Hello everyone!
Could you please help me with a plot. When I plot a mean v, it plots differently as 0. I need to mean line to be at the same angle as my fluctuations.
My code is:
clear all; close all;
N=100;
t = (0:N);
t1=(0:N-1);
t2=(0:N-1);
tt = rand(1,N) * 2 * pi;
ang = rand(1,N) * 2 * pi;
amp_u = 2;
amp_v = 1;
amp_w = 0.5;
u_mean=2;
v_mean=1;
w_mean=1;
u1 = amp_u.*sin(100*t2)+u_mean;
v1 = sin(ang) .* amp_v+v_mean;
w1=sin(ang) .* amp_w+w_mean;
u = cumsum([2, u1]);
x(1:N)=mean(u1);
v = cumsum([2, v1]);
w = cumsum([2, w1]);
figure(1); plot(t,u);
hold on
plot(t1,x);

1 Commento

Mean will be a scalar, constant value and it plots accordingly as well.
"When I plot a mean v ..."
You are not plotting v in your code above.
"I need to mean line to be at the same angle as my fluctuations."
Can you explain what do you mean by this? Perhaps a figure will be more helpful.

Accedi per commentare.

 Risposta accettata

‘I need to mean line to be at the same angle as my fluctuations.’
You apparently want the linear regression of ‘u1’ as a function of ‘t1’ not the mean of ‘u1’.
Try this —
% clear all; close all;
N=100;
t = (0:N);
t1=(0:N-1);
t2=(0:N-1);
tt = rand(1,N) * 2 * pi;
ang = rand(1,N) * 2 * pi;
amp_u = 2;
amp_v = 1;
amp_w = 0.5;
u_mean=2;
v_mean=1;
w_mean=1;
u1 = amp_u.*sin(100*t2)+u_mean;
v1 = sin(ang) .* amp_v+v_mean;
w1=sin(ang) .* amp_w+w_mean;
u = cumsum([2 u1]);
% x(1:N)=mean(u1);
B = [t(:) ones(size(t(:)))] \ u(:); % Linear Regression Parameters
x = [t2(1) 1; t2(end) 1] * B; % Linear Regression Evaluation
v = cumsum([2, v1]);
w = cumsum([2, w1]);
figure(1)
plot(t,u)
hold on
plot(t1([1 end]),x)
ylim([0 max(ylim)])
While I am thinking about it, do you have any thoughts on my Answer to plot a graph from user input?
.

Più risposte (2)

Your question is not veary clear, but I think you want to plot:
plot(t,mean(u1)*t)

Categorie

Scopri di più su 2-D and 3-D Plots in Centro assistenza e File Exchange

Prodotti

Release

R2022b

Tag

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by