Principles and applications of electrical engineering 6th edition rizzoni solutions manual by alexandra.mitchell765

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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 – 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., MATLAB® or Electronics Workbench® ) 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.

6.1 PROPRIETARY MATERIAL. © 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 6.1 Determine the frequency response 𝑉𝑜𝑢𝑡 (𝑗𝜔)/𝑉𝑖𝑛 (𝑗𝑤) for the circuit of Figure P6.. Assume: 𝐿 = 0.5 𝐻 and 𝑅 = 200 𝑘Ω.

Known quantities: 𝐿 = 0.5 𝐻, 𝑅 = 200 𝑘Ω

Find: (a) Determine the frequency response 𝑉𝑜𝑢𝑡 (𝑗𝜔)/𝑉𝑖𝑛 (𝑗𝜔) 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 𝑽𝒐𝒖𝒕 (𝒋𝝎)/𝑽𝒊𝒏 (𝒋𝝎) for the circuit of Figure P6.1. 𝑉𝑜𝑢𝑡 (𝑗𝜔)/𝑉𝑖𝑛 (𝑗𝜔) can be determine by the voltage division: 𝑉𝑜𝑢𝑡 = 𝑉𝑖𝑛

𝐿 𝐿+𝑅

Convert to the frequency domain: 𝑉𝑜𝑢𝑡 (𝑗𝜔) 𝑉𝑖𝑛 (𝑗𝜔)

𝑗𝜔𝐿 𝑗𝜔𝐿 + 𝑅

Substitute known values: 𝒋𝝎𝟎. 𝟓 𝟐𝟎𝟎 𝒌 + 𝒋𝝎𝟎. 𝟓

6.2 PROPRIETARY MATERIAL. © 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 (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. Convert the result from (a) into a manageable format using the complex conjugate of the denominator:

𝑉𝑜𝑢𝑡 (𝑗𝜔) 𝑉𝑖𝑛 (𝑗𝜔)

(0.5𝜔)2 + 𝑗𝜔100 𝑘 4𝑒10 + (0.5𝜔)2

Magnitude: 𝑉𝑜𝑢𝑡 (𝑗𝜔)

0.5𝜔2

𝑉𝑖𝑛 (𝑗𝜔)

4𝑒10 + (0.5𝜔)2

| = √(

) +(

100 𝑘 𝜔 4𝑒10 + (0.5𝜔)2

)

Phase: ∠

𝑉𝑜𝑢𝑡 (𝑗𝜔) 100 𝑘 𝜔 = atan ( ) (0.5𝜔)2 𝑉𝑖𝑛 (𝑗𝜔) 100𝑘 = atan ( ) 0.25𝜔

Plots: Magnitude 1.4

1.2

Magnitude

0.8

0.6

0.4

0.2

4 5 6 Frequency (rad/s)

10 6

x 10

6.3 PROPRIETARY MATERIAL. © 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) 90 80 70

Phase

60 50 40 30 20 10 0

4 5 6 Frequency (rad/s)

10 6

x 10

1.2

Magnitude

0.8

0.6

0.4

0.2

0 1 10

10 10 Frequency (rad/s)

6.4 PROPRIETARY MATERIAL. © 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) 90 80 70

Phase

60 50 40 30 20 10 0 1 10

10 10 Frequency (rad/s)

(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

10 10 Frequency (rad/s)

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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 𝑉𝑜𝑢𝑡 (𝑗𝜔)/𝑉𝑖𝑛 (𝑗𝜔) 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 𝑽𝒐𝒖𝒕 (𝒋𝝎)/𝑽𝒊𝒏 (𝒋𝝎) for the circuit of Figure P6.2. Determine 𝑉𝑜𝑢𝑡 (𝑗𝑤) and 𝑉𝑖𝑛 (𝑗𝑤) by using voltage division: 𝑅𝐶 𝑅 +𝐶 𝑉𝑜𝑢𝑡 = 𝑉𝑖𝑛 𝑅𝐶 𝑅+ 𝑅+𝐶 Convert to frequency domain: 1 𝑗𝜔𝐶 1 𝑅+ 𝑗𝜔𝐶 𝑉𝑜𝑢𝑡 (𝑗𝜔) = 𝑉𝑖𝑛 (𝑗𝜔) 1 𝑅 𝑗𝜔𝐶 𝑅+ 1 𝑅+ 𝑗𝜔𝐶 𝑅

