
Concept explainers
For the waveforms in Fig. 26.32, determine whether the following will be present in the Fourier series representation:
a. dc term
b. cosine terms
c. sine terms
d. even-ordered harmonics
e. odd-ordered harmonics

(a)
Whether dc term, cosine terms, sine terms, even ordered harmonics and odd ordered harmonics will be present in the given Fourier Series representation or not.
Answer to Problem 1P
The D.C. term is not present in the waveform, cosine terms are not present in the waveform, sine terms are present in the waveform, even ordered harmonics are not present in the waveform and odd ordered harmonics are present in the waveform.
Explanation of Solution
Calculation:
The waveform for Fourier series representation is drawn as shown in Figure 1.
Case 1:
D.C. value is the average value of the waveform. And as the area above the horizontal axis and below the horizontal axis is equal, so the value of area and the average is zero for the waveform.
Therefore, D.C. term is not present in the waveform.
Case 2:
If waveform is symmetric about vertical axis, it has an even symmetry. And for an even symmetry, the wave Fourier series contains D.C. and cosine term.
As waveform is not symmetric about the vertical axis, it means cosine terms are not present in the waveform.
Case 3:
If waveform is symmetric about origin it has odd symmetry. And for odd symmetry wave Fourier series contain D.C. and sine term.
As waveform is symmetric about origin, it means sine terms are present in the waveform.
Case 4:
Cosine term is even function and sine terms are odd function.
As waveform does not contain cosine term, therefore, even ordered harmonics are not present in the waveform.
Case 5:
Cosine term is even function and sine terms are odd function.
As the waveform contains sine term, therefore, odd ordered harmonics are present in the waveform.
Conclusion:
Thus, D.C. term is not present in the waveform, cosine terms are not present in the waveform, sine terms are present in the waveform, even ordered harmonics are not present in the waveform and odd ordered harmonics are present in the waveform.

(b)
Whether dc term, cosine terms, sine terms, even ordered harmonics and odd ordered harmonics will be present in the given Fourier Series representation or not.
Answer to Problem 1P
The D.C. term is present in the waveform, cosine terms are not present in the waveform, sine terms are not present in the waveform, even ordered harmonics are not present in the waveform and odd ordered harmonics are not present in the waveform.
Explanation of Solution
Calculation:
The waveform for Fourier series representation is drawn as shown in Figure 2.
Case 1:
D.C. value is the average value of the waveform. And as area above the horizontal axis and below the horizontal axis is not equal so value of area and average is not zero for the waveform.
Therefore D.C. term is present in the waveform.
Case 2:
If waveform is symmetric about vertical axis it has even symmetry. And for even symmetry wave Fourier series contain D.C. and cosine term.
As waveform is not symmetric about vertical axis it means cosine terms are not present in the waveform.
Case 3:
If waveform is symmetric about origin it has odd symmetry. And for odd symmetry wave Fourier series contain D.C. and sine term.
As waveform is not symmetric about origin it means sine terms are not present in the waveform.
Case 4:
Cosine term is even function and sine terms are odd function.
As waveform does not contain cosine term therefore even ordered harmonics are not present in the waveform.
Case 5:
Cosine term is even function and sine terms are odd function.
As waveform does not contains sine term therefore odd ordered harmonics are not present in the waveform.
Conclusion:
Thus, D.C. term is present in the waveform, cosine terms are not present in the waveform, sine terms are not present in the waveform, even ordered harmonics are not present in the waveform and odd ordered harmonics are not present in the waveform.

(c)
Whether dc term, cosine terms, sine terms, even ordered harmonics and odd ordered harmonics will be present in the given Fourier Series representation or not.
Answer to Problem 1P
The D.C. term is present in the waveform, cosine terms are present in the waveform, sine terms are not present in the waveform, even ordered harmonics are present in the waveform and odd ordered harmonics are not present in the waveform.
Explanation of Solution
Calculation:
The waveform for Fourier series representation is drawn as shown in Figure 3.
Case 1:
D.C. value is the average value of the waveform. And as area above the horizontal axis and below the horizontal axis is not equal so value of area and average is not zero for the waveform.
Therefore D.C. term is present in the waveform.
Case 2:
If waveform is symmetric about vertical axis it has even symmetry. And for even symmetry wave Fourier series contain D.C. and cosine term.
As waveform is symmetric about vertical axis it means cosine terms are present in the waveform.
Case 3:
If waveform is symmetric about origin it has odd symmetry. And for odd symmetry wave Fourier series contain D.C. and sine term.
