Chap 3
3.1 A continuous-time periodic signal x(t) is real value and has a fundamental period T=8. The nonzero Fourier series coefficients for x(t) are
j a a a a 4,2*
3311====--.
Express x(t) in the form
)cos()(0
k k k k t A t x φω+=∑∞
=
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∞
----=-∞
--=
=+++=++-=-∑
A discrete-time periodic signal x[n] is real valued and has a
fundamental period N=5.The nonzero Fourier series coefficients for x[n] are
10=a ,4/2πj e a --=,4/2πj e a =,3
/*4
42πj e a a ==- Express x[n] in the form
)sin(][1
0k k k k n A A n x φω++=∑∞
=
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
For the continuous-time periodic signal
)3
5sin(4)32cos(
2)(t t t x ππ++= Determine the fundamental frequency 0ω and the Fourier series
coefficients k a such that
t
jk k k
e
a t x 0)(ω∑∞
-∞
==
.
Solution:
for the period of )32cos(
t πis 3=T , the period of )3
5sin(t πis 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 π
π++=
)5sin(4)2cos(21
200t t ωω++=
000022551
2()2()2
j t j t j t j t e e j e e ωωωω--=++-- then, 20=a , 2
1
22==-a a , j a 25=-, j a 25-=
3.5 Let 1()x t be a continuous-time periodic signal with fundamental frequency
1ω and Fourier coefficients k a . Given that
211()(1)(1)x t x t x t =-+-
How is the fundamental frequency
2ω of 2()x t related to? Also,
find a relationship between the Fourier series coefficients k b of
2()x t and the coefficients k a You may use the properties listed in
Table 3.1. 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 -----+=+=
Suppose given the following information about a signal x(t): 1. x(t) is real and odd.
2. x(t) is periodic with period T=2 and has Fourier coefficients k a .
3. 0=k a for 1||>k .
4 1|)(|2120
2
=⎰dt t x .
Specify two different signals that satisfy these conditions. Solution:
0()j kt k k x t a e ω∞
=-∞
=
∑
while: )(t x is real and odd, then k a is purely imaginary and odd , 00=a , k k a a --=,.
2=T , then 02/2ωππ==
and
0=k a for 1>k
so
0()j kt k k x t a e ω∞
=-∞
=
∑
00011j t j t a a e a e ωω--=++
)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
∴ j a 2/21±=
∴
)sin(2)(t t x π±=
3 Consider a continuous-time LTI system whose frequency response is
⎰∞
∞
--=
=ω
ωωω)
4sin()()(dt e t h j H t j
If the input to this system is a periodic signal
⎩⎨
⎧<≤-<≤=8
4,14
0,1)(t t t x With period T=8,determine the corresponding system output y(t). Solution:
Fundamental period 8T =.02/8/4ωππ==
0()j kt k k x t a e ω∞
=-∞=
∑
∴ 00()()jk t k k y t a H jk e ωω∞
=-∞
=
∑
0004, 0
sin(4)()0, 0
k k H jk k k ωωω=⎧=
=⎨
≠⎩ ∴ 000()()4jkw t k k y t a H jk e a ω∞
=-∞
=
=∑
Because 48
004
111()1(1)088T a x t dt dt dt T ==+-=⎰⎰⎰
另:x(t)为实奇信号,则a k 为纯虚奇函数,也可以得到a 0为0。 So ()0y t =.
Consider a continuous-time ideal lowpass filter S whose frequency
response is
1, (100)
()0,.......100H j ωωω⎧≤⎪=⎨>⎪⎩
When the input to this filter is a signal x(t) with fundamental period
6/π=T and Fourier series coefficients k a , it is found that
)()()(t x t y t x S
=→.
For what values of k is it guaranteed that 0=k a ? Solution:
for
0()j kt k k x t a e ω∞
=-∞=
∑
∴ 00()()jk t k k y t a H jk e ωω∞
=-∞
=
∑
即对于所有的k ,1)(0=ωjk H
for
1, (100)
()0, (100)
H j ωωω⎧≤⎪=⎨
>⎪⎩ 也就是说1000<ωk , 126/0=⇒=ωπT 即12k<100,k<=8,故当k>8时,a k =0。
.Consider a continuous-time LTI system S whose frequency response
is
1,||250
()0,H j otherwise
ωω≥⎧=⎨
⎩
When the input to this system is a signal x(t) with fundamental period
/7T π= and Fourier series coefficients k a ,it is found that the
output y(t) is identical to x(t).
