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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Chapter 6:
Frequency Response and System Concepts โ Instructor Notes
Chapter 6 can be covered immediately following Chapter 4, or after completing Chapter 5. There is no direct dependence of Chapter 6 on Chapter 5. Modularity is a recurrent feature of this book, and we shall draw attention to it throughout these Instructor Notes. Section 6.1 introduces the notion of sinusoidal frequency response and motivates the use of sinusoidal signals; the Fourier Series method of representing signals is described in detail in Section 6.2. Further, the text and examples also illustrate the effect of a multi-components signal propagating through a linear system. Four examples accompany this presentation. Section 6.3 introduces filters, and outlines the basic characteristics of low-, high- and band-pass filters. The concept of resonance is treated in greater depth than in the previous edition, and a connection is made with the natural response of second order circuits, which may be useful to those instructors who have already covered transient response of second-order circuits. Four detailed examples are included in this section, Further, the boxes Focus on Measurements: Wheatstone bridge filter (pp. 315-317), Focus on Measurements: AC line interference filter (pp. 317319), and Focus on Measurements: Seismic displacement transducer (pp. 319-322) touch on additional application examples. The first and last of these boxes can be linked to related material in Chapters 2, 3, and 4. The instructor who has already introduced the operational amplifier as a circuit element will find that section 8.3, on active filters, is an excellent vehicle to reinforce both the op-amp concept and the frequency response ideas. Another alternative (employed by this author) consists of introducing the op-amp at this stage, covering sections 8.1 through 8.3. Finally, Section 6.4 covers Bode plots, and illustrates how to create approximate Bode plots using the straight-line asymptotic approximation. The box Focus on Methodology: Bode Plots (p. 327) clearly outlines the method, which is further explained in two examples. The homework problems present several frequency response, Fourier Series, filter and Bode plot exercises of varying difficulty. The instructor who wishes to use one of the many available software aids (e.g., MATLABยฎ or Electronics Workbenchยฎ ) to analyze the frequency response of more complex circuits and to exploit more advanced graphics capabilities, will find that several advanced problems lend themselves nicely to such usage. A number of new application oriented problems have been introduced in the 5 th Edition, including problems related to loudspeaker crossover networks (6.64, 6.66 and 6.69), and 60-Hz line noise filtering (6.68). The 5th Edition of this book includes 7 new problems; some of the 4 th Edition problems were removed, increasing the end-of-chapter problem count from 76 to 81.
Learning Objectives for Chapter 6 1. 2. 3. 4.
Understand the physical significance of frequency domain analysis, and compute the frequency response of circuits using AC circuit analysis tools. Compute the Fourier spectrum of periodic signals using the Fourier series representation, and use this representation in connection with frequency response ideas to compute the response of circuits to periodic inputs. Analyze simple first- and second-order electrical filters, and determine their frequency response and filtering properties. Compute the frequency response of a circuit and its graphical representation in the form of a Bode plot.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.1 Determine the frequency response ๐๐๐ข๐ก (๐๐)/๐๐๐ (๐๐ค) for the circuit of Figure P6.. Assume: ๐ฟ = 0.5 ๐ป and ๐ = 200 ๐ฮฉ.
Known quantities: ๐ฟ = 0.5 ๐ป, ๐ = 200 ๐ฮฉ
Find: (a) Determine the frequency response ๐๐๐ข๐ก (๐๐)/๐๐๐ (๐๐) for the circuit of Figure P6.1. (b) Plot the magnitude and phase of the circuit for frequencies between 10 and 107 rad/s on graph paper, with a linear scale for frequency. (c) Repeat part (b), using semilog paper. Place the frequency on the logarithmic axis. (d) Plot the magnitude response on semilog paper with magnitude in decibels.
Analysis: (a) Determine the frequency response ๐ฝ๐๐๐ (๐๐)/๐ฝ๐๐ (๐๐) for the circuit of Figure P6.1. ๐๐๐ข๐ก (๐๐)/๐๐๐ (๐๐) can be determine by the voltage division: ๐๐๐ข๐ก = ๐๐๐
๐ฟ ๐ฟ+๐
Convert to the frequency domain: ๐๐๐ข๐ก (๐๐) ๐๐๐ (๐๐)
=
๐๐๐ฟ ๐๐๐ฟ + ๐
Substitute known values: ๐๐๐. ๐ ๐๐๐ ๐ + ๐๐๐. ๐
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6 (b) Plot the magnitude and phase of the circuit for frequencies between ๐๐ and ๐๐๐ rad/s on graph paper, with a linear scale for frequency. Convert the result from (a) into a manageable format using the complex conjugate of the denominator:
๐๐๐ข๐ก (๐๐) ๐๐๐ (๐๐)
=
(0.5๐)2 + ๐๐100 ๐ 4๐10 + (0.5๐)2
Magnitude: ๐๐๐ข๐ก (๐๐)
0.5๐2
๐๐๐ (๐๐)
4๐10 + (0.5๐)2
|
| = โ(
2
) +(
2
100 ๐ ๐ 4๐10 + (0.5๐)2
)
Phase: โ
๐๐๐ข๐ก (๐๐) 100 ๐ ๐ = atan ( ) (0.5๐)2 ๐๐๐ (๐๐) 100๐ = atan ( ) 0.25๐
Plots: Magnitude 1.4
1.2
Magnitude
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4 5 6 Frequency (rad/s)
7
8
9
10 6
x 10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Phase (degrees) 90 80 70
Phase
60 50 40 30 20 10 0
0
1
2
3
4 5 6 Frequency (rad/s)
7
8
9
10 6
x 10
(c) Repeat part (b), using semilog paper. Place the frequency on the logarithmic axis. Magnitude 1.4
1.2
Magnitude
1
0.8
0.6
0.4
0.2
0 1 10
2
10
3
10
4
5
10 10 Frequency (rad/s)
6
10
7
10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Phase (degrees) 90 80 70
Phase
60 50 40 30 20 10 0 1 10
2
10
3
10
4
5
10 10 Frequency (rad/s)
6
10
7
10
(d) Plot the magnitude response on semilog paper with magnitude in decibels. Magnitude 50 45 40
Magnitude
35 30 25 20 15 10 5 0 1 10
2
10
3
10
4
5
10 10 Frequency (rad/s)
6
10
7
10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.2 Repeat the instructions of Problem 6.1 for the circuit of Figure P6.2.
Known quantities: Given in figure.
Find: (a) Determine the frequency response ๐๐๐ข๐ก (๐๐)/๐๐๐ (๐๐) for the circuit of Figure P6.2. (b) Plot the magnitude and phase of the circuit for frequencies between 10 and 107 rad/s on graph paper, with a linear scale for frequency. (c) Repeat part (b), using semilog paper. Place the frequency on the logarithmic axis. (d) Plot the magnitude response on semilog paper with magnitude in decibels.
