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(2
1t u t t u t --+=ππ )}()(){4cos(2
1t u t u t -+=π )4cos(2
1t 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=ππm m 16*8/12=, it ’s not rational number.
(e). Periodic. N=16 Solution as follow:
)62cos(2)8sin()4cos(
2][ππππ+-+=n n n n x in this equation,
)4cos(2n π
, it ’s period is N=2π*m/(π/4)=8, m=1.
)8sin(n π
, it ’s period is N=2π*m/(π/8)=16, m=1.
)62cos(
2ππ+-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(0222n 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, (140)
0N 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, (1145)
5N N k N
k =0, then 5<N ∴4=N
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=
01/421/4sin(
)12()jk t jk t
k T k a x t e dt e dt T
k ωππ
π
---=
==⎰⎰
2
8sin(
)2()[16(2)]
k k k b a H jk k k ππππ==+ 3.35 Solution: T= /7π,
02/14T ωπ==.
kt jw k k e a t x 0)(∑
∞
-∞==
∴t jkw k k e jkw H a t y 0)()(0∑
∞
-∞
==
∴0()k k b a H jkw =
for
⎩⎨⎧≥=otherwise w jw H ,.......
0250,.......1)(,01,. (17)
()0,.......k H jkw otherwise ⎧≥⎪=⎨⎪⎩
that is 0250
250, (14)
k k ω<<
, and k is integer, so 18....17k or k <≤. Let ()()y t x t =,k k b a =, it needs 0=k a ,for 18....17k or k <≤.
3.37 Solution:
1
1()[]()2
1
2()213
12
411511cos 2
24
n
j j n
j n n n n j n
n j n
n n j j j H e h n e
e e
e e e e ω
ωωωωωω
ωω∞
∞
--=-∞
=-∞
-∞
--=-∞
=-==
=
+=+=---∑
∑
∑
∑
A periodic sequence has Fourier Series:2(
)[]jk n N
k k N x n a e
π
=<>
=∑
.
N is the fundamental period of []x n .
The output of LTI system with inputed []x n is 22(
)[]()j
k jk n N
N
k k N y n a H e
e
ππ=<>
=
∑
.
Its coefficients of Fourier Series: 2()j
k N
k k b a H e
π=
(a)[][4]k x n n k δ∞
=-∞
=
-∑
.N=4, 1
4
k a =
.So 23
1
4
()524cos()44
j k N
k k b a H e k ππ==
-
316
5cos()42
k b k π=
-
3.40 Solution: According to the property of fourier series: (a). )2cos(2)cos(2000000
0t T
k
a t kw a e a e
a a k k t jkw k t jkw k k π==+='
- (b). Because 2
)
()()}({t x t x t x E v -+=
}{2
k v k k k a E a a a =+='-
(c). Because 2
)
(*)()}({t x t x t x R e +=
2
*
k
k k a a a -+='
(d). k k k a T
jk
a jkw a 2
2
0)2()(π=='
(e). first, the period of )13(-t x is 3
T T =
'
then 3
)(1)13(13
1
21
31
20dm
e m x T dt e t x T a m T jk T t T jk T k +'--'-'
-'⎰⎰'=-'='
ππ
T
jk
k m T jk T T jk T jk m T jk T e
a dm e m x T e dm e e m x T ππππ
π22112221
1
)(1)(1---------=⎥⎦
⎤⎢⎣⎡==⎰⎰
3.43 (a) Proof:
（i ）Because ()x t is odd harmonic ,(2/)()jk T t k k x t a e π∞
=-∞
=
∑
,where 0k a = for every
non-zero even k.
(2/)()2
(2/)(2/)()2T jk T t k k jk jk T t
k k jk T t
k k T x t a e
a e e a e ππππ∞+=-∞
∞
=-∞
∞
=-∞
+==
=-∑∑
∑
It is noticed that k is odd integers or k=0.That means
()()2
T
x t x t =-+
（ii ）Because of ()()2
T
x t x t =-+,we get the coefficients of Fourier Series
222/200/2
22(/2)/2/2
00
22/2/200111
()()()11()(/2)11()()(1)jk t jk t jk t T T T T T T k T jk t jk t T T T T T
jk t jk t T T k T
T a x t e dt x t e dt x t e dt
T T T x t e dt x t T e dt T T x t e dt x t e dt T T πππππππ-----+--==+=++=--⎰⎰⎰⎰⎰⎰⎰ 2/2
1[1(1)]()jk t T k
T x t e dt T π
-=--⎰
It is obvious that 0k a = for every non-zero even k. So ()x t is odd harmonic ,
(b)
Extra problems:
∑∞
-∞
=-=
k kT t t x )()(δ, π=T
(1). Consider )(t y , when )(jw H is
t
(2). Consider )(t y , when )(jw H is
Solution:
∑∞
-∞
=-=
k kT t t x )()(δ↔
π
11=T , 220==
T
w π
(1).
kt j k k t
jkw k k e k j H a e
jkw H a t y 20)2(1
)()(0∑
∑
∞
-∞
=∞
-∞
==
=
π
π
2
=
(for k can only has value 0)
(2).
kt j k k t
jkw k k e k j H a e jkw H a t y 20)2(1
)()(0∑
∑
∞
-∞
=∞
-∞
==
=
π
π
π
t
e e t j t j 2cos 2)(1
22=
+=
- (for k can only has value –1 and 1)。