As waveform is not symmetric about origin it means sine terms are not present in the waveform.
Case 4:
Cosine term is even function and sine terms are odd function.
As waveform contain cosine term therefore even ordered harmonics are present in the waveform.
Case 5:
Cosine term is even function and sine terms are odd function.
As waveform does not contains sine term therefore odd ordered harmonics are not present in the waveform.
Conclusion:
Thus, D.C. term is present in the waveform, cosine terms are present in the waveform, sine terms are not present in the waveform, even ordered harmonics are present in the waveform and odd ordered harmonics are not present in the waveform.

(d)
Whether dc term, cosine terms, sine terms, even ordered harmonics and odd ordered harmonics will be present in the given Fourier Series representation or not.
Answer to Problem 1P
The D.C. term is present in the waveform, cosine terms are not present in the waveform, sine terms are present in the waveform, even ordered harmonics are not present in the waveform and odd ordered harmonics are present in the waveform.
Explanation of Solution
Calculation:
The waveform for Fourier series representation is drawn as shown in Figure 4.
Case 1:
D.C. value is the average value of the waveform. And as area above the horizontal axis and below the horizontal axis is equal so value of area and average is zero for the waveform.
Therefore D.C. term is present in the waveform.
Case 2:
If waveform is symmetric about vertical axis it has even symmetry. And for even symmetry wave Fourier series contain D.C. and cosine term.
As waveform is not symmetric about vertical axis it means cosine terms are not present in the waveform.
Case 3:
If waveform is symmetric about origin it has odd symmetry. And for odd symmetry wave Fourier series contain D.C. and sine term.
As waveform is symmetric about origin it means sine terms are present in the waveform.
Case 4:
Cosine term is even function and sine terms are odd function.
As waveform does not contain cosine term therefore even ordered harmonics are not present in the waveform.
Case 5:
Cosine term is even function and sine terms are odd function.
As waveform contains sine term therefore odd ordered harmonics are present in the waveform.
Conclusion:
Thus, D.C. term is present in the waveform, cosine terms are not present in the waveform, sine terms are present in the waveform, even ordered harmonics are not present in the waveform and odd ordered harmonics are present in the waveform.
Want to see more full solutions like this?
Chapter 26 Solutions
Laboratory Manual for Introductory Circuit Analysis
- Q3. a) The frequency response method enables the study of the steady-state response of a system G(s). What type of inputs are used for frequency response? If the system is linear and stable, how does the output differ from the input? Compare the main characteristics of two types frequency response plots. b) Consider the control system shown in Figure Q3. Controller E(s) R(s) Desired output C(s) Plant G(s) Y(s) Actual output 3(s + 3) C(s) = k G(s) = = s(s - 1)(s + 10) Figure Q3. Closed-loop system. (i) Considering definitions in the study of bounded-input bounded-output stability, is G(s) stable? Classify the poles and zeros of G(s). (ii) G(s) defined in Figure Q3 is a system completely characterised by its transfer function. Explain why this is the case. (iii) Obtain the closed-loop transfer function P(s) = Y(s)/R(s) of the system. (iv) Based on your result for the previous question [Question 3b)-(iii)], use the Routh-Hurwitz stability criterion to determine suitable values of gain K…arrow_forwardPlease, I want the solution in two ways: Method 1 (without the Smith chart): Method 2 (using the Smith chart): A short circuit stub of length 0.04λ is used to match a 50 Ω lossless line to a load ZL = RL + j30 Ω. Use Smith chart to find:(a) The distance between the stub and the load.(b) The value of RL .arrow_forwardTHE FIRST PAGE OF THIS QUESTION SECTION BELOW IS THE FIRST IMAGE UPLOADED, WHICH SHOWS A digital synchronous sequential circuit and then comes the questions below:1B) Suppose the flip-flops are 74F74 devices and the AND gates are 74F08 devices. Let maxtpd,D=9ns, maxtsu,D=3ns, and maxtpd,AND=6ns. What is the maximum clock frequency at which the circuit can operate reliably? 2) Compare serial transmission and parallel transmission and discuss their advantages and disadvantages. 3) Explain briefly how the slave can protect itself from being overwhelmed by the master in I2 4) A hypothetical logic family has the following specifications. VOH=4.6V VIH=4.0V VOL=0.5V VIL=1.0V IOH=-1mA IIH=50μA IOL=8mA IIL=-0.6mA (4a) What are the noise margins? (4b) What is the fan-out capability?…arrow_forward
