代码搜索：continuous
找到约 2,697 项符合「continuous」的源代码
代码结果 2,697
www.eeworm.com/read/355157/10290133
m ackley.m
% ackley.m
% Ackley's function, from http://www.cs.vu.nl/~gusz/ecbook/slides/16
% and further shown at:
% http://clerc.maurice.free.fr/pso/Semi-continuous_challenge/Semi-continuous_challenge.htm
www.eeworm.com/read/355157/10290177
m griewank.m
% Griewank.m
% Griewank function
% described by Clerc in ...
% http://clerc.maurice.free.fr/pso/Semi-continuous_challenge/Semi-continuous_challenge.htm
%
% used to test optimization/global minimi
www.eeworm.com/read/355153/10290718
m griewank.m
% Griewank.m
% Griewank function
% described by Clerc in ...
% http://clerc.maurice.free.fr/pso/Semi-continuous_challenge/Semi-continuous_challenge.htm
%
% used to test optimization/global minimi
www.eeworm.com/read/353873/10410744
m display.m
function display(cdv)
% DISPLAY -
%
for k=1:length(cdv)
disp(sprintf('\n'))
disp('Continuous design variable')
disp('--------------------------')
display(cdv(k).gen)
txt=sprintf(
www.eeworm.com/read/353723/10425708
m ackley.m
% ackley.m
% Ackley's function, from http://www.cs.vu.nl/~gusz/ecbook/slides/16
% and further shown at:
% http://clerc.maurice.free.fr/pso/Semi-continuous_challenge/Semi-continuous_challenge.htm
www.eeworm.com/read/353723/10425753
m griewank.m
% Griewank.m
% Griewank function
% described by Clerc in ...
% http://clerc.maurice.free.fr/pso/Semi-continuous_challenge/Semi-continuous_challenge.htm
%
% used to test optimization/global minimi
www.eeworm.com/read/424063/10500325
m ben3bdat.m
T1 = 1/50;
T2 = 1/200;
T3 = 1/1000;
T4 = 1/2000;
offT1 = 1.0e-8;
offT2 = 0.0;
offT3 = -1.0e-8;
offT4 = -2.0e-8;
offZOH = 2.0e-8;
%***continuous transfers
gn1 = -15*[1,1.5];
gd1 = [1,6
www.eeworm.com/read/424063/10500625
m ben3adat.m
T1 = 1/50;
T2 = 1/200;
offT1 = 0.0;
offT2 = -1.0e-8;
offZOH = 1.0e-8;
%***continuous transfers
gn1 = -15*[1,1.5];
gd1 = [1,6,13];
gn2 = [-0.1,9];
gd2 = [1,6,13];
%***SIMO Transfer f
www.eeworm.com/read/160223/10555459
m ex0310.m
% Analog Signal
Dt = 0.00005;
t = -0.005:Dt:0.005;
xa = exp(-1000*abs(t));
%
% Continuous-time Fourier Transform
Wmax = 2*pi*2000;
K = 500; k = 0:1:K;
W = k*Wmax/K;
Xa = xa * exp(-j*t'*W) * Dt;
Xa = r
www.eeworm.com/read/278064/10576968
m ackley.m
% ackley.m
% Ackley's function, from http://www.cs.vu.nl/~gusz/ecbook/slides/16
% and further shown at:
% http://clerc.maurice.free.fr/pso/Semi-continuous_challenge/Semi-continuous_challenge.htm