how to find data points of function passed through a distorted "pipe"?
1 view (last 30 days)
Show older comments
This is not going to be your ordinary question, but hopefully somebody can assist me.
I want to take a function of data points, for instance a random set of points ("rnpi" in my code), and fit it to an upper and lower bounding function, such as a pair of sine waves (AS1 and AS2 in my code). The data point locations will be distorted by the waves, so that a straight line will come out curvy, etc. I am including a diagram showing what I want. How do I do this?
iP = [0, 88];
iB = [30, 74];
npi = 200; % SHOULD BE EQUAL TO LENGTH OF RANDOM PATTERN BELOW
aa1 = iP(1); % MIN OF RANGE
aa2 = iP(2); % MAX OF RANGE
ab1 = iB(1); % MIN OF RANGE
ab2 = iB(2); % MAX OF RANGE
nz1 = 0:pi/npi:4*pi;
nz2 = pi + (0:pi/npi:4*pi);
nr1 = 3/2; % NUMBER OF REPEATS /2
nr2 = 5/2; % NUMBER OF REPEATS /2
AS1 = (1 + cos(pi + nz1*nr1))/2; % 0-1
AS1 = aa1 + (ab1 * AS1); % aa1-ab1
AS2 = (1 + cos((pi*nr2)-(nz2*nr2)))/2; % 0-1
AS2 = ab2 + ((aa2-ab2) * AS2); % ab1-ab2
plot(AS1)
hold on
plot(AS2)
hold off
% NOW TEST WITH A RANDOM PATTERN
rnpi = round(88 * rand(npi,1));
% THE REST GOES HERE.....................
0 Comments
Accepted Answer
John D'Errico
on 7 Nov 2023
A difficult question to answer, since so much is left to guess. And the vaguesness of your question suggests all you want is a result that looks qualitatively as you have drawn. Mathematics is not good at subjective things. Sorry.
But is is not difficult to do something that looks vaguely as you have drawn, as simply an interpolation between two curves.
ULim = [74 88];
LLim = [0 30];
XLim = [0 2*pi];
x = linspace(XLim(1),XLim(2));
UpperFcn = @(x) (cos(x)+1)/2*diff(ULim) + ULim(1);
LowerFcn = @(x) (1-(cos(x))/2)*diff(LLim) + LLim(1);
fplot(UpperFcn,XLim)
hold on
fplot(LowerFcn,XLim)
Now just create an interpolated function that trades off the two boundaries.
TFcn = @(x) (x-XLim(1))./diff(XLim);
MidFcn = @(x) (1-TFcn(x)).*UpperFcn(x) + TFcn(x).*LowerFcn(x);
fplot(MidFcn,XLim)
Again, totally subjective. Just a picture that looks vaguely like what you asked about.
2 Comments
More Answers (0)
See Also
Categories
Find more on Logical in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!