Fundamentals Of Applied Electromagnetics
Fundamentals of Applied Electromagnetics 6ebyFawwaz T. Ulaby, Eric Michielssen, and Umberto RavaioliSolved ProblemsFawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
ChaptersChapter 1 Introduction: Waves and PhasorsChapter 2 Transmission LinesChapter 3 Vector AnalysisChapter 4 ElectrostaticsChapter 5 MagnetostaticsChapter 6 Maxwell’s Equations for Time-Varying FieldsChapter 7 Plane-Wave PropagationChapter 8 Wave Reflection and TransmissionChapter 9 Radiation and AntennasChapter 10 Satellite Communication Systems and Radar SensorsFawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Chapter 1 Solved ProblemsProblem 1-4Problem 1-7Problem 1-15Problem 1-18Problem 1-20Problem 1-21Problem 1-24Problem 1-26Problem 1-27Problem 1-29Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 1.4A wave traveling along a string is given byy(x,t) 2 sin(4πt 10πx) (cm),where x is the distance along the string in meters and y is the vertical displacement. Determine: (a) the direction of wavetravel, (b) the reference phase φ0 , (c) the frequency, (d) the wavelength, and (e) the phase velocity.Solution:(a) We start by converting the given expression into a cosine function of the form given by (1.17): π y(x,t) 2 cos 4πt 10πx (cm).2Since the coefficients of t and x both have the same sign, the wave is traveling in the negative x-direction.(b) From the cosine expression, φ0 π/2.(c) ω 2π f 4π,f 4π/2π 2 Hz.(d) 2π/λ 10π,λ 2π/10π 0.2 m.(e) up f λ 2 0.2 0.4 (m/s).Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 1.7A wave traveling along a string in the x-direction is given byy1 (x,t) A cos(ωt β x),where x 0 is the end of the string, which is tied rigidly to a wall, as shown in Fig. (P1.7). When wave y1 (x,t) arrives at thewall, a reflected wave y2 (x,t) is generated. Hence, at any location on the string, the vertical displacement ys will be the sumof the incident and reflected waves:ys (x,t) y1 (x,t) y2 (x,t).(a) Write down an expression for y2 (x,t), keeping in mind its direction of travel and the fact that the end of the stringcannot move.(b) Generate plots of y1 (x,t), y2 (x,t) and ys (x,t) versus x over the range 2λ x 0 at ωt π/4 and at ωt π/2.Figure P1.7: Wave on a string tied to a wall at x 0 (Problem 1.7).Solution:(a) Since wave y2 (x,t) was caused by wave y1 (x,t), the two waves must have the same angular frequency ω, and sincey2 (x,t) is traveling on the same string as y1 (x,t), the two waves must have the same phase constant β . Hence, with itsdirection being in the negative x-direction, y2 (x,t) is given by the general formy2 (x,t) B cos(ωt β x φ0 ),(1.1)where B and φ0 are yet-to-be-determined constants. The total displacement isys (x,t) y1 (x,t) y2 (x,t) A cos(ωt β x) B cos(ωt β x φ0 ).Since the string cannot move at x 0, the point at which it is attached to the wall, ys (0,t) 0 for all t. Thus,ys (0,t) A cos ωt B cos(ωt φ0 ) 0.(1.2)(i) Easy Solution: The physics of the problem suggests that a possible solution for (1.2) is B A and φ0 0, in which casewe havey2 (x,t) A cos(ωt β x).(1.3)(ii) Rigorous Solution: By expanding the second term in (1.2), we haveA cos ωt B(cos ωt cos φ0 sin ωt sin φ0 ) 0,Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
or(A B cos φ0 ) cos ωt (B sin φ0 ) sin ωt 0.(1.4)This equation has to be satisfied for all values of t. At t 0, it givesA B cos φ0 0,(1.5)B sin φ0 0.(1.6)and at ωt π/2, (1.4) givesEquations (1.5) and (1.6) can be satisfied simultaneously only ifA B 0(1.7)orA Band φ0 0.(1.8)Clearly (1.7) is not an acceptable solution because it means that y1 (x,t) 0, which is contrary to the statement of the problem.The solution given by (1.8) leads to (1.3).(b) At ωt π/4, π 2πxy1 (x,t) A cos(π/4 β x) A cos ,4λ π 2πxy2 (x,t) A cos(ωt β x) A cos .4λPlots of y1 , y2 , and y3 are shown in Fig. P1.7(b).Figure P1.7: (b) Plots of y1 , y2 , and ys versus x at ωt π/4.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
