華中科大信號與系統(tǒng)chap2-講稿-2-r3

1、©HUST 2009 (1/35M. Zhang2009年3月27日張敏明School of Optoelectronics Science & Engineering ( 第二講第二章連續(xù)時間系統(tǒng)的時域分析©HUST 2009 (2/35M. ZhangSchool of Optoelectronics Science & Engineering 2.3 沖激響應與階躍響應©HUST 2009 (3/35M. ZhangSchool of Optoelectronics Science & Engineering 1. 沖激響應(Impu

2、lse response 。(h t (t 求解沖激響應,需要使用沖激函數(shù)匹配法。在作用下,等式右端將出現(xiàn)及其導數(shù)項(t (t (1011(1011(n n n n m mm m C r t C r t C r t C r t E t E t E t E t +=+©HUST 2009 (4/35M. ZhangSchool of Optoelectronics Science & Engineering 0t +強迫響應項為零,因而沖激的作用是決定初始條件即改變0+初始條件,而初始條件決定齊次解系數(shù)A 例2-9. 對例2-5所示電路i (t 求對的沖激響應(e t t =解

3、:(7(10(6(4(6(4(e t t i t i t i t e t e t e t t t t =+=+2512(t th i t h t A e A e =+齊次解:0t +不含0t =©HUST 2009 (5/35M. ZhangSchool of Optoelectronics Science & Engineering 令:(h t a t b t c t d u t =+則有:(h t a t b t c u t h t a t b u t =+=+代入:(7(10(6(4(a tb tc td u t a t b t c u t a t b u t t t

4、 t +=+解得:1, 1, 1, 3a b c d =( (0(01h t a t b t c u t h h c+=+=( (0(01h t a t b u t h h b +=+=1243, 13A A =©HUST 2009 (6/35M. ZhangSchool of Optoelectronics Science & Engineering 相互關系?+2541(33t t h i t h t e e =+(1(h t a t b u t t u t =+=(2(00t +<<在時并不為零(即含有項,而齊次解只是(h t 0t =(t 時的解,因此其非

5、零取值范圍為0t +0t 2541(33t t h t t e e u t =+©HUST 2009 (7/35M. ZhangSchool of Optoelectronics Science & Engineering 關于h (t 中沖激項的說明(h t (t (1(,(,.,(m n m n t t t 第四章將給出理論分析©HUST 2009 (8/35M. ZhangSchool of Optoelectronics Science & Engineering (t u dt d t =(t u H dt dt u dt dH t H =即:(t

6、 g dt dt h =。(g t (u t 求法與類似,但可能有強迫響應分量。(g t (h t©HUST 2009 (9/35M. Zhang School of Optoelectronics Science & Engineering 2.4 卷積*(The Convolution Integral©HUST 2009 (10/35M. ZhangSchool of Optoelectronics Science & Engineering 本節(jié)的主要內容(1 卷積的概念(2 卷積計算的三種方法(3 卷積的性質a. 圖形計算法b. 解析計算法c. 微

7、積分計算法©HUST 2009 (11/35M. ZhangSchool of Optoelectronics Science & Engineering 1. 卷積的定義數(shù)學定義任意兩個信號與的卷積運算定義為:(1t f (2t f (2121t f t f d t f f t f =(2-41 (p.62因為:(*(*(1221t f t f t f t f =1212(f f t d f t f d =故有:©HUST 2009 (12/35M. ZhangSchool of Optoelectronics Science & Engineering

8、零狀態(tài)響應求解的方法與求是一樣的(齊次解(t r zi (t r h (t r t r t r zsp zsh zs +=求解的經典法是(t r zs 求解的現(xiàn)代法:(zs r t (zs r t e t h t =©HUST 2009 (13/35M. ZhangSchool of Optoelectronics Science & Engineering =H (t (t h t H t r = H (t (=t h t r H (t e (t e H t r =(H e t d e H t d =(e h t d e t h t =1. 信號的分解(1-58 p.252

9、. 沖激信號抽樣特性+偶函數(shù)特性是關于t 的函數(shù)©HUST 2009 (14/35M. ZhangSchool of Optoelectronics Science & Engineering =這個面積是關于t 的函數(shù),且通常是一個分段函數(shù)。3. 卷積的圖形計算法(1特點:直觀、將抽象關系形象化,繁瑣。例:求(*(t h t e t r =12(t h t (*(r t e t h t e h t d =從卷積的定義出發(fā):解:2111(t e t 目的:加深對卷積積分過程及其物理意義的理解©HUST 2009 (15/35M. ZhangSchool of Opt

10、oelectronics Science & Engineering 3. 卷積的圖形計算法(2Step 1.的圖形;的關于自變量和畫出函數(shù) ( ( t h e 改換圖形橫坐標,由t 改為,為自變量;選定其中的一個信號進行反褶,這里是:;(h 把反褶后的信號做位移,位移量是t ,得到;(t h 1(t h t注意t > 0 圖形右移©HUST 2009 (16/35M. ZhangSchool of Optoelectronics Science & Engineering Step 2.積分。相乘并關于和函數(shù) ( ( t h e 3. 卷積的圖形計算法(3將從

11、到移動,對移動過程中的重疊部分相乘,(t h =t 并對相乘后圖形進行積分。(a21<t 1(t h t 211(e 0(=d t h e t r©HUST 2009 (17/35M. ZhangSchool of Optoelectronics Science & Engineering 3. 卷積的圖形計算法(4(b121t (t h t 211(e (16144 211 121( (2t 21-+=×=tt d t t d t h e t r (c231t (t h t 211(e (16343 211 231( (121-=×=td t t

