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/*PRACTICA 2 EJERCICIO 10a El n鷐ero Pi se puede aproximar por medio de la siguiente serie: Pi/4 = 2 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - 1/15 + ... a) Escribir un programa que le pregunte al us

function [norm]=norm(x) norm=1/sqrt(2*pi)*int((e.^((-(t.^2))/2),t,-inf,x);

#include #include void main(void) { double pi = 3.14159265; printf("Tangent of pi is %f\n", tan(pi)); printf("Tangent of pi/4 is %f\n", tan(pi / 4.0)); }

function Y = gensino(f,n,Ts) X = zeros(1,length(n));X(1) = 1; % generate the impulses A = [1 -2*cos(2*pi * f *Ts) 1]; B = [0 sin(2*pi * f *Ts) 0 ]; Y = filter(B,A,X);

var x,y,z,w; const a=1, t0=0, tmax=10, dt=0.001, pi=3.1416, eps=1e-20; system y'=a*x &1; x'=-a*y &1; w=pi*3; z=x*y; sysend.

var x,y,z,w; const a=1000, t0=0, tmax=10, dt=0.001, pi=3.1416, eps=1e-20; system y'=a*x &1; x'=-a*y &1; w=pi*3; z=x*y; sysend.

%% % $$e^{\pi i} + 1 = 0$$ %% % $$0 \leq x \leq 6\pi$$ %% % $$0 \leq x \leq 6\pi$$

%% N=8192; f0=0.1; f1=0.2; f2=0.3; f3=0.5; n=[0:8*pi]; x0=cos(2*pi*f0*n); x1=cos(2*pi*f1*n); x2=cos(2*pi*f2*n); x3=cos(2*pi*f3*n); X0=fft(x0,N); X1=fft(x1,N); X2=fft(x2,N); X3=fft(x3

%% Plots of Different Signals %% Continuous Time Sinusoid % $$x(t) = cos(2\pi ft) ,f=1,2 $$ t = [0:0.01:2]; x1 = cos(2*pi*t); x2 = cos(2*pi*2*t); x3 = t.*cos(2*pi*3*t); x4 = cos(2*pi*4*t.*t);

G=tf(1.5,[1,2,3]); t=0:0.1:2*pi; u=sin(t); y=lsim(G,u,t); plot(t,u,t,y) figure u=sin(2*t); y=lsim(G,u,t); plot(t,u,t,y)