Induction Machine Faults Detection and Localization by Neural Networks Methods

To generate the healthy and faulty states, we use two models: multi winding and three-phase of induction machine. As described below.

2.1 Multi winding model

The equations of model fault of induction motor can be written as [9, 10]:

$[\mathrm{L}] \frac{\mathrm{d} \mathrm{l}}{\mathrm{d} \mathrm{t}}=[\mathrm{V}]-[\mathrm{R}][1]$                 (1)

$\left[\begin{array}{ccccc}{\mathrm{L}_{\mathrm{sc}}} & {0} & {-\mathrm{N}_{\mathrm{r}} \mathrm{M}_{\mathrm{sr}} / 2} & {0} & {0} \\ {0} & {\mathrm{L}_{\mathrm{sc}}} & {0} & {\mathrm{N}_{\mathrm{r}} \mathrm{M}_{\mathrm{sr}} / 2} & {0} \\ {-3 \mathrm{M}_{\mathrm{sr}} / 2} & {0} & {\mathrm{L}_{\mathrm{rc}}} & {0} & {0} \\ {0} & {3 \mathrm{M}_{\mathrm{sr}} / 2} & {0} & {\mathrm{L}_{\mathrm{rc}}} & {0} \\ {0} & {0} & {0} & {0} & {\mathrm{L}_{\mathrm{e}}}\end{array}\right] \frac{\mathrm{d}}{\mathrm{d} t}\left[\begin{array}{l}{\mathrm{I}_{\mathrm{ds}}} \\ {\mathrm{I}_{\mathrm{qs}}} \\ {\mathrm{I}_{\mathrm{dr}}} \\ {\mathrm{I}_{\mathrm{qr}}} \\ {\mathrm{I}_{\mathrm{e}}}\end{array}\right]=\left[\begin{array}{c}{V_{\mathrm{ds}}} \\ {\mathrm{V}_{\mathrm{qs}}} \\ {0} \\ {0} \\ {0}\end{array}\right]-$ $\left[\begin{array}{ccccc}{\mathrm{R}_{\mathrm{s}}} & {-\mathrm{L}_{\mathrm{sc}} \omega_{\mathrm{r}}} & {0} & {\mathrm{N}_{\mathrm{r}} M_{\mathrm{sr}} \omega_{\mathrm{r}} / 2} & {0} \\ {\mathrm{L}_{\mathrm{sc}} \omega_{\mathrm{r}}} & {\mathrm{R}_{\mathrm{s}}} & {\mathrm{N}_{\mathrm{r}} \mathrm{M}_{\mathrm{sr}} \omega_{\mathrm{r}} / 2} & {0} & {0} \\ {0} & {0} & {\mathrm{R}_{\mathrm{r}}} & {0} & {0} \\ {0} & {0} & {0} & {\mathrm{R}_{\mathrm{r}}} & {0} \\ {0} & {0} & {0} & {0} & {\mathrm{R}_{\mathrm{e}}}\end{array}\right]\left[\begin{array}{l}{\mathrm{I}_{\mathrm{ds}}} \\ {\mathrm{I}_{\mathrm{qs}}} \\ {\mathrm{I}_{\mathrm{dr}}} \\ {\mathrm{I}_{\mathrm{qr}}} \\ {\mathrm{I}_{\mathrm{e}}}\end{array}\right]$

With defect model of the rotor in order to simulate the defect of rotor broken bars, a fault resistance R_RF is given:

$\left[\mathrm{R}_{\mathrm{RF}}\right]=\left[\mathrm{R}_{\mathrm{R}}\right]+\left[\begin{array}{ccccccc}{0} & {\cdots} & {0} & {0} & {0} & {\cdots} & {\cdots} \\ {\vdots} & {\ldots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\cdots} \\ {0} & {\cdots} & {0} & {0} & {0} & {0} & {\cdots} \\ {0} & {\cdots} & {0} & {\mathrm{R}_{\mathrm{bk}}} & {-\mathrm{R}_{\mathrm{bk}}} & {0} & {\cdots} \\ {0} & {\cdots} & {0} & {-\mathrm{R}_{\mathrm{bk}}} & {\mathrm{R}_{\mathrm{bk}}} & {0} & {\cdots} \\ {0} & {\cdots} & {0} & {0} & {0} & {0} & {\cdots} \\ {\vdots} & {\cdots} & {\vdots} & {\vdots} & {\vdots} & {\cdots}\end{array}\right]$

where, the four terms of this matrix are [11, 12]:

$\begin{aligned} \mathrm{R}_{\mathrm{rdd}}=& 2 \mathrm{R}_{\mathrm{b}}(1-\cos (\mathrm{a}))+2 \frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{N}_{\mathrm{r}}}+\frac{2}{\mathrm{N}_{\mathrm{r}}}(1-\cos (\mathrm{a})) \sum_{\mathrm{k}} \mathrm{R}_{\mathrm{bkf}}(1-\cos (2 \mathrm{k}-1) \mathrm{a}) \end{aligned}$      (2)  

$\left.\mathbf{R}_{\mathrm{rdq}}=-\frac{2}{\mathrm{N}_{\mathrm{r}}}(\mathbf{1}-\cos (\mathbf{a})) \sum_{\mathbf{k}} \mathbf{R}_{\mathrm{bkf}} \sin (2 \mathbf{k}-\mathbf{1}) \mathbf{a}\right)$    (3)

$\mathbf{R}_{\mathrm{rqd}}=-\frac{2}{\mathrm{N}_{\mathrm{r}}}(\mathbf{1}-\mathbf{c o s}(\mathbf{a})) \sum_{\mathbf{k}} \mathbf{R}_{\mathrm{bkf}} \sin (2 \mathbf{k}-\mathbf{1})$a)      (4)

And

$\begin{aligned} \mathrm{R}_{\mathrm{rqq}}=2 \mathrm{R}_{\mathrm{b}}(1-\cos (\mathrm{a})) &+2 \frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{N}_{\mathrm{r}}} +\frac{2}{\mathrm{N}_{\mathrm{r}}}(1-\cos (\mathrm{a})) \sum_{\mathrm{k}} \mathrm{R}_{\mathrm{bkf}}(1+\cos (2 \mathrm{k}-1) \mathrm{a}) \end{aligned}$

2.2 Three-phase model

Another method for modeling of the induction machine is presented. The defects of short-circuit between coils, taking into account the changing parameters such as resistors and inductors based on the classics assumptions. The machine can be modeled by the following equations [13]:

$\left[\mathrm{V}_{\mathrm{s}}\right]=\left[\mathrm{R}_{\mathrm{s}}\right]\left[\mathrm{I}_{\mathrm{s}}\right]+\frac{\mathrm{d}}{\mathrm{dt}}\left[\Phi_{\mathrm{s}}\right]$                   (5)

$[0]=\left[\mathrm{R}_{\mathrm{r}}\right]\left[\mathrm{I}_{\mathrm{r}}\right]+\frac{\mathrm{d}}{\mathrm{dt}}\left[\Phi_{\mathrm{r}}\right]$                         (6)

$\left.\left[\Phi_{s}\right]=\left(\left[M_{s s}\right]+\left[L_{s}\right]\right)\left[I_{s}\right]+\left[M_{s r}\right]\left[I_{r}\right]\right)$      (7)

$\left[\Phi_{r}\right]=\left[M_{r s}\right]\left[I_{s}\right]+\left(\left[M_{r r}\right]+\left[L_{r}\right]\right)\left[I_{r}\right]$        (8)

The following matrices represent voltages, currents, and fluxes at the stator:

$\left[V_{s}\right]=\left[\begin{array}{c}{V_{s a}} \\ {V_{s b}} \\ {V_{s c}}\end{array}\right] ;\left[I_{s}\right]=\left[\begin{array}{c}{I_{s a}} \\ {I_{s b}} \\ {I_{s c}}\end{array}\right] ;\left[\Phi_{s}\right]=\left[\begin{array}{c}{\Phi_{s a}} \\ {\Phi_{s b}} \\ {\Phi_{s c}}\end{array}\right]$

The matrices represent voltages, currents, and fluxes at the rotor, the coefficients of short–circuit as following:

Short-circuit coefficient relative to the 1^st stator phase:

$K_{s a}=\frac{N_{c c 1}}{N_{s}}$

Short-circuit coefficient relative to the 2^nd stator phase:

$K_{s b}=\frac{N_{c c 2}}{N_{s}}$

Short-circuit coefficient relative to the 3^nd stator phase:

$K_{s c}=\frac{N_{c c 3}}{N_{s}}$

With Ncc: The number of turns in short-circuit.

