How do i add a time vector to a existing table?

I have a set of given data but they do not include a time. I am trying to add a time (Tijd) vector but if i add a time vector, i get the messages that vectors must be the same length.

The data is a set amount of measurements (4001) but with a sample rate of 50Hz. I ofcourse want the time to be in seconds. So I thought I would make the steps in the graph steps of 81 because 4001/50 ≈ 81. But even without the steps it still gives me the error of vectors must be the same length.

Any and all help is appreciated!

clear all

close all

clc

load('ACE2-opdracht-2-persoonlijk.mat', 'meetdata');

A = meetdata;

Motortoerental = A (:,1);

Stuurspanning = A (:,2);

Tijd = 1:81:4001;

figure;

subplot(1,2,1)

plot(Tijd, Motortoerental); hold on

xlabel('Tijd(s)')

ylabel('Toerental(Omw/min)')

grid;

subplot(1,2,2)

plot(Tijd, Stuurspanning); hold on

xlabel('Tijd(s)')

ylabel('Stuurspanning(V)')

grid;

[Stephen23]

Avoiding the possibility of floating point issues with the last value:

(0:4000)./50

ans = 1×4001

0 0.0200 0.0400 0.0600 0.0800 0.1000 0.1200 0.1400 0.1600 0.1800 0.2000 0.2200 0.2400 0.2600 0.2800 0.3000 0.3200 0.3400 0.3600 0.3800 0.4000 0.4200 0.4400 0.4600 0.4800 0.5000 0.5200 0.5400 0.5600 0.5800

[Wouter]

This also works! Why would i do it this way?

Answer 1

[Torsten]

Shouldn't it be

Tijd = (0:0.02:4000*0.02).'

?

[Wouter]

Yes! this works. Can you explain it to me?

[Torsten]

Your vector Tijd must have 4001 elements in order to plot Tijd against the other two arrays Motortoerental and Stuurspanning.

You said the measurements are every 0.02 s - thus

Tijd = (0:0.02:4000*0.02).'

[Wouter]

Oh now i understand. Thanks for the help!