12、d t he t r t ,21 積分區(qū)間1 ,21 積分區(qū)間©HUST 2009 (18/35M. ZhangSchool of Optoelectronics Science & Engineering 3. 卷積的圖形計算法(5(d323t (t h t 211(e (4324 211 323( (212-t +=×=tt d t t d t h e t r 1 ,2 t 積分區(qū)間(e3t 1(t h t211(e 0(=d t h e t r©HUST 2009 (19/35M. ZhangSchool of Optoelectronics Sci

13、ence & Engineering 3. 卷積的圖形計算法(6做出的圖形。(請參考圖2-14, p.64(t r 卷積圖形計算法的關鍵點/難點©HUST 2009 (20/35M. ZhangSchool of Optoelectronics Science & Engineering 3. 卷積的圖形計算法(7221(t f (t h t t te *0011(t e t -111(t h t 00*5分鐘動筆思考時間補充例:用圖形計算法求卷積©HUST 2009 (21/35M. ZhangSchool of Optoelectronics Scien

14、ce & Engineering 01( ,1 t t 積分區(qū)間111 ( 1+=×=t d t r t10( ,1 t t t 積分區(qū)間111 ( 1=×=tt d t r 11-1(t h (e t t -1( =t r 無重疊區(qū)間(a11-10(t h (e 11-1(t h (e (c©HUST 2009 (22/35M. ZhangSchool of Optoelectronics Science & Engineering21( 1 ,1 t t 積分區(qū)間td t r t =×=211 (11t0(t r 11-1211-1(

15、t h (e 11-1(t h (e 0( =t r 無重疊區(qū)間(e(a(b(c(d(e©HUST 2009 (23/35M. ZhangSchool of Optoelectronics Science & Engineering4. 卷積的性質(1. 卷積代數(shù)(1 交換律(*(*(1221t f t f t f t f =(2 分配律(*(*(*(3121321t f t f t f t f t f t f t f +=+(*(21t h t h t e t r +=( 21t h t h t h +=并聯(lián)系統(tǒng)的(1t h (2t h (t e (t h ©HU

16、ST 2009 (24/35M. ZhangSchool of Optoelectronics Science & Engineering (3 結合律(*(*(*(*(321321t f t f t f t f t f t f =(*(*(21t h t h t e t r =(*( 21t h t h t h =串連系統(tǒng)的(1t h (2t h (t e (t h©HUST 2009 (25/35M. ZhangSchool of Optoelectronics Science & Engineering5. 卷積的解析計算法(1若待卷積的兩個信號能用解析函數(shù)式表

17、達,則可直接按照卷積的積分表達式,利用卷積性質進行計算。故有:(+=22121*1(21(t u t t u t t u t u t r 211112(t e (t h tt*解:(121(+=t u t u t e (221(=t u t u t t h ©HUST 2009 (26/35M. ZhangSchool of Optoelectronics Science & Engineering化簡得:(+=221*1(.21*21(t u t t u t u t t u t r = + + + 21t -1(t h (e d t u u 1(2(21=3(43243(

18、112121 2 =t u t t t u d t 上式1(*2(212(21*1(=t u t u t t u t t u©HUST 2009 (27/35M. ZhangSchool of Optoelectronics Science & Engineering = + + + (t r =+=. . . . 3(43241(4124 23(16154421(161442222t u t t t u t t t u t t t u t t (分區(qū)間化簡©HUST 2009 (28/35M. ZhangSchool of Optoelectronics Scie

19、nce & Engineering t10(t u (u 0( (t e h t d u t =©HUST 2009 (29/35M. ZhangSchool of Optoelectronics Science & Engineering =11(e tf t u t t =22(h t f t u t t =2(u t t 1(u t 1t 2t t 0©HUST 2009 (30/35M. ZhangSchool of Optoelectronics Science & Engineering (sin (,(,2(t x t t u t u

20、t h t e u t y t x t h t =補充例題:設:求:解:sin (2tt u t u t e u t sin (sin (2t t tu t e u t tu t e u t =(02sin (sin (2ttt t ed u te d u t =02sin (sin (2t t t e e d u t e d u t = sin e dt = sin (de 0 0 t t = sin e t 0 t 0 e cos d 0 t = sin e t t 0 cos e t 0 e sin d 0 t 0 t sin e d = 2 (sin e 0 1 cos e y (t

21、= t t t t e t et sin e cos e u (t sin e cos e u (t 0 0 2 2 2 2 2 1 1 = sin t cos t + e t u (t sin t e 2 e t cos t u (t 2 2 2 t<0 0 t = (sin t cos t + e 2 0 t < 2 e t (1 + e 2 2 t 2 HUST 2009 (31/35 School of Optoelectronics Science & Engineering M. Zhang 例:如圖所示系統(tǒng),試求當 e(t = u (t 時, 系統(tǒng)的零狀態(tài)響應. e (t 2 x ( t 3 x ( t 2 x (t r (t 解:設輔助函數(shù) x(t ,使得: x(t = e(t 3 x(t 2 x(t 5分鐘動筆思考時間 r (t = 2 x(t + x(t 模擬/方框圖對應的微分方程為: r (t + 3r (t + 2r (t = 2e(t + e(t HUST 2009 (32/35 School of Optoelectronics Science & Engineering M.