The number of useful turns for the three stator phases, is then given by:

$N_{1}=N_{s}-N_{c c 1}=\left(1-K_{s a}\right) N_{s}=f_{s a} N_{s}$

$N_{2}=N_{s}-N_{c c 2}=\left(1-K_{s b}\right) N_{s}=f_{s b} N_{s}$

$N_{3}=N_{s}-N_{c c 3}=\left(1-K_{s c}\right) N_{s}=f_{s c} N_{s}$

The matrixes $\left[R_{S}\right],\left[L_{s f}\right],\left[M_{s s}\right],\left[M_{s r}\right],\left[M_{r s}\right]$ depend of three coefficient $f_{s a}, f_{s b}, f_{s c}$ are given by the following terms:

$\left[L_{s f}\right]=\left[\begin{array}{ccc}{f_{s a}^{2} L_{s f}} & {0} & {0} \\ {0} & {f_{s a}^{2} L_{s f}} & {0} \\ {0} & {0} & {f_{s a}^{2} L_{s f}}\end{array}\right]$      (9)

$\left[M_{s s}\right]=M_{s}\left[\begin{array}{ccc}{f_{s a}^{2}} & {-\frac{f_{s a} f_{s b}}{2}} & {-\frac{f_{s a} f_{s c}}{2}} \\ {-\frac{f_{s a} f_{s b}}{2}} & {f_{s b}^{2}} & {-\frac{f_{s c} f_{s b}}{2}} \\ {-\frac{f_{s a} f_{s c}}{2}} & {-\frac{f_{s c} f_{s b}}{2}} & {f_{s c}^{2}}\end{array}\right]$      (10)

$\left[M_{s r}\right]=M\left[\begin{array}{ccc}{f_{s a} \cos \theta} & {f_{s a} \cos \left(\theta+\frac{2 \pi}{3}\right)} & {f_{s a} \cos \left(\theta-\frac{2 \pi}{3}\right)} \\ {f_{s b} \cos \left(\theta-\frac{2 \pi}{3}\right)} & {f_{s b} \cos \theta} & {f_{s b} \cos \left(\theta+\frac{2 \pi}{3}\right)} \\ {f_{s c} \cos \left(\theta+\frac{2 \pi}{3}\right)} & {f_{s c} \cos \left(\theta-\frac{2 \pi}{3}\right)} & {f_{s c} \cos \theta}\end{array}\right]$ 

With $\left[M_{s r}\right]=\left[M_{r s}\right]^{t}$

The matrix of stator resistances $\left[R_{s}\right]$ is given by:

$\left[R_{s}\right]=R s\left[\begin{array}{ccc}{f_{s a}} & {0} & {0} \\ {0} & {f_{s b}} & {0} \\ {0} & {0} & {f_{s c}}\end{array}\right]$         (11)

The set of rotor variables (flows and currents) can be changed to new variables with the same pulsation as the stator variables. Thus, all the parameters of the model will be independent of the angular position "θ" the transformation is given by the following matrix [14]

$[T]=\left[\begin{array}{ccc}{\cos \left(\theta+\frac{1}{2}\right)} & {\cos \left(\theta+\frac{2 \pi}{3}\right)+\frac{1}{2}} & {\cos \left(\theta-\frac{2 \pi}{3}\right)+\frac{1}{2}} \\ {\cos \left(\theta-\frac{2 \pi}{3}\right)+\frac{1}{2}} & {\cos \left(\theta+\frac{1}{2}\right)} & {\cos \left(\theta+\frac{2 \pi}{3}\right)+\frac{1}{2}} \\ {\cos \left(\theta+\frac{2 \pi}{3}\right)+\frac{1}{2}} & {\cos \left(\theta-\frac{2 \pi}{3}\right)+\frac{1}{2}} & {\cos \left(\theta+\frac{1}{2}\right)}\end{array}\right]$

It is easy to show that this matrix is orthogonal:

$[T]^{-1}=[T]^{T}$                               (12)

Considering the Eq. (7) using the matrices T as:

$\left[\Phi_{s}\right]=\left[M_{s}\right]\left[I_{s}\right]+\left[M_{s r}\right]\left[I_{r}\right]\quad=\left[M_{s}\right]\left[I_{s}\right]+\left[M_{s r}\right][T]^{-1}[T]\left[I_{r}\right]$         (13)

When Simplify the equation written:

