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1、10.1 利用fourier函数求下列信号的傅里叶变换F(jw),并用ezplot函数绘出其幅度频谱|F(jw)|和相位频谱。观察比较三个信号的幅频特性和相频特性，并用傅里叶的性质加以解释。syms t phase im re f=sin(2*pi*t)/(2*pi*t);F=fourier(f);subplot(3,1,1)ezplot(f);title(原图)axis(-pi pi -0.3 1.1)subplot(3,1,2)ezplot(abs(F)title(幅度谱)axis(-3*pi 3*pi 0.3 0.6)im=imag(F);re=real(F);phase=atan(im
2、/re);subplot(3,1,3)ezplot(phase)title(相位谱)axis(-3*pi 3*pi -0.5 0.5)10.2 试用ifourier函数求下列傅里叶变换的逆变换，并画出其时域波形。syms t wF=2*pi*cos(w/2)/(pi2-w2);f=ifourier(F,t)ezplot(f);axis(-0.8 0.8 -0.2 1.1);title(时域波形)10.3syms t wf1=sym(Heaviside(t+1)-Heaviside(t-1)subplot(2,2,1)ezplot(f1)hold onaxis(-2 2 0 1.1);plot(
3、-1 -1,0 1)hold onplot(1 1,0 1)hold ontitle(f1(t)Fw=fourier(f1,t,w)FFw=maple(convert,Fw,piecewise);FFP=abs(FFw)subplot(2,2,2)ezplot(FFP,-10*pi 10*pi);axis(-10*pi 10*pi 0 2.1)title(F(jw)f=Fw*Fw;subplot(2,2,3)ezplot(f)title(f(t)=f1(t)*f1(t)subplot(2,2,4)ezplot(abs(f),-10*pi 10*pi)axis(-3*pi 3*pi 0 5)title(F(jw)10.4b=0 0 2;a=0.04 0.4 2;h,w=freqs(b,a,100);h1=abs(h);h2=angle(h);subplot(2,1,1);plot(w,h1);hold onplot(7.0711 7.0711,0 0.707,:);plot(0 7.0711,0.707 0.707,:);axis(0 50 0 1.5);grid;title(H(jomega)d幅频特性);subplot(2,1,2);plot(w,h2*180/pi);axis(0 100 -200 0);gridtitle(H(jomega)相频特性);
