电子科技大学电子工程学院：《信号与系统》课程教学资源（PPT课件讲稿，英文版）Chapter 2 Linear Time-invariant Systems

Chapter 2 Linear Time-invariant Systems

1 Chapter 2 Linear Time-invariant Systems

Chapter 2 LTI SyStems Example 1 an LtI system fl y 2 t 012t ()=(0)-f(-2) y2t)=y()-y1(t-2) L 1 0 t

2 Chapter 2 LTI Systems Example 1 an LTI system 0 2 t f (t) 1 1 ( ) ( 2) = f 1 t − f 1 t − 0 1 2 t y (t) 1 L 1 ( ) ( 2) = y1 t − y1 t − 0 2 4 t 1 -1 0 2 4 t f (t) 2 1 L y (t) 2 −1

Chapter 2 LTI SyStems 52.1 Discrete-time LTI Systems The Convolution Sum 卷积和) 52.1.1 The Representation of Discrete- Time Signals in Terms of impulses Example 2 xn 1012 xkFn-k ...+x 15a+1]+x0]]+x[]n-1 +∞ x=∑xkn-k k=-00

3 Chapter 2 LTI Systems §2.1 Discrete-time LTI Systems : The Convolution Sum （卷积和） §2.1.1 The Representation of Discrete-Time Signals in Terms of impulses xn=+ x−1 n+1+ x0 n+ x1n−1+ xn xk n k k =  − + =−  xk n− k Example 2 −1 0 1 2 1  2  3 xn n

Chapter 2 LTI SyStems 52.1.2 The Discrete- Time Unit Impulse Responses and the Convolution-Sum Representation of LtI Systems 1. The Unit Impulse Responses 单位脉冲响应 L0,6|n 2. Convolution-Sum(卷积和) y]=∑xkhm-k]=xh k k时刻的脉冲在n时刻的响应 系统在n时刻的输出包含所有时刻输入脉冲的影响 4

4 Chapter 2 LTI Systems §2.1.2 The Discrete-Time Unit Impulse Responses and the Convolution-Sum Representation of LTI Systems 1. The Unit Impulse Responses 单位脉冲响应 h n L n   =  0 ,   2. Convolution-Sum （卷积和） yn xkhn k k =  − + =− 系统在n时刻的输出包含所有时刻输入脉冲的影响 k时刻的脉冲在n时刻的响应 = xnhn

Chapter 2 LTI SyStems 3.卷积和的计算 ①利用定义计算 例23x团]=a"以=x=? ②图解法 Example 2.4 1,0<n<4 a",0≤n≤6 h 0<a<1 0. otherwise 0. othe rwise Determine the output signal yl 5

5 Chapter 2 LTI Systems 3. 卷积和的计算 ① 利用定义计算 例2.3 xn a un n = hn= un xnhn= ? ② 图解法 Example 2.4         = 0 , otherwise 1 , 0 n 4 x n         = 0 , otherwise a , 0 6 n n h n 0  a 1 Determine the output signal yn

Chapter 2 LTI SyStems Summarizing, we obtain 0 n10 Ly=11 LX=5 Lh=7 LV=LX+Lh-I 6

6 Chapter 2 LTI Systems Summarizing ,we obtain yn= 0 0 4  n  6 n  0 0  n  4 a a n − − + 1 1 1 a a a n n − − − + 1 4 1 a a a n − − − 1 4 7 6  n 10 n 10 Ly=11 Lx=5 Lh=7 Ly=Lx+Lh-1

Chapter 2 LTI SyStems ③不带进位的普通乘法 适用于因果序列或有限长度序列之间的卷积 Example 3 x四]}={2,15}n=01,2 ={3,1,42}n=0,2,3 Determine yn=xn * hn ④多项式算法(适用于有限长度序列) 利用多项式算法求卷积和的逆运算 7

7 Chapter 2 LTI Systems ③ 不带进位的普通乘法 ——适用于因果序列或有限长度序列之间的卷积 Example 3 xn= 2,1,5 hn= 3,1,4,2 n = 0,1,2 n = 0,1,2,3 Determine yn= xnhn ④ 多项式算法（适用于有限长度序列） 利用多项式算法求卷积和的逆运算

Chapter 2 LTI SyStems 52.2 Continuous-Time LTI Systems: The Convolution Integral (卷积积分) 52.2.1 The Representation of Continuous-Time Signals in Terms of impulses x(t=x()5(t-id Sifting Property 52. 2.2 The Continuous-Time Unit Impulse Response and the Convolution Integral Representation of LtI Systems y()=x()*h() m x(h(t-r)dr

8 Chapter 2 LTI Systems §2.2 Continuous-Time LTI Systems : The Convolution Integral （卷积积分） §2.2.1 The Representation of Continuous-Time Signals in Terms of impulses x(t) = x( ) (t − )d  + − ——Sifting Property §2.2.2 The Continuous-Time Unit Impulse Response and the Convolution Integral Representation of LTI Systems y t x t h t x h t d ( ) ( ) ( ) (   ) ( ) + − =  = − 

Chapter 2 LTI SyStems 523卷积的计算 1.由定义计算卷积积分 例2.6x()=e"(ka>0h()=u()>y() 2.图解法 例27求下列两信号的卷积 1,0<t<T 0<t<2T x()= h()= 0,其余t 0,其余t 3.利用卷积积分的运算性质求解

9 Chapter 2 LTI Systems §2.3 卷积的计算 1. 由定义计算卷积积分 例2.6 ( ) = ( ),  0 − x t e u t a at h(t)= u(t) y(t) 2. 图解法 例2.7 求下列两信号的卷积 x(t) = 1, 0  t T 0, 其余t h(t) = t, 0  t  2T 0, 其余t 3. 利用卷积积分的运算性质求解

Chapter 2 LTI SyStems 52.3 Properties of LTI Systems x()厂()y()=x(0)*h() x*风n LT系统的特性可由单位冲激响应完全描述 Example 2.9 1n=0,1 h 0 otherwise ① LTI SySten ② Nonlinear System ③ Time-variant System (a)]=(x[]+x[n-1)2(a)y()=cos(3)x(t) 巩l=max(pun -1)(b)y()=ex()

10 Chapter 2 LTI Systems §2.3 Properties of LTI Systems x(t) h(t) y(t) = x(t)h(t) hn yn= xnhn xn LTI系统的特性可由单位冲激响应完全描述 Example 2.9       = 0 1 h n otherwise n = 0 ,1 ① LTI system ② Nonlinear System ( )   (    ) 2 a y n = x n + x n −1 (b) yn= max(xn, xn−1) ③ Time-variant System (a cos 3 ) y t t x t ( ) = ( ) ( ) (b ) ( ) ( ) t y t e x t =