信号与系统奥本海姆习题答案

Chapter 1 Answers

1.6 (a).No

Because when t<0, )(1t x =0.

(b).No

Because only if n=0, ][2n x has valuable.

(c).Yes

Because ∑∞

-∞

=--+--+=

+k k m n k m n m n x ]}414[]44[{]4[δδ

∑∞

-∞=------=

k m k n m k n )]}(41[)](4[{δδ ∑∞

-∞

=----=

k k n k n ]}41[]4[{δδ

N=4. 1.9 (a). T=π/5

Because 0w =10, T=2π/10=π/5.

(b). Not periodic.

Because jt t e e t x --=)(2, while t

e -is not periodic, )(2t x is not periodic.

(c). N=2

Because 0w =7π, N=(2π/0w )*m, and m=7.

(d). N=10

Because n j j e e

n x )5/3(10

/343)(ππ=, that is 0w =3π/5, N=(2π/0w )*m, and m=3.

(e). Not periodic.

Because 0w =3/5, N=(2π/0w )*m=10πm/3 , it ’s not a rational number.

1.14 A1=3, t1=0, A2=-3, t2=1 or -1

dt

t dx )

( is Solution: x(t) is

Because ∑∞

-∞

=-=k k t t g )2()(δ,

dt t dx )(=3g(t)-3g(t-1) or dt

t dx )

(=3g(t)-3g(t+1) 1.15. (a). y[n]=2x[n-2]+5x[n-3]+2x[n-4]

Solution:

]3[2

1

]2[][222-+

-=n x n x n y ]3[2

1

]2[11-+

-=n y n y ]}4[4]3[2{2

1

]}3[4]2[2{1111-+-+-+-=n x n x n x n x

]4[2]3[5]2[2111-+-+-=n x n x n x

Then, ]4[2]3[5]2[2][-+-+-=n x n x n x n y

(b).No. For it ’s linearity.

the relationship between ][1n y and ][2n x is the same in-out relationship with (a).

you can have a try.

1.16. (a). No. For example, when n=0, y[0]=x[0]x[-2]. So the system is memory. (b). y[n]=0. When the input is ][n A δ,

then, ]2[][][2

-=n n A

n y δδ, so y[n]=0.

(c). No.

For example, when x[n]=0, y[n]=0; when x[n]=][n A δ, y[n]=0.

So the system is not invertible. 1.17. (a). No.

For example, )0()(x y =-π. So it ’s not causal. (b). Yes.

Because : ))(sin()(11t x t y = ,

))(sin()(22t x t y =

))(sin())(sin()()(2121t bx t ax t by t ay +=+

1.21. Solution: We have known:

(a).

(b).

(c).

(d).

1.2

2. Solution:

We have known:

(a).

(b).

(e).

(g)

1.23. Solution:

For

)]()([2

1

)}({t x t x t x E v -+=

)]()([2

1

)}({t x t x t x O d --=

then, (a).

(b).

(c). 1.24.

For:

])[][(2

1

]}[{n x n x n x E v -+=

])[][(2

1

]}[{n x n x n x O d --=

then, (a).

(b).

1.25. (a). Periodic. T=π/

2. Solution: T=2π/4=π/2. (b). Periodic. T=2. Solution: T=2π/π=2. (d). Periodic. T=0.5.

Solution: )}()4{cos()(t u t E t x v π=

)}())(4cos()()4{cos(21

t u t t u t --+=ππ )}()(){4cos(21

t u t u t -+=π )4cos(2

1

t π= So, T=2π/4π=0.5 1.26. (a). Periodic. N=7

Solution: N=

m *7

/62ππ

=7, m=3. (b). Aperriodic.

Solution: N=

ππ

m m 16*8

/12=, it ’s not rational number. (e). Periodic. N=16

Solution as follow:

)6

2

cos(

2)8

sin(

)4

cos(

2][π

π

π

π

+

-+=n n n n x

in this equation,

)4

cos(

2n π

, it ’s period is N=2π*m/(π/4)=8, m=1.