At ωt π/2,2πx,λ2πxy2 (x,t) A cos(π/2 β x) A sin β x A sin.λy1 (x,t) A cos(π/2 β x) A sin β x A sinPlots of y1 , y2 , and y3 are shown in Fig. P1.7(c).Figure P1.7: (c) Plots of y1 , y2 , and ys versus x at ωt π/2.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 1.15A laser beam traveling through fog was observed to have an intensity of 1 (µW/m2 ) at a distance of 2 mfrom the laser gun and an intensity of 0.2 (µW/m2 ) at a distance of 3 m. Given that the intensity of an electromagnetic waveis proportional to the square of its electric-field amplitude, find the attenuation constant α of fog.Solution: If the electric field is of the formE(x,t) E0 e αx cos(ωt β x),then the intensity must have a formI(x,t) [E0 e αx cos(ωt β x)]2 E02 e 2αx cos2 (ωt β x)orI(x,t) I0 e 2αx cos2 (ωt β x)where we define I0 E02 . We observe that the magnitude of the intensity varies as I0 e 2αx . Hence,at x 2 m,I0 e 4α 1 10 6at x 3 m,I0 e 6α 0.2 10 6(W/m2 ),(W/m2 ).10 6I0 e 4α 5I0 e 6α0.2 10 6e 4α · e6α e2α 5α 0.8(NP/m).Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 1.18Complex numbers z1 and z2 are given byz1 3 j2z2 1 j2Determine (a) z1 z2 , (b) z1 /z 2 , (c) z21 , and (d) z1 z 1 , all all in polar form.Solution:(a) We first convert z1 and z2 to polar form: p 132 22 e j tan 2/3z1 (3 j2) 13 e j33.7 13 e j(180 33.7 ) 13 e j146.3 . 11 4 e j tan 2 5 e j63.4 .z2 1 j2 13 e j146.3 5 e j63.4 65 e j82.9 .z1 z2 (b)r 13 e j146.313 j82.9 z1 e. j63.4z255e(c) z21 ( 13)2 (e j146.3 )2 13e j292.6 13e j360 e j292.6 13e j67.4 .(c)z1 z 1 13 e j146.3 13 e j146.3 13.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 1.20Find complex numbers t z1 z2 and s z1 z2 , both in polar form, for each of the following pairs:(a) z1 2 j3, z2 1 j3,(b) z1 3, z2 j3,(c) z1 3 30 , z2 3 30 ,(d) z1 3 30 , z2 3 150 .Solution:(d)t z1 z2 3 30 3 150 (2.6 j1.5) ( 2.6 j1.5) 0, s z1 z2 (2.6 j1.5) ( 2.6 j1.5) 5.2 j3 6e j30 .Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 1.21Complex numbers z1 and z2 are given byz1 5 60 ,z2 4 45 .(a)(b)(c)(d)(e)Determine the product z1 z2 in polar form.Determine the product z1 z 2 in polar form.Determine the ratio z1 /z2 in polar form.Determine the ratio z 1 /z 2 in polar form. Determine z1 in polar form.Solution: z1 5e j60 1.25e j105 .(c) z24e j45Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 1.24If z 3e jπ/6 , find the value of ez .Solution:z 3e jπ/6 3 cos π/6 j3 sin π/6 2.6 j1.5ez e2.6 j1.5 e2.6 e j1.5 e2.6 (cos 1.5 j sin 1.5) 13.46(0.07 j0.98) 0.95 j13.43.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 1.26Find the phasors of the following time functions:(a) υ(t) 9 cos(ωt π/3) (V)(b) υ(t) 12 sin(ωt π/4) (V)(c) i(x,t) 5e 3x sin(ωt π/6) (A)(d) i(t) 2 cos(ωt 3π/4) (A)(e) i(t) 4 sin(ωt π/3) 3 cos(ωt π/6) (A)Solution:(d)i(t) 2 cos(ωt 3π/4),Ie 2e j3π/4 2e jπ e j3π/4 2e jπ/4 A.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 1.27Find the instantaneous time sinusoidal functions corresponding to the following phasors:(a) Ve 5e jπ/3 (V)(b) Ve j6e jπ/4 (V)(c) Ie (6 j8) (A)(d) I 3 j2 (A)(e) I j (A)(f) I 2e jπ/6 (A)Solution:(d) Ie 3 j2 3.61 e j146.31 , i(t) Re{3.61 e j146.31 e jωt } 3.61 cos(ωt 146.31 ) A.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 1.29The voltage source of the circuit shown in Fig. P1.29 is given byvs (t) 25 cos(4 104t 45 ) (V).Obtain an expression for iL (t), the current flowing through the inductor.Figure P1.29: Circuit for Problem 1.29.Solution: Based on the given voltage expression, the phasor source voltage is Ves 25e j45(V).(1.9)The voltage equation for the left-hand side loop isR1 i R2 iR2 vs(1.10)diL,dt(1.11)For the right-hand loop,R2 iR2 Land at node A,i iR2 iL .(1.12)R1 Ie R2 IeR2 VesR2 IeR2 jωLIeLIe IeR2 IeL(1.13)Next, we convert Eqs. (2)–(4) into phasor form:(1.14)(1.15)e we have:Upon combining (6) and (7) to solve for IeR2 in terms of I,IeR2 jωLI.R2 jωLFawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics(1.16)c 2010 Prentice Hall
