Integration of a function.
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
JACINTA ONWUKA
am 16 Aug. 2019
Kommentiert: JACINTA ONWUKA
am 16 Aug. 2019
I'm trying to integrate a function but is giving be this error below
Incorrect dimensions for matrix multiplication.
Check that the number of columns in the first matrix matches the number
of rows in the second matrix. To perform elementwise multiplication, use '.*'.
I cant identify to set the dimention right but is still giving error. Please help me.
L=1;
T=100;
r=0.03;
I1=0.1;
p=0.1;
epsilon=0.3;
rho=1000;
beta= 0.1;
Cr=3000;
RhoP=0;
Mycp = 0:10:100;
n = zeros(numel(Mycp),1 );
n2 = zeros(numel(Mycp),1 );
n3 = zeros(numel(Mycp),1 );
Jcp = zeros(numel(Mycp),1 );
for i = 1:numel(Mycp )
MycpCurrent=Mycp(i);
delta = 1-MycpCurrent/100;
tau = (1/(beta*(L+delta*p)))*log((L*(I1+delta*p))/(delta*p*(L-I1 )));
t05 =(1/(beta*(L+delta*p)))*log((L*(0.05*L+delta*p))/(delta*p*(L-0.05*L )));
I2= @(t)(L*delta*p*(exp (beta*(L+delta*p)*t)-1)) ./ (L + delta*p* exp(beta*(L+delta*p)*t ));
I3= @(t)(L*(I1+delta*p)*exp((epsilon*beta)*(L+delta*p)*(t-tau))-...
delta*p*(L -I1))./(L-I1+(I1+delta*p)*exp(epsilon*beta*(L+delta*p)*(t-tau)));
fun = @(t,MycpCurrent) MycpCurrent*L*exp(-r*t);
fun2=@(t) rho*I2(t)*I3(t).*exp(-r*t)+((Cr*L*exp(-r*tau)));
fun3=@(t)RhoP*I2(t)*I3(t).*exp(-r*t);
n(i) = integral(@(t)fun(t,MycpCurrent),0,100, 'ArrayValued',1);
n2(i)= integral(fun2,t05,tau); %this is giving me error.
n3(i)= integral(fun3,tau,100);
end
0 Kommentare
Akzeptierte Antwort
infinity
am 16 Aug. 2019
Hello,
You just need to remove "." in the formulation of I2, I3, fun2 and fun3. The code will run withour errors.
Weitere Antworten (0)
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!