For what values of k is it guaranteed that 0k a =? 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
=k a ,for
18....17k or k <≤.
Chap 4
4.1 Use the Fourier transform analysis equation(4.9)to calculate the Fourier transforms of;
(a))1()1(2---t u e t (b)|
1|2--t e
Sketch and label the magnitude of each Fourier transform. Solution: (a). ()()j t X j x t e dt ωω∞
--∞
=⎰
2(1)(1)t j t e u t e dt ω∞
----∞=-⎰
2(1)(2)21
1
t j t j t e e dt e e dt ωω∞
∞-----=
=⎰
⎰
(2)(2)
221222j t j j e e e e e j j j ωωωωωω
∞
------===----+
(b).
()()j t X j x t e dt ωω∞
--∞
=⎰
2
1)2(21)2(2
1
221
22124422|21|21ωωωω
ωω
ωωωωωωω+=
-++=-++-=+==---∞
---∞
+--∞
-+--∞
+--∞
∞
---⎰⎰⎰j j j t j t j t j t t
j t t
j t e j e j e e j e e j e dt
e e dt e
e
dt e
e
4.2 Use the Fourier transform analysis equation(4.9) to calculate the Fourier transforms of: (a))1()1(-++t t δδ (b)
)}2()2({-+--t u t u dt
d
Sketch and label the magnitude of each Fourier transform. Solution: (a). ()()j t X j x t e dt ωω∞
--∞
=⎰
-j t - [(t 1)(t-1)]e dt ωδδ∞
∞=++⎰
-j t
-j t -
(t 1)e (t-1)e dt dt ωωδδ∞
∞
∞
-∞
=++⎰
⎰
2cos j j e e
ωω
ω-=+= (b). ()()j t X j x t e dt ωω∞
--∞=⎰
{(2)(2)}j t d
u t u t e dt dt
ω∞
--∞
=
--+-⎰
{(2)(2)}j t t t e dt ωδδ∞
--∞=---+-⎰ (2)(2)j t
j t t e
dt t e dt ωωδδ∞
∞
---∞
-∞
=
---+-⎰
⎰
222sin 2j j e e j ωωω-=-+=-
4.5 Use the Fourier transform synthesis equation(4.8) to determine the inverse Fourier transform of )()()(ωωωj X j e j X j X ∠=,where
|()|2{(3)(3)}X j u u ωωω=+-- πωω+-=∠2
3
)(j X
Use your answer to determine the values of t for which x(t)=0. Solution:
dw e e jw X dw e jw X t x jwt jw X j jwt )()(21)(21)(<∞
∞
-∞
∞
-⎰
⎰
=
=
π
π
dw e e
w u w u jwt w j )2
3
()}3()3({221ππ+-∞∞
---+=⎰
dw e
e dw e
e w
t j j jwt
w j )2
3(3
3
)2
3
(33
1
--+--⎰
⎰
=
=
π
ππ
π
)
2
3(1)
2
3(1)
2
3
(3)
2
3(33
3
)2
3(--⋅
-=
-⋅
-=
-----t j e e t j e t j t j w
t j π
π
)2
3()23
(3sin 2)23()23(3sin 21---=--⋅
-=t t j t j t j ππ If x(t)=0 then ⎪⎪⎩
⎪⎪⎨⎧≠-±±==-0)23(,...2,1,0,.........)2
3(3t k k t π
That is 0., (2)
3
3≠+=
k k t π Given that x(t) has the Fourier transform ()X j ω, express the Fourier transforms of the signals listed below in the terms of
()X j ω.You may find useful the Fourier transform properties listed
in Table4.1.
(a))1()1()(1t x t x t x --+-= (b))63()(2-=t x t x
(c) )1()(22
3-=t x dt
d t x
Solution: Accorrding to the properties of the Fourier transform, we ’ll get: (a). )(t x −→
←FT
()X j ω ∴
)1(t x - −→←FT
()j X j e ωω--
)1(t x -- −→←FT ()j X j e ωω-
then
)1()1()(1t x t x t x --+-= −→
←FT
1()()()2()cos j j X j X j e X j e X j ωωωωωωω--=-+-=-
(b). )(t x −→←FT
()X j ω
)(b at x + −→←FT
1()b j a X j e a a
ωω
∴ )63()(2-=t x t x −→←FT
221()()33
j X j X j e ωω
ω-= (c).