Substituting (8) in (5) and then solving for Ie leads to:jR2 ωL e eR1 Ie I VsR2 jωL jRωL2Ie R1 VesR2 jωL R1 R2 jR1 ωL jR2 ωLe VesIR2 jωL R jωL2Ie Ves .R1 R2 jωL(R1 R2 )(1.17)Combining (6) and (7) to solve for IeL in terms of Ie givesIeL R2eI.R2 jωL(1.18)Combining (9) and (10) leads to R2R2 jωLVesR2 jωLR1 R2 jωL(R1 R2 )R2Ves . R1 R2 jωL(R1 R2 )IeL Using (1) for Ves and replacing R1 , R2 , L and ω with their numerical values, we haveIeL 30j4 104 0.4 10 3 (20 30)20 30 30 25 e j45600 j800 7.5 j45 7.5e j45 j98.1 e 0.75ej53.16 j810e 25e j45(A).Finally,iL (t) Re[IeL e jωt ] 0.75 cos(4 104t 98.1 ) (A).Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Chapter 2 Solved ProblemsProblem 2-5Problem 2-16Problem 2-34Problem 2-45Problem 2-48Problem 2-64Problem 2-75Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 2.5For the parallel-plate transmission line of Problem 2.4, the line parameters are given by: R0 1 Ω/m,0L 167 nH/m, G0 0, and C0 172 pF/m. Find α, β , up , and Z0 at 1 GHz.Solution: At 1 GHz, ω 2π f 2π 109 rad/s. Application of (2.22) gives:pγ (R0 jωL0 )(G0 jωC0 ) [(1 j2π 109 167 10 9 )(0 j2π 109 172 10 12 )]1/2 [(1 j1049)( j1.1)]1/2 1/2 qj tan 1 1049j90 21 (1049) e 1.1e, ( j e j90 )hi 1/2 1049e j89.95 1.1e j90hi 1/2 1154e j179.95 34e j89.97 34 cos 89.97 j34 sin 89.97 0.016 j34.Hence,α 0.016 Np/m,β 34 rad/m.ω2π f2π 109 1.85 108 m/s.ββ34 0 R jωL0 1/2Z0 G0 jωC0 1/21049e j89.95 1.1e j90hi 1/2 954e j0.05up 31e j0.025 ' (31 j0.01) Ω.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 2.16 A transmission line operating at 125 MHz has Z0 40 Ω, α 0.02 (Np/m), and β 0.75 rad/m. Find theline parameters R0 , L0 , G0 , and C 0 .Solution: Given an arbitrary transmission line, f 125 MHz, Z0 40p and β 0.75 rad/m. Since Z0 Ω, α 0.02 Np/m,is real and α 6 0, the line is distortionless. From Problem 2.13, β ω L0C 0 and Z0 L0 /C 0 , therefore,L0 Then, from Z0 β Z00.75 40 38.2 nH/m.ω2π 125 106pL0 /C 0 ,C0 From α R0 G0 and R0C 0 L0 G0 , R0 R0 G0rL038.2 nH/m 23.9 pF/m.2402Z0R0 0 0 RGG0andG0 rL0 αZ0 0.02 Np/m 40 Ω 0.6 Ω/mC0α 2 (0.02 Np/m)2 0.5 mS/m.R00.8 Ω/mFawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 2.34A 50-Ω lossless line is terminated in a load impedance ZL (30 j20) Ω.Figure P2.34: Circuit for Problem 2.34.(a) Calculate Γ and S.(b) It has been proposed that by placing an appropriately selected resistor across the line at a distance dmax from the load(as shown in Fig. P2.34(b)), where dmax is the distance from the load of a voltage maximum, then it is possible torender Zi Z0 , thereby eliminating reflection back to the end. Show that the proposed approach is valid and find thevalue of the shunt resistance.Solution:(a) ZL Z0 30 j20 50 20 j20 (20 j20) 0.34e j121 .ZL Z0 30 j20 5080 j2080 j201 Γ 1 0.34S 2.1 Γ 1 0.34Γ (b) We start by finding dmax , the distance of the voltage maximum nearest to the load. Using (2.70) with n 1, θr λ λ 121 π λλdmax 0.33λ .4π2180 4π 2Applying (2.79) at d dmax 0.33λ , for which β l (2π/λ ) 0.33λ 2.07 radians, the value of Zin before adding theFawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