Simplify and solve for 𝑉𝑜𝑢𝑡 (𝑗𝜔)/𝑉𝑖𝑛 (𝑗𝜔): 6.6 PROPRIETARY MATERIAL. © 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 𝑉𝑜𝑢𝑡 (𝑗𝜔) 1 = 𝑉𝑖𝑛 (𝑗𝜔) 2 + 𝑗𝜔𝑅𝐶 Substitute known values: 𝑽𝒐𝒖𝒕 (𝒋𝝎) 𝟏 = 𝑽𝒊𝒏 (𝒋𝝎) 𝟐 + 𝒋𝝎𝟎. 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. Determine the equations for magnitude in phase. First, simplify 𝑉𝑜𝑢𝑡 (𝑗𝜔)/𝑉𝑖𝑛 (𝑗𝜔) by multiplying by the complex conjugate of the denominator: 𝑉𝑜𝑢𝑡 (𝑗𝜔) 2 − 0.1𝑗𝜔 = 𝑉𝑖𝑛 (𝑗𝜔) 4 + 0.01𝜔 2 Magnitude:

2 2 𝑉𝑜𝑢𝑡 (𝑗𝜔) 2 0.1𝜔 | = √( ) + ( ) 𝑉𝑖𝑛 (𝑗𝜔) 4 + 0.01𝜔 2 4 + 0.01𝜔 2

Phase: ∠

𝑉𝑜𝑢𝑡 (𝑗𝜔) −0.1𝜔 = arctan( ) 𝑉𝑖𝑛 (𝑗𝜔) 2

Plots: Magnitude 0.45 0.4 0.35

Magnitude

0.3 0.25 0.2 0.15 0.1 0.05 0

4 5 6 Frequency (rad/s)

10 6

x 10

6.7 PROPRIETARY MATERIAL. © 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) -20 -30 -40

Phase

-50 -60 -70 -80 -90 -100

4 5 6 Frequency (rad/s)

10 6

x 10

Magnitude

0.3 0.25 0.2 0.15 0.1 0.05 0 1 10

10 10 Frequency (rad/s)

6.8 PROPRIETARY MATERIAL. © 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) -20 -30 -40

Phase

-50 -60 -70 -80 -90 -100 1 10

10 10 Frequency (rad/s)

(d) Plot the magnitude response on semilog paper with magnitude in decibels. Magnitude 0

-20

Magnitude

-40

-60

-80

-100

-120 1 10

10 10 Frequency (rad/s)

6.9 PROPRIETARY MATERIAL. © 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 6.3 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 𝑉𝑜𝑢𝑡 (𝑗𝜔)/𝑉𝑖𝑛 (𝑗𝜔) 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 𝑽𝒐𝒖𝒕 (𝒋𝝎)/𝑽𝒊𝒏 (𝒋𝝎) for the circuit of Figure P6.3. To determine the frequency response, place an imaginary voltage source at 𝑉𝑖𝑛 and determine 𝑉𝑜𝑢𝑡 . For this problem, 𝑉𝑜𝑢𝑡 may be determined from the voltage division of the top node: 𝑉𝑜𝑢𝑡 = 𝑉1

