信號(hào)與系統(tǒng)課件分析信息處理教學(xué)中心

SIGNALSAND信號(hào)與系第六章離散信號(hào)與系統(tǒng)的變換域分析第六章離散信號(hào)與系統(tǒng)的變換域分析郵電大信號(hào)分析與信息處理教學(xué)中《信號(hào)與系統(tǒng)》SIGNALSAND 離散信號(hào)與系統(tǒng)的變換域分析概6.1Z6.2Z6.6散系統(tǒng)函數(shù)與系統(tǒng)特6.7離散信號(hào)與系統(tǒng)的頻域分本章要作返回《信號(hào)與系統(tǒng)》SIGNALSAND 時(shí)域分析：跟連續(xù)信號(hào)與系統(tǒng) 離散時(shí)間傅里葉變換分析、Z返回《信號(hào)與系統(tǒng)》SIGNALSAND Z變 返Z變換可以從 ZZ定義：離散信號(hào)（序列）ZF(z)?f(1)z1f(0)z0f(1)z1?f(k)zkf(k)zkk F(z)Zf(k)相應(yīng) f(k)Z1F(z)記 f(k)F(z

ZZ《信號(hào)與系統(tǒng)》SIGNALSAND Z fs(t)f(t)T(t)f(t)(tkT)f(kT)(t f(t0tf(t0tfS(t?? t

FS(s)則fs(t的拉氏1

f(kTzesTsT

lnz，F(xiàn)s(s

sT

ln

f(kT)zkF(zk當(dāng)f(k)為因果序列k0的部當(dāng)f(k)為因果序列k0的部分時(shí)，F(xiàn)(z)k0f(kkZ k k

（a為正實(shí)數(shù)的Z變換 F(z)f(kz-kakzk(azk kaz-11zaa0SANDjIma0SANDZ變換的收斂域?yàn)閳A心在原點(diǎn)，半徑為a

雙邊Z變

k

jImzf(k) k（a，b為正實(shí)數(shù)

F(z)akzk

bkzkk

az-

z-

azbab，則無(wú)收斂域，Z有限序列ZZ《信號(hào)與系統(tǒng)》SIGNALSAND 常見(jiàn)序列的單邊Z單位函數(shù)Z[(k)] (kz-kz- kk單位階躍序列1z 1z- - Z[(k)](k)zk k單邊指數(shù)序列ak

1z- z Z[a(k)]ak

1az- z-當(dāng)aeT時(shí)

SIGNALSAND aej0T

sin0kTε(k)cos0kTε(k

z(zcos0T)jzsinz-e

z-cos

z22zcosT P276P2766-Z0ZcoskTε(k)0

z(z-cos0T z2-2zcosTZsinkTε(k)

zsin0T z2-2zcosT《信號(hào)與系統(tǒng)》SIGNALSAD Z 返例F(z

kF(z)21.5z-11.25z-21.125z-3f(k)2,1.5,1.25,1.125 f(k)的解析式(閉式解) 《信號(hào)與系統(tǒng)》SIGNALSAND N(z) bzm

zm1?bzF(z)

10D(z) azn 10

zn1?az ，通常將F(z)展開(kāi)z 然后兩邊同時(shí)乘以z，得到典型序列的Z變換式，再進(jìn)行《信號(hào)與系統(tǒng)》SIGNALSAND 例用部分分式展開(kāi)法求的原函數(shù)f(k)。

F(z)

2z2-1.5z解F(z)

(z1)(z F(z)

z0 z- zz f(k(0.5)k《信號(hào)與系統(tǒng)》SIGNALSAND 例62 已知F(z)

4z(z1)(z

解F(z) 4z z(z1)(z1 z z (zz101

C 11 1 (1F(Z)18z7z z z (zf(k)(k)8(k)7(2)k(k)3k(2)k(k《信號(hào)與系統(tǒng)》SIGNALSAND ?例62 求F(z) ?

