x(t)=sin(2*pi*f1*t),f1=0.02,这个信号如何用matlab进行模拟?

matlab中问题：x = A1 * sin(2 * pi * f1 * t) + A2 * sin(2 * pi * f2 * t) + rand(size(t))f1=364.21 ,f2= 124.7,A1 x(t)=sin(2*pi*f1*t),f1=0.02,这个信号如何用matlab进行模拟? Matlab u(x,t)=sin(5*pi*x)cos(5*pi*t)+2sin(7*pi*x)cos(7*pi*t) 其中0 关于matlab对于信号的FT,DTFT,DFT的提问x(t)=sin(2pi*f1*t)+sin(2pi*f2*t)+sin(2pi*f3*t)f1=2HZ f2=2.05HZ f3=1.9hz确定合适的采样频率,对信号采样画出 FT,DTFT,DFT,并进行对比 x_t=sin(2*pi*f1*t)+sin(2*pi*f2*t)+sin(2*pi*f3*t);n_t=wgn(1,M,0),r_t=x_t+n_t,用matlab求 r_t方差均值均方值 Matlab程序编写和出图u(x,t) = 10 + 5*cos(2*pi*t/T) + (1 + cos(2*pi*t/T))*x/5 + sum{n=1 to n=N} ( (10/pi*n [(T/2*pi) * sin(2*pi*t/T) - (25/(pi*n))^2 cos(2*pi*t/T)] * sin(pi*n*x/25) )I need it once with T=1 and once with T=365.Also for N you can matlab rotate程序求大神修改t=0:pi/1000:3.5*pi;k=fix(2*t/pi)+1;x=0.5*sqrt(2)*cos(0.5*k*pi+0.25*pi)+k.*cos(t);y=0.5*sqrt(2)*sin(0.5*k*pi+0.25*pi)+k.*sin(t);h=plot(x,y,'r');m=pi:pi/20:4.5*pi;n=fix(2*m/pi)-1;x0=2*cos(1.25*pi);y0=2*sin(1.25*pi);x2= MATLAB中dsolve解出的t是什么含义?我使用了dsolve（’Dy = sin(pi*x*y)‘）；然后结果中：atan(2/(1+exp(2*t*pi*x+2*C1*pi*x))*exp(t*pi*x+C1*pi*x),-(exp(2*t*pi*x+2*C1*pi*x)-1)/(1+exp(2*t*pi*x+2*C1*pi*x)))/x/pi有t和c1两个量,c1想 x=4*sin(2*pi*0.01*t).*sin(2*pi*3*t)+2*cos(pi*t*t/4);在matlab中怎样表达运行出现错误：Error using ==> mtimesInner matrix dimensions must agree. matlab参数方程绘图问题t=0:pi/1000:3*pi k=fix(2*t/pi)+1; y=0.5*sqrt(2)*sin(0.5*k*pi+0.25*pi)+k*sin(t); x=0.5*sqrt(2)*cos(0.5*k*pi+0.25*pi)+k*cos(t); plot(x,y) axis equal 为什么不能运行啊, matlab中t=linspace(0,2*pi,100);x=[t;t]';y=[sin(t);cos(t)]';plot(x,y), matlab求函数曲线的积分 请问,怎么弄才能出真确结果syms x tf1=sqrt(1+(155*2*pi/t*cos(2*pi/t*(x-350))).^2)int(f1,x)f1 =((96100*pi^2*cos((2*pi*(x - 350))/t)^2)/t^2 + 1)^(1/2)Warning:Explicit integral could not be found.ans =int(((96100 在matlab中,如何画出一个信号的频谱图?例如：这个信号 x=sin(2*pi*f1*t)+0.5*cos(2*pi*f2*t);能不用FFT 求他的频谱图吗?求代码!最好讲解下,f1=50Hz,f2=27Hz matlab 中如何表示一个元素m是集合B元素,也就是下面这段程序错误在哪里,f1=49.8;t=0:0.001:1;A=2:2:49;B=3:2:49;for m=0:49;if m==0x(t)=0.2;elseif m==1x(t)=5*sin(2*pi*f1*t+45);elseif m==Ax(t)=(2/m)*sin(2*pi*f1*t+30);elseif m==Bx(t matlab中x(20:41)=0.2*sin(2*pi*50*t(20: matlab问题,>> t=0:0.2:3*pi;x=exp(-t).*sin(t);plot(t,x) >> t=0:0.2:3*pi;x=exp(-t)*sin(t);plot(t,x)>> t=0:0.2:3*pi;x=exp(-t).*sin(t);plot(t,x)>> t=0:0.2:3*pi;x=exp(-t)*sin(t);plot(t,x)这两个为什么一个可以执行,一个却不可以 用MATLAB研究莫尔斯码的幅度调制与解调.本题研究莫尔斯码的幅度调制与解调.本题中信号的形式为(1)x（t)=m1(t)cos(2*pi*f1*t)+m2(t)cos(2*pi*f2*t)+m1(t)cos(2*pi*f2*t)其中信号x(t)由文件ctftmod.mat定义,可用命 MATLAB求助,用FFT直接计算x=cos(2*pi*40*t)+2sin(2*pi*120*t)+w(t)的功率谱.用FFT直接计算x=cos(2*pi*40*t)+2sin(2*pi*120*t)+w(t)的功率谱.最好带上注释,