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完整word版,《信号与系统》综合复习资料

完整word版,《信号与系统》综合复习资料
完整word版,《信号与系统》综合复习资料

《信号与系统》综合复习资料

一、简答题:

1、dt

t df t f x e t y t

)

()

()0()(+=-其中x(0)是初始状态,为全响应,为激励,)()(t y t f 试回答该系统是否是线性的? 。

解答:由于无法区分零输入响应和零状态响应,因而系统为非线性的。 2、

4sin()()?6t t dt π

δ+∞

-∞-?=?__________________________。

解:根据冲激函数的性质:

2)60sin(4)()6sin(4--=-=-?∞

∞π

δπdt t t

3、()*(4)?k k εδ-=_答:)4(-k ε_____________________________________________。

4、已知系统的零状态响应和输入之间的关系为:()(1)zs y k f k =-,其中,激励为()f ?,

零状态响应为()zs y ?,试判断此系统是否是时不变的? 。 解:设)()(0k k f k f -=,若系统为时不变的则有:

)1())(1()(000k k y k k y k k y zs zs zs +-=--=-

根据)1()(k f k y zs -=,则将题设代入,可得:)1()(0k k f k y zs --= 很明显,)()(0k k y k y zs zs -≠ 因而系统为时变的。

5、(2)t δ=___________________________。 答案:)(2

1

)2(t t δδ=

6、已知描述系统的微分方程为'()sin ()()y t ty t f t +=

其中()()f t y t 为激励,为响应,试判断此系统是否为时不变的? 解:系统是时变的。 7、

()?t e t dt δ+∞

--∞

=?

____________________________。

解:1 。

8、已知信号3

()sin cos 62

f k k k π

π=+,则,该信号的周期为? 解:设k k f 6

sin )(1π

=,其周期为121=T ;

设k k f 2

3sin

)(2π=,其周期为342=T ;

二者的最小公倍数为12,因而信号为周期信号,其周期为12=T .

9、线性是不变系统传输信号不失真的频域条件为:___________________________。 答:)()(0t t K t h -=δ

10、 设系统的激励为()f t ,系统的零状态响应)(t y zs 与激励之间的关系为:

)()(t f t y zs -=,判断该系统是否是线性的,并说明理由。

解:若系统为线性的,则应满足齐次性和可加性。 (1)齐次性。

设)()(1t af t f =,且)()(11t y t f zs =-若系统满足齐次性,必有:)()(1t ay t y zs zs =下面看结论是否成立。根据输入与输出之间的关系可得)()(t f t y zs -=,将题设代入可得到:

)()()()(11t ay t af t f t y zs zs =-=-=所以结论成立,从而系统满足齐次性。

(2)可加性。

设)()()(21t f t f t f +=, 其中,)()(11t y t f zs =-,)()(22t y t f zs =-,若系统满足可加性,则必有结论)()()(21t y t y t y zs zs +=。下面证明这一结论。

根据输入与输出之间的关系可得)()(t f t y zs -=,将题设代入可得到:

)()()()()]()([)()(212121t y t y t f t f t f t f t f t y zs zs zs +=-+-=-+-=-=

所以系统满足可加性。 综合(1)(2)可得,系统为线性的。

11、已知描述LTI 连续系统的框图如图所示,请写出描述系统的微分方程。

解:由于输入输入之间无直接联系,设中间变量)(t x 如图所示,则各积分器的的输入信号分别如图所示。由加法器的输入输出列些方程:

左边加法器:)(3)(2)()(t x t x t f t x '--='' (1) 右边加法器:)(2)()(t x t x t y '-''= (2) 由(1)式整理得到:)()(2)(3)(t f t x t x t x =+'+'' (3) 消去中间变量)(t x : )](2)([2)(2t x t x t y '-''= (4) )]'(2)([3)(3t x t x t y '-''=' (5)

])(2)([)('''-''=''t x t x t y (6)

将(4)(5)(6)左右两边同时相加可得:

)](2)([2])(2)([3])('2)([)(2)(3)(t x t x t x t x t x t x t y t y t y '-''+''-''+''-''=+'+''

整理可得到:

)(2)()(2)(3)(t f t f t y t y t y '-''=+'+''

12、已知一信号()f k 如图所示,请用单位阶跃序列()k ε及其移位序列表示()f k 。

答案: )4()1()(---=k k k f εε

1 ()f k

k

4 3 2 1 0

二、作图题:

1、已知信号()f k 的波形如图所示,画出信号(2)(2)f k k ε+?--的波形。

解:

2?

左移个单位

???

1

2

3

1

0 k

()k ε

1

???

-4 -3 -2 0 k

(2)k ε--

2?

右移个单位

1

???

2

3

4

0 k

(2)k ε-

?

翻转

再根据信号乘积,可以得到(2)(2)f k k ε+?--的波形:

2、已知

()()12f t f t 、的波形如下图,求()

()()12f t f t f t =*(可直接画出图形)

解:本题可以利用图解的方法,也可以利用卷积公式法来进行计算。

卷积公式法: 1()()(2)f t t t εε=--

2()()(1)f t t t εε=--

1212()()*()()()f t f t f t f f t d τττ+∞

-∞

==-?

12()()()[()(2)][()(1)]f t f f t d t t d τττετετετεττ+∞+∞-∞

-∞

=-=--?----?

?

()()()()(1)(2)()(2)(1)f t t d t d t d t d ετεττετεττ

ετεττετεττ

+∞

+∞-∞

-∞

+∞+∞

-∞

-∞

=-------+---????

利用阶跃函数的性质对上面的式子进行化简:

11

2

2

()()(1)(1)(2)(2)(3)(3)

t t t t f t d d d d t t t t t t t t ττττ

εεεε--=--+=------+--????

()[()(1)][(1)(2)](3)[(2)(3)]f t t t t t t t t t εεεεεε=--+--------

根据上面的表达式,可以画出图形:

3、已知信号()f t 的波形如图所示,画出信号1

(1)2

f t -

的波形。

解:

?

向左移动一个单位

?

翻转

4、已知信号)(t f 的波形如图所示,请画出函数)21(t f - 的波形。

解:

1

向左平移?

)

1(+t f ()

t f ?

横坐标展缩

翻转

?

/2

1压缩到原来的?

三、综合题目:(请写明步骤,否则不得分)

1、某LTI 系统的冲激响应()()2()h t t t δδ'=+,若激励信号为()f t 时,其零状态响应

()()t zs y t e t ε-=,求输入信号()f t 。

解:()()2()h t t t δδ'=+转换到s 域,可得:

2)(+=s s H

零状态响应为:()()t

zs y t e t ε-=,转换到s 域可得:

1

1

)(+=

s s Y zs ,则在s 域输入的象函数为: 2

111)2)(1(1211

)()()(+-+=++=++==s s s s s s s H s Y s F zs

取其拉氏反变换可得:

)()()(2t e e t f t t ε---=

2、某离散系统的输出()y k 与输入()f k 之间的关系为: 0

()2

()i

i y k f k i ∞

==

-∑

求系统的单位序列响应()h k 。 解:根据单位序列响应的概念可得:

∑∞

--=0

)(2)(i i i k k h δ

则:Λ+-+=)1(2)(2)(1

0k k k h δδ 观察规律可得:)(2)(k k h k

ε=

3、已知因果系统的差分方程为:()3(1)2(2)()y k y k y k f k +-+-=,其中,

()2()k f k k ε=。若已知(0)0,(1)2y y ==,求系统的全响应()y k 。

解:系统的齐次方程为:()3(1)2(2)0y k y k y k +-+-=

特征方程为:2

320λλ++= 所以特征根分别为:1212λλ=-=-, 所以系统的齐次解可以表示为:()1(1)2(2)k

k

yh k c c =-+-

已知系统的输入为()2()k

f k k ε=,则系统的特解可以表示为:()2k

yp k p =,将其代入到原差分方程,可得:13

p = 所以特解1()23

k

yp k =

所以系统的全解可表示为:

1

()()()1(1)2(2)23

k k k y k yh k yp k c c =+=-+-+

将初始条件(0)0,(1)2y y ==代入,可得待定系数:

2

1,213

c c ==-

所以系统的全响应为:21()(1)(2)(2),033

k k

k y k k =---+≥

4、图示离散系统有三个子系统组成,已知)4

cos(

2)(1π

k k h =, )()(2k a k h k ε=,激励)1()()(--=k a k k f δδ,求:零状态响应()zs y k 。

解:由题意可知,该系统为子系统的串联,则:

12()()*()h k h k h k =

()()*()zs y k f k h k =

所以12()()*[()*()]zs y k f k h k h k = 将已知条件代入有:

()[()(1)]*2cos(

)*()4

k zs k y k k a k a k π

δδε=-- 整理可得:

1()[()(1)]*()*2cos(

)4

[()(1)]*2cos()

4

k zs k k k y k k a k a k k a k aa k πδδεπ

εε-=--=--

[()(1)]*2cos(

)4

k k a k k π

εε=-- ()*2cos()4

k k a k πδ= ()*2cos(

)4

k k π

δ= 2cos(

)4

k π= 5、已知一个因果LTI 系统的输出()y t 与输入()f t 有下列微分方程来描述: ''()6'()8()2()y t f t y t f t ++= (1)确定系统的冲激响应()h t ;

(2)若2()()t

f t e t ε-=,求系统的零状态响应)(t y zs 。 解:(1)冲激响应()h t 满足方程

''()6'()8()2()h t h t h t t δ++=

及初始状态(0)(0)0h h --'== 对方程两边同时取拉氏变换:

2

()6()8()2s H s sH s H s ++= 整理得:2()211

()()6842

Y s H s F s s s s s -=

==+++++ 所以系统的冲激响应为: 24()()()t

t h t e e t ε--=-

(2)21

()()2

t

f t e

t s ε-=?

+ 零状态响应可以表示为:

222

()()()()

68

212

(2)(4)2(2)(4)

zs Y s H s F s F s s s s s s s s ==

++==

+++++

利用部分分式展开可得:

22

11

21

22()(2)(4)(2)24

zs Y s s s s s s ==-++++++ 取其逆变换可得: 所以 4211

()(())()2

2

t

t zs y t e t e t ε--=+-

6、已知某线性时不变系统对输入()f t 的零状态响应为:

()(1)t

t zs y t e f d τττ--∞

=-?,求该系统的单位冲激响应()h t 和频率响应()H j ω。

解: τττd f e t y t

t zs ?

---=

)1()(

自变量的范围:t <<∞-τ即:0>-τt 该范围可用阶跃函数表示:)(τε-t 原方程可变为: ττεττd t f e t y t zs )()1()(--=

?

+∞

--

利用单位冲激响应的定义可得:ττετδτd t e t h t )()1()(--=

?

+∞

--

根据冲激函数的取样性质可得:)1()()(11-=-=-=-t e t e t h t t ετεττ 因为:1

1

)(+?

-ωεj t e t

所以利用时移特性有:ωωεj t

e j t e --+?

