ch2_符號(hào)計(jì)算

1、第 2 章 符號(hào)計(jì)算所謂符號(hào)計(jì)算是指：解算數(shù)學(xué)表達(dá)式、方程不是在離散化的數(shù)值點(diǎn)上進(jìn)行，而是憑借一系列恒等式，數(shù)學(xué)定理，通過(guò)推理和演繹，力求獲得解析結(jié)果。這種計(jì)算建立在數(shù)值完全準(zhǔn)確表達(dá)和推演嚴(yán)格解析的基礎(chǔ)之上，因此所得結(jié)果是完全準(zhǔn)確的。2.1 符號(hào)對(duì)象和符號(hào)表達(dá)式2.1.1 基本符號(hào)對(duì)象和運(yùn)算算符 1 生成符號(hào)對(duì)象的基本規(guī)則生成符號(hào)對(duì)象的MATLAB規(guī)則：（1）（2） 2 精準(zhǔn)符號(hào)數(shù)字和符號(hào)常數(shù)sym(Num)sc=sym(Num)sym('Num')sc=sym('Num')說(shuō)明【例2.1-1】演示：精準(zhǔn)符號(hào)數(shù)字或數(shù)字表達(dá)式的創(chuàng)建。1）clear allR1=s

2、in(sym(0.3)%R2=sin(sym(3e-1)%R3=sin(sym(3/10)%R4=sin(sym('3/10')%disp('R1屬于什么類別？ 答：',class(R1)disp('R1與R4是否相等？（是為1，否為0） 答：',int2str(logical(R1=R4)R1 =sin(3/10)R2 =sin(3/10)R3 =sin(3/10)R4 =sin(3/10)R1屬于什么類別？ 答：symR1與R4是否相等？（是為1，否為0） 答：1 2）S1=sin(sym('0.3')%S2=sin(sym(

3、'3e-1')%eRS=vpa(abs(R1-S1),64);disp('S1屬于什么類別？ 答：',class(S1)disp('S1與R1是否相同？ 答: ',int2str(logical(R1=S1)disp('S1與R1的誤差為')disp(double(eRS)S1 =0.29552020666133957510532074568503S2 =0.29552020666133957510532074568503S1屬于什么類別？ 答：symS1與R1是否相同？ 答: 0S1與R1的誤差為 6.3494e-413）F1=

4、sym(sin(3/10)%F2=sym(sin(0.3)%eFS=vpa(abs(F1-S1),32);disp('F1屬于什么類別？ 答：',class(F1)disp('S1與F1是否相同？ 答: ',int2str(logical(F1=S1)disp('F1與S1的誤差為')disp(double(eFS) F1 =5323618770401843/18014398509481984F2 =5323618770401843/18014398509481984F1屬于什么類別？ 答：symS1與F1是否相同？ 答: 0F1與S1的誤差為

5、2.8922e-17 3 基本符號(hào)變量說(shuō)明表 2.1-1 sym和syms指令功能異同比較sym指令的功用syms指令的功用能不能例2.1-1，2.1-2，2.2-1能不能例2.1-1，2.1-2，2.1-5，2.2-3，能能例2.1-5，2.2-3，2.5-4，2.5-5，2.6-1；表2.1-3例2.1-6，2.2-2，2.3-4；表2.1-3不能能例2.1-2，2.1-3，2.1-4，2.7-1，2.8-3等；表2.1-3不能能能不能例2.1-2，2.1-3，2.3-5能能例2.1-6，2.5-4例2.1-6，2.5-4不能能例2.5-5 4 符號(hào)計(jì)算中的各種算符2.1.2 符號(hào)計(jì)算中的

6、函數(shù)指令表2.1-2 MATLAB中可調(diào)用的符號(hào)計(jì)算函數(shù)指令類別情 況 描 述符號(hào)工具包函數(shù)三角函數(shù)、雙曲函數(shù)及反函數(shù)；除atan2外指數(shù)、對(duì)數(shù)函數(shù)（如exp, expm）復(fù)數(shù)函數(shù)（如abs, angle）矩陣分解函數(shù)（如eig, svd）方程求解函數(shù)solve微積分函數(shù)（如diff, int）積分變換和反變換函數(shù)（如laplace, ilaplace）繪圖函數(shù)（如ezplot, ezsurf）特殊函數(shù)單位脈沖和階躍函數(shù)（如dirac, heaviside） 函數(shù)（如beta, gamma）橢圓積分（如ellipke）貝塞爾函數(shù)（如besseli, besselj）MuPAD庫(kù)函數(shù)借助eval

