How to perform summation with double subscript notation?
古いコメントを表示
Hello everyone, I have to perform some analysis based on the following equation, which contains summation operator with double subscript notation.
I have written a code. Can anyone please look at it and confim whether I am wrong or right?
alpha =0;
for i=1:2
for j=1:2
alpha=alpha +(C(i)*A(j)*((((E-Eg(j)+Ep(i))^2)/(exp(Ep(i)/(k*T))-1))+(((E-Eg(j)-Ep(i))^2)/(1-exp(-Ep(i)/(k*T))))));
end
end
Alpha=(alpha+(Ad*((E-Egd)^(1/2))));
採用された回答
First, provide the ‘C’ and the other missing vectors, then save ‘alpha’ as a matrix, then use the sum function to sum its elements.
alpha =0;
for i=1:2
for j=1:2
alpha(i,j) = C(i)*A(j)*((((E-Eg(j)+Ep(i))^2)/(exp(Ep(i)/(k*T))-1)));
end
end
Alpha=(sum(alpha(:))+(Ad*((E-Egd).^(1/2))))
Try that with the vectors to see if the result is as desired.
.
4 件のコメント
Md. Golam Zakaria
2022 年 3 月 6 日
h=4.136*10^-15; % Planck's Constant
k=8.617*10^-5; % Boltzmann's Constant
c=3*10^8; % speed of light
T=300; % Ambient Temparature
beta=7.021*10^-4;
gamma=1108;
Eg0_1=1.1557;
Eg0_2=2.5;
Egd0=3.2;
Eg1=Eg0_1-((beta*(T^2))/(T+gamma));
Eg2=Eg0_2-((beta*(T^2))/(T+gamma));
Egd=Egd0-((beta*(T^2))/(T+gamma));
Eg=[Eg1 Eg2];
Ep=[1.827*10^-2 5.773*10^-2];
C=[5.5 4.0];
A=[3.231*10^2 7.237*10^3];
Ad=1.052*10^6;
walenength=(.2*10^-6):(.0001*10^-6):(1.2*10^-6);
num=numel(walenength);
Alpha=nan(1,num);
for t=1:num
lambda=walenength(t);
E=((h*c)/lambda);
alpha =0;
for i=1:2
for j=1:2
alpha=alpha +(C(i)*A(j)*((((E-Eg(j)+Ep(i))^2)./(exp(Ep(i)/(k.*T))-1))+(((E-Eg(j)-Ep(i))^2)/(1-exp(-Ep(i)/(k*T))))));
end
end
Alpha(t)=(alpha+(Ad*((E-Egd)^(1/2))));
end
figure
plot((walenength./10^-6),real(Alpha))
hold on
plot((walenength./10^-6),imag(Alpha))
plot((walenength./10^-6),abs(Alpha),'--')
hold off
set(gca,'YScale','log')
ylim([(10^0) (10^9)])
xlabel('Wavelength \lambda ,(\mum)')
ylabel('Absorption Coefficient, \alpha(m^{-1})')
legend('Re(\alpha(T))','Im(\alpha(T))','|\alpha(T)|', 'Location','best')
This is difficult to follow. I do not understand taking the square root in the second term of the ‘Alpha(t)’ assignment, since the square root is not in the posted symbolic code.
It would likely be best for you to go through this to be certain there are no coding errors, since this is not an area of my expertise. I have no idea what the ‘expected’ result is, so I have nothing to compare it to.
.
Md. Golam Zakaria
2022 年 3 月 6 日
@Star Strider actually the square root is there, I mistyped the equation. You just tell me if there is any coding error regarding the summation.
I do not see anything wrong with it. The easiest way to troubleshoot it is to see what the individual terms evaluate to, and then see if those are correct —
h=4.136*10^-15; % Planck's Constant
k=8.617*10^-5; % Boltzmann's Constant
c=3*10^8; % speed of light
T=300; % Ambient Temparature
beta=7.021*10^-4;
gamma=1108;
Eg0_1=1.1557;
Eg0_2=2.5;
Egd0=3.2;
Eg1=Eg0_1-((beta*(T^2))/(T+gamma));
Eg2=Eg0_2-((beta*(T^2))/(T+gamma));
Egd=Egd0-((beta*(T^2))/(T+gamma));
Eg=[Eg1 Eg2];
Ep=[1.827*10^-2 5.773*10^-2];
C=[5.5 4.0];
A=[3.231*10^2 7.237*10^3];
Ad=1.052*10^6;
walenength=(.2*10^-6):(.0001*10^-6):(1.2*10^-6);
num=numel(walenength);
Alpha=nan(1,num);
for t=1:(num/1000)
lambda=walenength(t);
E=((h*c)/lambda);
alpha =0;
for i=1:2
for j=1:2
fprintf([repmat('—',1, 20) '\nt = %4d\ti = %d\tj = %d\n'],t,i,j)
Term_1(i,j) = (((E-Eg(j)+Ep(i))^2)./(exp(Ep(i)/(k.*T))-1))
Term_2(i,j) = (((E-Eg(j)-Ep(i))^2)/(1-exp(-Ep(i)/(k*T))))
alpha=alpha +(C(i)*A(j)*((((E-Eg(j)+Ep(i))^2)./(exp(Ep(i)/(k.*T))-1))+(((E-Eg(j)-Ep(i))^2)/(1-exp(-Ep(i)/(k*T))))));
end
end
Alpha(t)=(alpha+(Ad*((E-Egd)^(1/2))));
end
figure
plot((walenength./10^-6),real(Alpha))
hold on
plot((walenength./10^-6),imag(Alpha))
plot((walenength./10^-6),abs(Alpha),'--')
hold off
set(gca,'YScale','log')
ylim([(10^0) (10^9)])
xlabel('Wavelength \lambda ,(\mum)')
ylabel('Absorption Coefficient, \alpha(m^{-1})')
legend('Re(\alpha(T))','Im(\alpha(T))','|\alpha(T)|', 'Location','best')
See if the intermediate values appear to be correct.
.
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