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 25⁄ .
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%

7. Electronusa Mechanical System [Research Center for Electronic and Mechanical]
7 | P a g e
For 10Hz,
(%) =
( ) − ( )
( )
× 100%
(%) =
1.3 − 4.37
1.3
× 100%
(%) =
1.3000 − 0.00437
1.3000
× 100%
(%) =
1.29563
1.3000
× 100%
(%) = 99.66%
For 100Hz,
(%) =
( ) − ( )
( )
× 100%
(%) =
1.3 − 38.9
1.3
× 100%
(%) =
1.3000 − 0.0389
1.3000
× 100%
(%) =
1.2611
1.3000
× 100%
(%) = 97%
For 1kHz,
(%) =
( ) − ( )
( )
× 100%
(%) =
1.3 − 83.3
1.3
× 100%
(%) =
1.3000 − 0.0833
1.3000
× 100%
(%) =
1.2167
1.3000
× 100%
(%) = 93.59%
For 10kHz,
(%) =
( ) − ( )
( )
× 100%
(%) =
1.3 − 84.8
1.3
× 100%
(%) =
1.3000 − 0.0848
1.3000
× 100%
(%) =
1.2152
1.3000
× 100%

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%

9. Electronusa Mechanical System [Research Center for Electronic and Mechanical]
9 | P a g e
For 10Hz,
(%) =
( ) − ( )
( )
× 100%
(%) =
993.2 − 24.6
993.2
× 100%
(%) =
968.6
993.2
× 100%
(%) = 97.52%
For 100Hz,
(%) =
( ) − ( )
( )
× 100%
(%) =
993.2 − 218
993.2
× 100%
(%) =
775.2
993.2
× 100%
(%) = 78.05%
For 1kHz,
(%) =
( ) − ( )
( )
× 100%
(%) =
993.2 − 466
993.2
× 100%
(%) =
527.2
993.2
× 100%
(%) = 53.08%
For 10kHz,
(%) =
( ) − ( )
( )
× 100%
(%) =
993.2 − 475
993.2
× 100%
(%) =
518.2
993.2
× 100%
(%) = 52.17%
For 16kHz,
(%) =
( ) − ( )
( )
× 100%
(%) =
993.2 − 475
993.2
× 100%
(%) =
518.2
993.2
× 100%

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