Analysis: (a) Determine the frequency response ๐ฝ๐๐๐ (๐๐)/๐ฝ๐๐ (๐๐) for the circuit of Figure P6.2. Determine ๐๐๐ข๐ก (๐๐ค) and ๐๐๐ (๐๐ค) by using voltage division: ๐ ๐ถ ๐ +๐ถ ๐๐๐ข๐ก = ๐๐๐ ๐ ๐ถ ๐ + ๐ +๐ถ Convert to frequency domain: 1 ๐๐๐ถ 1 ๐ + ๐๐๐ถ ๐๐๐ข๐ก (๐๐) = ๐๐๐ (๐๐) 1 ๐ ๐๐๐ถ ๐ + 1 ๐ + ๐๐๐ถ ๐
Simplify and solve for ๐๐๐ข๐ก (๐๐)/๐๐๐ (๐๐): 6.6 PROPRIETARY MATERIAL. ยฉ The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6 ๐๐๐ข๐ก (๐๐) 1 = ๐๐๐ (๐๐) 2 + ๐๐๐ ๐ถ Substitute known values: ๐ฝ๐๐๐ (๐๐) ๐ = ๐ฝ๐๐ (๐๐) ๐ + ๐๐๐. ๐ (b) Plot the magnitude and phase of the circuit for frequencies between ๐๐ and ๐๐๐ rad/s on graph paper, with a linear scale for frequency. Determine the equations for magnitude in phase. First, simplify ๐๐๐ข๐ก (๐๐)/๐๐๐ (๐๐) by multiplying by the complex conjugate of the denominator: ๐๐๐ข๐ก (๐๐) 2 โ 0.1๐๐ = ๐๐๐ (๐๐) 4 + 0.01๐ 2 Magnitude:
|
2 2 ๐๐๐ข๐ก (๐๐) 2 0.1๐ | = โ( ) + ( ) ๐๐๐ (๐๐) 4 + 0.01๐ 2 4 + 0.01๐ 2
Phase: โ
๐๐๐ข๐ก (๐๐) โ0.1๐ = arctan( ) ๐๐๐ (๐๐) 2
Plots: Magnitude 0.45 0.4 0.35
Magnitude
0.3 0.25 0.2 0.15 0.1 0.05 0
0
1
2
3
4 5 6 Frequency (rad/s)
7
8
9
10 6
x 10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Phase (degrees) -20 -30 -40
Phase
-50 -60 -70 -80 -90 -100
0
1
2
3
4 5 6 Frequency (rad/s)
7
8
9
10 6
x 10
(c) Repeat part (b), using semilog paper. Place the frequency on the logarithmic axis. Magnitude 0.45 0.4 0.35
Magnitude
0.3 0.25 0.2 0.15 0.1 0.05 0 1 10
2
10
3
10
4
5
10 10 Frequency (rad/s)
6
10
7
10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Phase (degrees) -20 -30 -40
Phase
-50 -60 -70 -80 -90 -100 1 10
2
10
3
10
4
5
10 10 Frequency (rad/s)
6
10
7
10
(d) Plot the magnitude response on semilog paper with magnitude in decibels. Magnitude 0
-20
Magnitude
-40
-60
-80
-100
-120 1 10
2
10
3
10
4
5
10 10 Frequency (rad/s)
6
10
7
10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.3 Repeat the instructions of Problem 6.1 for the circuit of Figure P6.3.
Known quantities: Values in Figure P6.3.
Find: (a) Determine the frequency response ๐๐๐ข๐ก (๐๐)/๐๐๐ (๐๐) for the circuit of Figure P6.3. (b) Plot the magnitude and phase of the circuit for frequencies between 10 and 107 rad/s on graph paper, with a linear scale for frequency. (c) Repeat part (b), using semilog paper. Place the frequency on the logarithmic axis. (d) Plot the magnitude response on semilog paper with magnitude in decibels.
Analysis: (a) Determine the frequency response ๐ฝ๐๐๐ (๐๐)/๐ฝ๐๐ (๐๐) for the circuit of Figure P6.3. To determine the frequency response, place an imaginary voltage source at ๐๐๐ and determine ๐๐๐ข๐ก . For this problem, ๐๐๐ข๐ก may be determined from the voltage division of the top node: ๐๐๐ข๐ก = ๐1
C 1000 ฮฉ + ๐ถ
where ๐1 is: ๐1 = ๐๐๐
๐๐๐ 2000 ฮฉ + ๐๐๐
and ๐๐๐ is: ๐๐๐ =
2000 ฮฉ โ (1000 ฮฉ โ ๐ถ) 2000 ฮฉ + (1000 ฮฉ + ๐ถ) 6.10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
=
2๐6 ฮฉ โ ๐ถ 3000 ฮฉ + ๐ถ
Put ๐๐๐ into the frequency domain: 1 ๐๐๐ถ ๐๐๐ (๐๐) = 1 3000 ฮฉ + ๐๐๐ถ 2๐6 ฮฉ โ
Simplify: 1 ๐๐๐ถ ๐๐๐ถ ๐๐๐ (๐๐) = โ 1 ๐๐๐ถ 3000 + ๐๐๐ถ 2๐6 โ
2๐6 โ =
๐๐๐ถ ๐๐๐ถ
3000 ฮฉ โ jฯC + =
๐๐๐ถ ๐๐๐ถ
2๐6 1 + 3000 ๐๐๐ถ
Substitute ๐๐๐ (๐๐) into ๐1 (๐๐): 2๐6 1 + 3000 ๐๐๐ถ ๐1 (๐๐) = ๐๐๐ (๐๐) 2๐6 2000 + 1 + 3000 ๐๐๐ถ Simplify: 2๐6 1 + 3000 ๐๐๐ถ 1 + 3000 ๐๐๐ถ ๐1 (๐๐) = ๐๐๐ (๐๐) โ 2๐6 1 + 3000 ๐๐๐ถ 2000 + 1 + 3000 ๐๐๐ถ 1 + 3000 ๐๐๐ถ 1 + 3000 ๐๐๐ถ = ๐๐๐ (๐๐) 1 + 3000 ๐๐๐ถ 2000 โ (1 + 3000 ๐๐๐ถ) + 2๐6 โ 1 + 3000 ๐๐๐ถ 2๐6 โ
1 + 3000 ๐๐๐ถ 1 + 3000 ๐๐๐ถ = ๐๐๐ (๐๐) 1 + 3000 ๐๐๐ถ 2000 โ (1 + 3000 ๐๐๐ถ) + 2๐6 โ 1 + 3000 ๐๐๐ถ 2๐6 โ
= ๐๐๐ (๐๐)
2๐6 2000 + ๐๐6๐6 โ ๐ถ + 2๐6
โ ๐๐๐ (๐๐)
2๐6 2๐6 + ๐๐60 6.11
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Substitute ๐1 (๐๐) into ๐๐๐ข๐ก (๐๐):
๐๐๐ข๐ก (๐๐) = ๐1 (๐๐)
1 ๐๐๐ถ 1000 +
1 ๐๐๐ถ
Simplify:
๐๐๐ข๐ก (๐๐) = ๐1 (๐๐)
= ๐1 (๐๐)
1 ๐๐๐ถ 1 1000 + ๐๐๐ถ
โ
๐๐๐ถ ๐๐๐ถ 1000 โ ๐๐๐ถ +
= ๐1 (๐๐)
๐๐๐ถ ๐๐๐ถ
๐๐๐ถ ๐๐๐ถ
1 1 + 0.01๐๐
Plug ๐1 (๐๐) into this equation: ๐๐๐ข๐ก (๐๐) = ๐๐๐ (๐๐)
2๐6 1 2๐6 + ๐๐60 1 + 0.01๐๐
