How to measure phase speed in Hovmoller

I have this Hovmoller plot of OLR anomalies with longitude x time, only showing the negative anomalies:

I want to know if theres a way to calculate the phase speed of this anomaly. That is, following the OLR minima over the time and then do a linear least square fit and obtain the quantity. Also, show the points and the fit in the same plot :)

Thank you, Ive been stucked with this for weeks

[Sam Chak]

@Luis Jesús Olvera Lazcano, could you please provide both the Hovmöller diagram code and the mathematical formula for the phase speed? This information will assist us in measuring the phase speed in MATLAB.

[Luis Jesús Olvera Lazcano]

Hi im so sorry for taking so long to answer. I finally discover how to do it I will put my code for anyone who wants it in the future.

Thank you for your interest in helping!

[Sam Chak]

Thank you, @Luis Jesús Olvera Lazcano. I'm looking forward to seeing your code in your answer. By the way, it is meaningful to help others, and it is absolutely fine to 'click-accept' your self-discovered solution.

[Luis Jesús Olvera Lazcano]

Im sorry for the wait @Sam Chak, Ive been busy these days. I will post my solution down below, thank you for your attention. Besides my own answer, @Shubham also brings another solution that might work for anyone else in the future.

Answer 1

[Shubham]

Hey @Luis Jesús Olvera Lazcano,

I understand that you want to calculate the phase speed of the anomalies.

You can try finding the minima points from the data. You can refer to the “islocalmin” function. The function returns a logical array representing the indices for the local minimum.

https://www.mathworks.com/help/matlab/ref/islocalmin.html

After extracting the minima points you can fit a linear curve between them. You can use the “polyfit” function to get a linear fit.

https://www.mathworks.com/help/matlab/ref/polyfit.html

You can use the coefficients returned by the “polyfit” function and use them to show the points and the fit in the same plot.

For the phase speed you can try using the coefficient of the linear fit itself.

Hope this helps!!

[Luis Jesús Olvera Lazcano]

Wow, this sounds like a more elegant solution!

I actually found myself the method to obtain the minima, as it requires some thresholds to calculate the phase speed correctly.

Nevertheless I appreciate your answer, Im looking forward to use it in the future!

Answer 2

[Luis Jesús Olvera Lazcano]

My solution to obtain the phase speed of a wave in the OLR anomaly-field Hovmoller is as it follows:

for x=2:3 
    caso1 = hov_indico{x}(:,:); % cell containing x Hovmollers, i.e., 180 x 51: longitude x days
    % longitude goes 2° per position, meaning: 180°, 178°W,...
     long_step = 31:2:91; % doing a 2 step longitude. Starting point of the wave is 31: 60°W
    for i=long_step 
        j = squeeze(caso1(i,:));
         if min(j)>-5 % if the minimum of the vector is higher than -5 W/m^2
             break
         else
        [k(long_step==i), idx(long_step==i)] = min(j); % k Obtains the longitudinal minimum value
        % and idx gets the position of the minimum 
        longitude(long_step==i)=lon(i); % captures the position of the longitudinal minimums
%             if idx(long_step==i)-idx(long_step==(i-5))<=-2 % condicional para evitar que el ciclo
            if idx(long_step==i)<idx(long_step==(i-10))
                % condition to avoid that the cycle captures minimum values
                % that are days below the i-step, i.e., avoid capturing
                % minimum of previous days
                idx(long_step==i)=[]; % if the condition is true, then the last input of the cycle
                % is empty and the cycle breaks
                longitude(long_step==i)=[];
                k(long_step==i)=[];
               break
            end
        end
    end
    tiempos = t_in(1,idx); % vector containing only the days that have minimum, same length as longitude vector
    
    p = polyfit(longitude,tiempos,1); % polyfit to find the slope, i.e., the phase speed
    phase_speed{2:3==x,1} = (1/p(1))*(111319.44/86400); % as the slope is y/x: days/longitude, divide 1 over the slope
    % then multiply per 111319.44 meters/86400 seconds to obtain the phase speed in m/s
    
    pv{2:3==x,1} = polyval(p,longitude); % evaluate the polyfit coefficients to obtain the fit
end

The resulting plot including the OLR anomalies and the polyval fit is:

having a phase speed of 4.7 m/s