shunt resistance is: ZL jZ0 tan β lZin Z0Z0 jZL tan β l (30 j20) j50 tan 2.07 50 (102 j0) Ω.50 j(30 j20) tan 2.07Thus, at the location A (at a distance dmax from the load), the input impedance is purely real. If we add a shunt resistor R inparallel such that the combination is equal to Z0 , then the new Zin at any point to the left of that location will be equal to Z0 .Hence, we need to select R such that111 R 102 50or R 98 Ω.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 2.45The circuit shown in Fig. P2.45 consists of a 100-Ω lossless transmission line terminated in a load withZL (50 j100) Ω. If the peak value of the load voltage was measured to be VeL 12 V, determine:(a) the time-average power dissipated in the load,(b) the time-average power incident on the line,(c) the time-average power reflected by the load.Figure P2.45: Circuit for Problem 2.45.Solution:(a)Γ ZL Z0 50 j100 100 50 j100 0.62e j82.9 .ZL Z0 50 j100 100150 j100The time average power dissipated in the load is:1Pav IeL 2 RL221 VeL RL2 ZL 5011 VeL 2RL 122 2 0.29 W.22 ZL 250 1002(b)iPav Pav(1 Γ 2 )Hence,iPav Pav0.29 0.47 W.21 Γ 1 0.622(c)riPav Γ 2 Pav (0.62)2 0.47 0.18 W.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 2.48Repeat Problem 2.47 using CD Module 2.6.Solution:Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 2.64to a 50-Ω line.Use CD Module 2.7 to design a quarter-wavelength transformer to match a load with ZL (100 j200) ΩSolution: Figure P2.64(a) displays the first solution of Module 2.7 where a λ /4 section of Z02 15.5015 Ω is inserted atdistance d1 0.21829λ from the load.Figure P2.64(b) displays a summary of the two possible solutions for matching the load to the feedline with a λ /4transformer.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 2.75 Generate a bounce diagram for the voltage V (z,t) for a 1-m–long lossless line characterized by Z0 50 Ωand up 2c/3 (where c is the velocity of light) if the line is fed by a step voltage applied at t 0 by a generator circuit withVg 60 V and Rg 100 Ω. The line is terminated in a load RL 25 Ω. Use the bounce diagram to plot V (t) at a pointmidway along the length of the line from t 0 to t 25 ns.Solution:Rg Z0 100 50501 ,Rg Z0 100 50 150 3ZL Z0 25 50 25 1 ΓL .ZL Z0 25 50753Γg From Eq. (2.149b),V1 Vg Z060 50 20 V.Rg Z0 100 50Also,T ll3 5 ns. up 2c/3 2 3 108The bounce diagram is shown in Fig. P2.75(a) and the plot of V (t) in Fig. P2.75(b).Figure P2.75: (a) Bounce diagram for Problem 2.75.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Figure P2.75: (b) Time response of voltage.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Chapter 3 Solved ProblemsProblem 3-9Problem 3-17Problem 3-19Problem 3-20Problem 3-22Problem 3-23Problem 3-25Problem 3-33Problem 3-35Problem 3-36Problem 3-41Problem 3-50Problem 3-55Problem 3-57Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 3.9 Find an expression for the unit vector directed toward the origin from an arbitrary point on the line describedby x 1 and z 3.Solution: An arbitrary point on the given line is (1, y, 3). The vector from this point to (0, 0, 0) is:A x̂(0 1) ŷ(0 y) ẑ(0 3) x̂ ŷy 3ẑ,pp A 1 y2 9 10 y2 ,A x̂ ŷy ẑ 3.â p A 10 y2Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 3.17 Find a vector G whose magnitude is 4 and whose direction is perpendicular to both vectors E and F, whereE x̂ ŷ 2 ẑ 2 and F ŷ 3 ẑ 6.Solution: The cross product of two vectors produces a third vector which is perpendicular to both of the original vectors. F and another along the opposite direction. Hence,Two vectors exist that satisfy the stated conditions, one along E G 4 (ŷ 3 ẑ 6) F(x̂ ŷ 2 ẑ 2) E 4 (ŷ 3 ẑ 6) E F (x̂ ŷ 2 ẑ 2) ( x̂ 6 ŷ 6 ẑ 3) 4 36 36 9 8844 ( x̂ 6 ŷ 6 ẑ 3) x̂ ŷ ẑ.9333Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 3.19Vector field E is given byE R̂ 5R cos θ θ̂θ12sin θ cos φ φ̂φ 3 sin φ .RDetermine the component of E tangential to the spherical surface R 2 at point P (2, 30 , 60 ).Solution: At P, E is given by12sin 30 cos 60 φ̂φ 3 sin 60 2 R̂ 8.67 θ̂θ 1.5 φ̂φ 2.6.E R̂ 5 2 cos 30 θ̂θThe R̂ component is normal to the spherical surface while the other two are tangential. Hence,Et θ̂θ 1.5 φ̂φ 2.6.Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Problem 3.20When sketching or demonstrating the spatial variation of a vector field, we often use arrows, as inFig. P3.20, wherein the length of the arrow is made to be proportional to the strength of the field and the direction ofthe arrow is the same as that of the field’s. The sketch shown in Fig. P3.20, which represents the vector field E r̂r, consistsof arrows pointing radially away from the origin and their lengths increase linearly in proportion to their distance away fromthe origin. Using this arrow representation, sketch each of the following vector fields:(a) E1 x̂y,(b) E2 ŷx,(c) E3 x̂x ŷy,(d) E4 x̂x ŷ2y,(e) E5 φ̂φ r,(f) E6 r̂ sin φ .Figure P3.20: Arrow representation for vector field E r̂r (Problem 3.20).Solution:(b)Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice Hall
Figure P3.20(b): E2 ŷ x(e)Figure P3.20(e): E5 φ̂φ rFawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Pre
(ii) Rigorous Solution: By expanding the second term in (1.2), we have Acoswt B(coswtcosf 0 sinwtsinf 0) 0; Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics c 2010 Prentice Hall
surveys given to the students at the end of each MATLAB assignment. The Electromagnetics Concept Inventory was also used. KEYWORDS computers and education, computer-assisted instruction and learning, computer exploration in electromagnetics, electromagnetics teac
Chapter 1 Electromagnetics and Optics 1.1 Introduction In this chapter, we will review the basics of electromagnetics and optics. We will briefly discuss various laws of electromagnetics leading to Maxwell's equations. The Maxwell's equations will be used to derive the wave equation which forms the basis for the study of optical fibers in .
1. Numericals based on solving Engg. Electromagnetics problems using MATLAB for tutorials are available in CD accompanied with the book of “Fundamentals of Engineering Electromagnetics by Sunil Bhooshan” 2. Matlab Experiments manual for Electromagnetics by Dr. M.H. Bakr SR. NO. LIST OF PRACTICALS 1. INTR
Chapter 1 Exercise Solutions Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Exercise 1.6 Exercise 1.7 Exercise 1.8 Exercise 1.9 Exercise 1.10 Exercise 1.11 Exercise 1.12 Fawwaz T. Ulaby and Umberto Ravaioli, Fundamentals of Applied Electromagnetics c 2019 Prentice Hall
The total electric field in a dielectric material is the sum of the applied elec-tric field and of an induced electric field, resulting from the polarization of the material.As a simple example, aperfect electric conductor-1. FUNDAMENTALS OF ELECTROMAGNETICS.,, .
electromagnetics education and surveys and experiences in teaching/learning electromagnetic fields and waves are pre-sented in [1]–[6]. Generally, there is a great diversity in the teaching of undergraduate electromagnetics courses, in content, scope, and pedagogical philosophy. Some electromagnetics courses
Electromagnetics Inductors and the Wave Equation Inductors Capacitors deal with voltage and store energy in an electric field. Inductors, on the other hand, deal with . To see how the wave equation behaves, let's use Matlab to simulate an LC circuit. To start with, let's
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