Simplify and solve for ๐๐๐ข๐ก (๐๐)/๐๐๐ (๐๐): ๐๐๐ข๐ก (๐๐) 2๐6 โ ๐๐๐ (๐๐) 2๐6 + 2๐4๐๐ Simplify: ๐ฝ๐๐๐ (๐๐) ๐ โ ๐ฝ๐๐ (๐๐) ๐ + ๐. ๐๐๐๐ (b) Plot the magnitude and phase of the circuit for frequencies between ๐๐ and ๐๐๐ rad/s on graph paper, with a linear scale for frequency. First, get ๐๐๐ข๐ก (๐๐)/๐๐๐ (๐๐) into a manageable form: ๐๐๐ข๐ก (๐๐) 1 1 โ 0.01๐๐ = โ ๐๐๐ (๐๐) 1 + 0.01๐๐ 1 โ 0.01๐๐ โ
1 โ 0.01 ๐๐ 1 + 0.0001๐ 2
Calculate magnitude of complex expression: 6.12 PROPRIETARY MATERIAL. ยฉ The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
|
2 2 ๐๐๐ข๐ก (๐๐) 1 0.01๐ | = โ( ) +( ) 2 2 ๐๐๐ (๐๐) 1 + 0.0001๐ 1 + 0.0001๐
Phase response: โ0.01๐ ๐๐๐ข๐ก (๐๐) 1 โ 0.0001๐ 2 ) โ = arctan ( 1 ๐๐๐ (๐๐) 1 โ 0.0001๐ 2 = arctan(โ0.01๐) Plots: Magnitude 1 0.9 0.8
Magnitude
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
0
1
2
3
4 5 6 Frequency (rad/s)
7
8
9
10 6
x 10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Phase (degrees) 0 -10 -20
Phase
-30 -40 -50 -60 -70 -80 -90
0
1
2
3
4 5 6 Frequency (rad/s)
7
8
9
10 6
x 10
(c) Repeat part (b), using semilog paper. Place the frequency on the logarithmic axis. Magnitude 1 0.9 0.8
Magnitude
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 10
2
10
3
10
4
5
10 10 Frequency (rad/s)
6
10
7
10
6.14 PROPRIETARY MATERIAL. ยฉ The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Phase (degrees) 0 -10 -20
Phase
-30 -40 -50 -60 -70 -80 -90 1 10
2
10
3
10
4
5
10 10 Frequency (rad/s)
6
7
10
10
(d) Plot the magnitude response on semilog paper with magnitude in decibels. Magnitude 0
-20
Magnitude
-40
-60
-80
-100
-120 1 10
2
10
3
10
4
5
10 10 Frequency (rad/s)
6
10
7
10
6.15 PROPRIETARY MATERIAL. ยฉ The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.4 Repeat Problem 6.1 for the circuit of Figure P6.4. R1 = 300ohm R2 = R3 = 500ohm, L = 4H, C1 = 40 ฮผF, C2 = 160 ฮผF.
Solution: Known quantities: Resistance, inductance and capacitance values, in the circuit of Figure P6.4.
Find: a) b) c) d)
The frequency response for the circuit of Figure P6.4. Plot magnitude and phase of the circuit using a linear scale for frequency. Repeat part b., using semilog paper. Plot the magnitude response using semilog paper with magnitude in dB.
Analysis:
1 Vout 1 โ 0.0008๏ท 2 + j (0.05)๏ท j๏ท (C1 + C2 ) ( j๏ท ) = = 1 Vin 1 โ 0.0008๏ท 2 + j (0.11)๏ท R1 + R2 || R3 + j๏ทL + j๏ท (C1 + C2 ) R2 || R3 + j๏ทL +
a) Vout Vin
( j๏ท ) =
(1 โ ๏ท 2 0.0008) 2 + ((0.05)๏ท ) 2 (1 โ ๏ท 2 0.0008) 2 + ((0.11)๏ท ) 2
V ๏ฆ (0.05)๏ท ๏ถ ๏ฆ (0.11)๏ท ๏ถ ๏ out ( j๏ท ) = arctan๏ง ๏ท โ arctan๏ง ๏ท 2 2 Vin ๏จ 1 โ ๏ท 0.0008 ๏ธ ๏จ 1 โ ๏ท 0.0008 ๏ธ
The plots obtained using Matlab are shown below: b)
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
c)
d)
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.5 Determine the frequency response of the circuit of Figure P6.5 and generate frequency response plots. ๐ 1 = 20 ๐ฮฉ, ๐ 2 = 100 ๐ฮฉ, ๐ฟ = 1 ๐ป, and ๐ถ = 100 ๐๐น.
Assume:
Known quantities: ๐ 1 = 20 ๐ฮฉ, ๐ 2 = 100 ๐ฮฉ, ๐ฟ = 1 ๐ป, ๐ถ = 100 ๐๐น
Find: The frequency response for the circuit and generate frequency response plots.
Analysis: The output voltage, ๐ฃ๐๐ข๐ก , is the voltage across the capacitor, ๐ฃ๐ . The capacitorโs voltage may be calculated as a voltage division of the top node voltage, ๐ฃ1 : ๐ฃ๐๐ข๐ก = ๐ฃ1
๐ง๐ถ ๐ 2 + ๐ง๐ถ
To determine ๐ฃ๐๐ข๐ก , find ๐ฃ1 . First, calculate ๐ฃ1 as a voltage division of ๐ฃ๐๐ : ๐ฃ1 = ๐ฃ๐๐
๐ 1 โฅ (๐ 2 + ๐ง๐ถ ) ๐ฟ + ๐ 1 โฅ (๐ 2 + ๐ง๐ถ )
Convert to frequency domain: 1 ) ๐๐๐ถ ๐ฃ1 (๐๐) = ๐ฃ๐๐ (๐๐) 1 ๐๐๐ฟ + ๐ 1 โฅ (๐ 2 + ) ๐๐๐ถ ๐ 1 โฅ (๐ 2 +
๐ 1 (๐ 2 +
= ๐ฃ๐๐ (๐๐)
๐๐๐ฟ (๐ 1 + ๐ 2 + = ๐ฃ๐๐ (๐๐) =
1 ) ๐๐๐ถ
1 1 ) + ๐ 1 (๐ 2 + ) ๐๐๐ถ ๐๐๐ถ
๐๐๐ถ๐ 1 ๐ 2 + ๐ 1 ๐ 1 โ ๐ 2 ๐ฟ๐ถ(๐ 1 + ๐ 2 ) + ๐๐(๐ฟ + ๐ถ๐ 1 ๐ 2 )
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6 Convert ๐ฃ๐๐ข๐ก to the frequency domain: 1 ๐๐๐ถ
๐ฃ๐๐ข๐ก (๐๐) = ๐ฃ1 (๐๐)
๐ 2 +
1 ๐๐๐ถ
Plug in ๐ฃ1 (๐๐): 1 ๐ฃ๐๐ข๐ก (๐๐) ๐๐๐ถ๐ 1 ๐ 2 + ๐ 1 ๐๐๐ถ = ๐ฃ๐๐ (๐๐) ๐ 1 โ ๐ 2 ๐ฟ๐ถ(๐ 1 + ๐ 2 ) + ๐๐(๐ฟ + ๐ถ๐ 1 ๐ 2 ) ๐ + 1 2 ๐๐๐ถ =
=
