復(fù)試真題庫信號與系統(tǒng)ch7

1、ch7用離散時間復(fù)指數(shù)信號表示信號：Z變換(the Laplace Transform)郝曉莉陳后金北京交通大學(xué)電子信息工程學(xué)院Ch7.1Introduction1、從離散時間傅里葉變換到Z變換傅里葉分析具有清晰的物理意義，但某些信號的傅里葉變換不存在。引入Z變換，從而也可以對這些信號進行分析。 Z變換實質(zhì)是將信號xn乘以衰減因子r-n 的傅里葉分析。Topics（一）使用Z變換分析信號 The z-Transform（ Z變換） Properties of z-Transform Inversion of z-Transform（二）使用Z變換分析系統(tǒng)（Z變換

2、的性質(zhì)）（Z反變換） Solving Differential Equations With Initial Conditions（系統(tǒng)響應(yīng)求解） The Transfer Function （系統(tǒng)函數(shù)） Computational Structures（系統(tǒng)結(jié)構(gòu)）Ch7.2 the z-Transform（拉普拉斯變換）Definitions （定義）Regions of Convergence Z plane （Z平面）（收斂域）Zeros and Poles （零極點）雙邊Z變換1. Definitions雙邊z變換X (z) =n=-xnz -nxn = 1X (z)zn-1dzz反變

3、換2jcC為X(z) 的收斂域(ROC )中的一閉合曲線符號表示正變換：X(z)=Zxn反變換： xk =Z-1X(z)xnz X (z)或物理意義：將離散信號分解為不同頻率復(fù)指數(shù)esTk的線性組合Regions of Convergence（雙邊z變換的收斂域）收斂域:z變換存在的條件對任意信號xn，若滿足上式，則xn應(yīng)滿足 lim xnr-n = 0nRegions of Convergence （ROC）:使上式成立的所有r值。xnr -n aROC1- az-1(1)x n = anun;(2)x n = -anu-n -1;Example:12(3)x n = anun - bnu-

4、n -13Determine X1(z), X2(z) and X3(z)xn = -bnu-n -1Solutions:(2)左邊序列-1X (z) =n=- anu-n -1z -n =n=- an z-nIm(z)= -(az -1)-n= 1- (az -1)nRe(z)n=1n=011= 1-=z aROC :1- a-1z1- az-1(1)x n = anun;(2)x n = -anu-n -1;Example:12(3)x n = anun - bnu-n -13Determine X1(z), X2(z) and X3(z)xn = anun - bnu-n -1Solut

5、ions:(3)雙邊序列Im(z)11- az -111- bz -1X (z) =+Re(z)a z |a|的條件下，序列的Z變換才存在?；拘蛄械腪變換Zd n = 1,Za nun =z 01)11-a z-1a2)z1un3)1- z -1z-1nun(1- z-1)24)az-1naunn-(1- az)12Ch7.4 properties ofz-Transform（Z變換的主要性質(zhì)）重點看以下幾個：1.線性(linearity)ax1n + bx2n aX1(z) + bX2 (z)properties ofz-Transform（Z變換的主要性質(zhì)）2時移（Time Shift）

6、-nxn - n = zX (z)00Example ：xn = un - un - 5z-51- z-51X (z) =1- z-1-1- z-1=-11- zz 0properties ofz-Transform（Z變換的主要性質(zhì)）Example：X(z)=1/(z-a)|z| a ， 求 xn。11- az -1X (z) = z -1xn = an-1un -1properties ofz-Transform（Z變換的主要性質(zhì)）3. 序列卷積 (convolution)x1n* x2n X1(z)X2 (z) X (z)1- z -1nExample ：Z xk = Zxn* un =

7、k =0properties ofz-Transform（Z變換的主要性質(zhì)）4.序列指數(shù)加權(quán)（multiplication by exponential sequence）xnZ X (z)an xnZ X (z / a)properties ofz-Transform（Z變換的主要性質(zhì)）5.序列線性加權(quán)-Z域微分 （differentiation in the z-domain）nxn -z dX (z)dz1Example ：un 1- z -1(-1)z-2d1nun - z-1 ) = -z(-11- z(1- z)12dz=(1- z-1)2Ch7.5 Inversion ofZ -

