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Ch6.1 引言( Introduction )

Ch6 Representation Signals by Using Continuous-Time Complex Exponentials: The Laplace Transform (用连续时间复指数信号表示信号:拉普拉斯变换). Ch6.1 引言( Introduction ). (一)使用拉普拉斯变换分析信号 The Laplace Transform (拉普拉斯变换) Properties of Laplace Transform (拉普拉斯变换的性质) Inversion of the Laplace Transform (拉普拉斯反变换)

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Ch6.1 引言( Introduction )

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  1. Ch6 Representation Signals by Using Continuous-Time Complex Exponentials: The Laplace Transform(用连续时间复指数信号表示信号:拉普拉斯变换)

  2. Ch6.1 引言(Introduction) • (一)使用拉普拉斯变换分析信号 • The Laplace Transform (拉普拉斯变换) • Properties of Laplace Transform (拉普拉斯变换的性质) • Inversion of the Laplace Transform (拉普拉斯反变换) • (二)使用拉普拉斯变换分析系统 • Solving Differential Equations With Initial Conditions • (系统响应求解) • The Transfer Function (系统函数)

  3. From Fourier transform to Laplace transform从傅里叶变换到拉普拉斯变换 x(t)=eatu(t)a >0的傅里叶变换? 不存在! 将 x(t) 乘以衰减因子e-t 若

  4. From Fourier transform to Laplace transform 从傅里叶变换到拉普拉斯变换 推广到一般情况 令s=+j 拉普拉斯正变换 定义: 对 x(t)e-t求傅里叶反变换可得 拉普拉斯反变换

  5. The Laplace transform applies to more general signals than the Fourier transform does. (a) Signal for which the Fourier transform does not exist. (b) Attenuating factor associated with Laplace transform. (c) The modified signal x(t)e-t is absolutely integrable for  > 1.

  6. Real and imaginary parts of the complex exponential est, where s =  + j.

  7. Ch6.2 the Laplace Transform(拉普拉斯变换) • Definitions (定义) • Regions of Convergence (收敛域) • S plane (S平面) • Zeros and Poles (零极点)

  8. Bilateral Laplace Transform(双边拉普拉斯变换) 信号x(t)可分解成复指数est的加权叠加,权重正比于X(s) 。 符号表示: s是复数称为复频率,F(s)称复频谱。

  9. Regions of Convergence (双边拉普拉斯变换的收敛域) 收敛域: 双边拉普拉斯变换存在的条件 对任意信号x(t) ,若满足上式,则 x(t)应满足 Regions of Convergence (ROC):使上式成立的所有值。

  10. j 收 敛 区 0  Regions of Convergence(收敛域) Ex: x (t) = et u(t) 的傅里叶变换?拉氏变换? S平面 左半平面 右半平面 X(s)存在 X(s)不存在

  11. Re(s)>-a Re(s)< -a 不同的信号,虽然具有相同形式的拉氏变换,但收敛域不同。

  12. The ROC for x(t) = eatu(t) is depicted by the shaded region. A pole is located at s = a.

  13. The ROC for y(t) = –eatu(–t) is depicted by the shaded region. A pole is located at s = a.

  14. 拉普拉斯变换与傅里叶变换的关系 1)当收敛域包含j轴时,拉普拉斯变换和傅里叶变换均存在。 2)当收敛域不包含j轴时,拉普拉斯变换存在而傅里叶变换均不存在。

  15. Zeros and Poles(零极点) Zeros poles

  16. MATLAB code: b=[1];a=[1 3]; Splane(b,a); w=linspace(0,5,256); subplot(2,1,1); splane(b,a); h=freqs(b,a,w); subplot(2,1,2); plot(w,abs(h));

  17. Ch6.3 The Unilateral Laplace Transform(单边拉普拉斯) Definitions:

  18. 基本信号的拉普拉斯变换

  19. 基本信号的拉普拉斯变换

  20. 基本信号的拉普拉斯变换

  21. 基本信号的拉普拉斯变换

  22. Ch6.4 properties of Laplace Transform(单边拉普拉斯的性质) 1. Linearity(线性特性) 2. scaling (展缩特性)

  23. 3. Time Shift(时移特性) if then 4. s-domain shift(指数加权性质---s域位移) Ex:

  24. 5. convolution(卷积特性) Example:Find the unitial Laplace transform of y(t)=e-3 tu(t)*e-tu(t)

  25. 6. Differentiation in the S-domain(S域微分特性) Ex:

  26. 7.Differentiation in the time-domain(时域微分特性) 证明:

  27. 8. Integration Property(积分特性) Where, Ex:

  28. 9. Initial- and Final-Value Theorems(初值定理和终值定理)