C 1000 Ω + 𝐶

where 𝑉1 is: 𝑉1 = 𝑉𝑖𝑛

𝑍𝑒𝑞 2000 Ω + 𝑍𝑒𝑞

and 𝑍𝑒𝑞 is: 𝑍𝑒𝑞 =

2000 Ω ∗ (1000 Ω ∗ 𝐶) 2000 Ω + (1000 Ω + 𝐶) 6.10

PROPRIETARY MATERIAL. © 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

2𝑒6 Ω ∗ 𝐶 3000 Ω + 𝐶

Put 𝑍𝑒𝑞 into the frequency domain: 1 𝑗𝜔𝐶 𝑍𝑒𝑞 (𝑗𝜔) = 1 3000 Ω + 𝑗𝜔𝐶 2𝑒6 Ω ∗

Simplify: 1 𝑗𝜔𝐶 𝑗𝜔𝐶 𝑍𝑒𝑞 (𝑗𝜔) = ∗ 1 𝑗𝜔𝐶 3000 + 𝑗𝜔𝐶 2𝑒6 ∗

2𝑒6 ∗ =

𝑗𝜔𝐶 𝑗𝜔𝐶

3000 Ω ∗ jωC + =

𝑗𝜔𝐶 𝑗𝜔𝐶

2𝑒6 1 + 3000 𝑗𝜔𝐶

Substitute 𝑍𝑒𝑞 (𝑗𝜔) into 𝑉1 (𝑗𝜔): 2𝑒6 1 + 3000 𝑗𝜔𝐶 𝑉1 (𝑗𝜔) = 𝑉𝑖𝑛 (𝑗𝜔) 2𝑒6 2000 + 1 + 3000 𝑗𝜔𝐶 Simplify: 2𝑒6 1 + 3000 𝑗𝜔𝐶 1 + 3000 𝑗𝜔𝐶 𝑉1 (𝑗𝜔) = 𝑉𝑖𝑛 (𝑗𝜔) ∗ 2𝑒6 1 + 3000 𝑗𝜔𝐶 2000 + 1 + 3000 𝑗𝜔𝐶 1 + 3000 𝑗𝜔𝐶 1 + 3000 𝑗𝜔𝐶 = 𝑉𝑖𝑛 (𝑗𝜔) 1 + 3000 𝑗𝜔𝐶 2000 ∗ (1 + 3000 𝑗𝜔𝐶) + 2𝑒6 ∗ 1 + 3000 𝑗𝜔𝐶 2𝑒6 ∗

1 + 3000 𝑗𝜔𝐶 1 + 3000 𝑗𝜔𝐶 = 𝑉𝑖𝑛 (𝑗𝜔) 1 + 3000 𝑗𝜔𝐶 2000 ∗ (1 + 3000 𝑗𝜔𝐶) + 2𝑒6 ∗ 1 + 3000 𝑗𝜔𝐶 2𝑒6 ∗

= 𝑉𝑖𝑛 (𝑗𝜔)

2𝑒6 2000 + 𝑗𝜔6𝑒6 ∗ 𝐶 + 2𝑒6

≈ 𝑉𝑖𝑛 (𝑗𝜔)

2𝑒6 2𝑒6 + 𝑗𝜔60 6.11

G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6

Substitute 𝑉1 (𝑗𝜔) into 𝑉𝑜𝑢𝑡 (𝑗𝜔):

𝑉𝑜𝑢𝑡 (𝑗𝜔) = 𝑉1 (𝑗𝜔)

1 𝑗𝜔𝐶 1000 +

1 𝑗𝜔𝐶

Simplify:

𝑉𝑜𝑢𝑡 (𝑗𝜔) = 𝑉1 (𝑗𝜔)

= 𝑉1 (𝑗𝜔)

1 𝑗𝜔𝐶 1 1000 + 𝑗𝜔𝐶

∗

𝑗𝜔𝐶 𝑗𝜔𝐶 1000 ∗ 𝑗𝜔𝐶 +

= 𝑉1 (𝑗𝜔)

𝑗𝜔𝐶 𝑗𝜔𝐶

1 1 + 0.01𝑗𝜔

Plug 𝑉1 (𝑗𝜔) into this equation: 𝑉𝑜𝑢𝑡 (𝑗𝜔) = 𝑉𝑖𝑛 (𝑗𝜔)

2𝑒6 1 2𝑒6 + 𝑗𝜔60 1 + 0.01𝑗𝜔

Simplify and solve for 𝑉𝑜𝑢𝑡 (𝑗𝜔)/𝑉𝑖𝑛 (𝑗𝜔): 𝑉𝑜𝑢𝑡 (𝑗𝜔) 2𝑒6 ≈ 𝑉𝑖𝑛 (𝑗𝜔) 2𝑒6 + 2𝑒4𝑗𝜔 Simplify: 𝑽𝒐𝒖𝒕 (𝒋𝝎) 𝟏 ≈ 𝑽𝒊𝒏 (𝒋𝝎) 𝟏 + 𝟎. 𝟎𝟏𝒋𝝎 (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. First, get 𝑉𝑜𝑢𝑡 (𝑗𝜔)/𝑉𝑖𝑛 (𝑗𝜔) into a manageable form: 𝑉𝑜𝑢𝑡 (𝑗𝜔) 1 1 − 0.01𝑗𝜔 = ∗ 𝑉𝑖𝑛 (𝑗𝜔) 1 + 0.01𝑗𝜔 1 − 0.01𝑗𝜔 ≈

1 − 0.01 𝑗𝜔 1 + 0.0001𝜔 2

Calculate magnitude of complex expression: 6.12 PROPRIETARY MATERIAL. © 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 𝑉𝑜𝑢𝑡 (𝑗𝜔) 1 0.01𝜔 | = √( ) +( ) 2 2 𝑉𝑖𝑛 (𝑗𝜔) 1 + 0.0001𝜔 1 + 0.0001𝜔

