I have everything figured out except I cant seem to get the fplot for my anonymous variable A to project over figure(2). This issue may be complicated by the fact that I have a subplot going on in figure(1). The entire code is below, but the very bottom is where Im struggling. I dont get an error messgae, it just simply doesnt plot the function.
clear
clc
clf
%Origonal mass of decaying Radium-226
Anot= 20; %(mg)
%Decay rate
r=-0.000436;%(per year)
%Time elapsed, 20 evely spaced values
t=linspace(1,4000,20); %(years)
%Equation for mass as a function of time
original_data_A=Anot.*(exp(r.*t)); %(mg)
%Creating array the same size as origonal_data_A
numberOfSamples=length(original_data_A);
%Adds noise to original_data_A (+ or – 0.5, then scales each value to 10%)
addSampleNoise=(rand([1 numberOfSamples])-0.5).*(original_data_A.*0.1);
A_noisy=original_data_A+addSampleNoise;
fprintf('First value of A + noise: %0.04f mg \n',A_noisy(1));
fprintf('Last value of A + noise: %0.04f mg \n', A_noisy(end));
figure(1)
%plot linear x-axis vs linear y-axis
subplot(2,2,1)
plot(t,A_noisy,'*')
title('Linear-Linear Fit')
xlabel('Time (years)')
ylabel('Mass (mg)')
axis([0 4000 0 25])
%plot ln(x) vs ln(y)
subplot(2,2,2)
%transform the data from x -> ln(x) and y -> ln(y)
plot(log(t),log(A_noisy),'*')
title('Log-Log Fit')
xlabel('ln(Time (years))')
ylabel('ln(Mass (mg))')
axis([4 10 1 3.5])
%x linear, y-> ln(y)
subplot(2,2,3)
plot(t,log(A_noisy),'*')
title('Linear-Log Fit')
xlabel('Time (years)')
ylabel('ln(Mass (mg))')
axis([0 4000 1 3.5])
%x->ln(x), y linear
subplot(2,2,4)
plot(log(t),A_noisy,'*')
title('Log-Linear Fit')
xlabel('ln(Time (years))')
ylabel('Mass (mg)')
axis([4 10 0 25])
fprintf('The function resulting in a plot, closest to linear is EXPONENTIAL.')
fprintf('\nOtherwise, shown as the plot with linear x and natural log of y.\n')
%coeffitiants represent m (slope) and b (y intercept) of linearized
%equation
p=polyfit(t,log(A_noisy),1);
fprintf('First order polynomial fit returns \n m= %0.06f \n b= %0.06f',p(1),p(2))
%to get the true m and b out of polyfit, we revurse the form it is given
%in. m is linear in the exponential function, so we leave m the way it is.
b=exp(p(2));
m=p(1);
fprintf('\nThe resulting Anot = %0.06f \n The resulting r= %0.06f',b,m)
%Original equation reconstructed
A=@(t) m.*exp(b.*t);
figure(2);
hold on
plot(t,A_noisy,'*r');
title('Mass vs Time, Fitted Polynomial')
xlabel('Time (years)')
ylabel('Mass (mg)')
figure(2);
hold on
fplot(A,[min(t), max(t)],'-b')
Best Answer