计算卷积的方法

*计算卷积的方法1.用图解法计算卷积2.用函数式计算卷积3.利用性质计算卷积4.数值解法分段时限卷积积分限方法一.用图解法计算卷积dthfthtfty)()()()()()())(()()(ththhht平移翻转1）将f(t)和h(t)中的自变量由t改为，成为函数的自变量;2）把其中一个信号翻转、平移;3）将f()与h(t)相乘；对乘积后的图形积分。)(tft)(tht)(h)()(thft)()(),()(),(*)(tuethtutfthtft计算)(f)(h01)(*)(0)(tedethtfttt例*.积分限的确定:dedthetrtt0)()()()(方法一:t*0)(th)(e若两个函数的左边界分别为tl1,tl2,右边界分别为tr1,tr2,积分的下限为max[tl1,tl2];积分的上限为min[tr1,tr2].0)()(非零值下限是非零值下限是－uth卷积分下限是零非零值上限是非零值上限是)()(uttht卷积分上限是0.1:t解重合面积为零:f1(t)*f2(t)=010.2tifa0t-21dtffff)()(2121*0a1f1(t)tdtffff)()(2121计算a1tt-20t10t-20t-21tf2(t)t02btttabdtba020)(4)(224tab21.3tifa1tt-20t10t-21021021)(4)(2tabdtbaff)12(4tab32.4tif0t-21tdtbafft)(21221)23(4)(4212-t2ttabtab03.521fftif012tt12300.25ab=42abtab21t32t10t)23(42ttab24tab21ffab43结语:若f1(t)与f2(t)为有限宽度的脉冲,f1*f2的面积为f1和f2面积之积,f1*f2的宽度为f1和f2宽度之和.方法二.利用门函数直接计算卷积分)()(])[(jiijtkttututttG)(jttu1jtt0jtt])[(ijtktttG10jitttitjtt*.表达式的推导0itjtt])[(ijtktttGit)(itu)(ijtttujttti)(itu1ti01.将被卷积的两个函数f(t)和h(t)都表示成单位阶跃u(t)移位加权之和.1.).........()()(1ipiittutftf2)......()()(1jqjjttuthth其中fi(t)和hj(t)分别是f(t)的第i段和h(t)的第j段数学表达式.ti和tj分别是fi(t)和hj(t)的起点.2.将(1)和(2)代入卷积公式:dttuthttufhfjqjjipii)]()()][()([11dttututhtfjijpiqji)]()()[()(11由以上讨论可知:得出卷积积分的上下限和定义域如下:)()()(11jijpiqjtttitttudthfhfji例题:设)()(2tuetht)1(4)()(ttutuetft求系统的z.s.r.解:ttttftudetudeedthfhfty1)(2)(20)1(4)()()()()1()12()()()1(22tutetueettt设系统是因果的,但激励是非因果的,)()(tuetht)()(2tuetft求yf(t).u(-t))()()(tututu)()()(02)(2tudeetudeehfyttttf)(31)(31)(31)]()([3122tuetuetuetutuetttt解:)]1()([1tutuaf)]1()1()[1(22tututbf)()()(12121212jiijttttttudtffffjiti=1;(tj=0,tj=1)ti=-1;(tj=0,tj=1)dtutuuuabff)]1()()][1()1()[1(212111111]}1)][2()1()()1({[2ttttdtutututuab21ff计算01*f1f2-11ab)(f)(fj1i2tftf为为)]2()1()1)(3()()1([4)4()1(222tututttutuabttt1)13()12()3()1()2()1()]3()2([*)1(*2131ttdddtuudtuututututt下式错在哪里？)4()4()3()3()13()12(2131tuttuttudtudtt确的解法为错在忽略了定义域，正错是防止疏忽的好措施。分上限减去下限因子在每一段积分后乘上积一般情况下,,*.*.快速定限表若参与卷积的两个函数fs(t)和fl(t)都是只有一个定义段,它们的时限长度分别为TS和TL,并且TSTL,长函数fl(t)的左右时限分别为LL和RL,而短函数fs(t)的的左右时限分别为LS和RS,并规定积分号内括号统一只表示即只反转时限长的函数.)()(tfflst0lsltld.ssrld.slrrtd.0定义域和卷积结果例:f(t)h(t)4521131.求出关于的不定积分212)()(ddthfls2.将两函数的时限值两两相加,得出定义域1+4=5;1+5=6;3+4=7;3+5=85678005453t14t3.确定积分限62214tt2254tt216253005876hf=26t72t-65t6关键：1.卷积结果各分段时限的确定.2.各分段内卷积积分限的确定。0t516-2t7t80t8hft5867分解成单位阶跃分量之和)0(f)(1tf)(11ttf1t1ttegralDaHarmatgtuln)()(*.Duharmalintegraldttdgtethtetr)()()()()()(*)(tgdttdetdtgtgetre0)()()()()('0)()()()())()(()()()()(ttetudttdedttuteddttdetutete),()(,)()(,tutgtute响应为时当由系统之因果性tdtgtgedtutgeuddegdtdetre0)()()()()()()]()0()()([*)('0题－面，20286.p)(1th)(1th)(3th)(2th)(te)(tr:202:可知－由题图解)()]()([)()(3121ththththth)]([)]()1([)(ttuttu)]([)]1([)(ttutu)1()(tutu).(y,)(f(t)2.h(t)..1)2()1()(),()(h,.zs'21tttututhtut响应求该复合系统的零状态时当系统的输入求该系统的冲激响应分别为两个子系统的冲激响应某复合系统如图所示)(h1t)(h2tcostf(t))(y1t)(yt)(f2t的输出为子系统时当输入解)(h,(t)f(t)1.:1t)()()()()()(y11tututthtft的输入为子系统由图可知)(h,2t)(coscos)()(f12ttuttyt为故复合系统的冲激响应)]2()1([)]([cos)()(fh(t)22tututtutht)2()2sin()1()1sin()]2()1([)]([sin)]2()1([]cos[tuttutttttuttdt的输出为，子系统当系统输入)(h)(f(t).21'tt)()()()()()(y'11zsttutthtft:)(h,2的输入为子系统此时t)(cos)(cos)()(f12tttttytzs:应为故复合系统的零状态响)2()1()]2()1([)()()()(y22zstututututthtft故不能利用性系统由于该复合系统为非线点评,:)()()(yzstftht。计算零状态响应的条件系统、零初态是用卷积。请记住线性时不变来直接计算零状态响应)(,,2,21,,)2()1()]2()()[121()(.tiFCRRCbatuttututte响应电流求且初态为零＝设串连电路中所示的加于图所示如图激励电压)(te11223340ta图＋－)(tib图求系统的冲激响应解:ttdictRi)()(1)()(2)(2)(tuettit＝)(2)(dtdi(t)'tti)(2)(2tt‘)(2t)(2tu)2(21)()121()(:ttututte激励电压化简为tdtheti0)()()(:计算积分dtuetuutt)](2)(2)][2(21)()121[()(0＝dtuuedtuttt)]()()2()()()2[()(00＝dtuuedtuttt)()2()()2(0)(0＋:,2,则上式成为到以及后两个积分限是从取样性质考虑到单位冲激函数的t)2(][)2()(][)()2()(2)()(0)()(tueettutueetuttitttttt:,整理得带入积分上下限)2(]1[)()1()()2(tuetuetitt)(ti0t11234523预习§2.7作业p862-21(1)(2)