)8

sin(

n π

, it ’s period is N=2π*m/(π/8)=16, m=1.

)6

2

cos(

2π

π

+

-n , it ’s period is N=2π*m/(π/2)=4, m=1.

So, the fundamental period of ][n x is N=(8,16,4)=16.

1.31. Solution

Because )()1()(),2()()(113112t x t x t x t x t x t x ++=--=. According to LTI property ,

)()1()(),2()()(113112t y t y t y t y t y t y ++=--=

Extra problems:

Sketch ?

∞

-=

t

dt t x t y )()(.

1. Suppose Solution:

2. Suppose Sketch:

(1). )]1(2)1()3()[(--+++t t t t g δδδ

(2). ∑∞

-∞

=-k k t t g )2()

(δ

(2).

Chapter 2

2.1 Solution:

Because x[n]=(1 2 0 –1)0, h[n]=(2 0 2)1-, then (a).

So, ]4[2]2[2]1[2][4]1[2][1---+-+++=n n n n n n y δδδδδ

(b). according to the property of convolutioin:

]2[][12+=n y n y

(c). ]2[][13+=n y n y

][*][][n h n x n y =

][][k n h k x k -=

∑∞

-∞=

∑∞

-∞=-+--=

k k k n u k u ]2[]2[)2

1(2 ][2

11)21

()21(][)21(1

2)2(02

22n u n u n n k k --==+-++=-∑

][])2

1

(1[21n u n +-=

the figure of the y[n] is:

2.5 Solution:

We have known: ??

?≤≤=elsewhere n n x ....090....1][，，, ???≤≤=elsewhere N n n h ....00....1][，

，

,(9≤N )

Then, ]10[][][--=n u n u n x , ]1[][][---=N n u n u n h

∑∞

-∞

=-=

=k k n u k h n h n x n y ][][][*][][

∑∞

-∞

=-------=

k k n u k n u N k u k u ])10[][])(1[][(

So, y[4] ∑∞

-∞

=-------=

k k u k u N k u k u ])6[]4[])(1[][(

???????≥≤=∑∑==4,...14, (14)

N N k N

k =5, then 4≥N

And y[14] ∑∞

-∞

=------=

k k u k u N k u k u ])

4[]14[])(1[][(

???????≥≤=∑∑==14,...114, (114)

5

5

N N k N

k =0, then 5

2.7 Solution:

[][][2]k y n x k g n k ∞

=-∞

=

-∑

（a ）[][1]x n n δ=-,[][][2][1][2][2]k k y n x k g n k k g n k g n δ∞

∞

=-∞

=-∞

=

-=--=-∑∑

(b) [][2]x n n δ=-,[][][2][2][2][4]k k y n x k g n k k g n k g n δ∞∞

=-∞

=-∞=-=--=-∑∑

(c) S is not LTI system.. (d) [][]x n u n =,0

[][][2][][2][2]k k k y n x k g n k u k g n k g n k ∞

∞

∞

=-∞

=-∞

==-=-=-∑∑∑

2.8 Solution:

)]1(2)2([*)()(*)()(+++==t t t x t h t x t y δδ )1(2)2(+++=t x t x

Then,

That is, ?????

????≤<-≤<-+-=-<<-+=others t t t t t t t t y ,........

010,....2201,.....41..,.........412,.....3)(

2.10 Solution:

(a). We know:

Then,

)()()(αδδ--='t t t h

)]()([*)()(*)()(αδδ--='='t t t x t h t x t y )()(α--=t x t x

that is,

So, ????

???+≤≤-+≤≤≤≤=others t t t t t t y ,.....

011,.....11,....0,.....)(ααααα

(b). From the figure of )(t y ', only if 1=α, )(t y ' would contain merely there

discontinuities.