Phase response: −0.01𝜔 𝑉𝑜𝑢𝑡 (𝑗𝜔) 1 − 0.0001𝜔 2 ) ∠ = arctan ( 1 𝑉𝑖𝑛 (𝑗𝜔) 1 − 0.0001𝜔 2 = arctan(−0.01𝜔) Plots: Magnitude 1 0.9 0.8

Magnitude

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

4 5 6 Frequency (rad/s)

10 6

x 10

6.13 PROPRIETARY MATERIAL. © 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

4 5 6 Frequency (rad/s)

10 6

x 10

Magnitude

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 10

10 10 Frequency (rad/s)

6.14 PROPRIETARY MATERIAL. © 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

10 10 Frequency (rad/s)

(d) Plot the magnitude response on semilog paper with magnitude in decibels. Magnitude 0

-20

Magnitude

-40

-60

-80

-100

-120 1 10

10 10 Frequency (rad/s)

6.15 PROPRIETARY MATERIAL. © 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 μF, C2 = 160 μF.

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 − 0.0008 2 + j (0.05) j (C1 + C2 ) ( j ) = = 1 Vin 1 − 0.0008 2 + j (0.11) R1 + R2 || R3 + jL + j (C1 + C2 ) R2 || R3 + jL +

a) Vout Vin

( j ) =

(1 −  2 0.0008) 2 + ((0.05) ) 2 (1 −  2 0.0008) 2 + ((0.11) ) 2

V  (0.05)   (0.11)   out ( j ) = arctan  − arctan  2 2 Vin  1 −  0.0008   1 −  0.0008 

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

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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. 𝑅1 = 20 𝑘Ω, 𝑅2 = 100 𝑘Ω, 𝐿 = 1 𝐻, and 𝐶 = 100 𝜇𝐹.

Assume:

Known quantities: 𝑅1 = 20 𝑘Ω, 𝑅2 = 100 𝑘Ω, 𝐿 = 1 𝐻, 𝐶 = 100 𝜇𝐹

Find: The frequency response for the circuit and generate frequency response plots.

Analysis: The output voltage, 𝑣𝑜𝑢𝑡 , is the voltage across the capacitor, 𝑣𝑐 . The capacitor’s voltage may be calculated as a voltage division of the top node voltage, 𝑣1 : 𝑣𝑜𝑢𝑡 = 𝑣1

𝑧𝐶 𝑅2 + 𝑧𝐶

To determine 𝑣𝑜𝑢𝑡 , find 𝑣1 . First, calculate 𝑣1 as a voltage division of 𝑣𝑖𝑛 : 𝑣1 = 𝑣𝑖𝑛

𝑅1 ∥ (𝑅2 + 𝑧𝐶 ) 𝐿 + 𝑅1 ∥ (𝑅2 + 𝑧𝐶 )

Convert to frequency domain: 1 ) 𝑗𝜔𝐶 𝑣1 (𝑗𝜔) = 𝑣𝑖𝑛 (𝑗𝜔) 1 𝑗𝜔𝐿 + 𝑅1 ∥ (𝑅2 + ) 𝑗𝜔𝐶 𝑅1 ∥ (𝑅2 +

𝑅1 (𝑅2 +

= 𝑣𝑖𝑛 (𝑗𝜔)

𝑗𝜔𝐿 (𝑅1 + 𝑅2 + = 𝑣𝑖𝑛 (𝑗𝜔) =

1 ) 𝑗𝜔𝐶

1 1 ) + 𝑅1 (𝑅2 + ) 𝑗𝜔𝐶 𝑗𝜔𝐶

𝑗𝜔𝐶𝑅1 𝑅2 + 𝑅1 𝑅1 − 𝜔 2 𝐿𝐶(𝑅1 + 𝑅2 ) + 𝑗𝜔(𝐿 + 𝐶𝑅1 𝑅2 )

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6 Convert 𝑣𝑜𝑢𝑡 to the frequency domain: 1 𝑗𝜔𝐶

𝑣𝑜𝑢𝑡 (𝑗𝜔) = 𝑣1 (𝑗𝜔)

𝑅2 +

1 𝑗𝜔𝐶

Plug in 𝑣1 (𝑗𝜔): 1 𝑣𝑜𝑢𝑡 (𝑗𝜔) 𝑗𝜔𝐶𝑅1 𝑅2 + 𝑅1 𝑗𝜔𝐶 = 𝑣𝑖𝑛 (𝑗𝜔) 𝑅1 − 𝜔 2 𝐿𝐶(𝑅1 + 𝑅2 ) + 𝑗𝜔(𝐿 + 𝐶𝑅1 𝑅2 ) 𝑅 + 1 2 𝑗𝜔𝐶 =

