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I am simulating an impulse signal of a resonant bearing fault with this formula: $$ y(k)=\sum_r A_r \sin\left(\displaystyle \frac{2\pi f \left(k - rF/f_m\right)}{F}\right)\cdot e^{\displaystyle-\beta(k−rF/f_m)/F} $$

where:

  • Amplitude signal $A_r =1.5$
  • Resonant frequency $f =2000$
  • Sampling frequency $F=10\cdot 10\cdot 3$
  • Characteristic fault frequency $f_m =50$
  • Decay factor $\beta =500$

What I can't figure out is the interpretation of $r$ in this equation and why is the sum of this term. If someone understands it please help and also what value of this $r$ should I consider.

The signal should look like:

Transient signal

EDIT:

I am adding the Python code for the function, it goes right for value of r=0 also I'm not adding any r. I cant understand the interpretation of this.

#Parameters:
Ar= 1.5   #Amplitude
f = 2000    #Resonant Frequency in Hz
b=500       #Exponential decay factor
fm=50      #Fault characteristic frequency
F=10*10**3  #Freq. Sampling
Tot_samples=20000
r=0

def y(k):
    t=(k-r*F/fm)/F
    exp=np.exp(-b*t)
    sin=np.sin(2*pi*f*t)
    return (Ar*sin*exp)   
#Create the summatory of r
_y=[]
y_signal=[]
for j in np.linspace(0,200,200,endpoint=False):  #only first 200 points of 
the impulse
    _y.append(y(j))
plt.plot(_y)
plt.show()
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    $\begingroup$ It's you who can know what that r is, by carefully reading the paper that gives you this sum. On the other hand my guess is that this summation is adding M terms of damped sinusoidals (which are probably solutions to a linear differential equation of the oscillating bearing under resonant) hence r represents each of those terms. $\endgroup$
    – Fat32
    Commented Jun 20, 2017 at 23:29
  • $\begingroup$ the paper dont mention what r means $\endgroup$
    – tom zko
    Commented Jun 21, 2017 at 0:00
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    $\begingroup$ probably it does but you cannot see... otherwise it's a shame :-) $\endgroup$
    – Fat32
    Commented Jun 21, 2017 at 0:24
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    $\begingroup$ Please link the paper if you expect better answer. $\endgroup$
    – jojeck
    Commented Jun 21, 2017 at 6:51

1 Answer 1

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In the formula, the idea behind $r$ is to set the amount of sinusoidal with exponential decaying amplitude terms that you want.

Maybe a more intuitive way to see it is to write the formula as:

\begin{equation} y(k)=\sum_{r}A_r\sin\left(2\pi f\left[\dfrac{k}{F}-\dfrac{r}{f_m}\right]\right)\cdot e^{-\beta\left(\dfrac{k}{F}-\dfrac{r}{f_m}\right)} \end{equation}

Notice that $\dfrac{k}{F}$ is $k$ (integer) times the sampling period and as $f_m$ characteristic fault frequency $\dfrac{r}{f_m}$ is $r$ (integer) times the period between impacts due to failure in the bearing.

Then the general idea is that each term in the summation represents a sinusoidal with exponential decaying amplitude shifted in time with periodicity $(1/f_m)$.

But there is a problem with the formula, neither $\sin(x)$ or $e^x$ are $0$ for all $x<0$. So, with that formula, you'll have non-zero values in terms for $r>0$ in samples previous to the impact that the term represents. This can be fixed adding a Heaviside step function $u(x)$:

\begin{equation} y(k)=\sum_{r}A_r\sin\left(2\pi f\left[\dfrac{k}{F}-\dfrac{r}{f_m}\right]\right)\cdot e^{-\beta\left(\dfrac{k}{F}-\dfrac{r}{f_m}\right)}\cdot u\left(\dfrac{k}{F}-\dfrac{r}{f_m}\right) \end{equation}

In python something like this could be use:

Ar= 1.5   #Amplitude
f = 2000    #Resonant Frequency in Hz
b=500       #Exponential decay factor
fm=50      #Fault characteristic frequency
F=10*10**3  #Freq. Sampling
Tot_samples=600
k=np.arange(0,Tot_samples)  
Sx=np.zeros((Tot_samples,))

for r in range(3):
    Sx+=np.sin(2*np.pi*f*(k/F-r/fm))*np.exp(-b*(k/F-r/fm))*np.heaviside(k/F-r/fm, 1)
    
plt.plot(k/F,Sx)
plt.grid(True)
plt.xlabel('Time[s]')
plt.ylabel('Amplitude')
plt.show()

enter image description here

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