-1

1

)1(1 即:ωωωj e j j H -+=1

1

)(

7、已知某线性时不变连续系统的阶跃响应为3()(1.50.5)()t

t g t e

e t ε--=-;当系统的激励为

()(2)()f t t t ε=+,系统的初始值为(0)3,(0)9,y y ++'==-求系统的完全响应。

解:由于系统的阶跃响应为3()(1.50.5)()t

t g t e

e t ε--=-,根据阶跃响应与冲激响应)(t h 的

关系 可得:

)

(5.0)(5.4)()()5.05.4()()5.05.1()()(333t e t e t t e e t e e t g t h t t t t t t εεδεδ------+-=+-+-='=

将其转化到s 域,可得:4

315.035.41)(22

++=+++-=s s s s s s H 则描述系统的方程为:)()(3)(4)(t f t y t y t y ''=+'+'' 并将已知输入转化到s 域: 212)(s

s s F +=

则,系统的零状态响应的象函数为:)

3)(1(1

)3)(1(2)(2+++++=s s s s s s s Y zs

整理可得:3

1

251121)(++

+-

=s s s Y zs 取拉式反变换可得:3()(0.5 2.5)()t t

zs y t e e t ε--=-+

333()(0.57.5)()(0.5 2.5)() (0.57.5)()2()

t t t t

zs

t t

y t e e t e e t e e t t εδεδ------'=-+-+=-+

从而:(0)2,(0)5zs zs y y '+=+=- 所以:

(0)(0)(0)(0)321,

(0)(0)(0)(0)9(5)4

zi zi zs zi zi zs y y y y y y y y +=-=+-+=-=''''+=-=+-+=---=

因为描述系统的微分方程为:()4()3()()y t y t y t f t '''''++=

所以(0)(0)4(0)8 3.5 2.5()(1)(3)(1)(3)13

zi zi zi zi sy y y s Y s s s s s s s '-+-+-+-=

==+++++++

所以3()(3.5 2.5)()t t

zi y t e e t ε--=-

所以系统的全响应为:

()()()3()t zi zs y t y t y t e t ε-=+=

8、已知某LTI 连续系统的系统函数()2

31

22++++=s s s s s H ,求:

(1)系统的冲激响应()t h ;

(2)当激励)()(t t f ε=,初始状态()'

(0) 1 , 01y y --==时系统的零输入响应() zi y t 和

零状态响应()zs y t 。

解(1)因为)()(s H t h ?而2()()(3)()t

t

h t t e e

t δε--=+-

两边同时取拉普拉斯变换,可得:

)

2)(1()1(3)2()2)(1(23111)(+++-++++=+-++

=s s s s s s s s s H 整理可得:2

31

)2)(1()1(3)2()2)(1()(22++++=+++-++++=s s s s s s s s s s s H

(2)根据系统函数的定义:)()

()(s F s Y s H =而231)(22++++=s s s s s H

所以:231

)()(22++++=s s s s s F s Y ? )()1()()23(22s F s s s Y s s ++=++

两边同时取拉普拉斯逆变换,可得描述系统的微分方程为:

)()()()(2)(3)(t f t f t f t y t y t y +'+''=+'+''

而零输入响应)(t y zi 满足如下方程

0)(2)(3)('''=++t y t y t y zi zi zi

和初始状态:)0()0(--=y y zi )0()0('

--'=y y zi

对方程两边同时取拉普拉斯变换,可得:

0)(2_))0()((3_)0(_)0()(('

2=+-+--s Y y s sY y sy s Y s zi zi zi zi zi zi

整理可得:

2

3)

0(3)0()0()(2

'

++++=---s s y y sy s Y zi zi zi zi 将初始状态代入可得:1

3

22234)(2

+++-=+++=

s s s s s s Y zi 取拉普拉斯逆变换,可得系统的零输入响应为:)()32()(2t e e

t y t t

zi ε--+-=

)(*)()(t f t h t y zs =,所以:

)

2)(3(3)1)(3(13131)23111()()()(++-++++=++-++

==s s s s s s s s s F s H s Y zs

整理可得:

2

31213127332312132131)(+-

+++=+++-+++-++=s s s s s s s s s Y zs 取拉普拉斯逆变换可得系统的零状态响应为:)()2

7

32

1()(32t e e e t y t t t

zs ε---+-=

完整word版,2018年12月四级真题第一套

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完整word版2018年6月大学生英语四级真题试卷及答案第三套

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(完整版)2018年6月大学英语四级真题试卷一及答案

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