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2. Both DC and AC analyses are performed. In DC analysis, the Thevenin voltage, output voltage, and output resistance are calculated. In AC analysis, input impedance, voltage gain, and output impedance are derived.
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MATHEMATICAL MODELING OF COMPLEX REDUNDANT SYSTEM UNDER HEAD-OF-LINE REPAIREditor IJMTER
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This method is based on the fact that calculation in float numbers will remove accuracy. By reducing the number of calculation the accuracy of result increase significantly.
Umar Sidik
(BEng) Electrical and Electronic Engineering, Universitas Sumatera Utara, Indonesia
(MSc) Mechanical Engineering, National Defence University of Malaysia, Malaysia
in this presentation content different types of interpolation formulas which is used for many applications,and give accurate answer of big calculation in short time.
Neural Network Back Propagation AlgorithmMartin Opdam
Explanation of the back propagation algorithm and gradient descent using a simple multilayer perceptron neural network.
The C# code for this example can be found on https://github.com/mpopdam/NeuraNet
MATHEMATICAL MODELING OF COMPLEX REDUNDANT SYSTEM UNDER HEAD-OF-LINE REPAIREditor IJMTER
Suppose a composite system consisting of two subsystems designated as ‘P’ and
‘Q’ connected in series. Subsystem ‘P’ consists of N non-identical units in series, while the
subsystem ‘Q’ consists of three identical components in parallel redundancy.
In this paper we represent a modified Generalized Regression Artificial Neural net that can recognize all breast cancer of Wisconsin Diagnostic Breast Cancer and Wisconsin Prognostic Breast Cancer correctly. In this method the modified Neural Net trained with 50% of data & 50% for test. But the result is the ability of classify with 100% accuracy. The all 50% train & test data chosen randomly.
This method is based on the fact that calculation in float numbers will remove accuracy. By reducing the number of calculation the accuracy of result increase significantly.
Umar Sidik
(BEng) Electrical and Electronic Engineering, Universitas Sumatera Utara, Indonesia
(MSc) Mechanical Engineering, National Defence University of Malaysia, Malaysia
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Electronusa Mechanical System
1. Electronusa Mechanical System [Research Center for Electronic and Mechanical]
1 | P a g e
The Impedance Matching in The Audio Signal Processing
Umar Sidik.BEng.MSc*
Director of Engineering
Electronusa Mechanical System (CTRONICS)
*umar.sidik@engineer.com
1. Introduction
Commonly, impedance is obstruction to transfer energy in the electronic circuit. Therefore, the
impedance matching is required to achieve the maximum power transfer. Furthermore, the
impedance matching equalizes the source impedance and load impedance. In other hand, the
emitter-follower (common-collector) provides the impedance matching delivered from the base
(input) to the emitter (output). The emitter-follower has high input resistance and low output
resistance. In the emitter-follower, the input resistance depends on the load resistance, while the
output resistance depends on the source resistance. In addition, this study implements the radial
electrolytic capacitor 100 63⁄ .
2. Analytical Work
In this study, and form the Thevenin voltage, while and deliver ac signal as and
(figure 1).
(a) (b)
Figure 1. (a). The concept of circuit analyzed in the study
(b). The equivalent circuit
2.1 Analysis of dc
First step, we have to calculate the Thevenin’s voltage in figure 1:
=
+
×
For this circuit, is 5 , then:
=
24 Ω
10 Ω + 24 Ω
× 5
24 Ω
34 Ω
× 5
= (0.71) × 5
= 3.55
2. Electronusa Mechanical System [Research Center for Electronic and Mechanical]
2 | P a g e
Actually, in this circuit = , so = 3.55 .
The second step, we have to calculate :
= −
= 3.55 − 0.7
= 2.85
The third step, we have to calculate :
=
=
2.85
150Ω
= 19
2.2 Analysis of ac
In the analysis of ac, we involve the capacitor to pass the ac signal and we also involve the internal
resistance of emitter known as (figure 2).
(a) (b)
Figure 2. (a). The ac circuit
(b). The equivalent circuit for ac analysis
The first step, we have to calculate in the figure 2:
=
25
=
25
19
= 1.32Ω
The second step, we have to calculate ( ):
( ) = ( + 1) ( + )‖
( ) = (200 + 1) (150Ω + 8.2Ω)‖1.32Ω
3. Electronusa Mechanical System [Research Center for Electronic and Mechanical]
3 | P a g e
( ) = (201) (158.2Ω)‖1.32Ω
( ) = (201)
1
158.2Ω
+
1
1.32Ω
( ) = (201)
1.32
208.824Ω
+
158.2
208.824Ω
( ) = (201)
159.52
208.824Ω
( ) = (201)(0.764Ω)
( ) = 153.564Ω
The third step is to calculate :
=
( )
=
1
153.564Ω
= 0.0065
= 6.5
The fourth step is to calculate :
=
= (200)(0.0065 )
= 1.3
The last step is to calculate :
=
= (1.3 )(0.764Ω)
= 0.9932
= 993.2
3. Simulation Work
The simulation work can be classified into the dc analysis and the ac analysis.
