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


  • 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


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.

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

def y(k):
    return (Ar*sin*exp)   
#Create the summatory of r
for j in np.linspace(0,200,200,endpoint=False):  #only first 200 points of 
the impulse
  • 1
    $\begingroup$ 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 Jun 20 '17 at 23:29
  • $\begingroup$ the paper dont mention what r means $\endgroup$ – tom zko Jun 21 '17 at 0:00
  • 1
    $\begingroup$ probably it does but you cannot see... otherwise it's a shame :-) $\endgroup$ – Fat32 Jun 21 '17 at 0:24
  • 1
    $\begingroup$ Please link the paper if you expect better answer. $\endgroup$ – jojek Jun 21 '17 at 6:51

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