𝑗𝜔𝐶𝑅1 𝑅2 + 𝑅1 2 [𝑅1 − 𝜔 𝐿𝐶(𝑅1 + 𝑅2 ) + 𝑗𝜔(𝐿 + 𝐶𝑅1 𝑅2 )](1 + 𝑗𝜔𝑅2 𝐶) 𝑅1 + 𝑗𝜔𝐶𝑅1 𝑅2 𝐶𝑅 𝑅2 𝑅1 − 𝜔 2 𝐿𝐶 (𝑅1 + 2𝑅2 + 1 2 ) + 𝑗𝜔[𝐿 + 2𝐶𝑅1 𝑅2 − 𝜔 2 𝑅2 𝐿𝐶 2 (𝑅1 + 𝑅2 )] 𝐿

Substitute known values: 𝒗𝒐𝒖𝒕 (𝒋𝝎) 𝟐𝑬𝟒 + 𝒋𝝎𝟐𝑬𝟓 = 𝒗𝒊𝒏 (𝒋𝝎) 𝟐𝑬𝟒 − 𝟖𝑬𝟔 𝝎𝟐 + 𝒋𝝎(𝟒𝑬𝟓 − 𝟏𝟐𝟎𝝎𝟐 ) To generate plots, determine the magnitude and the phase. First, convert 𝑣𝑜𝑢𝑡 (𝑗𝜔)/𝑣𝑖𝑛 (𝑗𝜔) to a manageable format using the complex conjugate: 𝑣𝑜𝑢𝑡 (𝑗𝜔) 4𝐸 8 − 𝜔2 (8𝐸10 + 2.4𝐸 7 𝜔2 ) − 𝑗𝜔(4𝐸 9 + 1.6𝐸12 𝜔2 ) = (2𝐸 4 + 8𝐸 6 𝜔 2 )2 + 𝜔 2 (4𝐸 5 − 120𝜔 2 )2 𝑣𝑖𝑛 (𝑗𝜔) Magnitude: 2

𝑣𝑜𝑢𝑡 (𝑗𝜔) 4𝐸 8 − 𝜔 2 (8𝐸10 + 2.4𝐸 7 𝜔 2 ) 𝜔(4𝐸 9 + 1.6𝐸12 𝜔 2 ) | | = √( ) +( ) 4 6 2 2 2 5 2 2 4 (2𝐸 + 8𝐸 𝜔 ) + 𝜔 (4𝐸 − 120𝜔 ) (2𝐸 + 8𝐸 6 𝜔 2 )2 + 𝜔 2 (4𝐸 5 − 120𝜔 2 )2 𝑣𝑖𝑛 (𝑗𝜔)

Phase: ∠

𝑣𝑜𝑢𝑡 (𝑗𝜔) −𝜔(4𝐸 9 + 1.6𝐸12 𝜔2 ) = arctan ( 8 ) 𝑣𝑖𝑛 (𝑗𝜔) 4𝐸 − 𝜔 2 (8𝐸10 + 2.4𝐸 7 𝜔 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

1.5

0.5

0 1 10

10 10 Frequency (rad/s)

Phase (degrees) 90 80 70

Phase

60 50 40 30 20 10 0 1 10

10 10 Frequency (rad/s)

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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 μF and R = 2 kohm, a. Determine how the input impedance Z( jω) = Vi ( jω)/Ii ( jω) 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( jω) = R[1+j(1/ ω RC)] d. Determine the frequency ω = ωC 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(jω) at ω = 10 rad/s and ω = 10^5 rad/s.

Solution: Known quantities: Figure P6.6.

Find:

V i ( j ) , behaves at extremely high or low frequencies. I i ( j ) b) An expression for the input (or driving point) impedance.  1  c) Show that this expression can be manipulated into the form: Z ( j) = R1+ j  RC  d) Determine the frequency  = c for which the imaginary part of the expression in c) is equal to 1. e) Estimate the magnitude and angle of Z [j] at  = 10 rad/s and 100,000 rad/s. a)

How the input impedance, Z ( j )=

Analysis: a)

As  → ,



 Z C → 0  Short Z → R

As  → 0, Z C →   Open  Z →  b) and c) KVL : - V i + I i Z C + I i Z R = 0  1 1   Z ( j ) = V i = Z + Z = R1− j  C R = R+  j  C  C Ii 1 1 1 rad d) = 1  c = = = 1000 −6 RC RC s c [ 2000 ] [ 0.510 ]  rad  Z 10   200 k  − 90  s  e)  rad  Z 100 k   2 k 0  s  more precisely,