๐๐๐ถ๐ 1 ๐ 2 + ๐ 1 2 [๐ 1 โ ๐ ๐ฟ๐ถ(๐ 1 + ๐ 2 ) + ๐๐(๐ฟ + ๐ถ๐ 1 ๐ 2 )](1 + ๐๐๐ 2 ๐ถ) ๐ 1 + ๐๐๐ถ๐ 1 ๐ 2 ๐ถ๐ ๐ 2 ๐ 1 โ ๐ 2 ๐ฟ๐ถ (๐ 1 + 2๐ 2 + 1 2 ) + ๐๐[๐ฟ + 2๐ถ๐ 1 ๐ 2 โ ๐ 2 ๐ 2 ๐ฟ๐ถ 2 (๐ 1 + ๐ 2 )] ๐ฟ
Substitute known values: ๐๐๐๐ (๐๐) ๐๐ฌ๐ + ๐๐๐๐ฌ๐ = ๐๐๐ (๐๐) ๐๐ฌ๐ โ ๐๐ฌ๐ ๐๐ + ๐๐(๐๐ฌ๐ โ ๐๐๐๐๐ ) To generate plots, determine the magnitude and the phase. First, convert ๐ฃ๐๐ข๐ก (๐๐)/๐ฃ๐๐ (๐๐) to a manageable format using the complex conjugate: ๐ฃ๐๐ข๐ก (๐๐) 4๐ธ 8 โ ๐2 (8๐ธ10 + 2.4๐ธ 7 ๐2 ) โ ๐๐(4๐ธ 9 + 1.6๐ธ12 ๐2 ) = (2๐ธ 4 + 8๐ธ 6 ๐ 2 )2 + ๐ 2 (4๐ธ 5 โ 120๐ 2 )2 ๐ฃ๐๐ (๐๐) Magnitude: 2
๐ฃ๐๐ข๐ก (๐๐) 4๐ธ 8 โ ๐ 2 (8๐ธ10 + 2.4๐ธ 7 ๐ 2 ) ๐(4๐ธ 9 + 1.6๐ธ12 ๐ 2 ) | | = โ( ) +( ) 4 6 2 2 2 5 2 2 4 (2๐ธ + 8๐ธ ๐ ) + ๐ (4๐ธ โ 120๐ ) (2๐ธ + 8๐ธ 6 ๐ 2 )2 + ๐ 2 (4๐ธ 5 โ 120๐ 2 )2 ๐ฃ๐๐ (๐๐)
2
Phase: โ
๐ฃ๐๐ข๐ก (๐๐) โ๐(4๐ธ 9 + 1.6๐ธ12 ๐2 ) = arctan ( 8 ) ๐ฃ๐๐ (๐๐) 4๐ธ โ ๐ 2 (8๐ธ10 + 2.4๐ธ 7 ๐ 2 )
Plots:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6 -3
2.5
Magnitude
x 10
Magnitude
2
1.5
1
0.5
0 1 10
2
10
3
10
4
5
10 10 Frequency (rad/s)
6
10
7
10
Phase (degrees) 90 80 70
Phase
60 50 40 30 20 10 0 1 10
2
10
3
10
4
5
10 10 Frequency (rad/s)
6
10
7
10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.6 In the circuit shown in Figure P6.6, where C = 0.5 ฮผF and R = 2 kohm, a. Determine how the input impedance Z( jฯ) = Vi ( jฯ)/Ii ( jฯ) behaves at extremely high and low frequencies. b. Find an expression for the impedance. c. Show that this expression can be manipulated into the form Z( jฯ) = R[1+j(1/ ฯ RC)] d. Determine the frequency ฯ = ฯC for which the imaginary part of the expression in part c is equal to 1. e. Estimate (without computing it) the magnitude and phase angle of Z(jฯ) at ฯ = 10 rad/s and ฯ = 10^5 rad/s.
Solution: Known quantities: Figure P6.6.
Find:
V i ( j๏ท ) , behaves at extremely high or low frequencies. I i ( j๏ท ) b) An expression for the input (or driving point) impedance. ๏ฉ 1 ๏น c) Show that this expression can be manipulated into the form: Z ( j๏ท) = R๏ช1+ j ๏ซ ๏ทRC ๏บ๏ป d) Determine the frequency ๏ท = ๏ทc for which the imaginary part of the expression in c) is equal to 1. e) Estimate the magnitude and angle of Z [j๏ท] at ๏ท = 10 rad/s and 100,000 rad/s. a)
How the input impedance, Z ( j๏ท )=
Analysis: a)
As ๏ท โ ๏ฅ,
๏
๏ Z C โ 0 ๏ Short ๏Z โ R
As ๏ท โ 0, Z C โ ๏ฅ ๏ Open ๏ Z โ ๏ฅ b) and c) KVL : - V i + I i Z C + I i Z R = 0 ๏ฆ 1 1 ๏ถ ๏ Z ( j๏ท ) = V i = Z + Z = R๏ง1โ j ๏ท C R = R+ ๏จ j ๏ท C ๏ท C๏ธ Ii 1 1 1 rad d) = 1 ๏ ๏ทc = = = 1000 โ6 RC RC s ๏ทc [ 2000 ] [ 0.5๏10 ] ๏ฆ rad ๏ถ Z ๏ง10 ๏ท ๏ป 200 k๏ ๏ โ 90๏ฐ ๏จ s ๏ธ ๏ e) ๏ฆ rad ๏ถ Z ๏ง100 k ๏ท ๏ป 2 k๏ ๏0๏ฐ ๏จ s ๏ธ more precisely,
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
๏ฆ ๏ถ ๏ฆ rad ๏ถ 1 Z ๏ง10 ๏ท = 20.001 k๏ ๏ โ 89.43๏ฐ ๏ท = R ๏ง1โ j โ6 ๏จ s ๏ธ [ 10 ] [ 2000 ] [ 0.5๏10 ] ๏ธ ๏จ ๏ฆ ๏ถ ๏ฆ rad ๏ถ 1 Z ๏ง10 5 ๏ท =2 k๏ ๏ โ 0.06๏ฐ ๏ท = R ๏ง1โ j ๏จ s ๏ธ [ 10 5 ] [ 2000 ] [ 0.5๏10โ6 ] ๏ธ ๏จ
๏
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.7 In the circuit shown in Figure P6.7, where L = 2mHand R = 2 kohm, a. Determine how the input impedance Z( jฯ) = Vi ( jฯ)/Ii ( jฯ) behaves at extremely high and low frequencies. b. Find an expression for the impedance. c. Show that this expression can be manipulated into the form Z( jฯ) = R[1+j(1/ ฯ RC)] d. Determine the frequency ฯ = ฯC for which the imaginary part of the expression in part c is equal to 1. e. Estimate (without computing it) the magnitude and phase angle of Z(jฯ) at ฯ = 10^5 rad/s, ฯ = 10^6 rad/s ฯ = 10^7 rad/s
Solution: Known quantities: Figure P6.7.
Find: a)
How the input impedance,
Z ( j๏ท) =
V i ( j๏ท) , I i ( j๏ท)
behaves at extremely high or low frequencies. b) An expression for the input (or driving point) impedance.
๏ทL ๏น ๏ฉ Show that this expression ๏ can be manipulated into the form: Z ( j๏ท )= R ๏ช1 + j R ๏บ๏ป ๏ซ d) Determine the frequency ๏ท = ๏ท C for which the imaginary part of the expression in c) is equal to 1. c)
e)
Estimate the magnitude and angle of Z [j๏ท] at ๏ท = 105, 106, 107 rad/s.