Zex1x

x 2！得Fz)

z＝1

a

azz

?

azz a z a a z

a

zk k！ k k！

與Z變換的定義式比較，得f k《信號(hào)與系統(tǒng)》SIGNALSANDSTM 例試求Fz)

z4z6(z-

Z解F(z

z7z4

Z7(z

zz4z5Z6z F(z)z3z4z5zf(k)(k3)(k4)(k5)(k(k3)(k《信號(hào)與系統(tǒng)》SIGNALSAND 例Z）F(z)解法一

z32z2z31.5z2F(z)13.5z14.75z26.375z3f(k)解法二部分分式展開(kāi)法F(z) z32z2

26 z2(z1)(z

z z1 zF(z)26 z1 zf(k)2(k1)6(k)[813(0.5)k](k)《信號(hào)與系統(tǒng)》SIGNALSAND （2）F(z)

z3(z解F(z)zz 2 (z (z

(z

z1F(z)

(z (z z(k) k(k)

k2(k)

z2z1 zz(z

(z2

(z (z (z

k)(k) F(z)(k2k3k1)(k)(k1)2(k)《信號(hào)與系統(tǒng)》SIGNALSAND

z2(z1)(z2解F(z) z

Bz

(z1)(z2

z z2z0

111

C等式兩邊同乘z令z， 01 B故F(z)

變換 z

z2

F

z2

2

(k)z2zf(k)(1

z2k)(k)

sin2

(k)

z21《信號(hào)與系統(tǒng)》SIGNALSANDSYSTEMS 返 若Zf1(k)1z)，Zf2(k)F2(z則Za1f1(ka2f2(k)a1F1(za2F2(z移序（移位）性若Zf(k)F(z則Zf(k1)zF(zzfZf(k1)z1F(z)f(- 證明f(k1)z-k

zf(k1)z-(k1)zf(i)z-k - z

f

f

m

-k推廣Zf(km)zF(zf

f(k1)z-kz-1f(k1)z-

z-1fiz-iz-1[F(z)f(1)z推 -m kZf(km)zF(z)f(k)z k f(1f(2f(m0Zf(km)z-mF(z Zf(km)(km)z-mF(z于F(zL統(tǒng)TM大為B例631 試求離散信號(hào)f(k)(k1)(k1)的Z變換式解Zf(k)Z(k1)(k1)

，分別求ak1，ak1和ak1(k1的Z變換式f(kak，則f(k1ak1，根據(jù)右移序性質(zhì)， z1 z a(z由于是單邊Z變換，有Zak1

a(z z 《信號(hào)與系統(tǒng)》SIGNALSANDZB 已知F(z)f(k

，試求對(duì)應(yīng)于F(z解這類(lèi)題目用一般求Z反變換的方法求解是 F(z)1

z

f(k)(1)k10(k10)(1)k(k《信號(hào)與系統(tǒng)》SIGNALSAND 已知單邊周期序列f(k)f1(kmNm，n為整數(shù)，NZf1(k)(k)F1(z)f(kZF(z解:f(kf1(k)(kf1(kN)(kN -

-mN1z

1z1z1z 若z-F

z

(z)F1(z)1z- 《信號(hào)與系統(tǒng)》SIGNALSAND 若Zf(k)F(z

則Zakf(k)za z

z證明

f(k)akf(k)zkf

k k

a a也稱(chēng)為序列的指 11 2

k(kZZ k

z21

Z2sin2k(k)(2z)2

z2號(hào)與系統(tǒng)》SIGNALSADYSTEMSZ若Zf(k)F(z

則Zkf(k)

dF(z)d

f(k)zk

f(k)