7、in和feval指令可調(diào)用比mfunlist所列范圍更廣泛的MuPAD庫(kù)函數(shù)；需要具備MuPAD語(yǔ)言知識(shí)。說(shuō)明2.1.3 符號(hào)表達(dá)式和符號(hào)函數(shù) 1 符號(hào)表達(dá)式和符號(hào)函數(shù)（1）（2） 2 自由符號(hào)變量說(shuō)明【例2.1-2】1）clearsyms a b c x y u v%syms F(X,Y,Z)%k=sym(3)%G=sym('p*sqrt(q)+r*sin(t)')%EXPR=a*G*u+(b*x2+k)*v%f(x,y)=a*x2+b*y2-c%disp(F)% k =3G =p*q(1/2) + r*sin(t)EXPR =v*(b*x2 + 3) + a*u*(p*q(

8、1/2) + r*sin(t)f(x, y) =a*x2 + b*y2 - cF(X, Y, Z)symbolic function inputs: X, Y, Z 2）symvar(EXPR)%ans = a, b, p, q, r, t, u, v, x symvar(EXPR,20) %ans = x, v, u, t, r, q, p, b, a symvar(EXPR,1)% ans =x 3）disp(symvar(f)% a, b, c, x, y disp(symvar(f,2)% x, y 【例2.1-3】用符號(hào)法求方程的解。1）clearsyms u v w z%Eq=u*

9、w2+z*w-v%g(z)=u*w2+z*w=v%Eq =u*w2 + z*w - vg(z) =u*w2 + z*w = v 2）symvar(Eq,1)%<5>ans =w symvar(g(z),1)%<6> ans =w 3）R1=solve(Eq)%<7>R2=solve(g)%<8> R1 = -(z + (z2 + 4*u*v)(1/2)/(2*u) -(z - (z2 + 4*u*v)(1/2)/(2*u)R2 = -(z + (z2 + 4*u*v)(1/2)/(2*u) -(z - (z2 + 4*u*v)(1/2)/(2*u

10、) 4）S1=solve(Eq,z)%<9>S2=solve(g(z),z)% <10> S1 =(- u*w2 + v)/wS2 =(- u*w2 + v)/w 5）disp(simplify(subs(Eq,z,S1)% 0 disp(simplify(g(S2)% TRUE 說(shuō)明【例2.1-4】syms a b t u v x yA=a+b*x,sin(t)+u;x*exp(-t),log(y)+v%symvar(A,1)% A = a + b*x, u + sin(t) x*exp(-t), v + log(y)ans =x 2.1.4 符號(hào)對(duì)象的識(shí)別在MATL

11、AB中，函數(shù)指令很多。有的函數(shù)指令適用于多種數(shù)據(jù)對(duì)象（如數(shù)值、符號(hào)等），但也有的函數(shù)指令只對(duì)某種數(shù)據(jù)對(duì)象適用。在數(shù)值計(jì)算和符號(hào)計(jì)算混合使用的情況下，由于函數(shù)指令與數(shù)據(jù)對(duì)象不適配引起的錯(cuò)誤容易發(fā)生。為了避免這種錯(cuò)誤，MATLAB提供了用于識(shí)別數(shù)據(jù)對(duì)象屬性的指令：class(var)給出變量var的數(shù)據(jù)類別（如double, sym等）isa(var,'Obj')若變量var是Obj代表的類別，給出1，表示“真”whos給出所有MATLAB內(nèi)存變量的屬性【例2.1-5】數(shù)據(jù)對(duì)象及其識(shí)別指令的使用。1）cleara=1;b=2;c=3;d=4;%Mn=a,b;c,d%Mc='