Analysis: a)
As ๏ท โ ๏ฅ,
Z L โ ๏ฅ ๏ Open ๏Z โ ๏ฅ
As ๏ท โ 0,
Z L โ 0 ๏ Short ๏ Z โ R
b) KVL : โ V i + I i Z R + I i Z L = 0
๏
๏
๏
Z ( j๏ท ) = V i = Z L + Z R = j๏ทL + R Ii
๏ฆ ๏ทL ๏ถ c) Z ( j๏ท) = R + j ๏ทL = R ๏ง 1+ j ๏ท ๏จ R ๏ธ R 2000 rad ๏ท L = 1000 k d) c = 1 ๏ ๏ท c = = โ3 R L s 2 ๏10 e)
๏
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
๏ฆ rad ๏ถ 2 ๏ 10 โ3 ๏ 10 5 ๏ถ๏ท ๏ฆ Z ๏ง100k = 2000(1 + j 0.1) = 2.01 k๏๏5.71๏ฐ ๏ท = R ๏ง1 + j ๏ง ๏ท s ๏ธ 2000 ๏จ ๏จ ๏ธ ๏ฆ rad ๏ถ 2 ๏ 10 -3 ๏ 10 6 ๏ถ๏ท ๏ฆ Z ๏ง1M = 2000(1 + j 1) = 2.82 k๏ ๏45.00๏ฐ ๏ท = R๏ง1 + j ๏ง ๏ท s ๏ธ 2000 ๏จ ๏จ ๏ธ ๏ฆ rad ๏ถ 2 ๏ 10 โ3 ๏ 10 7 ๏ถ๏ท ๏ฆ Z ๏ง10M = 2000(1 + j 10 ) = 20.10 k๏ ๏84.29๏ฐ ๏ท = R ๏ง1 + j ๏ง ๏ท s ๏ธ 2000 ๏จ ๏จ ๏ธ Note, in particular, the behavior of the impedance one decade below and one decade above the cutoff frequency.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.8
Solution: Known quantities: With reference to Figure P6.8
L = 190 mH R1 = 2.3 k๏ C = 55 nF R 2 = 1.1 k๏ Find:
V i ( j๏ท) , behaves at extremely high or low frequencies. I i ( j๏ท) ๏ b) An expression for the input impedance in the form: ๏ฆ 1+ j f (๏ท) ๏ถ L 1 Z ( j๏ท) = Z o ๏ง๏ง ๏ท๏ท Z o = R1 + f 2 (๏ท)๏ธ R2C ๏จ 1+ j ๏ a)
How the input impedance, Z ( j๏ท) =
f 1 (๏ท) =
๏
๏ท 2 R1 LC โ R1 โ R 2 ๏ท(R1 R 2 C + L )
f 2 (๏ท) =
๏ท 2 LC โ1 ๏ทC R 2
c) Determine the four cutoff frequencies at which f1[๏ท] = +1 or -1 d) Determine the resonant frequency of the circuit. e) Plot the impedance vs. frequency.
Analysis:
As ๏ท โ ๏ฅ,
a)
As ๏ท โ 0,
and f2[๏ท] = +1 or -1.
Z L โ ๏ฅ ๏ Open, Z C โ 0 ๏ Short ๏Z โ R1 Z C โ ๏ฅ ๏ Open, Z L โ 0 ๏ Short ๏ Z โ R1 + R2
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6 b)
๏
๏
1 ] [ R 2 + j๏ทL ] j๏ทC 1 + [ R 2 + j๏ทL ] j๏ทC
[
Z[j๏ท ] =
V[j๏ท ] [ + ZL ] = Z R1 + Z C Z R 2 = R1 + I[j๏ท ] ZC + [ Z R2 + Z L ]
= R1 +
( R 1 [ 1 - ๏ท 2 LC ] + R 2 ) + j ( ๏ท R 1 R 2 C + ๏ทL ) (โ j ) R 2 + j ๏ทL = ๏ ๏ 1 - ๏ท 2 LC + j ๏ท R 2 C [ 1 - ๏ท 2 LC ] + j ๏ท R 2 C (โ j )
๏ Z ๏ j๏ท๏ =
( ๏ทR C + j (๏ท LC โ1)
๏ท(R1R2C + L) + j ๏ท 2 R1LC โ R1 โ R2 2
2
)= R R C + L ๏ 1 2
R2C
1+ j
j๏ทC = j๏ทC
๏ท 2 R1LC โ R1 โ R2 ๏ท(R1R2C + L) 1+ j
๏ท 2 LC โ1 ๏ทR2C
c) Both f1[๏ท] and f2[๏ท] can be positive or negative, and therefore equal to plus or minus one depending on the frequency; therefore, both cases must be considered. ๏ท c [ R1 R 2 C + L ] = ๏ฑ 1 f 1 [๏ท c] = R 1 [ 1- ๏ท 2c LC ] + R 2 1 + R2 R ] ๏ท c - R1 = 0 ๏ท 2c ๏ฑ [ 2 + L R1 C R 1 LC 1100 1 rad R2 + 1 = + = 13.69 k -9 L 0.19 s R1 C [ 2300 ] [ 55๏10 ]
๏
3400 rad R1 + R 2 = = 141.46 M 2 -9 R 1 LC [ 2300] [ 0.19 ] [ 55๏10 ] s Where
๏ทc
1 1 = โ [ ๏ฑ 13.69 ๏10 3 ] ๏ฑ ( [ ๏ฑ 13.69 ๏10 3 ]2 - 4[1][ - 141.5๏10 6 ] )1/2 2 2
rad rad ๏ท c4 = 20.569 k s s only the positive answers are physically valid, i.e., a negative frequency is physically impossible. 1 ๏ท 2c LC - 1 = ๏ฑ 1 ๏ ๏ท 2c ๏ฑ [ R 2 ] ๏ท c = 0 f 2 [๏ท c] = L LC ๏ทc R2C 1 1 rad R 2 = 1100 = 5.79 k rad = = 95.69 M 2 -9 L 0.19 s LC [ 0.19 ] [ 55๏10 ] s 1 1 6 2 1/2 [ ๏ฑ 5790 ] ๏ฑ ( [ ๏ฑ 5790 ] + 4[1][ 95.69๏10 ] ) = ๏ฑ 2895 ๏ฑ 10201 ๏ทc = 2 2 rad rad ๏ ๏ท c2 = 7.31 k ๏ท c3 = 13.09 k s s Again, the negative roots were rejected because they are physically impossible. d) Magnitude and phase response in semilogarithmic frequency plots: = ๏ฑ 6.845๏10 3 ๏ฑ 13.724 ๏10 3
๏
๏
๏
๏ ๏ท c1 = 6.879 k
๏
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.9
Solution: Known quantities: In the circuit of Figure P6.9: R1 = 1.3 k๏ R2 = 1.9 k๏ C = 0.5182 ๏F
Find: How the voltage transfer function: V o [j๏ท ] behaves at extremes of high and H v [j๏ท ] = V i [j๏ท ] low frequencies. b) An expression for the voltage transfer function, showing that it can be manipulated into the form: C Ho R2 Where : H o = f[๏ท ] = ๏ทR1 R 2 H v [j๏ท ] = 1 + j f[๏ท ] R1 + R 2 R1 + R 2 c) The "cutoff" frequency at which f[๏ท] = 1 and the value of Ho in dB. a)
๏
๏
๏
Analysis: a)
As ๏ท โ ๏ฅ : VD :
0 Z C โ 0 ๏ โ 90 ๏ Short 0 H v โ 0 ๏ โ 90
As ๏ท โ 0 :
๏
b)
0 Z C โ ๏ฅ ๏ โ 90 ๏ Open R2 VD : H v โ ๏ 00 R1 + R 2 1 [ ] [ R2 ] j๏ทC Z C Z R2 = = Z eq 1 Z C + Z R2 + R2 j๏ทC
j๏ทC j๏ทC
=
R2 1 + j ๏ท R2C
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
VD :
R2 Z eq 1 + j๏ท R 2 C 1 + j ๏ทR 2 C V o [j๏ท ] = = = H v [j๏ท ] = R 1 + j๏ท R 2 C 2 V i [j๏ท ] Z R1 + Z eq + R1 1 + j ๏ท R2C 1 R2 R2 = = ๏ท + + j ๏ท C + R1 R2 R1 R 2 R1 R 2 1 + j R1 R 2 C R1 + R 2
c)
๏
f[๏ท c ] =
๏ท c R1 R 2 C = 1
Ho =
1900 R2 = 1300 + 1900 R1 + R 2
R1 + R 2
๏ทc =
1300 + 1900 [ 1300 ] [ 1900] [ 0.5182๏10
-6
= 2.5 k ]
rad s
= 0.5938 = 20๏ Log[0.5938] = - 4.527 dB
๏
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.10 6.10 The circuit shown in Figure P6.10 is a second-order circuit because it has two reactive components (L and C). A complete solution will not be attempted. However, determine: a. The behavior of the voltage frequency response at extremely high and low frequencies. b. The output voltage Vo if the input voltage has a frequency where: Vi = 7.07โ ฯ 4 V R1 = 2.2 kohm R2 = 3.8 kohm Xc = 5kohm XL = 1.25 kohm c. The output voltage if the frequency of the input voltage doubles so that XC = 2.5 kohm XL = 2.5kohm d. The output voltage if the frequency of the input voltage again doubles so that XC = 1.25 kohm XL = 5kohm
Solution: Known quantities: Figure P6.10.
