Repeating x-axis for rectangular signal

Question

youjarr 2019-1-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/441289-repeating-x-axis-for-rectangular-signal

评论： Star Strider 2019-1-26

Hey guys,

i have an rectangular signal i want to plot.

On the Y axis is the voltage from 0 to 3.3 V (24945x1 single) and on the X axis is the position of the motor (circle) from 0-360°. (24945x1 double)

I have both signals but like I exprected it when I do

plot(Angle, Voltage)

it is not working like i wish. How can i let the X axis repeat it self and not overlapping the other datapoints.

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Answer 1

Star Strider 2019-1-24

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https://ww2.mathworks.cn/matlabcentral/answers/441289-repeating-x-axis-for-rectangular-signal#answer_357856

在 MATLAB Online 中打开

Try something like this:

Angle = reshape(([1; 1; 1]*(0:359))', 1, []);                   % Create Data;
Voltage = reshape(sind([1; 10; 20]*(0:359))', 1, []);           % Create Data;
figure
plot(Angle, Voltage)                                            % Original Plot
contAngle = unwrap(Angle*pi/180)*180/pi;                        % Use ‘unwrap’ To Recover Serial Vector
              
figure
plot(contAngle, Voltage)                                        % ‘Unwrapped’ Plot
xt = get(gca, 'XTick');
xtn = min(xt) : 90 : max(xt);
set(gca, 'Xtick',xtn, 'XTickLabel',ceil(rem(xtn,360.1)))
xlim([min(contAngle)  max(contAngle)])

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youjarr 2019-1-25

Hey,

thanks for your fast response. This is exactly what i wanted but now I see that i maybe did a mistake somewhere.

To tell you the whole story:

The x-axis was for the time in seconds but i needed to change it into angles to see the rotation/position of the movement.

so i wrote this small algorithm to convert my time in seconds to angle in degree u see here.

            t = rot90(t);
            t = flipud(t);
            t_start = t(1,1);
            
            %Calculation of the dt of 1 cycle (0°-360°)
            dt360_H1 = abs(PosNeg_Hall_1(1,2)-PosNeg_Hall_1(5,2));
            dt360_H2 = abs(PosNeg_Hall_2(1,2)-PosNeg_Hall_2(5,2));
            dt360_H3 = abs(PosNeg_Hall_3(1,2)-PosNeg_Hall_3(5,2));
            
            %Take the mean value of this three dt´s
            dt360 = (dt360_H1+dt360_H2+dt360_H3)/3;
            
            for ii = 1:length(t_90)
                calc = abs((360*(t_start-t_90(ii,1)))/dt360);
                if(calc > 359.95 && calc <= 360)
                    app.sek2degp(ii,1) = 0;
                    t_start = t_90(ii,1);
                else
                     app.sek2degp(ii,1) = calc;
                end
            end

To explain it a little bit more... one cycle starts at positive flank and ends at the fith positive flank this is equal to 0°-360°.

You think it is correct or logical what i have done here?

sek2deg will then be put in your written code as ContAngle

youjarr 2019-1-25

Thanks for your fast response. I hope now it is more readable. Im not sure about the if condition

            
            %t = time vector
            t = rot90(t); 
            t = flipud(t);
            t_start = t(1,1); %My assumed start point for the time
            
            %Calculation of the dt of 1 cycle (0°-360°) between 
            %the first pos. flank and the fifth pos. flank of all three Hall Signals
            dt360_H1 = abs(PosNeg_Hall_1(1,2)-PosNeg_Hall_1(5,2));
            dt360_H2 = abs(PosNeg_Hall_2(1,2)-PosNeg_Hall_2(5,2));
            dt360_H3 = abs(PosNeg_Hall_3(1,2)-PosNeg_Hall_3(5,2));
            
            %Take the mean value of this three dt´s
            dt360 = (dt360_H1+dt360_H2+dt360_H3)/3;
            
            for ii = 1:length(t_90)
                %here I calculate the angle out of the time
                calc = abs((360*(t_start-t_90(ii,1)))/dt360); 
                %This works like a "%" (modulo) but i am not reaching exactly the value
                %360° so i can´t use modulo
                if(calc > 359.95 && calc <= 360)
                    app.sek2degp(ii,1) = 0;
                    %if the "modulo" condition is true i save a new start/zero point
                    t_start = t_90(ii,1);
                else
                    %Just save the converted value
                     app.sek2degp(ii,1) = calc;
                end
            end

youjarr 2019-1-26

I´m not understanding the option 'MinSeparation'?! Separation between what because i can type in what i want it seems the the plots always looks the same?

And How can I use islocalmax to identify Positive nd negative flanks?

It seems that it can work much better than im doing now with diff.

But i dont understand it perfectly and I think the noise is maby to difficult for that function.

Star Strider 2019-1-26

From the documentation on islocalmax:

‘Minimum separation between local maxima, specified as the comma-separated pair consisting of 'MinSeparation' and a nonnegative scalar. ’

Another option is the ischange (link) function. That might be more applicalbe to what you want to do.

Please understand that without a representative sample of your data, I cannot specifically answer your questions. I can only suggest solutions, not knowing if they are appropriate.

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