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6

   rad  1 Z 10  = 20.001 k  − 89.43  = R 1− j −6  s  [ 10 ] [ 2000 ] [ 0.510 ]      rad  1 Z 10 5  =2 k  − 0.06  = R 1− j  s  [ 10 5 ] [ 2000 ] [ 0.510−6 ]  



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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( jω) = Vi ( jω)/Ii ( jω) 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( jω) = R[1+j(1/ ω RC)] d. Determine the frequency ω = ωC 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(jω) at ω = 10^5 rad/s, ω = 10^6 rad/s ω = 10^7 rad/s

Solution: Known quantities: Figure P6.7.

Find: a)

How the input impedance,

Z ( j) =

V i ( j) , I i ( j)

behaves at extremely high or low frequencies. b) An expression for the input (or driving point) impedance.

L   Show that this expression  can be manipulated into the form: Z ( j )= R 1 + j R   d) Determine the frequency  =  C for which the imaginary part of the expression in c) is equal to 1. c)

Estimate the magnitude and angle of Z [j] at  = 105, 106, 107 rad/s.

Analysis: a)

As  → ,

Z L →   Open Z → 

As  → 0,

Z L → 0  Short  Z → R

b) KVL : − V i + I i Z R + I i Z L = 0



Z ( j ) = V i = Z L + Z R = jL + R Ii

 L  c) Z ( j) = R + j L = R  1+ j   R  R 2000 rad  L = 1000 k d) c = 1   c = = −3 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

 rad  2  10 −3  10 5   Z 100k = 2000(1 + j 0.1) = 2.01 k5.71  = R 1 + j   s  2000     rad  2  10 -3  10 6   Z 1M = 2000(1 + j 1) = 2.82 k 45.00  = R1 + j   s  2000     rad  2  10 −3  10 7   Z 10M = 2000(1 + j 10 ) = 20.10 k 84.29  = R 1 + j   s  2000    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) , behaves at extremely high or low frequencies. I i ( j)  b) An expression for the input impedance in the form:  1+ j f ()  L 1 Z ( j) = Z o   Z o = R1 + f 2 () R2C  1+ j  a)

How the input impedance, Z ( j) =

f 1 () =



 2 R1 LC − R1 − R 2 (R1 R 2 C + L )

f 2 () =

 2 LC −1 C R 2

c) Determine the four cutoff frequencies at which f1[] = +1 or -1 d) Determine the resonant frequency of the circuit. e) Plot the impedance vs. frequency.

Analysis:

As  → ,

As  → 0,

and f2[] = +1 or -1.

Z L →   Open, Z C → 0  Short Z → R1 Z C →   Open, Z L → 0  Short  Z → R1 + R2

 6.26 PROPRIETARY MATERIAL. © 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 b)



1 ] [ R 2 + jL ] jC 1 + [ R 2 + jL ] jC

[

Z[j ] =

V[j ] [ + ZL ] = Z R1 + Z C Z R 2 = R1 + I[j ] ZC + [ Z R2 + Z L ]

= R1 +

( R 1 [ 1 -  2 LC ] + R 2 ) + j (  R 1 R 2 C + L ) (− j ) R 2 + j L =   1 -  2 LC + j  R 2 C [ 1 -  2 LC ] + j  R 2 C (− j )

 Z  j =

( R C + j ( LC −1)

(R1R2C + L) + j  2 R1LC − R1 − R2 2

)= R R C + L  1 2

R2C

1+ j

jC = jC

 2 R1LC − R1 − R2 (R1R2C + L) 1+ j

 2 LC −1 R2C

c) Both f1[] and f2[] can be positive or negative, and therefore equal to plus or minus one depending on the frequency; therefore, both cases must be considered.  c [ R1 R 2 C + L ] =  1 f 1 [ c] = R 1 [ 1-  2c LC ] + R 2 1 + R2 R ]  c - R1 = 0  2c  [ 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

c

1 1 = − [  13.69 10 3 ]  ( [  13.69 10 3 ]2 - 4[1][ - 141.510 6 ] )1/2 2 2

rad rad  c4 = 20.569 k s s only the positive answers are physically valid, i.e., a negative frequency is physically impossible. 1  2c LC - 1 =  1   2c  [ R 2 ]  c = 0 f 2 [ c] = L LC c 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 [  5790 ]  ( [  5790 ] + 4[1][ 95.6910 ] ) =  2895  10201 c = 2 2 rad rad   c2 = 7.31 k  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: =  6.84510 3  13.724 10 3