3.1 Analysis of dc
In the simulation, is 3 (figure 3), while in the analytical work is 3.55 .
The different of the analytical work and the simulation work is:
(%) =
( ) − ( )
( )
× 100%
(%) =
3.55 − 3
3.55
× 100%
4. Electronusa Mechanical System [Research Center for Electronic and Mechanical]
4 | P a g e
(%) =
0.55
3.55
× 100%
(%) = 18.33%
Figure 3. in the simulation
In the simulation, is 2.25 (figure 4), while in the analytical work is 2.85 . The different of the
analytical work and the simulation work is:
(%) =
( ) − ( )
( )
× 100%
(%) =
2.85 − 2.25
2.85
× 100%
(%) =
0.6
2.85
× 100%
(%) = 21.05%
Figure 4. in the simulation
In the simulation, is 15 (figure 5), while in the analytical work is 19 . The difference is:
(%) =
( ) − ( )
( )
× 100%
(%) =
19 − 15
19
× 100%
(%) =
4
19
× 100%
(%) = 21.05%
5. Electronusa Mechanical System [Research Center for Electronic and Mechanical]
5 | P a g e
Figure 5. in the simulation
3.2 Analysis of ac
In the analytical is 6.5 (0.0065 ), while in the simulation is 0.07 (figure 6). The
difference is:
(%) =
( ) − ( )
( )
× 100%
(%) =
0.07 − 0.0065
0.07
× 100%
(%) =
0.0635
0.07
× 100%
(%) = 90.71%
(a) (b) (c)
(d) (e)
Figure 6. (a). in the simulation at 1Hz
(b). in the simulation at 10Hz
(c). in the simulation at 100Hz
(d). in the simulation at 1kHz
(e). in the simulation at 10kHz
6. Electronusa Mechanical System [Research Center for Electronic and Mechanical]
6 | P a g e
In the simulation, is 14.9 (figure 7), while in the analytical is 1.3 . The difference is:
(%) =
( ) − ( )
( )
× 100%
(%) =
14.9 − 1.3
14.9
× 100%
(%) =
13.6
14.9
× 100%
(%) = 91.275%
(a) (b) (c)
(d) (e)
Figure 7. (a). in the simulation at 1Hz
(b). in the simulation at 10Hz
(c). in the simulation at 100Hz
(d). in the simulation at 1kHz
(e). in the simulation at 10kHz
In the simulation, is 0 at 1Hz, is 0 at 10Hz, is 0.05 at 100Hz, is 0.94 at 1kHz, 9.61 at
10kHz, and 15.2 at 16kHz (figure 8). The difference is:
For 1Hz,
(%) =
( ) − ( )
( )
× 100%
(%) =
1.3 − 0.53
1.3
× 100%
(%) =
1.30000 − 0.00053
1.3
× 100%
(%) =
1.29947
1.3
× 100%
(%) = 99.959%
8. Electronusa Mechanical System [Research Center for Electronic and Mechanical]
8 | P a g e
(%) = 93.47%
For 16kHz,
(%) =
( ) − ( )
( )
× 100%
(%) =
1.3 − 84.8
1.3
× 100%
(%) =
1.3000 − 0.0848
1.3000
× 100%
(%) =
1.2152
1.3000
× 100%
(%) = 93.47%
(a) (b) (c)
(d) (e) (f)
Figure 8. (a). in the simulation at 1Hz
(b). in the simulation at 10Hz
(c). in the simulation at 100Hz
(d). in the simulation at 1kHz
(e). in the simulation at 10kHz
(f). in the simulation at 16kHz
In the simulation, is 0 at 1Hz, is 0 at 10Hz, is 0.32 at 100Hz, is 5.36 at 1kHz, is 53.8
at 10kHz, and 85.3 at 16kHz (figure 9). The difference is:
For 1Hz,
(%) =
( ) − ( )
( )
× 100%
(%) =
993.2 − 2.97
993.2
× 100%
(%) =
990.23
993.2
× 100%
(%) = 99.7%
10. Electronusa Mechanical System [Research Center for Electronic and Mechanical]
10 | P a g e
(%) = 52.17%
In this study, the simulation shows that the and became stable started at 1 kHz.
(a) (b) (c)
(d) (e) (f)
Figure 9. (a). in the simulation at 1Hz
(b). in the simulation at 10Hz
(c). in the simulation at 100Hz
(d). in the simulation at 1kHz
(e). in the simulation at 10kHz
(f). in the simulation at 16kHz