how to find time interval of peak value in a timeseries ?

5 vues (au cours des 30 derniers jours)
Harish Dhanasekaran Velayutha Rajan
Commenté : Star Strider le 4 Août 2017
i have a load profile data set.. Time vs Load Demand as a timeseries object. If I have a threshold for peak as 800KW How do I find at what time my value goes above 800 kW ? The time series object is for every 30 mins interval over 24 hrs.
I need to find the highlighted time interval in the picture. Any idea?

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Star Strider
Star Strider le 4 Août 2017
If you want to interpolate to find the ‘exact’ time, this works:
t = 0 : 30 : 1440;
LD = 500 + 400*sin(2*pi*t/2880);
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0); % Returns Approximate Zero-Crossing Indices Of Argument Vector
X800_idx = zci(LD - 800); % Subtract 800, Find Approximate Zero-Crossing
LD800_Rng = LD(X800_idx-1 : X800_idx+1)'; % Define Range For Interpolation
t800_Rng = t(X800_idx-1 : X800_idx+1)'; % Define Range For Interpolation
t800 = interp1(LD800_Rng, t800_Rng, 800, 'linear', 'extrap'); % Find ‘Exact’ Time (Minutes)
figure(1)
plot(t, LD)
hold on
plot(t800, 800, '*r')
hold off
grid
  1 commentaire
Star Strider
Star Strider le 4 Août 2017
A slightly more robust version that finds the points where ‘LD’ crosses 800:
t = 0 : 30 : 1440; % Create Data
LD = 500 + 400*sin(2*pi*t/2880); % Create Data
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0); % Returns Approximate Zero-Crossing Indices Of Argument Vector
X800_idx = zci(LD - 800); % Subtract 800, Find Approximate Zero-Crossing
for k1 = 1:numel(X800_idx)
LD800_Rng = LD(X800_idx(k1)-1 : X800_idx(k1)+1)'; % Define Range For Interpolation
t800_Rng = t(X800_idx(k1)-1 : X800_idx(k1)+1)'; % Define Range For Interpolation
t800(k1) = interp1(LD800_Rng, t800_Rng, 800, 'linear', 'extrap'); % Find ‘Exact’ Time (Minutes)
end
figure(1)
plot(t, LD) % Plot Load Curve
hold on
plot(t800, [1 1]*800, '*r') % Plot ‘Load = 800’ Points
hold off
grid

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Plus de réponses (1)

fbaillon
fbaillon le 4 Août 2017
You can use the find function:
T= ... % Time
LD= .... % Load Demand
timeYouWant=T(find(LD>800e3,1,'first))
Something like that...

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