2.11 Solution:

(a). )(*)]5()3([)(*)()(3t u e

t u t u t h t x t y t

----==

??∞

∞

---∞

∞

--------=ττττττττd t u e u d t u e

u t t )()5()()3()(3)

(3

??-------=t

t t t d e t u d e

t u 5

)(33

)

(3)5()3(ττττ

?????????

??≥+-=-<≤-=<=---------???5

,.......353,.....313.........,.........0315395)

(33

)(3393

)(3t e e d e d e t e d e t t

t t t t t t t t ττττττ

(b). )(*)]5()3([)(*)/)(()(3t u e t t t h dt t dx t g t ----==δδ

)5()3()5(3)3(3---=----t u e t u e t t

(c). It ’s obvious that dt t dy t g /)()(=.

2.12 Solution

∑∑∞

-∞

=-∞

-∞

=--=

-=k t

k t

k t t u e

k t t u e t y )]3(*)([)3(*)()(δδ

∑∞

-∞

=---=

k k t k t u e

)3()

3(

Considering for 30<≤t ,we can obtain

3

3311

])3([)(---∞

=-∞

-∞

=--==-=∑∑e

e e e

k t u e e t y t

k k t

k k

t

. (Because k must be negetive ,1)3(=-k t u for 30<≤t ).

2.19 Solution:

(a). We have known:

][]1[2

1

][n x n w n w +-=

(1) ][]1[][n w n y n y βα+-=

(2)

from (1), 21)(1-

=

E E

E H

from (2), α

β-=E E

E H )(2

then, 2

12

212

)21(1)

2

1

)(()()()(--++-=

--=

=E E E E E E H E H E H α

αβ

αβ

∴][]2[2

]1[)21(][n x n y n y n y βα

α=-+-+-

but, ][]1[4

3

]2[81][n x n y n y n y +-+--=

∴??

?

??=??? ??=+=143)21

(:....812βααor ∴?????==1

41βα

(b). from (a), we know )

2

1)(41()()()(2

21--==E E E E H E H E H

2

1241-

+

--=

E E

E E ∴][)41()2

1

(2][n u n h n n ??????-=

2.20 (a). 1

??∞

∞

-∞

∞

-===1)0cos()cos()()cos()(0dt t t dt t t u δ

(b). 0

dt t t )3()2sin(5

+?δπ has value only on 3-=t , but ]5,0[3?-

∴dt t t )3()2sin(5

+?δπ=0

(c). 0

??

---=-6

4

15

5

1)2cos()()2cos()1(dt t t u d u πτπττ

?-'-=6

4

)2cos()(dt t t πδ

0|)2(s co ='=t t π 0|)2sin(20=-==t t ππ

∑∞

-∞

=-=

=k t h kT t t h t x t y )(*)()(*)()(δ

∑∞

-∞

=-=

k kT t h )(

∴

2.27Solution

()y A y t dt ∞

-∞

=

?,()x

A x t dt ∞-∞

=?,()h

A h t dt ∞

-∞

=?.

()()*()()()y t x t h t x x t d τττ∞

-∞

==

-?

()()()()()()()()()(){()}

y x h

A y t dt x x t d dt

x x t dtd x x t dtd x x d d x d x d A A ττττττττττξξτττξξ∞

∞

∞-∞-∞

-∞

∞

∞

∞

∞

-∞-∞

-∞

-∞∞

∞

∞

∞

-∞

-∞

-∞

-∞

=

=-=-=-=

==???????????

(a) ()()(2)t

t y t e x d τττ---∞

=

-?

,Let ()()x t t δ=,then ()()y t h t =. So , 2

()

(2)(2)()(2)()(2)t t t t t h t e

d e d e u t τξδττδξξ---------∞

-∞

=

-=

=-??

(b) (2)()()*()[(1)(2)]*(2)t y t x t h t u t u t e u t --==+---

(2)

(2)(1)(2)(2)(2)t t u e

u t d u e u t d ττττττττ∞

∞

-------∞

-∞=

+------??