12、a,b;c,d'%Ms=sym(Mc)% Mn = 1 2 3 4Mc =a,b;c,dMs = a, b c, d 2）SizeMn=size(Mn)SizeMc=size(Mc)SizeMs=size(Ms) SizeMn = 2 2SizeMc = 1 9SizeMs = 2 2 3）CMn=class(Mn)CMc=class(Mc)CMs=class(Ms) CMn =doubleCMc =charCMs =sym 4）isa(Mn,'double')isa(Mc,'char')isa(Ms,'sym') ans = 1ans

13、= 1ans = 1 5）whos Mn Mc Ms% Name Size Bytes Class Attributes Mc 1x9 18 char Mn 2x2 32 double Ms 2x2 60 sym 2.1.5 符號(hào)運(yùn)算機(jī)理和變量假設(shè) 1 符號(hào)運(yùn)算的工作機(jī)理 2 對(duì)符號(hào)變量的限定性假設(shè)說(shuō)明 3 清除變量和撤銷假設(shè)說(shuō)明【例2.1-6】符號(hào)變量的默認(rèn)數(shù)域是復(fù)數(shù)域。1）clear all%syms x%f=x3+475*x/100+5/2;%r=solve(f,x)% r = -1/2 (79(1/2)*i)/4 + 1/4 1/4 - (79(1/2)*i)/4 assumption

14、s(x)% ans = empty sym 2）assume(x,'real')%r21=solve(f,x)% r21 =-1/2 syms x clear%assume(imag(x)=0)%r22=solve(f,x) r22 =-1/2 disp(assumptions(x)% imag(x) = 0 3）sym(x,'clear')%assume(real(x)>0)%r3=solve(f,x) ans =xr3 = (79(1/2)*i)/4 + 1/4 1/4 - (79(1/2)*i)/4 disp(assumptions(x)% 0 &l

15、t; real(x) 4）assumeAlso(imag(x)>0)%r4=solve(f,x) r4 =(79(1/2)*i)/4 + 1/4 disp(assumptions(x) 0 < imag(x), 0 < real(x) 2.2 符號(hào)數(shù)字及表達(dá)式的操作2.2.1 符號(hào)數(shù)字轉(zhuǎn)換成雙精度數(shù)字說(shuō)明2.2.2 符號(hào)數(shù)字的任意精度表達(dá)形式說(shuō)明【例2.2-1】1）reset(symengine)%<1> sa=sym('1/3+sqrt(2)')% sa =2(1/2) + 1/3 2）digits%<3> Digits = 32 f

16、ormat longa=1/3+sqrt(2)%sa_Plus_a=vpa(sa+a,20)%<6>sa_Minus_a=vpa(sa-a,20)%<7>% a = 1.747546895706428sa_Plus_a =3.4950937914128567869sa_Minus_a =-0.000000000000000022658064826339973669 3）sa32=vpa(sa)%<8>digits(48)%<9>sa5=vpa(sa,5)%<10>sa48=vpa(sa)%<11> sa32 =1.7475

17、46895706428382135022057543sa5 =1.7475sa48 =1.74754689570642838213502205754303141190300520871 說(shuō)明2.2.3 符號(hào)表達(dá)式的基本操作說(shuō)明【例2.2-2】簡(jiǎn)化。syms xf=(1/x3+6/x2+12/x+8)(1/3)g1=simplify(f)% f =(12/x + 6/x2 + 1/x3 + 8)(1/3)g1 =(2*x + 1)3/x3)(1/3) g2=simplify(f,'Steps',10,'IgnoreAnalyticConstraints', tru

18、e)% g2 =1/x + 2 2.2.4 表達(dá)式中的置換操作 1 公因子法簡(jiǎn)化表達(dá)說(shuō)明【例2.2-3】對(duì)符號(hào)矩陣進(jìn)行特征向量分解。1）clear%A=sym('a b;c d')%V,D=eig(A)%A = a, b c, dV =(a/2+d/2-a2-2*a*d+d2+4*b*c)(1/2)/2)/c-d/c, (a/2+d/2+(a2-2*a*d+ d2+4*b*c)(1/2)/2)/c-d/c 1, 1D =a/2 + d/2 - (a2 - 2*a*d + d2 + 4*b*c)(1/2)/2, 0 0, a/2 + d/2 + (a2 - 2*a*d + d2

