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Plot and interp using Spline method

Asked by Ammar Ahmed on 17 Jun 2019
Latest activity Answered by the cyclist
on 17 Jun 2019
I'm trying to correct this line ploting using interpolate method .
but seem blocky shape appears
the coding in the body
function [TI1,TI2,TI3] = ER_Function
er1=(1+1.8*10^14*1i);
er2=(2.5+2.5*10^-3*1i);
Hc=(10^6);
H1=(0);
P1= 0:0.02:0.30;
Pc=(0.33).*exp((-(H1)/Hc));
c1=(1-3*Pc).*(((P1)./(Pc)).^Pc).*(((1-P1)./(1-Pc)).^(1-Pc));
T1=(1./(2.*(2+c1))).*(((3*P1-1+c1)*er1+(2-3.*P1-c1)*er2)+(((((3*P1-1+c1)*er1+(2-3.*P1)*er2).^2)+(4.*(er1).*(er2).*(1-c1).*(2+c1))).^0.5));
PI1=[0,0.20,0.23,0.25,0.27,0.30];
TI1= spline (P1,T1,PI1);
P2= 0:0.02:0.30;
er1=(1+1.8*10^14*1i);
er2=(2.5+2.5*10^-3*1i);
Hc=(10^6);
H2=(10^6);
Pc=(0.33).*exp((-(H2)/Hc));
c2=(1-3*Pc).*(((P2)./(Pc)).^Pc).*(((1-P2)./(1-Pc)).^(1-Pc));
T2=(1./(2.*(2+c2))).*(((3*P2-1+c2)*er1+(2-3.*P2-c2)*er2)+(((((3*P2-1+c2)*er1+(2-3.*P2)*er2).^2)+(4.*(er1).*(er2).*(1-c2).*(2+c2))).^0.5));
PI2=[0,0.20,0.23,0.25,0.27,0.30];
TI2= spline (P2,T2,PI2);
P3= 0:0.02:0.30;
er1=(1+1.8*10^14*1i);
er2=(2.5+2.5*10^-3*1i);
Hc=(10^6);
H3=(2*10^6);
Pc=(0.33)*(exp(-H3/Hc));
c3=(1-3*Pc).*(((P3)./(Pc)).^Pc).*(((1-P3)./(1-Pc)).^(1-Pc));
T3=(1./(2.*(2+c3))).*(((3*P3-1+c3)*er1+(2-3.*P3-c3)*er2)+(((((3*P3-1+c3)*er1+(2-3.*P3)*er2).^2)+(4.*(er1).*(er2).*(1-c3).*(2+c3))).^0.5));
PI3=[0,0.20,0.23,0.25,0.27,0.30];
TI3= spline (P3,T3,PI3);
J=plot ( PI1, real(TI1),'r',PI2, real (TI2),'g',PI3, real (TI3) , 'b');
J(1).LineWidth = 2;
J(2).LineWidth = 2;
J(3).LineWidth = 2;
title('Magnetic Field' ),xlabel('(P) concentration Volume'),ylabel('\bf\epsilon^e','FontSize',20)
legend({'H1=0','H2=10^6','H3= 2*10^6'},'Location','northwest','NumColumns',3)
end

  2 Comments

darova
on 17 Jun 2019
What is the problem?
lines are not smooth ,even when i used the interpolating method .
How I could make the lines smoother?Is there Specific command?

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1 Answer

Answer by the cyclist
on 17 Jun 2019

It's presumably because you have not chosen your query points to be very close together. Do you get something closer to what you expect if you define something like
PI3=[0:0.002:0.30];

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