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#include #include #define zero 1e-2 #define PI 3.141592653 int main() { double l1, l2, l3; double a1, a2, a3; double x1, x2, x3, y1, y2, y3; while(scanf("%lf%lf%lf%lf

#include #include int r, c; char map[21][21]; int mark[21][21][4]; int pi, pj, bi, bj, ti, tj; int f, p; int Bi, Bj; struct node { int pi, pj; int bi, bj; char wa

%-------------------------------------------------------------------------- % Copyright (C) Darius Plausinaitis % Written by Darius Plausinaitis %---------------------------------------------------

%产生信号序列 n=linspace(-pi,pi,1024); x=2*sin(n*40)+cos(n*100); figure,plot(x); title('输入信号'); %计算序列的DFT y1=fft(x); figure,plot(abs(y1)); grid on title('频谱'); %计算序列的PSD y2=conj(y1); Ppw=y1.*y2/

#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)); }

%改进双向空间平滑music算法 7阵元等距阵波达方向估计（d=λ/2）,2信源， 4个子阵（4阵元） 最小范数 clear all clc d=1; lma=2; N=2000; %设定采样点数 SNR=[3 3 2]; fs=8;

%改进双向空间平滑music算法 7阵元等距阵波达方向估计（d=λ/2）,2信源， 4个子阵（4阵元） 最小范数 clear all clc d=1; lma=2; N=1024; %设定采样点数 SNR=[2 4 6]; fs=8;

%改进双向空间平滑music算法，10阵元 等距阵波达方向估计（d=λ/2）,2信源 5个子阵（6阵元） v=2; N=2000; %设定采样点数 SNR=[-5 8]; fs=8; f

#include #include using namespace std; int main() {int a; couta; cout

#include using namespace std; int main() {int a=21; cout.setf(ios::showbase); cout