信號(hào)系統(tǒng)-教案2003年春季

kkkk

kf

[k]z

a

f

[k]z

k

baf

[k]

bf

[k]z211

2因?yàn)?

R1

z

R2

)所以af1[k]

bf2[k]

aF1(z)

bF2

(z)6.3

z

變換的性質(zhì)6.3.1線性性質(zhì)(R11

z

R12

)(R21

z

R22

)

F1(z)

F2(z)如果

f1[k]f2[k]PDF

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Pro

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0(k

0)f

[k]

1

k為偶數(shù)k為奇數(shù)例：2f

[k]

1

[1

(1)k]u[k]

z2(

z

1)z

2

1

z

zz

1

z

1

(

z

1)11

z

2f

[k]zk

1

z2

z4

z6

...

k

又:PDF

createdwith

pdfFactory

Pro

trialversion

(

z

a

)(

z

a

)zz

aaz

aaku[k]

aku[k

1]

例：

1

(

z

0)z

a z

aak

u[k]

ak

u[k

1]

[k]z

aPDF

createdwith

pdfFactory

Pro

trialversion

6.3.2移序性質(zhì)如果

f

[k]

F

(z)

(

R1

z

R2

)因?yàn)?/p>

f

[k

n]zk

f

[m]z(mn)

zn

f

[k]zkk

m

k所以

f

[k

n]

znF

(z)

(R1

z

R2

)PDF

createdwith

pdfFactory

Pro

trialversion

那么

f

[k

n] (n

0)

的單邊

z

變換則為

f

[k

n]z

k

f

[n]

f

[n

1]z

1

f

[n

2]z

2

...k

0=

zn

{

f

[n]z

n

f

[n

1]z

(n

1)

f

[n

2]z

(n

2)

...

}全部)不包括原點(diǎn))不包括無(wú)窮遠(yuǎn)點(diǎn))

[k]

[k

1]

z1

[k

1]

z

1

(z

平面

(z

平面

(z

平面例：F

(

z)f

[k

]u[k

]

設(shè)：PDF

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Pro

trialversion

注意！雙向序列的單邊z

變換所具有的移序特性。zn{

f

[0]

f

[1]z1

f

[2]z2

...+

f

[n1]z(n1)}zn{

f

[n]zn

f

[n1]z(n1

f

[n

2]z(n

...}

zn{

f

[0]

f

[1]z1

f

[2]z2

...+

f

[n1]z(n1)}

f

[k

n]zkk0

)2)k

0znF

(z)

zn

n

1

f

[k

]z

k

znF(z)

znn1

f

[k]zkk

0則

f

[k

n]u[k]

若

f

[k]u[k]

F(z)增添部分PDF

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原有部分應(yīng)減去部分1knkf

[k]zf

[k

n]u[k]

znF(z)

zn那么

f

[k

n](n

0)的單邊

z

變換則為

f

[k

n]zkk0

f

[n]

f

[n

1]z1

f

[n

2]z2

...

f

[1]zn

1

f

[0]zn

f

[1]zn

1

f

[2]zn

2

...表示為和式

znF(z)

zn

1

f

[k]zkk

n若

f

[k]u[k]

F(z)PDF

createdwith

pdfFactory

Pro

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例6.3.21

1ke

z

k

0

k!

z

11k!1

u[k

]

e

z

(

z

0)1(k

2)!u[k]

z2(e

z

1z1)

z2e

z

z2

z

(0

z

)n1kf

[k

n]u[k]

zn

F(z)

f

[k]zk0若

f

[k]u[k]

F(z)11

1z(e

z

1)

ze

z

z1

1(k

1)!u[k]

(0

z

)PDF

createdwith

pdfFactory

Pro

trialversion

6.3.3

序列指數(shù)

性質(zhì)如果

f

[k]

F

(z)

(R1

z

R2

)a

z

kkk

f

[k]k

因?yàn)?/p>

a f

[k]zk

(

a

R1

z

a

R2

)那么

ak

f

[k

]

F

za

PDF

createdwith

pdfFactory

Pro

trialversion

(z

/

0.5)(z

/

0.5)

4那么(0.5)k

4ku[k]

zz

44k

u[k]

例：已知PDF

createdwith

pdfFactory

Pro

trialversion

(z

/

0.5e

j/6

)(z

/

0.5ej/6

)

4那么(0.5ej/6)k

4k

u[k]

2e

j/6PDF

createdwith

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Pro

trialversion

=a

a

z2

za

0z2

2za

cos

a2za

sin02

cos0

1

z

sin00ak

sin(

k)u[k]

=0a

z2

z0

a

a

z2

2za

cos0

a2z(z

a

cos

)2

a

cos0

1

z

zcosak

cos(0k)u[k]

例6.3.3sin(0k)u[k]

zsin0z2

2zcos10cos(0k)u[k]

z(z

cos0)z2

2zcos101

zz

/

a

1z

aPDF

createdwith

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meR1R2O6.3.4時(shí)域反褶性質(zhì)如果f

[k

]