19、+ 4*b*c)(1/2)/2 2）subexpr(V;D)%<4>who% sigma = (a2 - 2*a*d + d2 + 4*b*c)(1/2)ans = (a/2 + d/2 - sigma/2)/c - d/c, (a/2 + d/2 + sigma/2)/c - d/c 1, 1 a/2 + d/2 - sigma/2, 0 0, a/2 + d/2 + sigma/2Your variables are:A D V ans sigma 3）Dw=subexpr(D,'w')%w = (a2 - 2*a*d + d2 + 4*b*c)(1/2)Dw

20、= a/2 + d/2 - w/2, 0 0, a/2 + d/2 + w/2 4）RVD,w=subexpr(V;D,'w')%<7> RVD = (a/2 + d/2 - w/2)/c - d/c, (a/2 + d/2 + w/2)/c - d/c 1, 1 a/2 + d/2 - w/2, 0 0, a/2 + d/2 + w/2w =(a2 - 2*a*d + d2 + 4*b*c)(1/2) 說(shuō)明 2 通用置換指令說(shuō)明【例2.2-4】1）clearsyms a b x;f1=a*sin(x)+b f1 =b + a*sin(x) 2）f2=subs(f1

21、,sin(x),'log(y)')%<4>class(f2)% f2 =b + a*log(y)ans =sym 3）f3=subs(f1,a,sym(3.11)%<6>class(f3)% f3 =b + (311*sin(x)/100ans =sym 4）f4=subs(f1,x,0,pi/2,pi)%<8>class(f4) f4 = b, a + b, bans =sym 5）format%format compact%t=0:pi/10:2*pi;%f5=subs(f1,a,b,x,2,3,t);%<13>class(f

22、5)plot(t,f5,'r:','LineWidth',5)%<15>% ans =sym圖2.2-1 利用符號(hào)表達(dá)式變量置換產(chǎn)生的單根曲線6）k=0.6;0.8;1; %f6=subs(subs(f1,a,b,k,2),x,t);%<17>class(f6)plot(t,f6) %<19> ans =sym圖2.2-2 利用兩次subs置換產(chǎn)生的多根曲線說(shuō)明2.3 符號(hào)微積分可以毫不夸張地說(shuō)，大學(xué)本科高等數(shù)學(xué)中的大多數(shù)微積分問(wèn)題，都能用符號(hào)計(jì)算解決，手工筆算演繹的煩勞都可以由計(jì)算機(jī)完成。2.3.1 極限和導(dǎo)數(shù)的符號(hào)計(jì)算說(shuō)明

23、【例2.3-1】?jī)煞N重要極限和。syms t x ks=sin(k*t)/(k*t);f=(1-1/x)(k*x);Lsk=limit(s,0)%Ls1=subs(Lsk,k,1)%Lf=limit(f,x,inf)%Lf1=vpa(subs(Lf,k,sym('-1'),48)% Lsk =1Ls1 =1Lf =exp(-k)Lf1 =2.7182818284590452353602874713526624977572470937 說(shuō)明【例2.3-2】已知，求、 、。syms a t x;f=a,t3;t*cos(x), log(x);df=diff(f)%dfdt2=dif

24、f(f,t,2)%dfdxdt=diff(diff(f,x),t)% df = 0, 0 -t*sin(x), 1/xdfdt2 = 0, 6*t 0, 0dfdxdt = 0, 0 -sin(x), 0 【例2.3-3】求的Jacobian矩陣。syms x1 x2;f=x1*exp(x2);x2;cos(x1)*sin(x2);%v=x1;x2;%<3>Jf=jacobian(f,v)% Jf = exp(x2), x1*exp(x2) 0, 1 -sin(x1)*sin(x2), cos(x1)*cos(x2) 說(shuō)明l 指令<3>把自變量x1和x2寫成列向量v。但

25、若寫成v=x1,x2，所得Jacobian矩陣完全一樣?！纠?.3-4】，求，。1）clearsyms xsyms d positivef_p=sin(x);%df_p=limit(subs(f_p,x,x+d)-f_p)/d,d,0)%<5>df_p0=limit(subs(f_p,x,d)-subs(f_p,x,0)/d,d,0)%<6> df_p =cos(x)df_p0 =1 2）f_n=sin(-x);%df_n=limit(f_n-subs(f_n,x,x-d)/d,d,0)%<8>df_n0=limit(subs(f_n,x,0)-subs(f

