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Derivation of sin(x) gives false cos(x)

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Bruce Rogers
Bruce Rogers on 29 Mar 2021
Commented: Robert Brown on 29 Mar 2021
Hello,
I'm trying to derivate the sin(x) two times to get the cos(x) and the -sin(x). The plot shows a cosine figure, but it doesn't start at 1 and the amplitude is false too.
My code is the following:
clear;
t = 0.05;
x = [0:0.05:2*pi];
y = sin(x);
figure(1)
scatter(x,y,'.')
xlabel('time')
ylabel('distance')
%%
vel = diff(y);
g = 1:length(vel);
for i = 1:length(g)
vel_t(i) = t * g(i);
end
figure(2)
scatter(vel_t,vel,'.')
xlabel('time')
ylabel('velocity')
%%
acc = diff(vel);
h = 1:length(acc);
for i = 1:length(h)
acc_t(i) = t * h(i);
end
figure(3)
scatter(acc_t,acc,'.')
xlabel('time')
ylabel('acceleration')
Is it wrong to use x = [0:0.05:2*pi] for defining my period?
Thanks for your help.
  1 Comment
Stephen Cobeldick
Stephen Cobeldick on 29 Mar 2021
"Is it wrong to use x = [0:0.05:2*pi] for defining my period?"
No, but those square brackets are superfluous, get rid of them.
"...it doesn't start at 1 and the amplitude is false too."
Hint: take a look at the definition of differentiation. Where in your code do you divide by the change in the independent variable?

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Accepted Answer

KSSV
KSSV on 29 Mar 2021
t = 0.05;
x = 0:0.05:2*pi;
y = sin(x);
figure(1)
plot(x,y,'r')
xlabel('time')
ylabel('distance')
%%
vel = gradient(y)./gradient(x);
figure(2)
plot(x,vel,'b')
xlabel('time')
ylabel('velocity')
%%
acc = gradient(vel)./gradient(x);
figure(3)
plot(x,acc,'g')
xlabel('time')
ylabel('acceleration')
  2 Comments
Stephen Cobeldick
Stephen Cobeldick on 29 Mar 2021
"Can you explain why diff() gave me that wrong results?"
The problem is not DIFF, the problem is that you did not divide by the change in the independent variable (in your code x), which is required by the definition of differentiation. KSSV did that division using ./

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More Answers (2)

VBBV
VBBV on 29 Mar 2021
Edited: VBBV on 29 Mar 2021
Use gradient instead of diff
  4 Comments
Bruce Rogers
Bruce Rogers on 29 Mar 2021
This didn't work, now the plot is showing a linear function. KSSV's solution worked well to solve my problem. But thank you very much for your time VBBV.

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Robert Brown
Robert Brown on 29 Mar 2021
Technically, the gradient is extrapolating data!!!
When you have a series of position measurements..... you require 2 position measurements to get to a velocity estimate, and require another position measurement to get to an acceleration measurement.
Assume Time = T1, T2, T3, ...
Position 1 = only know position P1 and T1
Position 2 = know P1 and P2 and T1 and T2
and estimate V1 = (P2 - P1) / (T2-T1) = average velocity getting from P1 to P2,
and corresponding time TV1 = (T2-T1)/2
Position 3 = know P1 and P2 and estimate of V1 (above) and P3 and V2 = (P3 - P2) / (T3 - T2)
and corresponding time TV2 = (T3-T2)/2
and know acceleration 1 A1 = (V2-V1)/(TV2 - TV1) =
((P3 - P2) / (T3 - T2) - (P2 - P1) / (T2-T1)) / ((T3-T2)/2 - (T2-T1)/2)
etc....
  1 Comment
Robert Brown
Robert Brown on 29 Mar 2021
Maybe I should have called it V1.5 meaning the average (estimated) velocity between 1 and 2, and T1.5 meaning the time 1/2 of the way between V1 and V2. Then by using V1.5 and V2.5.... you can determine V2, the (estimated) velocity at T2.
If you look back at your code, the diff function caused the velocity measurements to have 1 less data point since you don't know velocity until you know 2 positions, and the diff function caused the acceleration measurements to have 2 less data points, since you don't know acceleration until you know 3 positions and 2 estimated velocities... :-)

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