，3可知，55,101111==⇒==-a a a N由4，∑∑∑==-==⇒==909552225050][101n k k kka a n x又511==-a a ⎩⎨⎧±==∴取其它值k N k a k ,01,5综上，)5cos(10][552)(2n ea ea n x k nNj k N k nNj k πππ===∑∑-==故有，0,5,10===C B A π3.22 解：(a). (a) 2T =，()x t Q 是实的奇函数，00a ∴=11111111111(1)||,(0)22(1)(1)()kjk t jk t jk t k k k jk t jk tk k j a te te e k jk jk k j j x t e e k k ππππππππππ-------∞=-∞=-⎡⎤-==-+=≠⎢⎥⎣⎦--∴=+⎰∑∑b). 6T =，012a ∴=00,()(1)jkw t k k k k k even a x t a e j k odd k π∞=-∞⎧⎪=∴=⎨-⎪⎩∑c). T=3, 01a ∴=02/3/3223sin(2/3)2sin(/3),(0)2()jk jk k jk tkk ja e k e k k k x t a eππωπππ∞=-∞⎡⎤=+≠⎣⎦∴=∑3.28 解：a). a) N=7, 472675sin()117[]77sin()7jk j knk n ek a x n ek ππππ--===∑ b). N=6, ∑∑=--=----•===503345033116161][61n k j k j n kn j kn j k e e e e n x a ππππ =k kek j 6sin 32sin612πππ- 51≤≤k ; 320=ac). ∑=-=----+++-==223233323]212[61][61n n k j k j k j k j kn j k e e e e e n x a πππππ=k k 32cos 313cos 3261ππ-+, 50≤≤k c). (c) )(2114sin 1][44nj n j e e jnn x πππ---=-=, (30≤≤n )∑∑∑=+-=--=-+-=80)21(280)21(2802818141n k n j n k n j n kn j k ej e j e a πππ=)21(2)21(2)21(2)21(222118111811141+-+---------•+--•---•k j k j k j k j k j kj e e j e e j e e ππππππ =22cos 22221114122----•--k e e k j k j πππ 即： ,423)21(4110-=+-=a ),2cos 21()1(411πk a k k +-=+ 3,2,1=k (d) ∑∑∑=+-=--=-+-=110)23(6110)23(61106241241121n k n j n k n j n kn j k e j e j e a πππ =)23(6)23(2)23(6)23(262112411124111121+-+---------•+--•---•k j k j k j k j k j kj e ej e e j e e ππππππ = 26cos 222611112162-•---•--k e e k j k j πππ 即： 1221122226110-=-•-=a ,3cos 16cos212126cos 22121kk k a k πππ+-=--= 111≤≤k 3.30 N=6, a). 0111,1/2,a a a -=== b). */411/2,j b b eπ--==c). 22k l k ll c a b-=-=∑,可求得：*/4*/401122cos(/4)/2,/2,/2j j c c c ec c e πππ----===== 3.34解：设0(),jk tkk y t b eω∞=-∞=∑则0();k k b a H k ω=其中k k a b 、分别是()x t 和()y t 的傅里叶级数系数。
428()16tj t H j ee dt ωωω∞---∞==+⎰ (c) ()(1)()nn x t t n δ∞=-∞=--∑; 02,;T ωπ==[]12120,11()(1)(1)221,jk tjk k k a t t e dt e k ππδδ--⎧=--=-=⎨⎩⎰偶奇,1,4k k b k jk π⎧⎪∴=⎨⎪+⎩偶奇 (d) 由图所示()x t 可得：01,2T ωπ== 011sin(/2),,1,2,22/2k k a a k k ππ===±±L L 00,01,sin(/2),8(42)k k k b b k k k j k πππ≠⎧⎪∴==⎨⎪+⎩偶，奇 3.36 解： ()j nj j n e H e e ωωω→，将此代入差分方程中可得：1()()4j nj j j n j j n eH e e e H e e ωωωωωω--=, 求得 1()114j j H e e ωω-=-, a). N=8, 信号中的谐波分量为正负3次谐波，可得*331/2,a a j -==输出信号中的傅立叶级数系数为3/43/43333(),()j j b a H eb a H e ππ---== b). N=8, 信号中的谐波分量为正负1次谐波与正负2次谐波，可得11221/2,1a a a a --====输出信号中的傅立叶级数系数为/4/4/2/211112222(),(),(),()j j j j b a H e b a H e b a H e b a H e ππππ------====3.43 解：(a) 222/200/211()()()T T T jk t jk t jk t T T Tk T a x t e dt x t e dt x t e dt T T πππ---⎡⎤==+⎢⎥⎣⎦⎰⎰⎰若()()2T x t x t =-+，则22/2/2001()(1)()T T jk t jk t k T T k a x t e dt x t e dt T πππ--⎡⎤=--⎢⎥⎣⎦⎰⎰当k 为奇数时， 2/22()T jk t T k a x t e dt T π-=⎰当k 为偶数时， 0k a = ∴只有奇次谐波(b) ():2,x t T =奇谐信号， ()()2T x t x t =-- ∴,01()(1),10t t x t t t <<⎧=⎨-+-<<⎩()x t 如下图所示。