)(t x −→←FT
()X j ω
)1(-t x −→←FT ()j X j e ω
ω-
)(22t x dt
d −→←FT 2
()()j X j ωω ∴)1()(223-=t x dt
d t x −→
←FT
23()()j X j X j e ωωωω-=- Given the relationships
)()()(t h t x t y *=,And )3()3()(t h t x t g *=
And given that x(t) has Fourier transform )(ωj X and h(t) has
Fourier transform )(ωj H ,use Fourier transform properties to show that g(t) has the form
)()(Bt Ay t g =
Determine the values of A and B. Solution: for )()()(t h t x t y *=
−→←FT
()()()Y j X j H j ωωω=
and
)3()3()(t h t x t g *=
)3(t x −→←FT
1()33X j ω )3(t h −→←FT
1()33
H j ω
then
11()()()3333G j X j H j ωωω=
⋅1()93
Y j ω
=
−→←FT
)3(3
1)(t y t g =
∴
3,3
1
==B A
Consider a signal x(t) with Fourier transform )(ωj X .Suppose we are given the following facts: 1. x(t) is real and nonnegative.
2. ),()}()1{(21t u Ae j X j F t --=+ωωwhere A is independent of t.
3.
⎰
∞
∞
-=πωω2|)(|d j X .
Determine a closed-form expression for x(t). Solution: From (1), we know )(t x is real and 0)(≥t x ;
From (2), we know : )(2t u Ae t
- −→←FT
(1)()j X j ωω+
And we also know )(2t u Ae t
- −→
←FT
2
A
j ω+
So (1)()j X j ωω+=
2
A
j ω+
That is ()(1)(2)12
A A A
X j j j j j ωωωωω-=
=+++++
−→←FT
)()()(2t u Ae t u Ae t x t t ---=
From (3), we know:
2
()2X j d ωωπ∞
-∞
=⎰
But
2
2
2
220
()2()2()t t X j d x t dt A e e dt
ωωππ∞
∞
∞
---∞
-∞
==-⎰
⎰
⎰
dt e e e A
t t t )2(24320
2
---∞
+-=⎰
π
∞
----+---=0
4322
)4322(2t
t t e e e A π 22
12
2)413221(2A A ππ=+-= So
2
12
2A π=π2, that is 122
=A
, 12±=A
While 0)(≥t x , 12=A
Then )()()(2t u Ae t u Ae t x t t ---=)()(122t u e e t t ---=
Chap 3 3.1 A continuous-time periodic signal x(t) is real value and has a fundamental period T=8. The nonzero Fourier series coefficients for x(t) are j a a a a 4,2* 3311====--. Express x(t) in the form )cos()(0 k k k k t A t x φω+=∑∞ = 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∞ ----=-∞ --= =+++=++-=-∑ A discrete-time periodic signal x[n] is real valued and has a fundamental period N=5.The nonzero Fourier series coefficients for x[n] are 10=a ,4/2πj e a --=,4/2πj e a =,3 /*4 42πj e a a ==- Express x[n] in the form )sin(][1 0k k k k n A A n x φω++=∑∞ = 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
第一章作业解答 解:(b )jt t t j e e e t x --+-==)1(2)( 由于)()(2)1()1())(1(2t x e e e T t x T j t j T t j ≠==++-+-++-,故不是周期信号; (或者:由于该函数的包络随t 增长衰减的指数信号,故其不是周期信号;) (c )n j e n x π73][= 则πω70= 7 2 20 = ωπ 是有理数,故其周期为N=2; 解:]4[1][1)1(] 1[1][4 3--=--==+---=∑∑∞ =∞ =n u m n m k k n n x m k δδ -3 –2 –1 0 1 2 3 4 5 6 n 1 … 减去: -3 –2 –1 0 1 2 3 4 5 6 n u[n-4] 等于: -3 –2 –1 0 1 2 3 4 5 6 n … 故:]3[+-n u 即:M=-1,n 0=-3。 解:x(t)的一个周期如图(a)所示,x(t)如图(b)所示:
而:g(t)如图(c)所示 …… dt t dx ) (如图(d )所示: ……故: )1(3)(3) (--=t g t g dt t dx 则:1t ,0t 3,3 2121==-==;A A 1.15 解:该系统如下图所示: 2[n]
(1) ] 4[2]3[5]2[2]}4[4]3[2{2 1 ]}3[4]2[2{]3[2 1 ]2[][][1111111222-+-+-=-+-+-+-=-+ -==n x n x n x n x n x n x n x n x n x n y n y 即:]4[2]3[5]2[2][-+-+-=n x n x n x n y (2)若系统级联顺序改变,该系统不会改变,因为该系统是线性时不变系统。(也可以通过改变顺序求取输入、输出关系,与前面做对比)。 解:(a )因果性:)(sin )(t x t y = 举一反例:当)0()y(,0int s x t =-=-=ππ则时输出与以后的输入有关,不是因果的; (b )线性:按照线性的证明过程(这里略),该系统是线性的。 1.20 解:(a ))(2 1)2cos()(221t j t j e e t t x -+= = 则:)(2 1 )}(21{)(33221t j t j t j t j e e e e T t y --+=+=; (b) t j j t j j t j t j e e e e e e t t x 2121)12()12(22 1 21)(21))21(2cos()(-----+=+=-= 则:)3 1(3cos )(212121)()3 1 (3)31 (331312-=+=+=-----t e e e e e e t y t j t j t j j t j j (注意:此系统不是时不变系统。) (b)x(2-t)
Chapter 1 An swers 1.1 1.2 Con vert ing from polar to Cartesia n coord in ates: con vert ing from Cartesia n to polar coord in ates: jo 5 =5e , 1 . 3 耳 J e ■ 2 2 j(1-j)二e 4 , 2, =3e 七 =2e 七 1.3. 旳& o e (b) X 2 (t )弋心 4) P 『砧 k (c) dt 1 j ^T e - 4 2 1-j 2 b 2 _2 一1一「3一 e P :.=0, because E:.::::: , X 2 (t )二1 |2 1 T X 2(t)|dt 斗汁亍.. 2 dt - dt-:: J -oO (d) -/'2T X 2(t)=cos(t). Therefore, P oo =lim — f T 存2T L 1 " Xl [n] u[n] 2 P :一=0, because x 2[n]=e a< ? 8) , .Therefore, E - Jx 2(t)| lim1 =1 T _ L : E :: = jx 3(t)「dt = :8S (t)2dt - : T 工 dt = 1 COS(2t1 1 dt =_ 2 2 .Therefore, ;cos(t 「dt 尹丰; '|x1[ n] | l 4 丿 2 =1. therefore, n u[n] jx i [ n] | 4 =3 X 2[n] _ 1 N P :: = lim - x 2 [n] N Y 2N +1 7以2屮」| n E ::' n] 2 二:: X 3[n]=cos1 . Therefore, 14丿 2 1 N lim -------- '、' N ;:2N E* Sjx 3[n]| 匚cosgn) ^cos(^j- N 1 亠cos( n) 1 7( 2 )二1 烛 2 2 2 n) 匚1 V 伍1 lim --------- \ cos — j2N 4 The signal x[n] is shifted by 3 to the right. The shifted signal lim -— =N :.:2N 1 1.4. (a) 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 sig nal x[n] is flipped sig nal 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 Sig nal will be zero for n v-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 sig nal 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.
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]