  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 ] behaves at extremes of high and H v [j ] = V i [j ] 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[ ] = R1 R 2 H v [j ] = 1 + j f[ ] R1 + R 2 R1 + R 2 c) The "cutoff" frequency at which f[] = 1 and the value of Ho in dB. a)



Analysis: a)

As  →  : VD :

0 Z C → 0  − 90  Short 0 H v → 0  − 90

As  → 0 :



0 Z C →   − 90  Open R2 VD : H v →  00 R1 + R 2 1 [ ] [ R2 ] jC Z C Z R2 = = Z eq 1 Z C + Z R2 + R2 jC

jC jC

R2 1 + j  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 R 2 C 1 + j R 2 C V o [j ] = = = H v [j ] = R 1 + j R 2 C 2 V i [j ] Z R1 + Z eq + R1 1 + j  R2C 1 R2 R2 = =  + + j  C + R1 R2 R1 R 2 R1 R 2 1 + j R1 R 2 C R1 + R 2



f[ c ] =

 c R1 R 2 C = 1

Ho =

1900 R2 = 1300 + 1900 R1 + R 2

R1 + R 2

c =

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 = 7.07∠π 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  → 0

Z C →   Open Z L → 0  Short V o → 0

As  → 

Z C → 0  Short

VD :

Vo=

→   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  − 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

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

+ j 25.2510

134.34 V 1350 29.17 59.94

= 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( jω) = Vo( jω)/Vi ( jω)= Hvo/(1±jf(  ))

Solution: Known quantities: Figure P6.11.

Find: a)

The voltage transfer function in the form:

H v [j ] =

V o [j ] H vo = . [j  ] 1  j f[  ] 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 F 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 ] =

Z R2 = Vi Z R1 + Z C + Z R2 R1 +

V o [j ] V i [j ]

R2 R1 + R 2 1 - j

R2 1 + R2 jC

1 1

C [ R1 + R 2 ]



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( jω)/Vin( jω).

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C 1 R2 + jC R2

jC R2 = jC 1 + j  R2C

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5 th Edition Problem solutions, Chapter 6

VD :

H v [j ] =

V o [j ] V i [j ]

Z RL Z R1 + Z eq + Z RL

R L [ 1 + j R 2 C ] R1 + R 2 + R L + j [ R1 + R L ]  R 2 C

RL R 2 + RL R1 + 1 + jR 2 C

1 + jR 2 C = 1 + jR 2 C

1 + j  R2C RL R1 + R 2 + R L 1 + j [ R1 + R L ]  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( jω)/Vin( jω) 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 τ

C = 100“F

Find: Compute and plot the frequency response function.

Analysis: (a)

Vout = Vin

V ( j ) 1/jC 1 1 = , out = R + 1/jC jRC +1 Vin ( j) 1 + j/10

Vout 1 ,() = -arctan(0.1) = Vin 1 + 0.01 2     (b) The responses are shown below: 

(c)

(d) 6.35 PROPRIETARY MATERIAL. © 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 6.14 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L 1000 + j (0.1)

Y vs  

(b)

V1 VS



(c)



V1 R 1000 = = VS R + jL 1000 + j(0.1) vs 

V2 jL j(0.1) = = VS R + jL 1000 + j(0.1)

V2 VS

vs 

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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( n ) bn = c n cos( n )

and using the trigonometric identities sin 2 ( n ) + cos 2 ( n ) = 1: 



 

a n2 + bn2 = c n2 sin 2 ( n ) + c n2 cos2 ( n ) = c n2 Finally,  bn c n cos( n ) = = cot( n ) = tan(n ) a n c n sin( n ) where,  n = −  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:  1 1  A (n − 4 )T  t  (n + 4 )T , n = 0,1,2,... x(t) =  0 (n + 1 )T  t  (n + 3 )T , n = 0,1,2,...  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 =  0T x(t)dt =  −T4 Adt = T T 2 4

 6.38 PROPRIETARY MATERIAL. © 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

an =

 2   2  2 T 2 T t dt =  −T4 A cos n t dt =  x(t) cos n  T   T  T 0 T 4 T

 n  2A   2  T  4 A   n  t =  = 0 sin n  sin  − sin−  2  T   T  2n −T n   2 

(n)

bn = 

 2   2  2 T 2 T t dt =  −T4 A sin n t dt =  0 x(t) sin n  T   T  T T 4 T

 2  T  4  n   n  2A  A  t =  = − cos n  − cos  + cos−  T  2n −T  2   2  T  n  4

 2A  n   n A  = −2 cos  =   2   n   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 𝑥(𝑡) = { 𝐴

0≤𝑡≤

𝑇 3

𝑇 ≤𝑡≤𝑇 3

Known quantities: None.