8、 Transform(Z 反變換)xn = 1X (z)zn-1dz2jcC為F(z) 的ROC中的一閉合曲線。計算方法：冪級數(shù)展開和長除法部分分式展開留數(shù)計算法Ch3.13.2 Inverse Discrete-Time Fourier Transform（由X(ejW)求x(n)）z- M+L+ b z-1 + bbX (z) = M10 z- N+L+ a z-1 + aaN10+ B(z)= c+ c z-1 +L+ cz-( M - N )M - N01A(z) 真分式 當MN時存在d (n - (M - N ) + Z -1 B(z) x(n) = c d (n) + c d (n

9、 -1) +L+ c A(z) M - N01Ch7.5 Inversion ofZ - Transform(Z 反變換)將真分式分解為部分分式之和，然后求解各部分分式對應(yīng)的z反變換zaanun 11- az-1Inversion ofZ - Transform將真分式分解為部分分式之和，然后求解各部分分式對應(yīng)的z反變換za(n +1)anun 1(1- az -1)21Ex： H (z) =, determinehn(1- 2z-1)(1- 3z-1)ABH (z) =+solution：1 - 2z -11 - 3z -1A = (1- 2z-1)H (z)= -2z-1 = 12B =

10、(1- 3z-1)H (z)= 3z -1 =13- 23H (z) =+1 - 2z -11 - 3z -11Ex： H (z) =, determinehn(1- 2z-1)(1- 3z-1)H1(z)H2(z)討論不同收斂域H (z) = - 2+3solution：1- 2z -11- 3z -1(1) |z|3 ，H1(z)和 H2(z)均對應(yīng)右邊序列hn = (-2n+1 + 3n+1)un(2) 2|z|3，H1(z)對應(yīng)右邊序列， H2(z) 對應(yīng)左邊序列hn = -2n+1un - 3n+1u-n -1(3) |z| 4 ,zDetermine xn(1 - 2z -1 )2

11、 (1 - 4z -1 )A1- 2z-1B(1- 2z-1)2C1- 4z-1solution：X (z) =+C = (1- 4z-1)X (z)= 4z=4B = (1- 2z-1)2 X (z)= -1z=21d X (z)(1- 2z-1)2A = -2z =2(-2) dz -1xn = (-2 2n - (n +1)2n + 4 4n )unCh7.6 The Transfer Function（系統(tǒng)函數(shù)）yzsn=xn*hnYzs(z)=X(z)H(z)xnX(z)系統(tǒng)函數(shù):系統(tǒng)在零狀態(tài)條件下，輸出的Z變換與輸入的Z變換之比，記為H(z)。H (z) = Z yzs (z) =

12、 Yzs (z)Zx(z)X (z)hnH(z)Ex: Find the transfer function of the LTI systemdescriped by the differential equationyn - 0.7 yn -1 + 0.1yn - 2 = xn -1,n 0Solution：對方程兩邊做Z變換ynzY (z)yn -1z z-1Y (z)yn - 2z z-2Y (z)xn -1z z-1X (z)Y (z)1- 0.7z-1 + 0.1z-2 = z-1X (z)H (z) =1- 0.7z-1 + 0.1z-2z-1Ex: Find theimpuls

13、e response of the LTI systemdescriped by the differential equationyn + 5 yn -1 + 6 yn - 2 = xn -1,n 0Solution：系統(tǒng)函數(shù)為z-111H (z) =1+ 5z-1 + 6z-2=1+ 2z -1-1+ 3z -1系統(tǒng)的脈沖響應(yīng)hn = (-2)n - (-3)nunCausality and Stability（因果性與穩(wěn)定性）離散時間LTI系統(tǒng)BIBO穩(wěn)定的充分必要條件是因果系統(tǒng)的穩(wěn)定條件系統(tǒng)函數(shù)H(z)的全部極點位于的 z平面單位圓內(nèi)。hn n=-Ex: A causal system

14、 has the transfer functionH (z) =1- 0.7z-1 + 0.1z-22 + z-1Find the impulse response. Is the system stable?Solution ：The system is stable（穩(wěn)定）。極點z= 0.2，在z平面單位圓內(nèi);極點z= 0.5， 在z平面單位圓內(nèi)。7.8 the unilateral z-transform (單邊Z變換)1、單邊Z變換：X (z) = n=0xnz -nTime Shift(單邊Z變換的時移特性)x(n)x(n -1)x(n - 2)nnn000Zxn -1un = n