系统(第二版)-学习说明(练习答案)系计算机工程系2005.12目录17第35章第62章第83章第109章第119章第132章第140章160章答案1.1从极坐标转换:1.2从笛卡尔极坐标转换:limlim dtdtdt=cos(t)。因此,信号翻转限制信号对,所以因此,我们知道(2)线性压缩,因为线性压缩。因此,基态周期奇信号,所有值为零时为零只有当周期复指数时。10 10复数指数乘以衰减指数。因此,周期信号。复指数基本周期信号。fundamentalperiod我们得到fundamentalperiod complexexponential=3/5。找不到任何整数整数。因此,定期1.10。x(t)=2cos(10t+1)-sin(4t-1)周期第一项第一项,整个信号周期至少有多个第二项。-3-1-1-2-3-3-3第一项第二项第二项整个信号周期,至少在35.1.12中有多个共同的三项。图1.12。翻转信号对,所以,no=-3.1.13其导数图1.14。因此[n-3]=2x[n-2]+4x[n-3]+4x[n-4])=2x[n-2]+5x输入输出关系y[n]=2x[n-2]+5x[n-3]2x[n-4]输入输出关系的连接序列是反向的。我们可以很容易地证明[n-3])+4(x输入-输出关系在y[n]=2x[n-2]+5x[n-3]2x[n-4]1.16无记忆性,因为过去值我们可能总是得出系统输出,因为有时可能取决于考虑两个任意输入(sin(t))的未来值,
让线性组合任意标量给系统相应的输出线性。1.18.(a)考虑两个任意输入线性组合任意标量。给定系统,相应的输出随机输入相应的输出。考虑第二个输入输出对应的Alsonote+1)B。因此+1)B.1.19考虑两个任意输入(t-1)让线性组合任意标量。给定系统,相应的输出为线性。(ii)考虑相应输出的任意输入。考虑第二输入输出相应的输出。考虑两个任意输入[n-2]。让线性组合任意标量。考虑任意输入对应的二次输入。给定系统,相应的输出为线性。(ii)考虑任意输入对应的输出。考虑第二输入输出相应的输出也不考虑两个任意输入线性组合的任意标量。给定系统,相应的输出是线性的。(ii)考虑相应输出的任意输入。考虑第二输入输出相应的系统线性,因此我们知道x2(t)=cos (2(t-1/2))=线性性质,我们可以再次写出x1 cos(3t-1),因此,x1(t)=cos(2(t-1/2))=cos(3t-1)1.21。信号图1.21。图1.21 21 1.22图1.22 1.22图1.22 1.23奇数图1.23 1/2-1/2-1 0.50.5 3/2-2-3/2 x(2t+1)x(4-t/2)1012 1/2-2-1/1/2-1/2-1/2-1/2-1/2-1/2 1/2-1/2 1/2 1/2-1-2图1.22-4-1-2-4-1-2-7 xo[n]xo[n]3t/2/2-3t/2-1/2-2-1/2-7图1.24-2-2图1.24-2图1.23 1.23 1.23 1/2 1/2 1/2 1 1/2-1 3/2-3/2-1/2 101.24零件图1.24
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 2.3 Solution: ][*][][n h n x n y = ][][k n h k x k -= ∑∞ -∞= ∑∞ -∞ =-+--= k k k n u k u ]2[]2[)21(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 第一章 1.3 解: (a). 2 40 1 lim (),04T t T T E x t dt e dt P ∞ -∞∞→∞ -====⎰ ⎰ (b) dt t x T P T T T ⎰-∞→∞=2)(21 lim 121 lim ==⎰ -∞→dt T T T T ∞===⎰⎰∞ ∞ --∞ →∞dt t x dt t x E T T T 2 2 )()(lim (c). 2 22 lim ()cos (), 111cos(2)1 lim ()lim 2222T T T T T T T T T E x t dt t dt t P x t dt dt T T ∞ ∞→∞ --∞ ∞→∞→∞--===∞+===⎰⎰⎰ ⎰ (d) 034121lim )21(121lim ][121lim 0 22 =⋅+=+=+=∞→=∞→-=∞→∞∑∑N N n x N P N N n n N N N n N 3 4) 2 1 ()(lim 202 = ==∑ ∑-∞ =∞ →∞n N N n N n x E (e). 2()1,x n E ∞==∞ 2 11lim []lim 112121N N N N n N n N P x n N N ∞→∞→∞=-=-===++∑∑ (f) ∑-=∞→∞=+=N N n N n x N P 21)(121lim 2 ∑ -=∞ →∞∞===N N n N n x E 2 )(lim 1.9. a). 00210,105 T ππ ω== =; 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 =-=- 第九章作业解答 9.21 解: (a) 2}Re{21 )}({2->+=-s s t u e L t 3}Re{31 )}({3->+=-s s t u e L t 2}Re{3)(2(5 s 23121)(->+++ =+++=s s s s s s X ) Re(s ) (c) 2}Re{21 )}({2<--=-s s t u e L t 3}Re{31 )}({3<--=-s s t u e L t 2}Re{3)(2(5 s 23121)(<----=----=s s s s s s X ) 9.22 (a )0}Re{91 )(2>+=s s s X 根据:0}Re{)(sin 20200>+→ s s t tu ωωω 则:)(3sin 31}9 1{21t tu s L =+- (c )根据:0}Re{)(cos 2020<+→ --s s s t tu ωω 以及:)(X )(00s s t x e t s -→ 则:)(3cos }3)1(1{221t tu e s s L t --=+++-- (e) 3 2216512+++-=+++s s s s s 根据收敛域:2}Re{3-<<-s 故:)(}21{ 21t u e s L t -=+--- )(2}3 2{31t u e s L t --=+ )(2)()(32t u e t u e t x t t --+-= (g) 2222222)1(3131) 1(3)1(31)1(31)1(3)1()1(1+++-=+-+-=+-=+-+=++-s s s s s s s s s s s s (须先转换为真分式) 则根据收敛域:)(3)(3)()(t u te t u e t t x t t --+-=δ 9.28解: 其所有可能的收敛域: (1)1}{R >s e ,收敛域不包括虚轴,不稳定;收敛域位于最右边极点的右半平面,因果; (2)2}{R - 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 s s c 0 2cos π ==,又对任意整数k ,??? ? ??+±=???? ??±n f kf f n f f s s s 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 ∴ 4 20ππ =s f f ∴ 125 8 1 0== 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()?? ?<≥=-0 0t t e t h at c 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: 1.对一个LTI 系统,我们已知如下信息:输入信号 2()4()t x 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 (),t x 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 t y t e e u d e τ+∞ ---τ--∞ =ττ=⎰ ; ()(1)t t y 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()()t y 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()t e 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 1答案 习题 1.1用笛卡儿坐标形式(x+yj)表示下列复数。 解:利用欧拉公式:和复平面性质 ,有: , , 1.2用极坐标形式(re jθ,-π<θ≤π)表示下列复数。 解:根据,有: 1.3对下列每一个信号求P∞和E∞。 解: (a) (b) (c) (d) (e) (f) 1.1设n<-2和n>4时x[n]=0,对以下每个信号确定其值保证为零的n值。 解: (a)x[n-3]=0,n-3<-2或n-3>4,即 x[n-3]=0,n<1或n>7 (b)x[n+4]=0,n+4<-2或n+4>4,即 x[n+4]=0,n<-6或,n>0 (c)x[-n]=0,-n<-2或-n>4,即 x[-n]=0,n<-4或n>2 (d)x[-n+2]=0,-n+2<-2或-n+2>4,即 x[-n+2]=0,n<-2或n>4 (e)x[-n-2]=0,-n-2<-2或-n-2>4,即 x[-n-2]=0,,n<-6或n>0 1.2设t<3时x(t)=0,确定以下每个信号的值保证为零的t值。 解:(a)x(1-t)=0,1-t<3,即 x(1-t)=0,t>-2 (b)x(1-t)+x(2-t)=0,1-t<3且2-t<3,即 x(1-t)+z(2-t)=0,t>-1 (c)x(1-t)x(2-t)=0,1-t<3或2-t<3,即 x(1-t)x(2-t)=0,t>-2 (d)x(3t)=0,3t<3,即 x(3t)=0,t<1 (e)x(t/3)=0,t/3<3,即 x(t/3)=0,t<9 1.3判断下列信号的周期性。 解: (a)由于 对于-∞<t<∞,x1(t)的值不具备重复性,所以x1(t)不是周期信号。(b)由于 所以x2[n]也不具备周期性。 (c)由于 所以x3[n]是基波周期为4的周期序列。 1.4对以下每个信号求信号的偶部保证为零的所有自变量值。 解: (a) 只有当|n|>3时, (b) 即对一切t, (c) Charpt 1 1.21—(a),(b),(c) 一连续时间信号x(t)如图original所示,请画出下列信号并给予标注:a)x(t-1) b)x(2-t) c)x(2t+1) d)x(4-t/2) e)[x(t)=x(-t)]u(t) f)x(t)[δ(t+3/2)-δ(t-3/2)] (d),(e),(f) 1.22 一离散时间信号x[n]如图original所示,请画出下列信号并给予标注。 a)x[n-4] b)x[3-n] c)x[3n] e)x[n]u[3-n] f)x[n-2]δ[n-2] 1.23 确定并画出图original信号的奇部和偶部,并给予标注。 1.25 判定下列连续时间信号的周期性,若是周期的,确定它的基波周期。 