复试真题库信号与系统ch7

1、ch7用离散时间复指数信号表示信号：Z变换(the Laplace Transform)郝晓莉陈后金北京交通大学电子信息工程学院Ch7.1Introduction1、从离散时间傅里叶变换到Z变换傅里叶分析具有清晰的物理意义，但某些信号的傅里叶变换不存在。引入Z变换，从而也可以对这些信号进行分析。 Z变换实质是将信号xn乘以衰减因子r-n 的傅里叶分析。Topics（一）使用Z变换分析信号 The z-Transform（ Z变换） Properties of z-Transform Inversion of z-Transform（二）使用Z变换分析系统（Z变换

2、的性质）（Z反变换） Solving Differential Equations With Initial Conditions（系统响应求解） The Transfer Function （系统函数） Computational Structures（系统结构）Ch7.2 the z-Transform（拉普拉斯变换）Definitions （定义）Regions of Convergence Z plane （Z平面）（收敛域）Zeros and Poles （零极点）双边Z变换1. Definitions双边z变换X (z) =n=-xnz -nxn = 1X (z)zn-1dzz反变

3、换2jcC为X(z) 的收敛域(ROC )中的一闭合曲线符号表示正变换：X(z)=Zxn反变换： xk =Z-1X(z)xnz X (z)或物理意义：将离散信号分解为不同频率复指数esTk的线性组合Regions of Convergence（双边z变换的收敛域）收敛域:z变换存在的条件对任意信号xn，若满足上式，则xn应满足 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)左边序列-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)双边序列Im(z)11- az -111- bz -1X (z) =+Re(z)a z |a|的条件下，序列的Z变换才存在。基本序列的Z变换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变换的主要性质）重点看以下几个：1.线性(linearity)ax1n + bx2n aX1(z) + bX2 (z)properties ofz-Transform（Z变换的主要性质）2时移（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变换的主要性质）Example：X(z)=1/(z-a)|z| a ， 求 xn。11- az -1X (z) = z -1xn = an-1un -1properties ofz-Transform（Z变换的主要性质）3. 序列卷积 (convolution)x1n* x2n X1(z)X2 (z) X (z)1- z -1nExample ：Z xk = Zxn* un =

7、k =0properties ofz-Transform（Z变换的主要性质）4.序列指数加权（multiplication by exponential sequence）xnZ X (z)an xnZ X (z / a)properties ofz-Transform（Z变换的主要性质）5.序列线性加权-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 反变换)xn = 1X (z)zn-1dz2jcC为F(z) 的ROC中的一闭合曲线。计算方法：幂级数展开和长除法部分分式展开留数计算法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 当MN时存在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 反变换)将真分式分解为部分分式之和，然后求解各部分分式对应的z反变换zaanun 11- az-1Inversion ofZ - Transform将真分式分解为部分分式之和，然后求解各部分分式对应的z反变换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)讨论不同收敛域H (z) = - 2+3solution：1- 2z -11- 3z -1(1) |z|3 ，H1(z)和 H2(z)均对应右边序列hn = (-2n+1 + 3n+1)un(2) 2|z|3，H1(z)对应右边序列， H2(z) 对应左边序列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（系统函数）yzsn=xn*hnYzs(z)=X(z)H(z)xnX(z)系统函数:系统在零状态条件下，输出的Z变换与输入的Z变换之比，记为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：对方程两边做Z变换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：系统函数为z-111H (z) =1+ 5z-1 + 6z-2=1+ 2z -1-1+ 3z -1系统的脉冲响应hn = (-2)n - (-3)nunCausality and Stability（因果性与稳定性）离散时间LTI系统BIBO稳定的充分必要条件是因果系统的稳定条件系统函数H(z)的全部极点位于的 z平面单位圆内。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（稳定）。极点z= 0.2，在z平面单位圆内;极点z= 0.5， 在z平面单位圆内。7.8 the unilateral z-transform (单边Z变换)1、单边Z变换：X (z) = n=0xnz -nTime Shift(单边Z变换的时移特性)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(单边Z变换的时移特性)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变换分析系统响应）解差分方

16、程 时域差分方程 时域响应yn反变换变换 Z域代数方程 Z域响应Y（z）解代数方程ZZSolving Differential Equations with InitialConditions（利用Z变换分析系统响应）yn + a1 yn -1 + a2 yn - 2 = b0 xn + b1xn -1初始状态为y-1, y-2对差分方程两边做z变换，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变换分析系统响应）- 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）求系统响应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 （单位脉冲响应）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系统的方框图表示）延迟（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