How to perform integration inside for loop? [matrix dimension must agree issue]
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Md. Golam Zakaria
am 25 Feb. 2022
Bearbeitet: Md. Golam Zakaria
am 26 Feb. 2022
Hello Everyone I am a beginner at matlab. I am trying to perform integration inside a for loop. But I am getting 'matrix dimension must agree' error. I have a feeling that this is a minor problem and any expert can solve this problem within minuites. Would anyone be kind enough to spare few minuites to solve this proble. The code is below-
clc
clear all
close all
h=6.582*10^-16;
k=8.617*10^-5;
T=300;
beta=7.021*10^-4;
gamma=1108;
C1=5.5;
C2=4;
A1=3.231*10^2;
A2=7.237*10^3;
Ad=1.052*10^6;
Ep1=1.82*10^-2;
Ep2=5.773*10^-2;
Egd=3.2;
Eg0_1=1.1557;
Eg0_2=2.5;
Eg1=Eg0_1-((beta*(T^2))/(T+gamma));
Eg2=Eg0_2-((beta*(T^2))/(T+gamma));
Fs= 2.16*10^-5*pi; % Geometrical Factor for sun
H= 4.136*10^-15; % Plancks Constant
c= 3*10^8; % Speed of light
K = 8.6173*10^-5; % Boltzmanns Constant
Ts=5760; % Temparature of the sun
q=1.6*10^-19;
A=((2*Fs)/((H^3)*(c^2)));
x=1:0.004:5;
num=numel(x);
output=nan(1,num);
for v=1:num
lambda=0.2:0.001:1.2;
Irradiance=(A.*(((H*c)./lambda).^3./(exp((((H*c)./lambda)./(K.*Ts)))-1))).*q;
alpha=C1*A1*(((((h.*((2*pi)./lambda))-Eg1+Ep1).^2)./(exp(Ep1/(k*T))-1))+((((h.*((2*pi)./lambda))-Eg1-Ep1).^2)./(1-exp(-Ep1/(k*T)))))+C2*A2*(((((h.*((2*pi)./lambda))-Eg2+Ep2).^2)./(exp(Ep2/(k*T))-1))+((((h.*((2*pi)./lambda))-Eg2-Ep2).^2)./(1-exp(-Ep2/(k*T)))))+Ad.*((2*pi.*(c./lambda))-Egd).^(1/2);
depth=v;
attenuation=@(depth) (depth.*alpha);
alpha_x=integral(attenuation,0,depth');
output(v)=alpha.*Irradiance.*exp(-alpha_x);
end
plot(x,output)
For further simplification, I should note that I triend to use nested for loop for lambda, but that complecated the problem for me. But this code should also work.
0 Kommentare
Akzeptierte Antwort
VBBV
am 26 Feb. 2022
Bearbeitet: VBBV
am 26 Feb. 2022
clc
clear all
close all
h=6.582*10^-16;
k=8.617*10^-5;
T=300;
beta=7.021*10^-4;
gamma=1108;
C1=5.5;
C2=4;
A1=3.231*10^2;
A2=7.237*10^3;
Ad=1.052*10^6;
Ep1=1.82*10^-2;
Ep2=5.773*10^-2;
Egd=3.2;
Eg0_1=1.1557;
Eg0_2=2.5;
Eg1=Eg0_1-((beta*(T^2))/(T+gamma));
Eg2=Eg0_2-((beta*(T^2))/(T+gamma));
Fs= 2.16*10^-5*pi; % Geometrical Factor for sun
H= 4.136*10^-15; % Plancks Constant
c= 3*10^8; % Speed of light
K = 8.6173*10^-5; % Boltzmanns Constant
Ts=5760; % Temparature of the sun
q=1.6*10^-19;
A=((2*Fs)/((H^3)*(c^2)));
x=1:0.04:5;
num=numel(x);
output=zeros(1,num);
lambda=0.2:0.01:1.2
1:1.2;
for v=1:num
Irradiance=(A.*(((H*c)./lambda).^3./(exp((((H*c)./lambda)./(K.*Ts)))-1))).*q;
alpha=C1*A1*(((((h.*((2*pi)./lambda))-Eg1+Ep1).^2)./(exp(Ep1/(k*T))-1))+((((h.*((2*pi)./lambda))-Eg1-Ep1).^2)./(1-exp(-Ep1/(k*T)))))+C2*A2*(((((h.*((2*pi)./lambda))-Eg2+Ep2).^2)./(exp(Ep2/(k*T))-1))+((((h.*((2*pi)./lambda))-Eg2-Ep2).^2)./(1-exp(-Ep2/(k*T)))))+Ad.*((2*pi.*(c./lambda))-Egd).^(1/2);
attenuation=@(depth) (depth.*alpha);
depth=v;
alpha_x(v,:)=integral(@(depth) attenuation(depth),0,depth,'ArrayValued',true);
output(v,:)=alpha_x(v,:).*Irradiance.*exp(-depth);
end
plot(x,output)
axis([1 2 0 2600])
2 Kommentare
VBBV
am 26 Feb. 2022
Bearbeitet: VBBV
am 26 Feb. 2022
When computing output , why do you use alpha instead of alpha_x ?
What is the use of evaluating alpha_x that is not used anywhere in the code ?
Put
depth = v
after
attenuation=@(depth) (depth.*alpha);
This allows the assigned value for depth to be used correctly, otherwise it uses only symbolic variable
Weitere Antworten (1)
Torsten
am 25 Feb. 2022
Use
alpha_x{v} = integral(attenuation,0,depth,'ArrayValued',true);
output{v} = alpha.*Irradiance.*exp(-depth);
instead of
alpha_x=integral(attenuation,0,depth');
output(v)=alpha.*Irradiance.*exp(-depth);
But the result of your integration will simply be
alpha_x{v} = v*alpha
Is this really what you want ?
3 Kommentare
Torsten
am 25 Feb. 2022
Bearbeitet: Torsten
am 25 Feb. 2022
Yes, you need curly brackets for alpha_x as well as for output.
The reason is that for each v, alpha_x is a vector of the same length as lambda. In order to save these "num" vectors, they have either to be saved in a cell array (as done above) or in a matrix
alpha_x(:,v) = integral(attenuation,0,depth,'ArrayValued',true);
output(:,v) = alpha.*Irradiance.*exp(-alpha_x);
But the main question is whether the integral really gives the answer you expect, since - as said above -
alpha_x{v} = v*alpha
So no integration is needed to get this result.
Md. Golam Zakaria
am 25 Feb. 2022
Bearbeitet: Md. Golam Zakaria
am 25 Feb. 2022
Siehe auch
Kategorien
Mehr zu Logical finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!