a) x(t)=3cos(4t+π/3) T=2π/4=π/2; b) x(t)=e )1(-t j π T=2π/π=2; c) x(t)=[cos(2t-π/3)]2 x(t)=1/2+cos[(cos(4t-2π/3))]/2, so T=2π/4=π/2; d) x(t)=E v {cos (4πt)u(t)} 定义x(0)=1/2,则T=1/2; e) E v {sin(4πt)u(t)} 非周期 f )x(t)=∑∞ -∞=--n n t e )2( 假设其周期为T 则 ∑∞-∞=--n n t e )2(=∑∞-∞=+--n T n t e )22(=∑∞-∞=---n T n t e ))2(2(=∑∞ -∞=--n n t e )2( 所以T=1/2(最小正周期); 1.26 判定下列离散时间信号的周期性;若是周期的,确定他们的基波周期。 (a) x[n]=sin(6π/7+1) N=7 (b) x[n]=cos(n/8-π) 不是周期信号 (c )x[n]=cos(πn 2 /8) 假设其周期为N ,则8/8/)(22n N n ππ=++πk 2 所以易得N=8 (d )x[n]=)4 cos()2cos(n n π π N=8 (e) x[n]=)62cos(2)8sin()4cos(2π ππ π+-+n n n N=16 1.31 在本题中将要说明线性时不变性质的最重要的结果之一,即一旦知道了一个线性系统或线性时不变系统对某单一输入的响应或者对若干个输入的响应,就能直接计算出对许多其他输入信号的响应。 (a ) 考虑一个LTI 系统它对(a )的信号x1(t )的响应y1(t )示于(b ),确定并画出 该系统对于图(c )的信号x2(t )的响应。 (b ) 确定并画出(a )中的系统对于(d )的信号x3(t )的响应。 信号与系统第二版课后答案 第一章简介 1.1 信号与系统的定义 1.1.1 信号的定义 信号是对某一现象或信息的描述,可以是物理量、采样值、传感器输出等。根据信号的不同特性,可以将其分为连续信号和离散信号。 1.1.2 系统的定义 系统是对信号加工与处理过程的描述。系统可以是硬件电路、算法或计算机软件。根据系统对信号的作用方式,可以将其分为线性系统和非线性系统。 1.2 信号的分类 1.2.1 连续信号与离散信号 连续信号是在时间上连续变化的信号,可以用数学函数进行描述。离散信号则是在时间上呈现离散变化的信号,通常通过采样离散化得到。 1.2.2 有限信号与无限信号 有限信号是在有限时间内存在的信号,其持续时间有限。无限信号则是在无限时间内存在的信号,持续时间可以是无限的。 1.3 系统的分类 1.3.1 线性系统与非线性系统 线性系统满足线性叠加原理,即将输入信号与线性系统的响应相加所得到的输出信号仍然是系统的响应。非线性系统则不满足线性叠加原理。 1.3.2 因果系统与非因果系统 因果系统的输出只与当前和过去的输入有关,不受未来输入的影响。非因果系统的输出则可能与未来的输入有关。 第二章离散信号与系统 2.1 离散信号的表示与性质 2.1.1 离散信号的表示 离散信号可以通过序列来表示,其中序列是一组按照一定顺序排列的数字。离散信号可以是有限序列或无限序列。 2.1.2 离散信号的性质 离散信号的性质包括幅度、相位、频率、周期性等。这些性质可以通过变换来描述和分析离散信号。 2.2 离散系统的表示与性质 2.2.1 离散系统的表示 离散系统可以通过差分方程来表示,其中差分方程描述了输入和输出之间的关系。离散系统也可以通过单位脉冲响应来描述,单位脉冲响应是当输入为单位脉冲序列时系统的输出。 2.2.2 离散系统的性质 离散系统的性质包括稳定性、因果性、线性性等。这些性质对系统的行为和性能有重要影响。 2.3 离散系统的频域分析 2.3.1 傅立叶变换 傅立叶变换是一种将信号从时域转换到频域的方法,可以将信号表示为频率的函数。离散系统的频域分析可以通过离散时间傅立叶变换进行。 Chapter 8 8.1 Solution: ))((2)(0w w j X jw Y -= ∴ t jw e t x t y 0)(2)(= that is t jw e t y t x 0)(2 1)(-= )()()(t m t y t x = ∴ t jw e t m 02 1)(-= 8.3 Solution: According to the conditions as given, we will see: Suppose )(jw X is: Then, if )(t p passes through the ideal lowpass filter whose frequency response ()H j ω with cutofffrequency π2000,we will get ()()()0Y jw P j H j ωω==, that is 0)(=t y . So 0)( t y . 8.22 Solution: According to the conditions as given, we will get : 8.28 Solution: the system about the implantation of single-sideband modulation is: (a). According to the conditions as given, we will get : (b). According to the conditions as given, we will get : From (a) and (b), we know only the upper sideband are retained. 