12111012()jk tjk t jk t k a tedt te e jk jk jk k πππππππ---⎡⎤==-+⎢⎥⎣⎦=-⎰ ，（k 为奇数）0k a =， （k 为偶数） 3.44 解：由T ＝6，可得320ππω==T由条件4可知，∑∑∞∞--∞∞--=)3(33t jk k t jk k ea ea ππ即，1)1(1=-+k ，所以k 为奇数由于当k ＝0和k>2 时，有0=k a 所以 当k ＝2时，02=a 且02=-a 因此，t j t j ea ea t x 3131)(ππ--+=又由 )(t x 为实信号可知，11*1a a a ==-由条件5，6 可知 212)(61212133212==+=--⎰a a a dt t x所以 2111==-a a 则，0,3,1===C B A π3.48 解：(a) ∑>=<-=-=N n n N jk k e n n x N a π20][1ˆ0222][1n N jk k N m n N jk m N jk e a e e m x N πππ->=<--∑=• (b) N jk kk N n n N jk k e a a e n x n x N a ππ22])1[][(1ˆ->=<--=--=∑ c). 201ˆ([][])22jk n jk N k k kn N kkeven N ax n x n e a a e a k oddNππ--=<>⎧=--=-=⎨⎩∑(d)4/22441/211200/244/21/212202ˆ([][])22[][][]222222jk n Nk n N N N N j k n jk n jk n NN N n n n N N N N jk m jk n N N k n m N ax n x n e NN x n e x n e x n e N N N N m n a x n e x m e N ππππππ-=<>------===⎛⎫---+- ⎪⎝⎭===++⎛⎫=++- ⎪⎝⎭⎛⎫⎡⎤⎡⎤=-=++-+ ⎪⎢⎥⎢⎥⎝⎭⎣⎦⎣⎦=∑∑∑∑∑∑令44/21/21220222222N N jk n jk n jk N N k n n kN N a x n e x n e e Na πππ-----==⎡⎤⎡⎤++-+⎢⎥⎢⎥⎣⎦⎣⎦=∑∑22k k a a ∴=)e). **]][[1][*1ˆ22k N n n N jk N n n N jk k a e n x N e n x N a =-=-=∑∑>=<>=<-ππ f) )2(2)2(2]][[1][)1(1ˆN k N n n N N k j N n n N jk n k a e n x N e n x N a ->=<-->=<-==-=∑∑ππ(k=0,1,2,……..N-1)g) ∑∑>=<->=<-•=-=N n n j n N jk N n n N jk n k e e n x N e n x N a 22][21][)1(21ˆπππ ]][][[211012)2(2)2(2∑∑-=-=----+=N n N Nn N k n N jNk n N j en x e n x N ππ]][][[21110)()2(2)2(2∑∑-=-=------•++=N n N n N k j N k n N jNk n N j e eN n x e n x N πππ)1(21)(2)(N k j N k e a ---+=π ⎪⎩⎪⎨⎧=-为偶数，为奇数，k 0k 2)(N k a(h) Θ ()1()()(1)()2ny n x n x n =+-对信号x(n)周期为偶数时，y(n)的周期大小不变，仍为N, 直接利用变换性质即可，212k k N k a a a -⎡⎤=+⎢⎥⎣⎦)对信号x(n)周期为奇数时，此时y(n)的周期性发生了变换，周期为2N , 傅立叶级数系数为2221ˆ(()(1)())2jkn nNk n N ax n x n eNπ-=<>=+-∑再分成前后两部分，221212201ˆ(()(1)())(()(1)())2N N jk n jk n n n N N k n n N a x n x n e x n x n e N ππ----==⎧⎫=+-++-⎨⎬⎩⎭∑∑经整理后得：221122001ˆ(()(1)())(()(1)())(1)2N N jk n jk n n n k N N k n n a x n x n e x n x n e N ππ----==⎧⎫=+-+---⎨⎬⎩⎭∑∑ ∴ 1ˆ2k k aa =3.52 解： (a) Θ x[n]是实信号，∴ [][]x n x n *=而 ==--=∑n NjkN k kean x π21*][*][*21n x ea n NjkN k k =∑-=-π＝n NjkN k k ea n x π21][*∑-==∴ *k k a a -=或写为k k a a -=*令 k k k jc b a +=，则有k k k jc b a ---+=，从而有 k k b b -=，k k c c --= (b) 当N 为偶数时，2N 为一整数。