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! ------求圆錐体积和表面积------ PROGRAM Example_1_2 REAL :: pi, r, h, v, s pi = 3.141593 PRINT *,'Input radius r and height h ?' READ *, r, h v = pi*h*r**2/3.0 s = pi*r*(r + sqrt(r**2 + h**2))

package ForTest; public class Exercise2_13 { /** * @param args */ public static void main(String[] args) { final double PI=3.14159; double pi=0; int n=1; // int sign=1; //

G=tf([5,5],[1,4,2,3,0]); t=[0:0.1:150]'; u=sin(t+pi/6); lsim(G,u,t) ; title('正弦信号输出响应曲线'); xlabel('t'); ylabel('sin(t+pi/6)'); grid

G=tf([5,5],[1,4,2,3,0]); t=[0:0.1:150]'; u=cos(t+pi/6); lsim(G,u,t) ; title('余弦信号输出响应曲线'); xlabel('t'); ylabel('cos(t+pi/6)'); grid

G=tf([5,5],[1,4,2,3,0]); t=[0:0.1:150]'; u=sin(t+pi/6); lsim(G,u,t) ; title('正弦信号输出响应曲线'); xlabel('t'); ylabel('sin(t+pi/6)'); grid

#ifndef max_h #define max_h 1 int Max(int, int, int); double pi=3.14; #endif

function q = IntQBXF2(func,n) format long; pi = 3.1415926535; q = 0; A = zeros(n,1); x = zeros(n,1); for i=1:n A(i) = sin((i*pi)/(n+1))*sin((i*pi)/(n+1))*pi/(n+1); x(i) = cos(pi*i/(n+1

function q = IntQBXF1(func,n) format long; pi = 3.1415926535; q = 0; A = zeros(n,1); x = zeros(n,1); for i=1:n A(i) = pi/n; x(i) = cos(pi*(2*i-1)/2/n); y(i) = subs(sym(func), find

function [A0,A,B]=FZZ(func,T, n) syms t; func = subs(sym(func), findsym(sym(func)),sym('t')); A0=int(sym(func),t,-T/2,T/2)/T; for(k=1:n) A(k)=int(func*cos(2*pi*k*t/T), t,-T/2,T/2)*2/T; A

function c = DFF(f,N) c(1:N)=0; for(m=1:N) for(n=1:N) c(m)=c(m)+f(n)*exp(-i*m*n*2*pi/N); end c(m)=c(m)/N; end