26、_n,x,-d)/d,d,0)%<9> df_n =-cos(x)df_n0 =-1 3）f=sin(abs(x);dfdx=diff(f,x)% <11>dfdx0=subs(dfdx,x,0)% <12> dfdx =cos(abs(x)*sign(x)dfdx0 = 0 4）clfxn=-3/2*pi:pi/50:0;xp=0:pi/50:3/2*pi;xnp=xn,xp(2:end);hold onplot(xnp,subs(f,x,xnp),'k','LineWidth',3)% <14>plot(xn,

27、subs(df_n,x,xn),'-r','LineWidth',3)plot(xp,subs(df_p,x,xp),':r','LineWidth',3)legend(char(f),char(df_n),char(df_p),'Location','NorthEast')% <17>grid onxlabel('x')hold off 圖 2.3-1 函數(shù)及其導(dǎo)函數(shù)說(shuō)明【例2.3-5】設(shè)，求。1）clearsyms x yf(x,y)=sym('cos(x+s

28、in(y(x)=sin(y(x)')%<3>dfdx=diff(f,x)%<4> f(x, y) =cos(x + sin(y(x) = sin(y(x)dfdx(x, y) =-sin(x + sin(y(x)*(D(y)(x)*cos(y(x) + 1) = D(y)(x)*cos(y(x) 2）dfdx1=subs(dfdx,'D(y)(x)','dydx')%<5> dfdx1(x, y) =-sin(x + sin(y(x)*(dydx*cos(y(x) + 1) = dydx*cos(y(x) 3）dydx

29、=simplify(solve(dfdx1,'dydx')% dydx =-sin(x + sin(y(x)/(cos(y(x)*(sin(x + sin(y(x) + 1) 說(shuō)明【例2.3-6】syms xr5=taylor(x*exp(x)%r8=taylor(x*exp(x),'Order',9)%pretty(r8)% r5 =x5/24 + x4/6 + x3/2 + x2 + xr8 =x8/5040 + x7/720 + x6/120 + x5/24 + x4/6 + x3/2 + x2 + x 8 7 6 5 4 3 x x x x x x 2

30、- + - + - + - + - + - + x + x 5040 720 120 24 6 2 說(shuō)明2.3.2 序列/級(jí)數(shù)的符號(hào)求和說(shuō)明【例2.3-7】1）syms n kf1=1/(k*(k+1);s1=symsum(f1,k,1,n)s1 =n/(n + 1) 2）syms xf2=x(2*k-1)/(2*k-1);s2=symsum(f2,k,1,inf) s2 =piecewise(abs(x) < 1, atanh(x) 3）f3=1/(2*k-1)2,(-1)k/k;s3=symsum(f3,k,1,inf) s3 = pi2/8, -log(2) 說(shuō)明2.3.3 符號(hào)積

31、分說(shuō)明【例2.3-8】clearsyms a b xf=x*log(x)s=int(f,x)f =x*log(x)s =(x2*(log(x) - 1/2)/2 【例2.3-9】syms a b x%f2=a*x,b*x2;1/x,sin(x)%INTf2=int(f2)%pretty(INTf2)% f2 = a*x, b*x2 1/x, sin(x)INTf2 = (a*x2)/2, (b*x3)/3 log(x), -cos(x)/ 2 3 | a x b x | -, - | 2 3 | | log(x), -cos(x) / 說(shuō)明【例2.3-10】。syms x y zF2=int(

32、int(int(x2+y2+z2,z,sqrt(x*y),x2*y),y,sqrt(x),x2),x,1,2)VF2=vpa(F2)% F2 =(14912*2(1/4)/4641 - (6072064*2(1/2)/348075 + (64*2(3/4)/225 + 1610027357/6563700VF2 =224.92153573331143159790710032805 【例2.3-11】求阿基米德（Archimedes）螺線在到間的曲線長(zhǎng)度函數(shù)，并求出時(shí)的曲線長(zhǎng)度。1）syms a r th pr=a*th;%x=r*cos(th);%y=r*sin(th);%dL=simplif