Find: a) The behavior of the voltage transfer function or gain at extremely high and low frequencies. b) The output voltage Vo if the input voltage has a frequency where: V i = 7.07 V ๏ 45o R 1 = 2.2 k๏ R 2 = 3.8 k๏ X C = 5 k๏ c) The output voltage if the frequency of the input voltage doubles so that: XC = 2.5 k๏ X L = 2.5 k๏ d) The output voltage if the frequency of the input voltage again doubles so that: XC = 1.25 k๏ X L = 5 k๏
๏
๏
X L = 1.25 k๏
Analysis: a)
๏
๏
As ๏ท โ 0
Z C โ ๏ฅ ๏ Open Z L โ 0 ๏ Short V o โ 0
As ๏ท โ ๏ฅ
Z C โ 0 ๏ Short
VD :
Vo=
ZL
โ ๏ฅ ๏ Open
V i R 2 = [ 7.07] [ 3800] = 4.478 V ๏450 2200 + 3800 R1 + R 2
b) Z eq1 = Z R1 + Z C = R1 - j X C
Z eq2 =
Z R2 Z L = Z R2 + Z L
[ R2 ] [ j X L ] R2 + j X L
๏
๏
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
j R2 X L R2 + j X L R2 + j X L VD : V o = ๏ j R2 X L R2 + j X L R1 - j X C + R2 + j X L j R2 X L ๏Vo = Vi [ R1 R 2 + X C X L ] + j [ X L ( R1 + R 2 ) - X C R 2 ] 0 0 6 0 V i ๏ [ j R 2 X L ] = [ 7.07 V ๏45 ] [ ( 3.8 k๏ ) ( 1.25 k๏ ) ๏90 ] = 33.58๏10 ๏135 V i Z eq2 = Vi Z eq1 + Z eq2
๏
6 R 1 R 2 + X C X L = [ 2200] [ 3800] + [ 5000] [ 1250] = 14.61๏10 6 X L [ R 1 + R 2 ] - X C R 2 = [ 1250] [ 6000] - [ 5000 ] [ 3800 ] = - 11.50 ๏10
V0 =
33.58 ๏10 6 ๏1350 14.61๏10 6 - j 11.50 ๏10 6
=
33.58 ๏1350
= 1.806V ๏173.2 0
18.59 ๏ โ 38.2 0
0 0 6 0 V i ๏ [ j R 2 X L ] = [ 7.07 V ๏45 ] [ ( 3800) ( 2500) ๏90 ] = 67.17๏10 ๏135
๏
6 R 1 R 2 + X C X L = [ 2200] [ 3800] + [ 2500] [ 2500] = 14.61๏10
c)
6 X L [ R 1 + R 2 ] + X C R 2 = [ 2500 ] [ 6000 ] - [ 2500 ] [ 3800] = 5.50๏10
Vo =
67.17 ๏10 6 ๏1350 14.61๏10 6 + j 5.50 ๏10 6
=
67.17 V ๏1350 15.61 ๏20.6 0
= 4.303 V ๏114.4 0
d) 0 0 6 0 V i ๏ [ j R 2 X L ] = [ 7.07 V ๏45 ] [ ( 3800) ( 5000)๏90 ] = 134.34 ๏10 ๏135 ๏ R 1 R 2 + X C X L = [ 2200] [ 3800] + [ 1250] [ 5000] = 14.61๏10 6 6 X L [ R 1 + R 2 ] + X C R 2 = [ 5000 ] [ 6000 ] - [ 1250 ] [ 3800] = 25.25๏10
Vo =
๏
134.34 ๏10 6 ๏1350 14.61๏10
6
+ j 25.25๏10
6
=
134.34 V ๏1350 29.17 ๏59.94
0
= 4.605 V ๏75.050
Problem 6.11 In the circuit shown in Figure P6.11, determine the frequency response function in the form: Hv( jฯ) = Vo( jฯ)/Vi ( jฯ)= Hvo/(1ยฑjf( ๏ท ))
Solution: Known quantities: Figure P6.11.
Find: a)
The voltage transfer function in the form:
H v [j๏ท ] =
V o [j๏ท ] H vo = . [j ๏ท ] 1 ๏ฑ j f[ ๏ท ] Vi
b) Plot the Bode diagram, i.e., a semilog plot where the magnitude [in dB] of the transfer function is plotted on a linear scale as a function of frequency on a log scale.
Assume: The values of the resistors and of the capacitor in the circuit of Figure P6.12: C = 0.47 ๏ญF R1 = 16 ๏ R2 = 16 ๏
Analysis: a)
๏
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
VD :
Vo = Vi
H v [j๏ท ] =
Z R2 = Vi Z R1 + Z C + Z R2 R1 +
V o [j๏ท ] V i [j๏ท ]
=
R2 R1 + R 2 1 - j
R2 1 + R2 j๏ทC
1 1
๏ทC [ R1 + R 2 ]
b)
๏
Problem 6.12 The circuit shown in Figure P6.12 has R1 = 100ohm Ro = 100ohm R2 = 50ohm C = 80 nF Determine the frequency response Vout( jฯ)/Vin( jฯ).
Solution: Known quantities: The values of the resistors and of the capacitor in the circuit of Figure P6.12:
R1 = 100 ๏
RL = 100 ๏ R2 = 50 ๏
C = 80 nF
Find: Compute and plot the frequency response function.
๏
Analysis: Using voltage division:
Z eq =
Z R2 Z C = Z R2 + Z C
1 j๏ทC 1 R2 + j๏ทC R2
j๏ทC R2 = j๏ทC 1 + j ๏ท R2C
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
VD :
H v [j๏ท ] =
=
V o [j๏ท ] V i [j๏ท ]
=
Z RL Z R1 + Z eq + Z RL
R L [ 1 + j ๏ทR 2 C ] R1 + R 2 + R L + j [ R1 + R L ] ๏ท R 2 C
=
=
RL R 2 + RL R1 + 1 + j๏ทR 2 C
1 + j๏ทR 2 C = 1 + j๏ทR 2 C
1 + j ๏ท R2C RL R1 + R 2 + R L 1 + j [ R1 + R L ] ๏ท R 2 C R1 + R 2 + R L
Plotting the response in a Bode Plot:
๏
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.13 a. Determine the frequency response Vout( jฯ)/Vin( jฯ) for the circuit of Figure P6.13. b. Plot the magnitude and phase of the circuit for frequencies between 1 and 100 rad/s on graph paper, with a linear scale for frequency. c. Repeat part b, using semilog paper. (Place the frequency on the logarithmic axis.) d. Plot the magnitude response on semilog paper with magnitude in dB.