$\left[\Phi_{s}\right]=\left[M_{s}\right]\left[I_{s}\right]+\left[M_{s r}^{s}\right]\left[I_{r}^{s}\right]$        (14)

where,

$\left[M_{s r}^{s}\right]=\left[M_{s r}^{s}\right][T]^{-1}$ and $\left[I_{r}^{s}\right]=[T]\left[I_{r}\right]$

With:

$\left[M_{s r}^{s}\right]=\left[\begin{array}{ccc}{f_{s a} M} & {-f_{s a} \frac{M}{2}} & {-f_{s c} \frac{M}{2}} \\ {-f_{s b} \frac{M}{2}} & {f_{s b} M} & {-f_{s c} \frac{M}{2}} \\ {-f_{s c} \frac{M}{2}} & {-f_{s c} \frac{M}{2}} & {f_{s c} M}\end{array}\right]$          (15)

The global model of the induction machine in the presence of stator failures.

Flux equations:

$\frac{d \Phi_{r a}}{d t}=\delta\left(f_{s a} i_{s a}-\frac{f_{s b}}{2} i_{s b}-\frac{f_{s c}}{2} i_{s c}\right)-\frac{R_{r} A}{C} \Phi_{r a}-\left(\frac{R_{r} B}{c}+\frac{\sqrt{3}}{3} \omega_{r}\right) \Phi_{r b}-\left(\frac{R_{r} B}{c}+\frac{\sqrt{3}}{3} \omega_{r}\right) \Phi_{r c}$        (16)

$\begin{aligned} \frac{d \Phi_{r b}}{d t} &=\delta\left(-\frac{f_{s a}}{2} i_{s a}+f_{s b} i_{s b}-\frac{f_{s c}}{2} i_{s c}\right) \\-\left(\frac{R_{r} B}{c}+\frac{\sqrt{3}}{3} \omega_{r}\right) & \Phi_{r a}-\frac{R_{r} A}{c} \Phi_{r b}-\left(\frac{R_{r} B}{c}+\frac{\sqrt{3}}{3} \omega_{r}\right) \Phi_{r c} \end{aligned}$   (17)

$\begin{aligned} \frac{d \Phi_{r c}}{d t} &=\delta\left(-\frac{f_{s a}}{2} i_{s a}-\frac{f_{s b}}{2} i_{s b}+f_{s c} i_{s c}\right) -\left(\frac{R_{r} B}{c}+\frac{\sqrt{3}}{3} \omega_{r}\right) \Phi_{r a}-\left(\frac{R_{r} B}{c}+\frac{\sqrt{3}}{3} \omega_{r}\right) \Phi_{r b}-\frac{R_{r} A}{c} \Phi_{r c} \end{aligned}$       (18)

Stator current equations:

$\frac{d i_{s a}}{d t}=V_{S A}+K_{A 1} i_{s a}+K_{A 2} i_{s b}+K_{A 3} i_{s c}+K f_{s a} f_{s b}^{2} f_{s c}^{2}\left(G \Phi_{r a}+\left(\frac{\sqrt{3}}{2} \omega_{r}-\frac{G}{2}\right) \Phi_{r b}-\left(\frac{\sqrt{3}}{2} \omega_{r}-\frac{G}{2}\right) \Phi_{r c}\right)$       (19)

$\begin{aligned} \frac{d i_{s b}}{d t} &=V_{S B}+K_{B 1} i_{s a}+K_{B 2} i_{s b}+K_{B 3} i_{s c}+K f_{s a} f_{s b}^{2} f_{s c}^{2}\left(-\left(\frac{\sqrt{3}}{2} \omega_{r}-\frac{G}{2}\right) \Phi_{r a}+G \Phi_{r b}+\left(\frac{\sqrt{3}}{2} \omega_{r}-\frac{G}{2}\right) \Phi_{r c}\right) \end{aligned}$          (20)

$\frac{d i_{s c}}{d t}=V_{S C}+K_{C 1} i_{s a}+K_{C 2} i_{s b}+K_{C 3} i_{s c}+K f_{s a} f_{s b}^{2} f_{s c}^{2}\left(\left(\frac{\sqrt{3}}{2} \omega_{r}-\frac{G}{2}\right) \Phi_{r a}-\left(\frac{\sqrt{3}}{2} \omega_{r}+\frac{G}{2}\right) \Phi_{r b}+\Phi_{r c}\right)$     (21)