Find: The Fourier series coefficient of the periodic function, 𝑥(𝑡).

Analysis: The function is not even or odd, so all of the 𝑎𝑛 and 𝑏𝑛 coefficients are needed. To accomplish this, use equations 6.34-6.36: 6.39 PROPRIETARY MATERIAL. © 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

𝑎0 =

𝑎𝑛 =

1 𝑇 ∫ 𝐴 𝑑𝑡 𝑇 𝑇/3 𝟐𝑨𝑻 𝟑

2 𝑇 2𝜋 ∫ 𝐴 cos (𝑛 𝑡) 𝑑𝑡 𝑇 𝑇 𝑇 3

𝑇 2 𝐴𝑇 2𝜋 sin (𝑛 𝑡) |𝑇 𝑇 2𝜋𝑛 𝑇 3

𝐴 2𝜋 (sin(𝑛 2𝜋𝑡) − sin (𝑛 𝑡)) 𝜋𝑛 3

=−

𝑏𝑛 =

𝑛 = 1, 2, 3, …

𝑨 𝟐𝝅 𝐬𝐢𝐧 (𝒏 𝒕) 𝒏 = 𝟏, 𝟐, 𝟑, … 𝝅𝒏 𝟑

2 𝑇 2𝜋 ∫ 𝐴 sin (𝑛 𝑡) 𝑑𝑡 𝑇 𝑇 𝑇 3

=−

2 𝐴𝑇 2𝜋 𝑇 cos (𝑛 𝑡)|𝑇 𝑇 2𝜋𝑛 𝑇 3

= − =

𝐴 2𝜋 (cos(𝑛 2𝜋 𝑡) − cos (𝑛 𝑡)) 𝑛 = 1, 2, 3, … 𝜋𝑛 3

𝑨 𝟐𝝅 𝐜𝐨𝐬 (𝒏 𝒕) 𝒏 = 𝟏, 𝟐, 𝟑, … 𝝅𝒏 𝟑

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:   2  T T  cos t − t x(t) =   T  4 4  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

 2  1 T 1 T 1   2  4 a 0 =  T2 x(t)dt =  T4 cos t dt = t  = sin T  T − 2 T − 4 2   T −T 4

   1 1    = sin  − sin−  =  2   2   2   2   2   2  2 T 2 T t dt =  T4 cos t  cos n t dt =  T2 x(t) cos n − −  T  T   T  T T 2 4  n   n 2 cos  (-1) 2 −1 (n even) 2  2  2 =− =   n −1  n 2 −1  (n odd)  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.

 2A  T t x(t ) =  2A − (t - T)  T

0t

T 2

T t T 2

Solution: Known quantities: The periodic function shown in Figure P6.19 and defined as:

 2A  T t x(t ) =  2A − (t - T)  T

0t

T 2

T t T 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   1 1 T 1  T 2 2A 2A  x ( t ) dt =  t dt + −   (t − T )dt  = A     T 0 0 T T T T  2  2

an =

T  − 2A  2 T 2  T 2 2A  2   2   2   x ( t ) cos n t dt =   t cos n t  dt + T   (t − T ) cos n t dt  =     T 0 T 0 T 2 T  T   T    T  

A 4A A 2A sin(n ) − sin(2n ) = sin(n ) − 2 2 cos(2n ) + 2 2 cos(n ) + n n n n 4A A = − 2 2 cos(2n ) + 2 2 cos(n ) n n =

bn =

T  2 T 2  T 2 2A 2A   2   2   2   x ( t ) sin n t dt =   t sin n t  dt + T  −  (t − T )sin n t dt  =     T 0 T 0 T T  2  T   T   T  

A A 2A cos(n ) − cos(2n ) + cos(n ) + 2 2 sin(n ) + n n n 2A A A A A − cos(2n ) + cos(n ) + 2 2 cos(n ) + 2 2 sin(2n ) − 2 2 sin(n ) n n n n n

=−

bn = −

A 2A cos(n ) − cos(2n ) − 2 A cos(2n ) + A cos(n ) + 2A 2 cos(n ) cos(n ) + n n n n n

Thus, the Fourier series expansion of the function is: 6.42 PROPRIETARY MATERIAL. © 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  A  2   A  2  x(t ) = +  a n cos n t +  bn sin n t 2 2 n=1 (n )  T  n=1 n  T 

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