  29. Ch6.5 Inversion of Unilateral Laplase transform (拉普拉斯反变换) 部分分式法求拉普拉斯反变换:由X(s)求x(t) 当MN时存在 真分式

  30. Inversion by Partial-Fraction Expansion(部分分式法求拉普拉斯反变换) 设 利用部分分式法: (1) A(s)=0有N个单极点 其中:

  31. Ex: Find the inverse Laplase transform of Solution: num=[1 2]; den=[1 4 3 0];[r,p]=residue(num,den) r = -1/6 -1/2 2/3 p = -3 -1 0

  32. Inversion by Partial-Fraction Expansion(部分分式法求拉普拉斯反变换) (2) A(s)=0有N重极点 其中: 则:

  33. Ex: Find the inverse Laplase transform of Solution: X(s)有1个 3阶重极点 [r,p]=residue([1 -2],[1 3 3 1 0]); r =2 2 3 -2 p = -1 -1 -1 0

  34. Ex: Find the inverse Laplase transform of Solution: • X(s)为有理假分式,将X(s)化为有理真分式 [r,p,k]=residue([1 0 2 -4],[1 4 -2])

  35. Inversion by Partial-Fraction Expansion(部分分式法求拉普拉斯反变换) (3) A(s)=0有复极点 若x0(t)为实信号,则A1与A2为共轭以保证 为实系数。

  36. Ex: Find the inverse Laplase transform of Solution:

  37. Ex: Find the inverse Laplase transform of Solution: 令s2=q,

  38. Ex: Find the inverse Laplase transform of Solution: k2, k3用待定 系数法求

  39. Ch6.6 Solving Diffrential Equations with Initial Conditions(利用拉普拉斯变换分析系统响应) 解微分方程 时域差分方程 时域响应y(t) 拉氏反变换 拉氏变换 S域代数方程 S域响应Y(s) 解代数方程

  40. Solving Diffrential Equations with Initial Conditions(利用拉普拉斯变换分析系统响应) 已知 x (t),y(0-),y'(0-) ,求y(t)。 求解步骤: 1) 经拉氏变换将域微分方程变换为域代数方程 2)求解s域代数方程,求出Yzi(s), Yzs (s) 3)拉氏反变换,求出响应的时域表示式

  41. Solving Diffrential Equations with Initial Conditions(利用拉普拉斯变换分析系统响应) a2y(t) y"(t) a1y'(t) Yzi(s) Yzs(s)

  42. Ex: Use the unilateral Laplace transform to determine the output of a system represented by the differential equation y''(t) + 5y'(t) + 6y(t) = 2x '(t) + 8x(t) in response to input x(t) = e-tu(t).Assume that the initial conditions on the system are y(0-)=3 and y'(0-)=2. Solution:Using the differential property and taking the unilateral Laplace transform of both sides of the differential equation, we obtain (对微分方程取拉氏变换可得)

  43. Ex: Use the unilateral Laplace transform to determine the output of a system represented by the differential equation y''(t) + 5y'(t) + 6y(t) = 2x '(t) + 8x(t) in response to input x(t) = e-tu(t).Assume that the initial conditions on the system are y(0-)=3 and y'(0-)=2. Solution:

  44. Ch6.7 Laplace Transform Methods in Circuit Analysis(利用拉普拉斯变换分析电路) Laplace Transform Circuit Models(电路的S域模型): 时域 复频域

  45. Laplace Transform Circuit Models(电路的S域模型) R、L、C串联形式的S域模型

  46. Ex: Use the Laplace transform circuit models to determing the voltage vc(t) in the circuit for an applied voltage EU(t). The voltage across the capacitor vc(0-)= -E. Solution:建立电路的s域模型,由s域模型写回路方程 求出回路电流 电容电压为

  47. h(t) H(s) Ch6.11 The Transfer Function(系统函数) yzs(t)=x(t)*h(t) x(t) X(s) Yzs(s)=X(s)H(s) 系统函数:系统在零状态条件下,输出的拉氏变换式 与输入的拉式变换式之比,记为H(s)。

  48. The Transfer function and Differential Equation(系统函数与微分方程) Ex: Find the transfer function of the LTI system descriped by the differential equation y''(t) + 7y'(t)+10y(t)= 2x '(t)+ x(t)

  49. Causality and Stability(因果性与稳定性) The relationship between the locations of poles and the impulse response in a causal system. (a) A pole in the left half of the s-plane corresponds to an exponentially decaying impulse response. (b) A pole in the right half of the s-plane corresponds to an exponentially increasing impulse response. The system is unstable in this case.

  50. Causality and Stability(因果性与稳定性) The relationship between the locations of poles and the impulse response in a stable system. (a) A pole in the left half of the s-plane corresponds to a right-sided impulse response. (b) A pole in the right half of the s-plane corresponds to an left-sided impulse response. In this case, the system is noncausal.

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