33、y(sqrt(diff(x,th)2+diff(y,th)2);%L=int(dL,th,0,p) L =(asinh(p) + p*(p2 + 1)(1/2)*(a2)(1/2)/2 2）L2pi=subs(L,a,p,1,2*pi)%L2pid=vpa(L2pi)% L2pi =asinh(2*pi)/2 + pi*(4*pi2 + 1)(1/2)L2pid =21.256294148209098800702512272566 3）L1=subs(L,a,1);%ezplot(L1*cos(p),L1*sin(p),0,2*pi)%grid on%hold on%x1=subs(x,a,1

34、);%y1=subs(y,a,1);%h1=ezplot(x1,y1,0,2*pi);%<15>set(h1,'Color','r','LineWidth',5)%title(' ')%<17>legend('螺線長(zhǎng)度-幅角曲線','阿基米德螺線')hold off% 圖 2.3-2 阿基米德螺線和螺線長(zhǎng)度函數(shù)說(shuō)明2.4 微分方程的符號(hào)解法2.4.1 符號(hào)解法和數(shù)值解法的互補(bǔ)作用2.4.2 求微分方程符號(hào)解的一般指令說(shuō)明2.4.3 微分方程符號(hào)解示例【例2.4-1】求的解。c

35、lear all%<1>S=dsolve('Dx=y,Dy=-x')%disp(' ')disp('微分方程的解',blanks(2),'x',blanks(22),'y')disp(S.x,S.y) S = x: 1x1 sym y: 1x1 sym 微分方程的解 x y C1*cos(t) + C2*sin(t), C2*cos(t) C1*sin(t) 說(shuō)明【例2.4-2】圖示微分方程1）clear all%<1>y=dsolve('(Dy)2-x*Dy+y=0',&#

36、39;x')%<2>y = x2/4 - C52 + x*C5 2）clf,hold on%hy1=ezplot(y(1),-6,6,-4,8,1);%<4>set(hy1,'Color','r','LineWidth',5)%Sv=symvar(y(2);%<6>for k=-2:0.5:2%<7>y2=subs(y(2),Sv(1),k);%<8>ezplot(y2,-6,6,-4,8,1)end% <10>hold off%box on%legend('

37、奇解','通解','Location','Best')ylabel('y')title('fontsize14微分方程',' (y '')2 xy '' + y = 0 ','的解') 圖 2.4-1 說(shuō)明【例2.4-3】 。1）y=dsolve('x*D2y-3*Dy=x2','y(1)=0,y(5)=0','x') y =(31*x4)/468 - x3/3 + 125/468 2）ezplo

38、t(y,-1,6)hold onplot(1,5,0,0,'.r','MarkerSize',20)text(1,1,'y(1)=0')text(4,1,'y(5)=0')title('x*D2y - 3*Dy = x2, ','y(1)=0,y(5)=0')hold off 圖 2.4-2 說(shuō)明2.5 符號(hào)變換和符號(hào)卷積2.5.1 Fourier變換及其反變換(2.5-1)(2.5-2)說(shuō)明【例 2.5-1】求單位階躍函數(shù)的Fourier變換。1）syms t wut=heaviside(t);U

39、T=fourier(ut) UT =pi*dirac(w) - i/w 2）Ut=ifourier(UT,w,t)%SUt=simplify(Ut)%<5> Ut =(pi + pi*(2*heaviside(t) - 1)/(2*pi)SUt =heaviside(t) 3）t=-2:0.01:2;ut=heaviside(t);kk=find(t=0);%<8>plot(t(kk),ut(kk),'.r','MarkerSize',30)%hold onut(kk)=NaN;%<10>plot(t,ut,'-r&#

40、39;,'LineWidth',3)plot(t(kk),t(kk),ut(kk-1),ut(kk+1),'or','MarkerSize',10)%hold offgrid onaxis(-2,2,-0.2,1.2)xlabel('fontsize14t'),ylabel('fontsize14ut')title('fontsize14Heaviside(t)') 圖 2.5-1 說(shuō)明【例2.5-2】矩形脈沖的Fourier變換。1）syms A t w taoyt=A*(heaviside(t+

