please help me with this

I plotted the graph as:

beta = 0;                % Pitch angle 
ind2 = 1;

for lambda=0.1:0.01:11.8
   
    lambdai(ind2) = (1./((1./(lambda-0.02.*beta)+ (0.003./(beta^3+1)))));
    Cp(ind2)=0.73.*(151./lambdai(ind2)-0.58.*beta-0.002.*beta^2.14-13.2).*(exp(-18.4./lambdai(ind2)));
    ind2=ind2+1;
end
tab_lambda=[0.1:0.01:11.8];

% Kopt for MPPT (maximum power point tracking)
Cp_max=0.44;
lambda_opt=7.2;
kopt = ((0.5*ro*pi*(Radius^5)*Cp_max)/(lambda_opt^3));

figure
subplot(1,3,3)
plot(tab_lambda,Cp,'linewidth',1.5)
xlabel('lambda','fontsize',14)
ylabel('Cp','fontsize',14)

Now, I want the graph to be like this:

What I must change in the code to plot this new graph???

Please help me.

[Image Analyst]

Doesn't run. What is ro and Radius?

[rami shaker]

Radius = 46;

ro = 1.225;

Answer 1

[Alan Stevens]

Simply insert

hold 
xvals = [0 7.2 7.2]; yvals = [0.36 0.36];
plot(xvals,yvals,'--')

after your plot command (though the peak of the first curve you showed is higher than 0.36).

[rami shaker]

the first curve the peak is y=0.44 at x=7.2

but i want the peak will be y=0.36 at x=7.2 (2nd graph)

[rami shaker]

it will draw a dot , not a curve

[rami shaker]

it will draw a dot not a curve

[rami shaker]

I want that the curve will be like this (if you understand the issue)

Answer 2

[Steven Lord]

This doesn't quite do exactly what you want but it's close.

% Define the curve
x = 0:0.25:10;
y = 50-(x-4).^2;

% Plot it
plot(x, y);
hold on

% Find the peak
[maxY, maxYLoc] = max(y);

% Plot the dashed lines and the "x marks the spot"
xline(x(maxYLoc), '--')
yline(maxY, '--')
plot(x(maxYLoc), maxY, 'x')

% Set the limits so there's room to see the horizontal dashed line
ylim([0 60])

[rami shaker]

there is an error with your code:

Undefined function or variable 'xline'.

Error in int3 (line 13)

xline(x(maxYLoc ), '--')

[Walter Roberson]

https://www.mathworks.com/help/matlab/ref/xline.html

R2018b or later

[rami shaker]

Cp(ind2)=0.73.*(151./lambdai(ind2)-0.58.*beta-0.002.*beta^2.14-13.2).*(exp(-18.4./lambdai(ind2)));

i want this equation to be drawn not this one: y = 50-(x-4).^2

[Image Analyst]

What are those things? Is lambdai() a function or a vector? What is the "x" value -- ind2 or beta? Are ind2 and beta vectors or scalars?

How about this:

plot(allInd2Values, Cp, 'b-', 'LineWidth', 2);
grid on;
xlabel('ind2');
ylabel('Cp');

[rami shaker]

beta = 0;                % Pitch angle 
ind2 = 1;
for lambda=0.1:0.01:11.8
   
    lambdai(ind2) = (1./((1./(lambda-0.02.*beta)+ (0.003./(beta^3+1)))));
    Cp(ind2)=0.73.*(151./lambdai(ind2)-0.58.*beta-0.002.*beta^2.14-13.2).*(exp(-18.4./lambdai(ind2)));
    ind2=ind2+1;
end
tab_lambda=[0.1:0.01:11.8];
% Kopt for MPPT (maximum power point tracking)
Cp_max=0.44;
lambda_opt=7.2;
kopt = ((0.5*ro*pi*(Radius^5)*Cp_max)/(lambda_opt^3));
figure
subplot(1,3,3)
plot(tab_lambda,Cp,'linewidth',1.5)
xlabel('lambda','fontsize',14)
ylabel('Cp','fontsize',14)

i want to modify this code to plot the second curve with peak of 0.36 at x=7.2