Note to instructor: the resistance in the figure should be 1000 ๏ .
Solution: Known quantities: The values of the resistors and of the capacitor in the circuit of Figure P6.13: R= 1000 ฯ
C = 100โF
Find: Compute and plot the frequency response function.
Analysis: (a)
Vout = Vin
V ( j๏ท ) 1/j๏ทC 1 1 = , out = R + 1/j๏ทC j๏ทRC +1 Vin ( j๏ท) 1 + j๏ท/10
Vout 1 ,๏ฆ(๏ท) = -arctan(0.1๏ท) = Vin 1 + 0.01๏ท 2 ๏ ๏ ๏ ๏ (b) The responses are shown below: ๏
(c)
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.14 Consider the circuit shown in Figure P6.14. a. Sketch the amplitude response of Y = I /VS . b. Sketch the amplitude response of V1/VS . c. Sketch the amplitude response of V2/VS .
Note to instructor: the resistance in the figure should be 1 k ๏ and the inductance 100 mH.
Solution: Known quantities: Circuit as shown in Figure P6.14:
Find: Compute and plot the frequency response function.
Analysis: Assume R = 1k๏ and L = 100mH . 1 1 1 = (a) Y = = Z R + j๏ทL 1000 + j (0.1)๏ท
Y vs ๏ท ๏
(b)
V1 VS
๏
(c)
๏
V1 R 1000 = = VS R + j๏ทL 1000 + j(0.1)๏ท vs ๏ท
V2 j๏ทL j(0.1)๏ท = = VS R + j๏ทL 1000 + j(0.1)๏ท
V2 VS
vs ๏ท
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Section 6.2:
Fourier Analysis
Problem 6.15 Use trigonometric identities to show that the equalities in equations 6.16 and 6.17 hold.
Solution: Find: Use trigonometric identities to show that the equalities in equations 6.16 and 6.17 hold.
Analysis: Looking at figure 6.8, we can write the following equations: a n = c n sin(๏ฑ n ) bn = c n cos(๏ฑ n )
and using the trigonometric identities sin 2 (๏ฑ n ) + cos 2 (๏ฑ n ) = 1: ๏
๏
๏ ๏
a n2 + bn2 = c n2 sin 2 (๏ฑ n ) + c n2 cos2 (๏ฑ n ) = c n2 Finally, ๏ bn c n cos(๏ฑ n ) = = cot(๏ฑ n ) = tan(๏นn ) a n c n sin(๏ฑ n ) where, ๏ฐ ๏นn = โ ๏ฑ n . 2
๏
c n = a n2 + bn2
๏
Problem 6.16 Derive a general expression for the Fourier series coefficients of the square wave of Figure 6.15(a) in the text.
Solution: Known quantities: The square wave of Figure 6.11(a) in the text.
Find: A general expression for the Fourier series coefficients.
Assume: None
Analysis:
๏
The square wave is a function of time as follows: ๏ฌ 1 1 ๏ฏ A (n โ 4 )T ๏ฃ t ๏ฃ (n + 4 )T , n = ๏ฑ0,๏ฑ1,๏ฑ2,... x(t) = ๏ญ ๏ฏ0 (n + 1 )T ๏ฃ t ๏ฃ (n + 3 )T , n = ๏ฑ0,๏ฑ1,๏ฑ2,... ๏ฏ๏ฎ 4 4 We can compute the Fourier series coefficient using the integrals in equations (6.20), (6.21) and (6.22): 1 1 T A a 0 = ๏ฒ 0T x(t)dt = ๏ฒ โT4 Adt = T T 2 4
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
an =
๏ฆ 2๏ฐ ๏ถ ๏ฆ 2๏ฐ ๏ถ 2 T 2 T t ๏ทdt = ๏ฒ โT4 A cos๏ง n t ๏ทdt = ๏ฒ x(t) cos๏ง n ๏จ T ๏ธ ๏จ T ๏ธ T 0 T 4 T
=
๏ฆ n๏ฐ ๏ถ๏น 2A ๏ฉ ๏ฆ 2๏ฐ ๏ถ T ๏น 4 A ๏ฉ ๏ฆ n๏ฐ ๏ถ t๏ท = ๏ท๏บ = 0 ๏ชsin๏ง n ๏บ ๏ชsin๏ง ๏ท โ sin๏งโ ๏จ 2 ๏ธ๏ป T ๏ซ ๏จ T ๏ธ 2n๏ฐ ๏ปโT n๏ฐ ๏ซ ๏จ 2 ๏ธ
(๏ขn)
4
bn = ๏
๏ฆ 2๏ฐ ๏ถ ๏ฆ 2๏ฐ ๏ถ 2 T 2 T t ๏ทdt = ๏ฒ โT4 A sin๏ง n t ๏ทdt = ๏ฒ 0 x(t) sin๏ง n ๏จ T ๏ธ ๏จ T ๏ธ T T 4 T
=
๏ฆ 2๏ฐ ๏ถ T ๏น 4 ๏ฆ n๏ฐ ๏ถ ๏ฆ n๏ฐ ๏ถ๏น 2A ๏ฉ A ๏ฉ t๏ท = ๏ท๏บ = ๏ชโ cos๏ง n ๏บ ๏ชโ cos๏ง ๏ท + cos๏งโ ๏จ T ๏ธ 2n๏ฐ ๏ปโT ๏จ 2 ๏ธ ๏จ 2 ๏ธ๏ป T ๏ซ n๏ฐ ๏ซ 4
๏ฌ 2A ๏ฆ n๏ฐ ๏ถ๏น ๏ฏ๏ฏ n๏ฐ A ๏ฉ = ๏ชโ2 cos๏ง ๏ท๏บ = ๏ญ ๏จ 2 ๏ธ๏ป ๏ฏ n๏ฐ ๏ซ ๏ฏ๏ฎ 0
๏
(n even) (n odd)
Problem 6.17 Compute the Fourier series coefficient of the periodic function shown in Figure P6.17 and defined as: 0 ๐ฅ(๐ก) = { ๐ด
0โค๐กโค
๐ 3
๐ โค๐กโค๐ 3
Known quantities: None.
Find: The Fourier series coefficient of the periodic function, ๐ฅ(๐ก).
