信號與系統(tǒng)版習(xí)題課

二.系統(tǒng)無失真的傳輸條件：時域表示：r(t)=ke(t

-t0

)頻域表示：R(jw)=kE(jw)e-jwt0

系統(tǒng)的頻率特性：H(jw)=ke-jwt0系統(tǒng)的沖激響應(yīng)：h(t)=kd(t

-t0

)一.LTIS的頻域特性:R(jw)=H(jw)E(jw)三.理想低通濾波器的響應(yīng)Hi

(

jw)

=0...w為其它值ce-

jwt

0

...w

￡

wdxx1y

sin

xuSi[w

c

(t

-

t0

)]sin

wc

(t

-

t0

)p

wc

(t

-

t0

)式中：Si(y)=0階躍響應(yīng)：r(t)=2

+沖激響應(yīng)：h(t)=wccrw=

2p四.t五.佩利—維納準(zhǔn)則￥-￥￥-￥dw

<

￥2

H

(

jw

)

dw

<

￥1

+

w

2Ln

H

(

jw

)六.希爾伯特變換￥-￥￥-￥1

R(l)X

(

jw

)

=

-

dlp

w

-

l1

X

(l)R(

jw

)

=

dlp

w

-

l七.調(diào)制與解調(diào)；PCM編碼；取樣保持同步解調(diào)和抑制載波調(diào)幅和解調(diào)頻率窗函數(shù)的運用理想取樣，零階保持和一階保持取樣信道復(fù)用，碼速與帶寬的關(guān)系*四種時頻對應(yīng)關(guān)系基本性質(zhì)。與抽樣定理有關(guān)的性質(zhì)。周期信號的頻譜:周期性-抽樣性抽樣信號的頻譜:抽樣性-周期性與單邊特性有關(guān)的性質(zhì)。

(希爾泊特交換）解析信號-單邊譜與功率譜有關(guān)的性質(zhì)。相關(guān)函數(shù)-功率譜P312.5-16證明:z(t)=F-[Z(w)]=F-[2F(w)u(w)]f

(t)=

2F-[F(w)]*F-[u(w)]

F

(w

)

?jt1

[pd(t)

-

1

]2p

jt\

u(w

)

?jwu(t)

?

pd(w

)

+

1

;\

[pd(t)

+

1

]

?

2pu(-w

)F

(t)

?

2pf

(-w

);

f

(-t)

?

F

(-w

)tf

(t)

*1\

z(t)

=

2

f

(t)

*{

1

[pd(t)

-

1

]}

=

f

(t)

-

12p

jt

jp^z(t)

=

f

(t)+

f

(t)a^*.解析信號

f

(t)

=

f

(t)

+

f

(t)Fa

(w

)

=

F(w

)

+

j[-

j

sgn

w

]F(w

)

=

F(w

)(1+

sgn

w

)0.....w.

<0=

2F(w)..w.

>0^1f

(t)

= *cosw

0tptf

(t)

=

cosw

0t000jw

tae

jw

0tt

=

ecosw

t

+

j

sin

wf

(t)

=解析性?

2pd(w

-w

0

)單邊性*.求y(t)=?cost1·

x

(t)1H

(

jw)·y1(t)2H

(

jw

)cos

100t2x

(t)y

(t

)cos10tH1(

jw)20-2020

wH2(jw)w100-1001解：x1

(t)=cost

cos10t21X

(w

)

=

p

[d(w

+11)

+d(w

-11)

+d(w

+

9)

+d(w

-

9)]p

[11d(w

+11)

+11d(w

-11)

+

9d(w

+

9)

+

9(w

-

9)]Y1

(w

)

=

X

1

(w

)H1

(w

)

=11

9cos

11

t

+

cos

9

t2

22y

1

(

t

)

=

F-

[Y

(w

)]

=2

211

cos11t

cos100t

+

9

cos

9t

cos100tx2

(t)

=

y1

(t)

cos100t

=4

42x

(t)

=

11[cos111t

+

cos

89t]

+

9

[cos109t

+

cos

91t]用2

cosa

cos

b

=cos(a

+b)+cos(a

-b)2X

(w)

=11

[d(w

+111)

+d(w

-111)]+11

[d(w

+89)

+d(w

-89)]4

p

4

p44114

411

9cos89t

+9cos91t\

y(t)

=

F

[Y(-w)]=p[d(w

+89)

+d(w

-89)]+

p[d(w

+91)

+d(w

-91)]+9p[d(w

+109)

+d(w

-109)]+9p[d(w

+91)

+d(w

-91)]4

4Y(w)

=

X2

(w)H2

(w)

=*.已知實信號f(t)的最高頻率為fm(Hz)，試計算對各信號f(2t)，f(t)*f(2t)，f(t)f(2t)抽樣不混疊的最小抽樣頻率。解：根據(jù)信號時域與頻域的對應(yīng)關(guān)系及抽樣定理得對信號f(2t)抽樣時，最小抽樣頻率為4fm(Hz)；對f(t)*f(2t)抽樣時，最小抽樣頻率為2fm(Hz)；對f(t)

f(2t)抽樣時，最小抽樣頻率為6fm(Hz)。*.