2

2

(2)(2)1

2

(1)(4)t t t t u t e d u t e d ττττ---------=---??

(2)2(2)2

12(1)[]|(4)[]|t t t t u t e e u t e

e ττ-------=--- (1)(4)[1](1)[1](4)t t e u t e u t ----=-----

2.46 Solution

Because

)]1([2)1(]2[)(33-+-=--t u dt

d

e t u e dt d t x dt d t t )1(2)(3)1(2)(333-+-=-+-=--t e t x t e t x t δδ.

From LTI property ,we know

)1(2)(3)(3-+-→-t h e t y t x dt

d

where )(t h is the impulse response of the system. So ,following equation can be derived.

)()1(223t u e t h e t --=-

Finally, )1(2

1)()1(23+=

+-t u e e t h t 2.47 Soliution

According to the property of the linear time-invariant system: (a). )(2)(*)(2)(*)()(000t y t h t x t h t x t y ===

(b). )(*)]2()([)(*)()(00t h t x t x t h t x t y --==

)(*)2()(*)(0000t h t x t h t x --=

01

2y(t)

t

4

)2()(00--=t y t y

(c). )1()1(*)(*)2()1(*)2()(*)()(00000-=+-=+-==t y t t h t x t h t x t h t x t y δ

(d). The condition is not enough.

(e). )(*)()(*)()(00t h t x t h t x t y --==

τττd t h x )()(00+--=?

∞

∞

-

)()()(000t y dm m t h m x -=--=?

∞

∞

-

(f). )()]([)](*)([)(*)()(*)()(000000t y t y t h t x t h t x t h t x t y "

='

'

='

--'

=-'

-'

==

Extra problems:

1. Solute h(t), h[n]

(1). )()(6)(5)(2

2t x t y t y dt d

t y dt

d =++ (2). ]1[][2]1[2]2[+=++++n x n y n y n y Solution:

(1). Because 3

1

21)3)(2(1651)(2+-++=++=++=

P P P P P P P H

so )()()()3

1

21(

)(32t u e e t P P t h t t ---=+-++=δ (2). Because )1)(1(1

)1(22)(2

2i E i E E

E E E E E E H -+++=++=++=

i

E E

i

i E E i -+-+

++=1212 so []

][)1()1(2][1212][n u i i i k i E E i i E E i n h n n +----=??

???

?

??-+-+

++=δ

Chapter 3

3.1 Solution:

Fundamental period 8T =.02/8/4ωππ==

00000000033113333()224434cos()8sin()

44

j kt j t j t j t j t

k k j t j t j t j t

x t a e a e a e a e a e e e je je t t ωωωωωωωωωππ∞

----=-∞

--=

=+++=++-=-∑

3.2 Solution:

for, 10=a , 4

/2πj e

a --= , 4

/2πj e

a = , 3

/42πj e

a --=, 3

/42πj e

a =

n N jk k N k e a n x )/2(][π∑

>

=<=

n j n j n j n j e a e a e a e a a )5/8(4)5/8(4)5/4(2)5/4(20ππππ----++++=

n j j n j j n j j n j j e e e e e e e e )5/8(3/)5/8(3/)5/4(4/)5/4(4/221ππππππππ----++++= )358cos(4)454cos(21π

πππ++++=n n

)6

558sin(4)4354sin(21π

πππ++++=n n

3.3 Solution: for the period of )3

2cos(

t π

is 3=T , the period of )3

5sin(

t π

is 6=T

so the period of )(t x is 6 , i.e. 3/6/20ππ==w

)3

5sin(4)32cos(

2)(t t t x ππ++= )5sin(4)2cos(21

200t w t w ++=

)(2)(2

1

200005522t w j t w j t w j t w j e e j e e ----++=

then, 20=a , 2

1

22==-a a , j a 25=-, j a 25-=

3.5 Solution:

(1). Because )1()1()(112-+-=t x t x t x , then )(2t x has the same period as )(1t x ,

that is 21T T T ==, 12w w =

(2). 212111()((1)(1))jkw t jkw t

k T T b x t e dt x t x t e dt T

--=

=-+-??