Used coefficients:

$V_{S A}=d_{1} f_{s a}^{2} f_{s c}^{2} u_{s a}+d_{2} f_{s a} f_{s b} f_{s c}^{2} u_{s b}+d_{2} f_{s a} f_{s b}^{2} f_{s c} u_{s c}$

$V_{S B}=d_{2} f_{s a} f_{s b} f_{s c}^{2} u_{s a}+d_{1} f_{s a}^{2} f_{s c}^{2} u_{s b}+d_{2} f_{s a}^{2} f_{s b} f_{s c} u_{s c}$

$V_{S A}=d_{2} f_{s a} f_{s b}^{2} f_{s c} u_{s a}+d_{2} f_{s a}^{2} f_{s b} f_{s c} u_{s b}+d_{1} f_{s a}^{2} f_{s b}^{2} u_{s c}$

$A=\left(l_{r f}+M_{r}\right)^{2}-\frac{M_{r}^{2}}{4} ; B=\frac{M_{r} l_{r f}}{2}+\frac{3 M_{r}^{2}}{4};$

$C=l_{r f}^{3}+3 l_{r f}^{2} M_{r}+\frac{9}{4} M_{r}^{2} l_{r f} ; \delta=\frac{M_{s r R}(A-B)}{c}; $

$T=\frac{M_{s r}^{2} R_{r}(A-B)^{2}}{c^{2}} ; z=M_{s r}-\frac{3 M_{s r}^{2} R_{r}(A-B)}{2 c};$

$\lambda=z+l_{s f} ; H=\lambda^{2}-\frac{z \lambda}{2}-\frac{z^{2}}{4}$

$\begin{aligned}|\Gamma| &=f_{s a}^{2} f_{s b}^{2} f_{s c}^{2}\left(\lambda^{3}-\frac{3 z^{2} \lambda}{2}-\frac{z^{3}}{4}\right) ; d_{1}=\left(z+l_{s f}\right)^{2}-\frac{z^{2}}{4} \\ d_{2} &=z\left(z+l_{s f}\right)^{2}+\frac{z^{2}}{4} ; K=\frac{M_{s r} H(A-B)}{c|\Gamma|} ; G=\frac{R_{r}(A-B)}{c} \end{aligned}$

$\left\{\begin{array}{l}{K_{A 1=}-\frac{3}{2}\left(d_{1}+d_{2}\right) T f_{s a}^{2} f_{s b}^{2} f_{s c}^{2}+R_{s} d_{1} f_{s a} f_{s b}^{2} f_{s c}^{2}} \\ {K_{A 2=}-\frac{3}{2} \frac{\left(d_{1}+3 d_{2}\right)}{2} T f_{s a} f_{s b}^{3} f_{s c}^{2}+R_{s} d_{2} f_{s a} f_{s b}^{2} f_{s c}^{2}} \\ {K_{A 3=}-\frac{3}{2} \frac{\left(d_{1}+3 d_{2}\right)}{2} T f_{s a} f_{s b}^{2} f_{s c}^{3}+R_{s} d_{2} f_{s a} f_{s b}^{2} f_{s c}^{2}} \\ {K_{B 1=}-\frac{3}{2} \frac{\left(d_{1}+3 d_{2}\right)}{2} T f_{s a}^{3} f_{s b} f_{s c}^{2}+R_{s} d_{2} f_{s a}^{2} f_{s b} f_{s c}^{2}} \\ {K_{B 2=}-\frac{3}{2}\left(d_{1}+d_{2}\right) T f_{s a}^{2} f_{s b}^{2} f_{s c}^{2}+R_{s} d_{1} f_{s a}^{2} f_{s b} f_{s c}^{2}} \\ {K_{B 3}=-\frac{3}{2} \frac{\left(d_{1}+3 d_{2}\right)}{2} T f_{s a}^{2} f_{s b} f_{s c}^{3}+R_{s} d_{2} f_{s a}^{2} f_{s b} f_{s c}^{2}}\\{K_{C 1=}-\frac{3}{2} \frac{\left(d_{1}+3 d_{2}\right)}{2} T f_{s a}^{3} f_{s b}^{2} f_{s c}+R_{s} d_{2} f_{s a}^{2} f_{s b}^{2} f_{s c}}\\{K_{C 2=}-\frac{3}{2} \frac{\left(d_{1}+3 d_{2}\right)}{2} T f_{s a}^{2} f_{s b}^{3} f_{s c}+R_{s} d_{2} f_{s a}^{2} f_{s b}^{2} f_{s c}}\\{K_{C 3=}-\frac{3}{2}\left(d_{1}+d_{2}\right) T f_{s a}^{2} f_{s b}^{2} f_{s c}^{2}+R_{s} d_{1} f_{s a}^{2} f_{s b}^{2} f_{s c}}\end{array}\right.$

