Error in integration of square wave using cumtrapz

Hi everyone,

I am currently plotting a square velocity profile for a moving platform of which I need to take the integral of to find the displacement. The code I have written lets the user input the time length of the waveform, the interval, velocity amplitude, and frequency and plots the corresponding square wave. Next, I use cumtrapz to calculate the integral of the square wave, which results in a triangular wave as expected.

After completing the integration, I center the triangular displacement around 0 by subtracting the mean to shift the waveform vertically, and then use a modified circshift (fshift from File Exchange) to horizontally shift the wave to it's closest y=0 value. This method seems to work well for certain frequency values (.6 to 1 Hz), however at lower frequency values (.1 to .5 Hz), the scale of the integration is incorrect. For a velocity of 15 cm/s, using a sine wave of or a square wave with frequency of 1 Hz, I calculate displacement values between 1.5 and 2 in. If I change the frequency to .5 Hz or less, the displacement then becomes between 2-3 in, which seems to be incorrect as amplitude of a function should not change with a frequency change.

I have been trying to figure out why at certain frequencies I am receiving this error, as in theory the frequency and amplitude of a function should be independent from one another. I would appreciate any help with figuring out this issue, thanks!

ft = 0:dt:time_length;                                  %time vector definition
amp = 15;                                               %cm/s velocity input
TV_ampin = amp/2.54;                                    %cm/s to in/s
TV_freq = 1;                                            %frequency in Hz
triV = TV_ampin.*square(2.*pi.*TV_freq.*ft,50);         %create square velocity
triV_p = cumtrapz(ft, triV);                            %calculate integral
if triV_p(1,1) ~= 0                                     %if triV_p doesn't start at 0, shift it to zero by subtracting initial value
    trivvy = triV_p - triV_p(1,1);
    downshift = trivvy - mean(triV_p);                  %shift down by subtracting mean
    zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);%define zero crosses function
    zV = zci(downshift);                                %find zero crosses
    triV_position = fshift(downshift, zV(1));           %horizontally shift by first zero cross
else                                                    %if it already starts at zero, do the same without subtracting initial
    trivvy = triV_p;
    downshift = trivvy - mean(triV_p);
    zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);
    zV = zci(downshift);
    triV_position = fshift(downshift, zV(1));
end
figure; plot(ft, triV)
title('Velocity'); xlabel('Time (s)'); ylabel('Velocity');
figure; plot(ft, triV_position)
title('Position'); xlabel('Time (s)'); ylabel('Displacement');

[Mathieu NOE]

hello

before jumping into code checking , I will come back on your assumptions :

If I change the frequency to .5 Hz or less, the displacement then becomes between 2-3 in, which seems to be incorrect as amplitude of a function should not change with a frequency change.

first , why do you speak of a function ? in my mind we are talking about amplitude (scalar) of velocity and displacement

maybe you are talking the integration function , which of course is frequency dependant

for a sine wave , amplitude of displacement D and velocity V are linked by V = D*(2*pi*freq), freq is frequency of input signal

so if you keep your input velocity constant at 15 cm/s and you halve the input frequency , D will double

there is no magic or error here

[Ethan]

Yes I was mistaken, the frequency is definitely related to the integral of a function or dataset. When I plot a sine wave displacement for a specified time, frequency, and ampltidue, I can take the derivative of the function to get displacement:

sineVelocity = SV_ampin*sin(2*pi*SV_freq*ft+(pi/2));
sineV_position = -1*(SV_ampin/(2*pi*SV_freq))*cos(2*pi*SV_freq*ft+(pi/2));

Where I think there is an error is when I plot a square wave velocity and take the integral using cumtrapz, with the same time, frequency, and amplitude from the sine wave velocity, I get a much larger max displacement value for the triangular wave. I tried calculating the displacement from the sine vector using cumtrapz and it matches the sineV_position that is calculated above. If the frequency and amplitude of the sine and square waves match, should the amplitude of the integrals also match?

[Mathieu NOE]

for me , integrating a signal is looking at the area under the curve

if I look at the area below the curve for a sine vs a square wave of same amplitude and frequency , the square wave area is only 1.5868 bigger than the sine wave area

this is the same factor you should see between displacement of sinus vs square velocity profile

[Mathieu NOE]

vice versa !

this is the same factor you should see between displacement of square vs sinus velocity profile

[Ethan]

Between velocity and displacement, yes there should be a common factor of difference. I figured that the amplitude of the integral of a sine wave and the amplitude of the integral of a square wave, with both original waves having the same amplitude, would also be equal. There are ways I can work around this however, as I have an amplitude limit at 2.5 inches. Thank you for your help!

[Mathieu NOE]

you can also think of the square as a series of sine waves (fourier decomposition), so each sine after its own integration will add some extra amount of amplitude

so a sine and a square wave of same amplitude and freq will not have the same peak displacement after integration