How to make decreasing quadratic curve?

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Kalasagarreddi Kottakota
回答： Image Analyst 2022-10-20
Hi, I am looking a decreaing quadratic of red colour in the image below. But when I script, I get the curve (in blue). Can someone help me to fix this?
clear all; close all; clc;
fs = 10200; % sampling frequency
dt = 1/fs;
T = 26; % total time
tVec = dt:dt:T; % time vector
rpm2 = 4500; f2 = rpm2/60;
rpm3 = 0; f3 = rpm3/60;
beta = (f3-f2)/T.^2; % slope rate;
finst = f2 + beta*tVec.^2.';
Speed = finst * 60;
figure()
plot(tVec, Speed)
grid on; box on;

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采纳的回答

DGM
DGM 2022-10-20
Ran in:
Are you after something like this?
fs = 10200; % sampling frequency
dt = 1/fs;
T = 26; % total time
rpm2 = 4500; f2 = rpm2/60;
rpm3 = 0; f3 = rpm3/60;
tVec = dt:dt:T; % time vector
beta = (f3-f2)/T.^2; % slope rate;
finst = f3 - beta*(T-tVec).^2.';
Speed = finst * 60;
plot(tVec, Speed)
grid on; box on;

更多回答（1 个）

Image Analyst
Image Analyst 2022-10-20
Ran in:
Well your formula just doesn't do that. But here's something closer. At least I have the colors of the lines changing.
all_rpm2 = linspace(4500, -1000, 10);
numCurves = numel(all_rpm2)
numCurves = 10
% Create red colormap
cmap = zeros(numCurves, 3);
cmap(:, 1) = linspace(0.2, 1, numCurves)';
fs = 10200; % sampling frequency
dt = 1/fs;
T = 26; % total time
tVec = dt:dt:T; % time vector
for k = 1 : length(all_rpm2)
rpm2 = all_rpm2(k)
f2 = rpm2/60;
rpm3 = 0;
f3 = rpm3/60;
beta = (f3-f2)/T.^2; % slope rate;
finst = f2 + beta * tVec.^2.';
Speed = finst * 60;
plot(tVec, Speed, 'Color', cmap(k, :), 'LineWidth', 3)
grid on;
box on;
hold on;
end
rpm2 = 4500
rpm2 = 3.8889e+03
rpm2 = 3.2778e+03
rpm2 = 2.6667e+03
rpm2 = 2.0556e+03
rpm2 = 1.4444e+03
rpm2 = 833.3333
rpm2 = 222.2222
rpm2 = -388.8889
rpm2 = -1000

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