zk

k k z1kf(k)zkz1Zkfk即Zkf(k)z 于Z域中對(duì)Z變換取導(dǎo)數(shù)并乘以-z。推廣

f(k)z

d

《信號(hào)與系》dz SIGNALSAND例已知Z(k

k(kZz1Zk(k)－zdZ(k)－zd

dzz1

(z1)Zk2(k)－zdZk(k)－zd

zd

d

dz z(z

dz

1dz

(z1)2

(zZkak(k

dZak(k

d

dzza《信號(hào)與系z(mì)2GASAND 若Zf1(k)F1(z)，Zf2(k)F2(z則Zf1(k)*f2(k)1z)F2(z) x(k)ak(k)，h(k)bk X(z)

z

，H(z)

z

bzY(z)X(z)H(z) (za)(z abz zby(k)x(k)*h(k)Z a《信號(hào)與系統(tǒng)》SIGNALSAND 若Zf(k)F(z則Z

f(n)

z

F(z) 證明利用時(shí)域卷積定理，可以得到序列求和的Z因?yàn)閒(k)(k)*(k)

fZ

f(n)

Zf(k)(k)*

z

F(z)《信號(hào)與系統(tǒng)》SIGNALSAND 1 已知f(k)1f(k的單邊Zk

22，f(k)kf1解f2(k

2mf(k)11

根據(jù)序列求和性質(zhì)，得F2(z

z1zz 根據(jù)序列指 性質(zhì)，得

根據(jù)序列線 z2(3zF(z)＝z F (z1)2(z

(z《信號(hào)與系統(tǒng)》SIGNALSAND 若Zf(k)F(z)limF(zf(kf(0)＝limF(z證明根據(jù)ZF(z)f(k)zkf(0)f(1)z1fk

?f(n)znf(kZ當(dāng)z，一其余各均趨于所以 ff(kZ f m (z)m1f(k)zk

若Zf(k)F(z)f(kf(則f(z1)F(z證明根據(jù)ZZf(k1)f(k)zF(z)zf(0)F(z)(z1)F(z)zf li（z1)F(z)f(0)limf(k1)f(k)zkf(k1)f(k z1k

klimf(1)f(0)f(2)f(1)?f(n1)flimf(n1)f故f(z1)F(z

條件：f(kF(zz＝1處《信號(hào)與系統(tǒng)》SIGNALSANDSYSTEMS sjz

eTe（縱軸）0（原點(diǎn)）s0（左半平面）0

z單位圓z（單位圓上的一點(diǎn)z單位圓內(nèi)jImz

《信號(hào)與系統(tǒng)》SIGNALSAND 例某序列的ZFz)f(k的終值f()。解F(zza

zaa1時(shí)，f()lim(z1)F(z)limz

za1z1f()lim(z1)F(z)lim(z f(k)ak(k

za1時(shí)，f(k1k(ka1f()《信號(hào)與系統(tǒng)》SIGNALSAND *9Z若Zf(k)F(z 則Z

f(k)

F(v)v1dv

1

f(k)vk

k

vk

f(k)v(k1)dvf(k) f(k) k

k

k

k 同理可證Z

f(k)

zaz

aP2886-2Z《信號(hào)與系統(tǒng)》P2886-2Z*

k

(k)的z變解：Z(k

zZZ

(k)z

dvz

(v1)v

(v

v v z 《信號(hào)與系統(tǒng)》SIGNALSAND 例已知f(k)f(kf(k)1k(k1)k 且f1(k)(k)，試求f2(k的表達(dá)式 Zf(k)Zf(k)Zf zZf z 2(z1(z1)z1 2(z Zf2 z

(z故f2(kk《信號(hào)與系統(tǒng)》SIGNALSAND 例求序列f(k)(1)n(kn的單邊Z變換解f(k(k1)(k1(k21)(k3?Zf(k)

z-

z- z-z- z z(1z-1z-2-z-3?)z

z

(z 《信號(hào)與系統(tǒng)》SIGNALSAND 例試求序列f(k1)k(k1)(k1Z解(1)k(k)

(1)kk(k)

d z Z (1)kk(k)

zz1z

f(k)(1)k1(k1)(k1)

《信號(hào)與系統(tǒng)》SIGNALSAND R(k)