F

(z)(

R1

z

R2

)nf

[n]zn

f

[k](z

1)kk

f

[k]z

k因?yàn)閗

所以

f

[k]

F

1

(

R1

z

1

R2

)z

PDF

createdwith

pdfFactory

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trialversion

(

z

2)1z

22k

1u[k

1]

222z

1z

z

1112

=2k

1u[k

1]

(

2

z

1即

z

1

)例6.3.4PDF

createdwith

pdfFactory

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trialversion

2

k

1

u[

k

1]

z

kk

1

1

(

2

k

1

)

z

k

1

2

z

kk

2

k

1

(

2

z

)

k

1

2

z

1

z2

k

1

2

1

2

z

2

1

z12

(2

z

1

即z

)2若用定義求2k

1u[k

1]的z

變換PDF

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例6.3.46.3.5時(shí)域卷積和性質(zhì)(

Rf1

z

Rf2

)(

Rh1

z

Rh2

)如果

f

[k]h[k]

F

(z)

H

(z)kk

i

k

i

f

[i]

h[k

i]

k

z

f

[i]h[k

i]

z那么R(z)

R(z)令

r[k]

f

[k]

h[k]

f

[i]h[k

i]i

PDF

createdwith

pdfFactory

Pro

trialversion

i

nini

n

h[n]z

i

f

[i]zf

[i]

h[n]zn

z

if

[i]

h[n]z

n

inR(z)

令n

k

i所以

f

[k]

h[k]

F

(z)H

(z

)

(

R1

z

R2

)21f2

h2f1

h1R

maxR,

R

,R

minR

,

RPDF

createdwith

pdfFactory

Pro

trialversion

6.3.6

z

域微分性質(zhì)如果

f

[k]

F

(z) (

R1

z

R2

)dz所以

kf

[k]

z

d

F

(z)

dz

kf

[k]z

k

1k

因?yàn)镕(z)

f

[k]z

k

,

d

F

(z)k

(

z

1)zn

1(z

1)n

1(k

+

n)(k

n

1)

...

(k

1)u[k]

1n

!利用移序性質(zhì)、z域微分性質(zhì)與單位階躍序的z變換，導(dǎo)出如下變換對(duì)：PDF

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dz(k

1)

f

[k

1]u[k

1]

d

F(z)由6.3.2

移序性質(zhì)d22F

(z)...

...

...

...

dz2(k

1)(k

2)

f

[k

2]u[k

2]u[k

1]

(1)由6.3.2

移序性質(zhì)dzkf

[k]u[k

]

z

d

F

(z)由6.3.6

z

域微分性質(zhì)

(1)2

z

d

2

F

(

z

)dz

2k

(k

1)

f

[k

1]u[k

1]u[k]由6.3.6

z

域微分性質(zhì)f

[k]u[k]

F

(z)如果有PDF

createdwith

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k(k

1)(k

2)

...

(k

n

1)

f

[k

n

1]u[k

n

1]...

u[k

1]u[k]

(1)n

z

dn

F(z)dznF(z)n

dndzn(k

1)(k2)...(k

n)f

[k

n]u[kn]...

u[k

1]

(1)現(xiàn)令f

[k]

u[k]dnndnnz

n!dzn

z

1

(z

1)n1(1)F

(z)

(1)dzn,d

z

(z

1)z

(1)(z

1)2

(z

1)2(1)

(1)(2)dz

z

1

d2dz

(z

1)2

(z

1)3dz

2

z

1

z

d...

...

...

...PDF

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n!令

g[k]

1

(k

1)(k

2)

...

(k

n)u[k

n]

...

u[k

1]kzn1n

g[k]zk

0所以

g[k

(n

1)]

z

n1(z

1)n1此項(xiàng)為零1(z

1)n11

(k

1)(k

2)

...

(k

n)u[k

n]...

u[k

1]

n!(k

n)(k

n

1)...(k

1)u[k

1]...

u[k

n]

zn1(z

1)n11n!PDF

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1

(k

n)(k

n

1)

...

(k

1)u[k

1]...

u[k

n]

zn1n!(z

1)n1(

z

1)1

(k

n)(k

n

1)

...

(k

1)

u[k]

n!zn1(z

1)n1n!(

z

a

)zn

1(z

a)n

11

(k

n)(k

n

1)

...

(k

1)aku[k]

再由6.3.3

序列指數(shù)

性質(zhì)所以(k

1)u[k

1]PDF

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6.3.7初值定理如果

f

[k]

0 (k

0)，有f

[k

]

F

(z

)因?yàn)镕(z)

f

[k]zk

f

[0]

f

[1]z1

f[2]z2

f[3]z3

...k0limz

limz

f

[k]z

k

f

[0]

k

0所以

F

(z)

zn

aF(z)

n2zn1

an1zn2

a

z

a1

0bmzm

bm1zm1

bm2zm2

b1z

b0如果

f

[k]

0 (k

0)，且

f

[0]

0，若那么

n

mPDF

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6.3.8終值定理如果f

[k]

0(k

0)，有f

[k

]

F

(z)

z

1F(z)那么

lim

f

[k]

l