The torque equation is given by:

$C_{e}=P \frac{M_{s r}}{L_{r}}\left[\left(I_{s b} \Phi_{r c}-I_{s c} \Phi_{r b}\right)-\left(I_{s a} \Phi_{r c}-I_{s c} \Phi_{r a}\right)+\right.\left.\left(I_{s a} \Phi_{r a}-I_{s b} \Phi_{r b}\right)\right]$               

In this case, the more frequent bar breaks at the rotor level, we will present the rotor defects at the instant: t=1s break a bar, t=2s break two bars, t=3s break three bars, and t=4s break four bars, we apply a load of Cr=3.5Nm at t=0.5s, the value of the resistance of the broken bar will be considered equal to eleven (11) times, the value of the initial resistance. The currents of the stator phases are always out of phase by 120°. The modulation of the envelope of the stator current after the breaks of the bars is also noted, the increase of the amplitude of proportional to the number of broken bars which appears in the figures (1-2). With regard to the simulation of short-circuit faults between coils in an induction machine, in a first case, we presented the shapes of the stator currents in the case operation and with short-circuit-type faults between 40 turns (25%), and the second case we apply short circuit faults between 20 turns (12.5%). As shown in Figures (3-4 -5-6).

1.png

Figure 1. Stator current in case broken four bar  

2.png

Figure 2. Stator phase current (a,b,c)

3.png

Figure 3. Stator current in case type short-circuit faults between 40 turns (25%)

4.png

Figure 4. Stator phase current (a,b,c) in case of short-circuit faults between 40 turns (25%)

5.png

Figure 5. Stator current in case of short-circuit faults between 20 turns (12.5%)

6.png

Figure 6. Stator phase current (a,b,c)  in case  of short-circuit faults between 20 turns (12.5%)

The effective value of a quantity is the square root of the sum of the squares of the constant term and the effective values of the various sinusoidal terms of the development in series:

$R M S=\sqrt{\frac{1}{t} \int_{0}^{t} u(t)^{2} d t}$             (23)

For a signal sampled by a sampling step Te. $u(t)^{2}$ will be known only at the sampling instants: $\int_{0}^{t} u(t)^{2} d t$ and can be approximated by the area between $u(t)^{2}$ discretized and the time axis.

For $\mathrm{N}$ samples: $\int_{0}^{t} u(t)^{2} d t \cong \sum_{i=0}^{N-1} u_{i}^{2} \times$ Te so: $R M S \cong\sqrt{\frac{1}{N \times T e} \sum_{i=0}^{N-1} u_{i}^{2} \times T e}$

$R M S \cong \sqrt{\frac{1}{N} \sum_{i=0}^{N-1} u_{i}^{2}}$

The results for the different types of input signals are detailed in the following graph, see Figure 8.

These graphs show the performance of the calculation method of RMS for square signal and sine signal.

8.png

Figure 8. Result of RMS calculation

The detection and early diagnosis allow reducing damage and maintaining other components of induction machine, through the study of defects influence and the behavior of the machine in case of operation fault. In this paper we presented the induction machine fault by using multi-winding model for simulation of broken bars and three-phase model for simulation of short- circuit between 20-40 turns. The new features are presented by multi layers neural network trained by retro-propagation algorithm. The RMS values of measured machine parameters are extracted by processing and monitoring of machine behavior in the presence of faults in order to obtain the indicators values

The proposed diagnosis method could be applied by artificial intelligence represented neural networks (ANN) on induction machine during several parametric studies (selection of the type of network, choice of inputs, and choice of outputs ...). With the data acquisition operation, this aims to establish the network learning base. To be reliable indicators for detection and location of fault broken bars and inter-turns short-circuit fault. These results clearly indicate that the proposed neural networks have a great importance for fault identification and capable to reduce the failure severity. Furthermore, it has been manifest that this approach is accurate and simple in the process implement diagnosis.