抽樣定理的工程應(yīng)用許多實際工程信號不滿足帶限條件f

(t)f1(t)h(t)抗混疊低通濾波器wF

(

jw

)10H

(

jw

)wm

w-wm10w1F

(

jw

)w

m-

wm10*混疊誤差與截斷誤差比較wF

(

jw

)101TwFs

(

jw

)w

mw

s-w

s-w

m0......Fs

(

jw

)1Tww

mw

s-w

s

-w

m0......wF1

(

jw

)w

m-

wm10￥-￥P314,5-23(短實傅立葉變換）￥-￥X

(t,w

)

=x(t)

g

(t

-t)e-

jwt

dtx(t

+t)

g

(t)e-

jwt

dtX

(t,w

)

=21t表示時窗函數(shù)的移動。X1表示信號不動，窗函數(shù)向右移動；X

2表示窗函數(shù)不動，信號向左移動。而X

1和X

2卻同時是頻率和時間的函數(shù)。X

(

jw

)x(t)

?aaaxx12p

￥-￥X

(w

)y

*(aw

)e

jbw

dw￥-￥WT

(a,

b)

=(

p313,5

-

22)x(t)j

*(t

-

b

)dtWT

(a,

b)

=：a,b含義的解釋b相當(dāng)于使鏡頭相對于目標(biāo)平行移動。a的作用相當(dāng)于鏡頭向目標(biāo)推進或遠(yuǎn)離。*.時頻分析入門：*

濾波器品質(zhì)

因數(shù)：即相對

帶寬（帶寬與

中心頻率之比）恒定的特點?；痉治鰡卧狿310.5-5解：i1

(t)1R2R1V(t))11

221

1=

2

1

2

1+s2

LC

+

s(R

+

R

)C

+1s2

R

LC

+

s(R

R

C

+

L)

+

RR

+1/

scI

R

+

sLv

1H(s)

=

1

=1/(1

22+

2

1

2

s

2

+

(R

+

R

)s

+1(1

-

R

2

)s

+

(R

+

R

)H

(s)

=

R

L

=1H

,C

=1F帶入H

(s)得：據(jù)無失真?zhèn)鬏斣跁r域的條件,系統(tǒng)的沖激\傳輸?shù)倪^程無失真,無延時。=

0s

2

+

(R

+

R

)s

+1(1

-

R

2

)s

+

(R

-

R

)2

1

2據(jù)此H

(s)只有常數(shù)項，\1

2\

R1

=

R2

=

1W

,

H

(s)

=

1,

h(t)

=

d(t)響應(yīng)為：h(t)=kd(t

-t0

)*理想高通濾波器,其w

>w

0H

(w

)

=

e-

jwtd0w

<

w

0求理想高通濾波器的沖激響應(yīng).當(dāng)信號f

(t)=2e-tu(t)通過高通濾波器,若要求輸出信號的能量為輸入信號能

量的百分之50,求高通濾波器的截止頻率.Fw

(w

)1

0w

0-w

0w

0Fw

(w

)

=1-

G2w

(w

)0

0解:

H

(w

)

=

e-

jwtd

F

(w

)00Sa(w

t)pw2w

0w

)

?

d(t)

-1-

G

(000Sa[w

(t

-

t

)]pwd(t

-

t

)

-(w

)]

=w

0h(t)

=

F

-1[e-

jwtd

F由時移定理可得:由:f

(t)*h(t)?F(w)H(w)42e-tu(t)

?1

+

w

2F

(w

)

2

=w0H

(w

)

=

e-

jwtd

F

(w

);(

p350,6

-140)2

21￥

￥w

=

f

(t

)

dt

=

2p

F

(w

)

dw-

￥ -

￥i21210222:

w

=

1

w￥-

￥按題意要求￥

F

(w

)

H

(w

)

dw-

￥輸出

:

w

0

=

2pdw

1

+

w輸入：

wi

=

2p-￥￥