- THE FIRST PAGE OF THIS QUESTION SECTION BELOW IS THE FIRST IMAGE UPLOADED, WHICH SHOWS A digital synchronous sequential circuit and then comes the questions below:1B) Suppose the flip-flops are 74F74 devices and the AND gates are 74F08 devices. Let maxtpd,D=9ns, maxtsu,D=3ns, and maxtpd,AND=6ns. What is the maximum clock frequency at which the circuit can operate reliably? 2) Compare serial transmission and parallel transmission and discuss their advantages and disadvantages. 3) Explain briefly how the slave can protect itself from being overwhelmed by the master in I2 4) A hypothetical logic family has the following specifications. VOH=4.6V VIH=4.0V VOL=0.5V VIL=1.0V IOH=-1mA IIH=50μA IOL=8mA IIL=-0.6mA (4a) What are the noise margins? (4b) What is the fan-out capability?…arrow_forwardI need help on this question a) Find y(t) =yh(t) +yp(t) in time domainIs the system over-damped, under-damped, or critical?arrow_forwardGiven f(t)=a sin(ßt) a = 10 & ß = 23 Find the Laplace Transform using the definition F(s) = ∫f(t)e-stdtarrow_forward
- = Calculate Avf, Zif, and Zof for the amplifier circuit,Assume he = 50, hie 1.1k2, and identical transistors? 150kQ Vs 5002 HH +25v 10k +6 · 47ΚΩ 47k2 4.7k0} 33 ΚΩ 4.7ΚΩ 10k w 4.7kQ HH Voarrow_forwardFor the four-pole filter in Fig. (2), determine the capacitance values required to produce a critical frequency of 2680 Hz if all the resistors in the RC low-pass circuits are 1.8 K. Also select values for the feedback resistors to get a Butterworth response. Note: For a Butterworth response, the damping factor must be 1.848 for the first stage and 0.765 for the second stage. (2) Re Res ww " = 11arrow_forwardFor the circuit shown in Fig. 2.20, the transistors are identica' and have the following parameters: hje=50, hie = 1.1K, hr =0, and hoe = 0. Calculate Auf, Rif and Rof. Ans: 45.4; 112 KN; 129N. HH 150k 47k R 25 V 10k 47k 4.7k 5μF 33k 4.7k 50µF 50µF 4.7k 4.7k R₁ Roj R1000arrow_forward
- A triangular wave is applied to the input of Fig. (3). Determine what the output should be and sketch its waveform in relation to the input. 10μs. 0 5μs 15 μs 0.001 μF R₁ w 2.2karrow_forwardA three-phase, 480-V, 60-Hz, 6-pole, Y-connected induction motor has its speed controlled by slip power. The circuit parameters are given: Rs=0.06 ohms, Rr=0.05 ohms, Xs=0.2 ohms, Xr=0.3 ohms and Xm=6 ohms. The turn ratio of the rotor to stator winding is n=0.8. The no-load losses of the motor are equal to 150 W. The rotor and stator cupper losses are equal to 249.21 W. The slip power losses are estimated to 8000W. The load torque is 173.61 N.m. at 700 rpm. The efficiency is equal to: Select one: a. 71.5% b. None of these c. 81.5% d. 91.5% Question 2 Consider a 3-phase, 460-V, 100-hp, 0.88 power factor lagging, 4-pole, 1728 RPM, 60 Hz, Y-connected induction motor. The operating slip is equal to: Select one: a. 0.05 b. 0.01 c. 0.04 d. None of these Question 3 A 3 phase, 10 kW, 1750 rpm, Y- connected 460 V, 60 Hz, 4 poles, Y-connected induction motor has the following parameters: Rs = 0.5 Ohms, Rr = 0.3 Ohms, Xs = 0.9 Ohms, Xr = 0.9 Ohms, Xm = 25 Ohms. The no load…arrow_forwardelectric plants do for hand writingarrow_forward
- Introductory Circuit Analysis (13th Edition)Electrical EngineeringISBN:9780133923605Author:Robert L. BoylestadPublisher:PEARSONDelmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage LearningProgrammable Logic ControllersElectrical EngineeringISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill Education
- Fundamentals of Electric CircuitsElectrical EngineeringISBN:9780078028229Author:Charles K Alexander, Matthew SadikuPublisher:McGraw-Hill EducationElectric Circuits. (11th Edition)Electrical EngineeringISBN:9780134746968Author:James W. Nilsson, Susan RiedelPublisher:PEARSONEngineering ElectromagneticsElectrical EngineeringISBN:9780078028151Author:Hayt, William H. (william Hart), Jr, BUCK, John A.Publisher:Mcgraw-hill Education,





