use findpeaks function for interpolation of signal peaks with damped sinusoidal oscillation

hi all, i'm trying to calculate the damping coefficient of an object by interpolating the peaks of the acceleration signal. I was able, using findpeaks, to perform the interpolation. the problem is that I can't interpolate all the peaks, what can I do? another problem is that I have to consider only the peaks in the positive part of the signal. i am trying to use for and if loops but still not getting good results. thank you

This is the script that I am using:

load 'prova.txt' '-ascii' % file with acceleration data

t = prova(:,1);

a = prova(:,4);

dt=mean(diff(t));

t1=0.18;

t2=1.11;

id1= round(t1/dt);

id2= round(t2/dt);

t_tmp=t(id1:id2);

a_tmp=a(id1:id2);

[peak_value, peak_location] = findpeaks(a_tmp,"DoubleSided");

figure

plot(t_tmp,a_tmp)

hold on

plot(t_tmp(peak_location),a_tmp(peak_location),'ro')

From the script, I obtain:

to consider only the positive part of the signal, I proceeded like this

k=1;

for i=1:length(a_tmp)

if a_tmp(i)>0

a_tmp1(k)=a_tmp(i)

t_tmp1(k)=k;

k=k+1;

end

end

t_tmp1=t_tmp(1:length(a_tmp1),1)

figure

plot(t_tmp1,a_tmp1)

hold on

plot(t_tmp1(peak_location),a_tmp1(peak_location),'ro')

[Jonas]

looks like you use Octave? in Matlab there is not "DoubleSided" option for findpeaks as far as i am aware.

do you know the frequency of the sinusoid? use the min peak distance parameter with a slightly temporal smaller value and use pak prominence with a value smaller the minimum peak height you want to catch

to catch positive and negative vakues you have to do the run either twice with flipped signal or with abs value, then you have to use half of the min peak value as before since you have twoce as much peaks per time as when not taking abs value

[Star Strider]

That signal is significantly undersampled, so it is unlikely that you can do anything with it other than repeat the experiment and sample it at a much higher sampling frequency. There are several examples of fitting a expnoentially decaying sinusoid, however none of them will work with that signal.

[dpb]

Same thing appeared to me -- there's naught to be seen here folks! Just move along...

[David Goodmanson]

Hi Federico,

could you post the arrays for a and t? (Since the arrays do not appear to be very large, a text file would work). And could you say a bit more about why you can only use the peaks in the positive parts of the signal?

[Federico Paolucci]

@Jonas

yes i am using octave, but also matlab. Thanks for advice!

@Star Strider

I know, you're right, the signal is quite inadequate. Unfortunately for the exam I am doing I am forced to use the linear accelerometer contained in the mobile app, with sampling frequency equal to 100 Hz. I am doing a dynamic calibration of the front wing of a Formula Student vehicle, I can only use a rubber hammer,to apply the pulses, which present this type of acceleration signal. I will repeat the experiment being careful to give the hammer perfectly vertically on the surface of the wing, making sure that the aileron is positioned on a perfectly horizontal plane.

@David Goodmanson

Hi, this is t vector between 0.18 e 1.11 s (the display range I am using)

1.8453e-01

1.9126e-01

2.0060e-01

2.1040e-01

2.2023e-01

2.3051e-01

2.4127e-01

2.5099e-01

2.6031e-01

2.7032e-01

2.8113e-01

2.9161e-01

3.0119e-01

3.1045e-01

3.2029e-01

3.3594e-01

3.4220e-01

3.5079e-01

3.6029e-01

3.7083e-01

3.8122e-01

3.9192e-01

4.0056e-01

4.1062e-01

4.2065e-01

4.3080e-01

4.4257e-01

4.5067e-01

4.6544e-01

4.7110e-01

4.8044e-01

4.9271e-01

5.0045e-01

5.1060e-01

5.2049e-01

5.3042e-01

5.4048e-01

5.5039e-01

5.6053e-01

5.7041e-01

5.8062e-01

5.9062e-01

6.0282e-01

6.1099e-01

6.2020e-01

6.3036e-01

6.4050e-01

6.5047e-01

6.6037e-01

6.7024e-01

6.8030e-01

6.9016e-01

7.0020e-01

7.1596e-01

7.2029e-01

7.3073e-01

7.4024e-01

7.5044e-01

7.6149e-01

7.7041e-01

7.8016e-01

7.9033e-01

8.0041e-01

8.1108e-01

8.2048e-01

8.3028e-01

8.4073e-01

8.5123e-01

8.6159e-01

8.7058e-01

8.8029e-01

8.9047e-01

9.0034e-01

9.1166e-01

9.2023e-01

9.3038e-01

9.4050e-01

9.5045e-01

9.6016e-01

9.7053e-01

9.8271e-01

9.9023e-01

1.0007e+00

1.0104e+00

1.0203e+00

1.0309e+00

1.0406e+00

1.0503e+00

1.0603e+00

1.0705e+00

1.0847e+00

1.0908e+00

1.1006e+00

1.1102e+00

a vector

-1.8670e-01

-4.0000e-04

8.7700e-02

-5.0000e-04

1.3680e-01

-4.0000e-04

-1.6700e-01

1.4670e-01

9.4000e-03

-1.1800e-01

4.8500e-02

7.7900e-02

-1.6710e-01

1.9100e-02

8.7700e-02

-6.9100e-02

-5.9200e-02

1.5650e-01

-1.2790e-01

-1.0200e-02

6.8200e-02

-1.0400e-02

-1.2790e-01

1.6620e-01

-6.9000e-02

-2.9700e-02

4.8600e-02

4.8600e-02

-1.4740e-01

1.0750e-01

-1.9900e-02

-6.9000e-02

9.4000e-03

9.7600e-02

-1.3760e-01

5.8400e-02

4.5700e-02

-2.9800e-02

1.0740e-01

-7.8800e-02

-4.0000e-04

6.8200e-02

-2.0000e-02

-4.9400e-02

8.7800e-02

-2.0000e-02

-5.9200e-02

7.8000e-02

-3.0000e-04

-2.9800e-02

3.8800e-02

2.9000e-02

-7.8800e-02

2.9000e-02

2.9100e-02

-3.9600e-02

1.9100e-02

6.8100e-02

-7.8900e-02

2.8900e-02

3.2200e-02

-2.0100e-02

-3.9800e-02

7.7800e-02

-4.9600e-02

9.1000e-03

4.8300e-02

-6.0000e-04

-3.0100e-02

4.8300e-02

-3.0100e-02

2.8700e-02

2.2900e-02

-3.9900e-02

2.8700e-02

1.8900e-02

-5.9400e-02

3.6800e-02

2.8800e-02

I consider the positive peaks for the calculation of the damping coefficient

Answer 1

[Image Analyst]

As the others have said, it's way undersampled so the peaks are not really where the peaks should be -- they're in between. If I was forced to come up with a ballpark estimate for the exponential decay, I'd find the first peak, and the last peak, and just say that a perfect exponential decay went through them.

The general equation (with no y offset to raise or lower the curve) is

y = a1 * exp(-a2 * x)

and so you have two of those equations - one at each point.

y1 = a1 * exp(-a2*x1)

y2 = a1 * exp(-a2*x2)

The x and y are known. The two exponential parameters are unknown.

Just solve as you normally would using high school math. Divide the equations, etc.

log(y1/y2) = -a2*(x1 - x2)

and so on.

[Federico Paolucci]

ok perfect, thanks for the advice. as I said maybe I will repeat the test, but I could also do so. As for the rest I understand, thanks!