利用matlab进行信号与系统分析实验

【实验1】利用matlab 进行信号的时域分析 （1）指数信号 >>A=1; >> a=-0.4;

>> t=0:0.01:10;

>> ft=A*exp(a*t); >> plot(t,ft); >> grid;

>> axis([0 10 -0.1 1.1]; >> xlabel('t') >> ylabel('ft')

（2）正弦信号 >> A=1; >> w0=2*pi; >> phi=pi/6; >> t=0:0.01:3; >> ft=A*sin(w0*t+phi); >> plot(t,ft); >> grid;

>> axis([0 3 -1.1 1.1]); >> xlabel('t') >> ylabel('ft')

()t f t Ae α=()sin()

f t A t ωϕ=+

>>x=linspace(-20,20); >> y=sinc(x/pi); >> plot(x,y);>> grid; >> axis([-21 21 -0.5 1.1]); >> xlabel('x') >> ylabel('y')

（4）矩形脉冲信号 >> t=0:0.001:4; >> T=1;

>> ft=rectpuls(t-2*T,2*T); >> plot(t,ft); >> grid;

>> axis([-1 5 -0.1 1.1]); >> xlabel('t') >>ylabel('ft')

t t t Sa t f )

sin()()(=

=)]

()([)()

(10τετετ+-+==t t A t G t f

function ft=heaviside(t) ft=(t>0); >> t=-1:0.001:3; >> ft=heaviside(t); >> plot(t,ft); >> grid;

>> axis([-1 3 -0.1 1.1]); >> xlabel('t') >>ylabel('ft')

(6)复指数信号的时域波形 >> t=0:0.1:60;

>> f=exp(-0.1*t).*sin(2/3*t); >> plot(t,f); >> grid;

>> axis([0 60 -1 1]); >> xlabel('Time(sec)') >>ylabel('f(t)')

⎩⎨

⎧<>=)0(0)

0(1)(t t t ε)

3

2sin()(1.0t e t f t -=

(7)加入随机噪声的正弦波>> t=0:0.001:50;

>> y=sin(2*pi*50*t);

>> s=y+randn(size(t)); >> subplot(2,1,1);

>> plot(t(1:100),y(1:100)); >> grid;

>> subplot(2,1,2);

>> plot(t(1:100),s(1:100)); >>grid;

（8）周期矩形波

>> A=1;

>> t=0:0.0001:5;

>> y=A*square(2*pi*t,20); >> plot(t,y);

>> grid;

>> axis([0 5 -1.5 1.5]);

（9）信号的基本运算

>> syms t;

>>f=sym('(t/2+1)*(heaviside(t+2)-heaviside(t-2))');

>>subplot(3,2,1),ezplot(f,[-3,3]);

>>grid;

>> y1=subs(f,t,t+2);

>> subplot(3,2,2),ezplot(y1,[-5,1]);

>> title('f(t+2)');

>> grid;

>> y2=subs(f,t,t-2);

>> subplot(3,2,3),ezplot(y2,[-1,5]);

>> title('f(t-2)');

>> grid;

>> y3=subs(f,t,-t);

>> subplot(3,2,4),ezplot(y3,[-3,3]);

>> title('f(-t)');

>> grid;

>> y4=subs(f,t,2*t);

>> subplot(3,2,5),ezplot(y4,[-2,2]);

>> title('f(2t)');

>> grid;

例1求系统y ”(t )+2y ’(t )+100y (t )=10f (t )的零状态响应，已知f (t )=(sin2πt ) ε(t )。

>> t=0:0.01:5;

>> sys=tf([10],[1 2 100]); >> f=10*sin(2*pi*t); >> y=lsim(sys,f,t); >> plot(t,y); >> grid;

>> xlabel('Time(sec)') >> ylabel('y(t)')

例2 求系统y ”(t )+2y ’(t )+100y (t )=10f (t )的零状态响应，已知f (t )=δ(t )。

>> t=0:0.01:5;

>> sys=tf([10],[1 2 100]); >> y=impulse(sys,t); >> plot(t,y); >> xlabel('Time(sec)') >> ylabel('h(t)') >> grid;

[例1]实用matlab 画出图示周期三角波信号的频谱

>> N=8; >> n1=-N:-1;

>> c1=-4*j*sin(n1*pi/2)/pi^2./n1.^2; >> c0=0; >> n2=1:N;

>>c2=-4*j*sin(n2*pi/2)/pi^2./n2.^2; >> cn=[c1 c0 c2]; >> n=-N:N; >> subplot(2,1,1); >> stem(n,abs(cn)); >> ylabel('Cn 的幅度'); >> subplot(2,1,2); >> stem(n,angle(cn)); >> ylabel('Cn 的相位'); >>xlabel('\omega/\omega0');

t

⎪

⎩⎪⎨⎧=≠-=000 )2

sin(422n n n n j

C n ππ