15、=0= x(-1) + -nxn -1zxn -1z -nn=1 z -1 = x(-1) + z-1X (z)= x(-1) + n=1xn -1z -(n-1)Zxn - 2un = x(-2) + z-1x(-1) + z-2 X (z)Time Shift(單邊Z變換的時移特性)x(n)x(n -1)x(n - 2)nnn000z-1X (z) + x(-1)xn -1zz-2 X (z) + z-1x(-1) + x(-2)xn - 2z2. Solving Differential Equations with Initial Conditions（利用Z變換分析系統(tǒng)響應(yīng)）解差分方

16、程 時域差分方程 時域響應(yīng)yn反變換變換 Z域代數(shù)方程 Z域響應(yīng)Y（z）解代數(shù)方程ZZSolving Differential Equations with InitialConditions（利用Z變換分析系統(tǒng)響應(yīng)）yn + a1 yn -1 + a2 yn - 2 = b0 xn + b1xn -1初始狀態(tài)為y-1, y-2對差分方程兩邊做z變換，利用yn -1z z-1Y (z) + y-1yn - 2z z-2Y (z) + y-1z-1 + y-2n 0Y (z) + a z-1Y (z) + a y-1 + az-2Y (z) + ay-2 + ay-1z-111222= b X

17、(z) + b z-1X (z)01Solving Differential Equations with InitialConditions（利用Z變換分析系統(tǒng)響應(yīng)）- a y-1 - ay-2 - ay-1z-1b+ b z-1Y (z) = 122+ 01X (z)1+ a z-1 + az-21+ a z-1 + az-21212 Yzi(z) Yzs (z)a y-1 + ay-2 + ay-1z-1Yzi (z) = - 1221+ a z-1 + az-212b+ b z -1Yzs (z) = 01X (z)1+ a z -1z-2+ a12yn = Z -1Y(z)(z) +

18、 YzizsExample: Use the unilateral z-transform to determine theoutput of a systemyn - 0.7 yn -1 + 0.1yn - 2 = 7xn - 2xn -1y-1 = -26, y-2 = -202, xn = unn 0Determine (1)yzin, yzsn, (2) hn 。Solution：1）求系統(tǒng)響應(yīng)yn -1z z-1Y (z) + y-1yn - 2z z-2Y (z) + y-1z-1 + y-2Y (z) - 0.7z-1Y (z) + y(-1)+ 0.1z-2Y (z) + y(

19、-2) + y(-1)z-1= 7 X (z) - 2z-1X (z)7 - 2z-1(0.7 - 0.1z-1) y(-1) - 0.1y(-2)1- 0.7z-1 + 0.1z-2Y (z) =1- 0.7z-1 + 0.1z-2X (z) +Example: Use the unilateral z-transform to determine theoutput of a systemyn - 0.7 yn -1 + 0.1yn - 2 = 7xn - 2xn -1y-1 = -26, y-2 = -202, xn = unn 0Determine (1) yn, yzin, yzsn

20、, (2) hn 。7 - 2z-1(0.7 - 0.1z-1) y(-1) - 0.1y(-2)1- 0.7z-1 + 0.1z-2Y (z) =1- 0.7z-1 + 0.1z-2X (z) +- 0.5-10= + 512.512-+1- 0.2z-11- z-1 1- 0.2z-11- 0.5z-1 1- 0.5z -1yzsn = -0.5(0.2)- 5(0.5)+12.5nnn 0yzin = -10(0.2)+12(0.5)nnyn = yzsn + yzinn 0Example: Use the unilateral z-transform to determine theo

21、utput of a systemyn - 0.7 yn -1 + 0.1yn - 2 = 7xn - 2xn -1y-1 = -26, y-2 = -202, xn = unn 0Determine (1) yn, yzin, yzsn ; (2)hn 。Solution：(2)Impulse response （單位脈沖響應(yīng)）7 - 2z-1H (z) =1- 0.7z-1 + 0.1z-2-11+ z -12.51+ 0.5z -1H (z) = -1+hn = -d n + (-1)n+1un + 2.5(-0.5)n unCh7.9 Block Diagram of LTI system（LTI系統(tǒng)的方框圖表示）延遲（time shift）yn -1z z-1Y (z)Example: Draw direct form I and direct form II implementationsof the LTIsystem described by the difference equationyn + a1 yn -1 + a2 yn - 2 = b0 xn + b1xn -1