0kNRN(k21N3?01RN(k21N3?012Nk0解GN(k(k(kN則RN(k)kGN(k) zNGN(z)z1zR(z)zdG

Z域微分 N1

RN(k)m(k zN1NzN NzN(z (z

NzzN1NzN1NzN

ZRN(k)mz(

》SIGNALSAND (1)(k1)2(k解設(shè)f1(k)k1)(k1)F(z)z (z1)2 (z而f(k)k1f1(k)F(z)zdF(z)F(z)

z

(zf(k)(k22k1)(k1)F(z)zd[zd )]2[zd )] z1 z z1z(z1)

z(z)3與系(kk

(3)(k1)[(k)(k3)][(k)(k （2）設(shè)f(k)1)k(k)，則F(z

zk而f(k)f1(n)kF(z) F(z) zz1 z21

3z

z22z

z z2(z

z2(z F2(z)

zzz1

z41z3(z

(z21)(zz3F(z)

(z)

(z21)(z1)(z22z 《信號(hào)與系統(tǒng)》SIGNALSAND 例試求序列f(k)0 k為奇數(shù)的Z變換。解f(k)1[(1)k1](k)2F(z)1 ] z2z z z2 k?4m0例f(k)0

Z解f(k)(k(k4(k8(k4mF(z)1z4z8 zz z 《信號(hào)與系統(tǒng)》SIGNALSAND 例已知F(z)

z32z2zz3z2

f(0),f(1f(2),f()解用長(zhǎng)除法，得F(z)1z12.5z22z2 f(0)1,f(1)1,f(2)2z3 z2

z32z2zF(z)

z3z2

z(z2z

1j2f()lim(z1)F(z)z z

(z1)(z32z2zz3z2 《信號(hào)與系統(tǒng)》SIGNALSAND 返與拉氏變換類(lèi)似，Z變換分析法可以分別求解零輸 y(k2)5y(k1)6y(k)x(k2)應(yīng)?！缎盘?hào)與系統(tǒng)》SIGNALSAND ZB返解x(k)0y(k2)5y(k1)6y(k)Z變換，并應(yīng)用移序性質(zhì)，可得z2Y(z)z2y(0)zy(1)5[zY(z)zy(0)]6Y(z)Y(z)

y(0)z2[y(1)5z25zi 2z2 3z iz)

5z

(k)3(2)k《信號(hào)與系統(tǒng)》SIGNALSAND y(k)5y(k1)6y(k2)x(k)3x(kyzi(0)2，yzi(13，試求其零輸入響應(yīng)。Z變換，并應(yīng)用移序性質(zhì)，可得Y(z5[z1Y(zy(16[z2Y(zz1y(1y(20

(z)

5yzi(1)6yzi(2)6yzi(1)z115z16z2所需初始條件yzi(1)yzi(2)，k1 ziIGNALSSYSTEi 將初始條件yzi(1)和yzi(2)代入27z 2z2 kYzi(z)15z16z z25z6z-2z-k

yzi(k)3(2)k《信號(hào)與系統(tǒng)》SIGNALSAND 時(shí)域分析法：yzs(k)x(kh(k)根據(jù)時(shí)域卷積定理：Yzs(zX(z)H(z)H(zZh(k)與連續(xù)系統(tǒng)類(lèi)似，a2y(k2)a1y(k1)a0y(k)b2x(k2)b1x(k1)a[z2Y(z)z2y(0)zy(1)]a[zY(z)zy(0)]aY(z) b[z2X(z)z2x(0)zx(1)]b[zX(z)zx(0)]b0X(z)《信號(hào)與系統(tǒng)》SIGNALSA1DSYSTEMS

(a2z2a1za0(bz2bzb)X(z)bz2x(0)b y(-1)=0y(-2)=0，以此代入原差分方程，有a2y(0)a2y(0)a2y(1)a2y(1)a1y(0)b2x(1)因此，差分方程的Z(az2aza)Y(z)(bz2bzb)X(z)