111111(1)(1)jkw t

jkw t T T

x t e dt x t e dt T T --=

-+-?? 111)(jkw k k jkw k jkw k e a a e a e a -----+=+=

3.8 Solution:

kt jw k k e a t x 0)(∑

∞

-∞

==

while:

)(t x is real and odd, then 00=a , k k a a --=

2=T , then ππ==2/20w

and

0=k a for 1>k

so kt jw k k e a t x 0)(∑

∞

-∞

==

t jw t jw e a e a a 00110++=--

)sin(2)(11t a e e a t j t j πππ=-=-

for

12)(212

1

2

12

1

20220

==++=-?a a a a dt t x

∴2/21±=a ∴)sin(2)(t t x π±=

3.13 Solution:

Fundamental period 8T =.02/8/4ωππ==

kt jw k k e a t x 0)(∑

∞

-∞

==

∴t jkw k k e jkw H a t y 0)()(0∑

∞

-∞

==

0004, 0

sin(4)()0, 0

k k H jk k k ωωω=?=

=?

≠? ∴000()()4jkw t k k y t a H jkw e a ∞

=-∞

=

=∑

Because 48

004

111()1(1)088T a x t dt dt dt T ==+-=???

So ()0y t =.

kt jw k k e a t x 0)(∑

∞

-∞==

∴t jkw k k e jkw H a t y 0)()(0∑∞

-∞

== ∴dt e jkw H t y T

a t jkw T

k 0)()(10-?

=

for

????

?>≤=100

, (0100)

,.......1)(w w jw H ∴

if 0=k a , it needs 1000>kw

that is 12

100

,........1006/2>

>k k

ππ

and k is integer, so 8>K

3.22 Solution:

021)(11

1

0===

??

-tdt dt t x T

a T

dt te dt te dt e t x T a t jk t jk t jkw T k ππ

-----???===1

122112121)(10

t jk tde jk ππ

--?

-

=1

1

21

???

?

?

??

?---

=----1

11

121π

πππjk e te jk t jk t

jk ??????---+-

=--πππππ

πjk e e e e jk jk jk jk jk )()(21

??

?

???-+-

=ππππ

jk k k jk )sin(2)cos(221

[]π

ππππk j

k k j k jk k

)1()cos()cos(221-==-=0............≠k

40

440

2

()()118

4416t

j t

j t t j t

t j t H j h t e

dt e

e dt

e e

dt e e dt

j j ωωωωωωωω∞

∞

----∞

-∞

∞

----∞

=

=

=+=

+=

-++?

?

?

?

A periodic continous-signal has Fourier Series:. 0()j kt k k x t a e ω∞

=-∞

=∑

T is the fundamental period of ()x t .02/T ωπ=

The output of LTI system with inputed ()x t is 00()()jk t k k y t a H jk e ωω∞

=-∞

=∑

Its coefficients of Fourier Series: 0()k k b a H jk ω= (a)()()n x t t n δ∞

=-∞

=

-∑

.T=1, 02ωπ=1

1k a T

=

=. 01/221/21()()1jkw t jk t

k T a x t e dt t e dt T

πδ---=

==?? （Note ：If ()()n x t t nT δ∞

=-∞

=

-∑

,1

k a T

=

） So 22

82

(2)16(2)4()

k k b a H jk k k πππ==

=++ (b)()(1)()n n x t t n δ∞

=-∞

=

--∑ .T=2, 0ωπ=,1

1k a T

=

= 01/23/21/21/2

111()()(1)(1)221

[1(1)]2

jkw t jk t

jk t k T k a x t e dt t e dt t e dt

T ππδδ----==+--=--???

So 2

4[1(1)]

()16()

k k k b a H jk k ππ--==+, (c) T=1,

02ωπ=