Analysis: The function is not even or odd, so all of the ๐๐ and ๐๐ coefficients are needed. To accomplish this, use equations 6.34-6.36: 6.39 PROPRIETARY MATERIAL. ยฉ The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
๐0 =
=
๐๐ =
1 ๐ โซ ๐ด ๐๐ก ๐ ๐/3 ๐๐จ๐ป ๐
2 ๐ 2๐ โซ ๐ด cos (๐ ๐ก) ๐๐ก ๐ ๐ ๐ 3
=
๐ 2 ๐ด๐ 2๐ sin (๐ ๐ก) |๐ ๐ 2๐๐ ๐ 3
=
๐ด 2๐ (sin(๐ 2๐๐ก) โ sin (๐ ๐ก)) ๐๐ 3
=โ
๐๐ =
๐ = 1, 2, 3, โฆ
๐จ ๐๐ ๐ฌ๐ข๐ง (๐ ๐) ๐ = ๐, ๐, ๐, โฆ ๐ ๐ ๐
2 ๐ 2๐ โซ ๐ด sin (๐ ๐ก) ๐๐ก ๐ ๐ ๐ 3
=โ
2 ๐ด๐ 2๐ ๐ cos (๐ ๐ก)|๐ ๐ 2๐๐ ๐ 3
= โ =
๐ด 2๐ (cos(๐ 2๐ ๐ก) โ cos (๐ ๐ก)) ๐ = 1, 2, 3, โฆ ๐๐ 3
๐จ ๐๐ ๐๐จ๐ฌ (๐ ๐) ๐ = ๐, ๐, ๐, โฆ ๐ ๐ ๐
Problem 6.18
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Solution: Known quantities: The periodic function shown in Figure P6.18 and defined as: ๏ฌ ๏ฆ 2๏ฐ ๏ถ T T ๏ฏ cos t โ ๏ฃt๏ฃ x(t) = ๏ญ ๏ง๏จ T ๏ท๏ธ 4 4 ๏ฏ๏ฎ 0 else
Find: A general expression for the Fourier series coefficients.
๏
Analysis: The function in Figure P6.18 is an even function. Thus, we only need to compute the
an coefficients.
We can compute the Fourier series coefficient using the integrals in equations (6.20) and (6.21): T
๏ฆ 2๏ฐ ๏ถ 1 T 1 T 1 ๏ฉ ๏ฆ 2๏ฐ ๏ถ๏น 4 a 0 = ๏ฒ T2 x(t)dt = ๏ฒ T4 cos๏ง t ๏ทdt = t ๏ท๏บ = ๏ชsin๏ง ๏จT ๏ธ T โ 2 T โ 4 2๏ฐ ๏ซ ๏จ T ๏ธ๏ปโT 4
๏ฆ ๏ฐ ๏ถ๏น 1 1 ๏ฉ ๏ฆ๏ฐ ๏ถ = ๏ชsin๏ง ๏ท โ sin๏งโ ๏ท๏บ = ๏จ 2 ๏ธ๏ป ๏ฐ 2๏ฐ ๏ซ ๏จ 2 ๏ธ ๏ฆ 2๏ฐ ๏ถ ๏ฆ 2๏ฐ ๏ถ ๏ฆ 2๏ฐ ๏ถ 2 T 2 T t ๏ทdt = ๏ฒ T4 cos๏ง t ๏ท cos๏ง n t ๏ทdt = ๏ฒ T2 x(t) cos๏ง n โ โ ๏จ T ๏ธ ๏จT ๏ธ ๏จ T ๏ธ T T 2 4 ๏ฆ n๏ฐ ๏ถ ๏ฌ n 2 cos๏ง ๏ท ๏ฏ(-1) 2 โ1 (n even) 2 ๏จ 2 ๏ธ 2 =โ = ๏ญ ๏ฐ n โ1 ๏ฐ n 2 โ1 ๏ฏ (n odd) ๏ฎ 0
an =
๏
(
)
๏
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
Problem 6.19 Compute the Fourier series expansion of the function shown in Figure P6.19, and express it in sine-cosine (an , bncoefficients) form.
๏ฌ 2A ๏ฏ T t x(t ) = ๏ญ 2A ๏ฏโ (t - T) ๏ฎ T
0๏ฃt๏ฃ
T 2
T ๏ฃt ๏ฃT 2
Solution: Known quantities: The periodic function shown in Figure P6.19 and defined as:
๏ฌ 2A ๏ฏ T t x(t ) = ๏ญ 2A ๏ฏโ (t - T) ๏ฎ T
0๏ฃt๏ฃ
T 2
T ๏ฃt ๏ฃT 2
Find: Compute the Fourier series expansion.
Analysis: We can compute the Fourier series coefficient using the integrals in equations (6.20), (6.21) and (6.22):
a0 =
T ๏ฆ ๏ถ 1 1 T 1 ๏ฆ T 2 2A 2A ๏ถ x ( t ) dt = ๏ง t dt + ๏งโ ๏ท ๏ (t โ T )dt ๏ท๏ท = A ๏ง ๏ฒ ๏ฒ ๏ฒ T 0 0 T T๏จ T T ๏ธ 2๏จ ๏ธ 2
an =
T ๏ฆ โ 2A ๏ถ 2 T 2 ๏ฆ T 2 2A ๏ฆ 2๏ฐ ๏ถ ๏ฆ 2๏ฐ ๏ถ ๏ฆ 2๏ฐ ๏ถ ๏ถ x ( t ) cos n t dt = ๏ง๏ง ๏ฒ t cos๏ง n t ๏ท dt + ๏ฒT ๏ง ๏ (t โ T ) cos๏ง n t ๏ทdt ๏ท๏ท = ๏ง ๏ท ๏ท ๏ฒ T 0 T๏จ 0 T 2๏จ T ๏จ T ๏ธ ๏จ T ๏ธ ๏ธ ๏จ T ๏ธ ๏ธ
A 4A A 2A ๏sin(n๏ฐ ) โ sin(2n๏ฐ )๏ = sin(n๏ฐ ) โ 2 2 cos(2n๏ฐ ) + 2 2 cos(n๏ฐ ) + n๏ฐ n๏ฐ n๏ฐ n๏ฐ 4A A = โ 2 2 cos(2n๏ฐ ) + 2 2 cos(n๏ฐ ) n๏ฐ n๏ฐ =
bn =
T ๏ฆ 2 T 2 ๏ฆ T 2 2A 2A ๏ถ ๏ฆ 2๏ฐ ๏ถ ๏ฆ 2๏ฐ ๏ถ ๏ฆ 2๏ฐ ๏ถ ๏ถ x ( t ) sin n t dt = ๏ง๏ง ๏ฒ t sin๏ง n t ๏ท dt + ๏ฒT ๏ง โ ๏ (t โ T )sin๏ง n t ๏ทdt ๏ท๏ท = ๏ง ๏ท ๏ท ๏ฒ T 0 T๏จ 0 T T ๏ธ 2๏จ ๏จ T ๏ธ ๏จ T ๏ธ ๏จ T ๏ธ ๏ธ
A A 2A ๏cos(n๏ฐ ) โ cos(2n๏ฐ )๏ + cos(n๏ฐ ) + 2 2 sin(n๏ฐ ) + n๏ฐ n๏ฐ n๏ฐ 2A A A A A โ cos(2n๏ฐ ) + cos(n๏ฐ ) + 2 2 cos(n๏ฐ ) + 2 2 sin(2n๏ฐ ) โ 2 2 sin(n๏ฐ ) n๏ฐ n๏ฐ n๏ฐ n๏ฐ n๏ฐ
=โ
bn = โ
A 2A ๏cos(n๏ฐ ) โ cos(2n๏ฐ )๏ โ 2 A cos(2n๏ฐ ) + A cos(n๏ฐ ) + 2A 2 cos(n๏ฐ ) cos(n๏ฐ ) + n๏ฐ n๏ฐ n๏ฐ n๏ฐ n๏ฐ
Thus, the Fourier series expansion of the function is: 6.42 PROPRIETARY MATERIAL. ยฉ The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6
A ๏ฅ A ๏ฆ 2๏ฐ ๏ถ ๏ฅ A ๏ฆ 2๏ฐ ๏ถ x(t ) = + ๏ฅ a n cos๏ง n t๏ท + ๏ฅ bn sin๏ง n t๏ท 2 2 n=1 (n๏ฐ ) ๏จ T ๏ธ n=1 n๏ฐ ๏จ T ๏ธ
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