AND y(k2)5y(k1)6y(k)x(k2)3x(k)H(z)和單位函數(shù)響應(yīng)h(kz2H(z)H

z25z2

(z)

2

2 z(z25z h(k)1(k)[2(3)k1(2)k](k (k[2(3)k(2)k1](k 《信號(hào)與系統(tǒng)》SIGNALSAND y(k)y(k2)x(k)x(kH(z)和單位函數(shù)響應(yīng)h(k)；x(k)(k)yzs(k。1 z2解H(z)1 z2h(k)cosk(k)sink(k) z2Yzs(z)X(z)H(z)z1z2

1X(z)H(z)

z2

(z1)(z2 z z2y(k)1sink)(k) 信號(hào)與系統(tǒng)》SIGNALSAND 應(yīng)），相減得零輸入響ALSANDSYSEM。 例已知離散系統(tǒng)的差分方程為y(k2y(k1x(k)，x(kek(k)y(00，試求全響應(yīng)y(k。解對(duì)差分方程取Z變換，得Y(z)2[z1Y(z)y(1)]X(z) Y(z)

X(z)2y(1)

X(z)12z 12z z

z y(0)2y(1)x(0)

y(1)2以X(z) ，y(1)代入，z Y(z)

(ze1)(z

z

Yzs(z)Yzi(z)《信號(hào)與系統(tǒng)》SIGNALSAND Y(z) (ze1)(z2)z 12ez 12ez2z y(k)

(2)k k 1 其中yzi(k)(2y(k)

(2)k 1 《信號(hào)與系統(tǒng)》SIGNALSAND yzi(02，yzi(1x(k(k)y(k12

52解y(k)14(2)k5(3)k 自然（固有）響應(yīng)分量為4(2)k5(3)k2yzi(k)A(2)k

kyzi(0AB2，yzi(12A3BA5，B《信號(hào)與系統(tǒng)》SIGNALSAND yzi(k5(2

ky(ky(ky(k)k

1(2)

1(3)k(k)Y(z)

z 2z1 z

2z

(z1)(z2)(zZX(z)

zH(z)Yzs(z) X(z) (z2)(z

z25zy(k2)5y(k1)6y(k)《信號(hào)與系統(tǒng)》SIGNALSAND 離散系統(tǒng)函數(shù)與系統(tǒng)特 返H(zmbzm?bz

(zzrnH(z)n

a

?az

H0r

(zpizr，極點(diǎn)i特征方程的根），標(biāo)量系 關(guān)系：z

eTe特征根p自然響應(yīng)的模式

特征根p自然響應(yīng)的模式《信號(hào)與系統(tǒng)》SIGNALSAND jIjIm000010000000《信號(hào)與系統(tǒng)》SIGNALSAND 例DD

DDD23D2解(1)y(kx(k1y(k10.5y(ky(k)y(k1)0.5y(k2)x(k 2 (2)H(z) z 2 1z1 z零點(diǎn)為z10，極點(diǎn)為p1,20.5

0

jImH(z)

z2z z0.5 z0.52 2h(k)j[(0.5j0.5)k(0.5j0.5)k](k) 《信號(hào)與系統(tǒng)》SIGNALSAND

k(k yzs(k)h(n)z h(n)z zh(n)z zH(z)k 了H(z)。條件:z應(yīng)位于H(z)的收斂域內(nèi)。例如：H(z) z

，如激勵(lì)為ejk ze

y(k)ejke

《信號(hào)與系統(tǒng)》SIGNALSAND 例求圖示離散系統(tǒng)的單位函數(shù)響應(yīng)和單位階躍響應(yīng)。XX Yzz14解設(shè)輔助函數(shù)Q(z)Q(z)X(z)2z2Q(z)Y(z)Q(z)消去Q(z)H(z)

Y(z) z2

X(z) z2

3z z z《信號(hào)與系統(tǒng)》SIGNALSAND x(k)(k)X(z)Y(z)H(z)X(z)

zz(z

(