本文档局部内容来源于网络,如有内容侵权请告知删除,感谢您的配合! Signals & Systems (Second Edition) —Learning Instructions (Exercises Answers) Department of Computer Engineering 2005.12 Contents Chapter 1 (2) Chapter 2 (17) Chapter 3 (53) Chapter 4 (80) Chapter 5 (101) Chapter 6 (127) Chapter 7 (137) Chapter 8 (150) Chapter 9 (158) Chapter 10 (178) Chapter 1 Answers 1.1 Converting from polar to Cartesian coordinates: 111cos 222 j e ππ==- 111cos()222 j e ππ-=-=- 2 cos()sin()22 j j j e πππ =+= 2 cos()sin()22 j j j e π ππ -=-=- 52 2 j j j e e π π == 4 )sin())144 j j j πππ=+=+ 94 4 1j j j π π=- 94 4 1j j j π π--=- 4 1j j π-=- 1.2 converting from Cartesian to polar coordinates: 55j =, 22j e π-=, 233j j e π --= 212j e π--=, 41j j π+=, () 2 2 21j j e π-=- 4(1)j j e π -=, 411j j e π +=- 12e π - = 1.3. (a) E ∞ =40 1 4 t dt e ∞ -= ⎰, P ∞=0, because E ∞<∞ (b) (2) 42()j t t x e π+=, 2 ()1t x =.Therefore, E ∞ =2 2()dt t x +∞ -∞ ⎰=dt +∞ -∞⎰=∞, P ∞=2 11lim lim 222()T T T T T T dt dt T T t x --→∞→∞==⎰⎰lim11T →∞= (c) 2()t x =cos(t). Therefore, E ∞ =2 3()dt t x +∞ -∞⎰=2 cos()dt t +∞-∞ ⎰=∞, P ∞=2 111(2)1lim lim 2222 cos()T T T T T T COS t dt dt T T t --→∞ →∞ +==⎰⎰ (d) 1 [][]12n n u n x =⎛⎫ ⎪⎝⎭ , 2 []11[]4n u n n x =⎛⎫ ⎪⎝⎭. Therefore, E ∞ =20 4 13 1[]4n n n x +∞∞ -∞ === ⎛⎫∑∑ ⎪⎝⎭ P ∞=0,because E ∞ <∞. (e) 2[]n x =( )2 8 n j e ππ -+, 2 2[]n x =1. therefore, E ∞ =2 2 []n x +∞-∞ ∑=∞, P ∞=2 1 1 lim lim 1122121[] N N N N n N n N N N n x →∞ →∞=-=-==++∑∑. (f). Therefore, E ∞ =2 3 []n x +∞-∞ ∑=2 cos()4 n π+∞-∞ ∑=2 cos()4 n π+∞-∞ ∑, P ∞= 1 lim cos 214 n N N n N N π→∞=-=+⎛⎫∑ ⎪⎝⎭ 1cos() 112lim ( )2122N N n N n 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 of 3. Therefore, x(3t) will be zero for t<1. ∑ {δ [n + 4m - 4k ] - δ [n + 4m - 1 - 4k ]} ∑ {δ [n - 4(k - m )] - δ [n - 1 - 4(k - m )]} ∑ {δ [n - 4k ] - δ [n - 1 - 4k ]} s Because g (t ) = ∑ δ (t - 2k ) , Chapter 1 Answers 1.6 (a).No Because when t<0, x (t ) =0. 1 (b).No Because only if n=0, x [n ] has valuable. 2 (c).Y es Because x[n + 4m ] = = = ∞ k =-∞ ∞ k =-∞ ∞ k =-∞ N=4. 1.9 (a). T=π /5 Because w =10, T=2π /10= π /5. (b). Not periodic. Because x (t ) = e -t e - jt , while e -t is not periodic, x (t ) is not periodic. 2 2 (c). N=2 Because w =7 π , N=(2 π / w )*m, and m=7. 0 0 (d). N =10 Because x (n) = 3e j 3π / 10 e j (3π / 5)n , that is w =3 π /5, N=(2 π / w )*m, and m=3. 4 0 (e). Not periodic. Because w =3/5, N=(2 π / w )*m=10π m/3 , it ’not a rational number . 1.14 A1=3, t1=0, A2=-3, t2=1 or -1 Solution: x(t) is dx(t ) dt is ∞ k =-∞ 1.15. (a). y[n]=2x[n-2]+5x[n-3]+2x[n-4] dx(t ) dx(t ) =3g(t)-3g(t -1) or =3g(t)-3g(t+1) d t dt信号与系统_奥本海姆_中文答案_全章节
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