41、tao/2)-heaviside(t-tao/2);%Yw=fourier(yt,t,w)%Yws=simplify(Yw)%<4> Yw =A*(cos(tao*w)/2)*i + sin(tao*w)/2)/w - (cos(tao*w)/2)*i - sin(tao*w)/2)/w)Yws =(2*A*sin(tao*w)/2)/w 2）Yt=ifourier(Yws,w,t)%Yts=simplify(Yt)%<6> Yt =-(A*(pi*heaviside(t - tao/2) - pi*heaviside(t + tao/2)/piYts =-A*(hea

42、viside(t - tao/2) - heaviside(t + tao/2) 3）T=3;tn=-3:0.1:T;%ytn=subs(yt,A,tao,t,1,T,tn);%kk=find(tn=-t3/2|tn=t3/2);%<11>plot(tn(kk),ytn(kk),'.r','MarkerSize',30)%ytn(kk)=NaN;%<13>hold onplot(tn,ytn,'-r','LineWidth',3)hold offgrid onaxis(-T,T,-0.5,1.5) yt13

43、 =heaviside(t + 3/2) - heaviside(t - 3/2)圖 2.5-2 矩形波4）頻域曲線繪制Ywn=subs(Yws,A,tao,1,T);subplot(2,1,1),ezplot(Ywn),grid onsubplot(2,1,2),ezplot(abs(Ywn),grid on 圖 2.5-3 矩形脈沖的頻率曲線和幅度頻譜說(shuō)明【例2.5-3】求的Fourier變換，在此是參數(shù)，是時(shí)間變量。1）clear allsyms t x wft=exp(-(t-x)*heaviside(t-x);%<3>gt=exp(-(t-x);%<4>2）

44、F1=simplify(fourier(ft,t,w)%G1=simplify(fourier(gt,t,w)% F1 =exp(-w*x*i)/(w*i + 1)G1 =exp(x)*fourier(exp(-t), t, w) 3）F2=simplify(fourier(ft,t)% F3=simplify(fourier(ft)% F2 =-exp(-t2*i)/(t*i - 1)F3 =(exp(-t*w*i)*(w*i + 1)/(w2 + 1) 說(shuō)明2.5.2 Laplace變換及其反變換(2.5-3)(2.5-4)說(shuō)明【例2.5-4】分別求, , , 的Laplace變換。1）s

45、yms t s a bf1=exp(-a*t)*sin(b*t)%<2>F1=laplace(f1,t,s) f1 =exp(-a*t)*sin(b*t)F1 =b/(a + s)2 + b2) 2）syms a clear%<4>f2=heaviside(t-a)F2=laplace(f2,t,s)% ans =af2 =heaviside(t - a)F2 =laplace(heaviside(t - a), t, s) syms a positive%<7>F3=laplace(f2) % F3 =exp(-a*s)/s 3）f4=dirac(t-b)

46、;%F4=laplace(f4,t,s)%F4 =piecewise(b < 0, 0, 0 <= b, exp(-b*s) f5=dirac(t-a);%<11>F5=laplace(f5,t,s) %ft_F5=ilaplace(F5,s,t) % F5 =exp(-a*s)ft_F5 =dirac(a - t) 4）n=sym('n','clear');%<14>F6=laplace(tn,t,s)% F6 =piecewise(-1 < real(n), gamma(n + 1)/s(n + 1) n=sym(&

47、#39;n','positive');%<16>F6=laplace(tn,t,s)%<17> F6 =gamma(n + 1)/s(n + 1) 說(shuō)明2.5.3 Z變換及其反變換(2.5-5)(2.5-6)說(shuō)明【例2.5-5】一組Z變換、反變換算例。1）clearsyms n z clear%<2>gn=6*(1-(1/2)n)%G=ztrans(gn,n,z);%pretty(G)% gn =6 - 6*(1/2)n 6 z 6 z- - -z - 1 1 z - - 2 2）syms n w T z clear%<6>fwn=sin(w*n*T);%FW=ztrans(fwn,n,z);%pretty(FW),disp(' ')inv_FW=iztrans(FW,z,n)% z sin(T w) - 2 z - 2 cos(T w) z + 1 inv_FW =sin(T*n*w) 3）syms n z clear%<11>f1=1;F1=ztrans(f1,n,z);pretty(F1) inv_F1=iztrans(F1,z,n