フィルターのクリア

can anyone check this code?

3 ビュー (過去 30 日間)
Atrolita afra
Atrolita afra 2020 年 4 月 2 日
コメント済み: Ameer Hamza 2020 年 4 月 2 日
function l=LCR_CF(k,c,sigma)
%% pdf of process of various values of system parameters
for x=0:.0625:10
temp=0;
LCR=0;
ro=(x.^2/sigma);
ro=20*log10(ro);
for p=0:10
for n=0:p+1
% closed term expansion
b=nchoosek(p+(1/2),n);
% gamma term calculation
f=factorial(p);
g=(gamma(p+1))*(gamma(c));
g_ma=((ro.^(p+0.5))*(k.^c)*(2*sqrt(2*pi)))/(g*f);
% bessel function term calculation
z=2*sqrt(k*(1+ro));
K = besselk(p-n+c,z);
% Exponential term calculation
exponential=exp(-ro);
% Last term
q=(p-n+c)/2;
last_term=((k/(1+ro)).^q);
% Final pdf function
temp1= b*g_ma*K*exponential*last_term;
temp=temp+temp1;
end
LCR=LCR+temp;
end
l(1,x/(.0625)+1)=LCR;
end
THIS IS THE FUNCTION I AM TRYING TO PLOT BUT I AM GATTING AN ERROR ,""Undefined function 'LCR_CF' for input arguments of type 'double'.""
HOW CAN I SOLVE THIS?
I HAVE CALLED THIS FUNCTION BY THE FOLLOWING CODE:
clc
clear all
close all
syms x ;
%% For different values of k and c. Change the value of k and c according to figure.
sig=1;
y1=LCR_CF(0.2,1.5,sig); % Put the values of k,c,sig accordingly
y2=LCR_CF(0.5,1.5,sig);
y3=LCR_CF(1,1.5,sig);
y4=LCR_CF(1.5,1.5,sig);
y5=LCR_CF(.2,2,sig);
y6=LCR_CF(.5,2,sig);
y7=LCR_CF(1,2,sig);
y8=LCR_CF(1.5,2,sig);
%% Ploting data
x=0:0.0625:10;
figure
% plot(y1,x,'b',y2,x,'b',y3,x,'b',y4,x,'b',y5,x,'g--',...
% y6,x,'g--',y7,x,'g--',y8,x,'g--');
plot(x,y1,'b',x,y2,'b',x,y3,'b',x,y4,'b',x,y5,'g--',...
x,y6,'g--',x,y7,'g--',x,y8,'g--');
xlim([0 10]);
ylim([0 1]);
title('Normalized LCR');
xlabel('x');
ylabel('N(x)');
txt = {'Sigma=1','C=1.5, Solid Line', 'C=2, Dotted Line'};
text(6,6,txt);
txt = {'k=1.5'};
text(1,6,txt);
  13 件のコメント
Atrolita afra
Atrolita afra 2020 年 4 月 2 日
Atrolita afra
Atrolita afra 2020 年 4 月 2 日
instead of infinity i was trying with 10

サインインしてコメントする。

採用された回答

Ameer Hamza
Ameer Hamza 2020 年 4 月 2 日
I have corrected the syntax issue in your code. The MATLAB function nchoosek does not work for fractional numbers. However, Now it gives infinity at 0 and also gives complex values. I am not sure whether this is correct or wrong.
function l=LCR_CF(k,c,sigma)
%% pdf of process of various values of system parameters
nck = @(n, k) gamma(n+1)./(gamma(k+1).*gamma(n-k+1));
X = 0:.0625:10;
l = zeros(size(X));
for i=1:numel(X)
x = X(i);
temp=0;
LCR=0;
ro=(x.^2/sigma);
ro=20*log10(ro);
for p=0:10
for n=0:p+1
% closed term expansion
b=nck(p+(1/2),n);
% gamma term calculation
f=factorial(p);
g=(gamma(p+1))*(gamma(c));
g_ma=((ro.^(p+0.5))*(k.^c)*(2*sqrt(2*pi)))/(g*f);
% bessel function term calculation
z=2*sqrt(k*(1+ro));
K = besselk(p-n+c,z);
% Exponential term calculation
exponential=exp(-ro);
% Last term
q=(p-n+c)/2;
last_term=((k/(1+ro)).^q);
% Final pdf function
temp1= b*g_ma*K*exponential*last_term;
temp=temp+temp1;
end
LCR=LCR+temp;
end
l(i)=LCR;
end
  3 件のコメント
Atrolita afra
Atrolita afra 2020 年 4 月 2 日
Oh i got your logic thanks for the infornation. though i don't know wheather it is the correct figure or not but i got that i will manage the rest !
Ameer Hamza
Ameer Hamza 2020 年 4 月 2 日
Glad to be of help. Best of luck for your project.

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeLogical についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by