# Impulse signal function

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:

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()

• 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. Commented Jun 20, 2017 at 23:29
• the paper dont mention what r means Commented Jun 21, 2017 at 0:00
• probably it does but you cannot see... otherwise it's a shame :-) Commented Jun 21, 2017 at 0:24
– jojeck
Commented Jun 21, 2017 at 6:51

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:

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

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 $$u(x)$$:

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

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()