how to find data points of function passed through a distorted "pipe"?
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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.....................
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John D'Errico
2023년 11월 7일
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.
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