Fundamentals of Applied Electromagnetics 6e Solution PDF

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Fundamentals of Applied Electromagnetics 6e problems solution...

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Fundamentals of Applied Electromagnetics 6e by Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli

Solved Problems

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Chapters Chapter 1 Introduction: Waves and Phasors Chapter 2 Transmission Lines Chapter 3 Vector Analysis Chapter 4 Electrostatics Chapter 5 Magnetostatics Chapter 6 Maxwell’s Equations for Time-Varying Fields Chapter 7 Plane-Wave Propagation Chapter 8 Wave Reflection and Transmission Chapter 9 Radiation and Antennas Chapter 10 Satellite Communication Systems and Radar Sensors

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Chapter 1 Solved Problems Problem 1-4 Problem 1-7 Problem 1-15 Problem 1-18 Problem 1-20 Problem 1-21 Problem 1-24 Problem 1-26 Problem 1-27 Problem 1-29

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Problem 1.4

A wave traveling along a string is given by y(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 wave travel, (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). 2

Since 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 Electromagnetics

c 2010 Prentice Hall

Problem 1.7

A wave traveling along a string in the +x-direction is given by y1 (x,t) = Acos(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 the wall, a reflected wave y2 (x,t) is generated. Hence, at any location on the string, the vertical displacement ys will be the sum of 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 string cannot 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 since y2 (x,t) is traveling on the same string as y1 (x,t ), the two waves must have the same phase constant β . Hence, with its direction being in the negative x-direction, y2 (x,t) is given by the general form y2 (x,t) = B cos(ωt + β x + φ0 ),

(1.1)

where B and φ0 are yet-to-be-determined constants. The total displacement is ys (x,t ) = y1 (x, t ) + y2 (x,t) = Acos(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) = Acos 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 case we have y2 (x,t) = −Acos(ωt + β x). (1.3) (ii) Rigorous Solution: By expanding the second term in (1.2), we have Acos ωt + B(cos ωt cos φ0 − sin ωt sin φ0 ) = 0,

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 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 gives A + B cos φ0 = 0,

(1.5)

B sin φ0 = 0.

(1.6)

and at ωt = π/2, (1.4) gives Equations (1.5) and (1.6) can be satisfied simultaneously only if A=B=0

(1.7)

or A = −B

and

φ 0 = 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 2πx , − y1 (x,t) = Acos(π/4 − β x) = Acos λ 4   π 2πx . + y2 (x,t) = −Acos(ωt + β x) = −Acos λ 4 Plots of y1 , y2 , and y3 are shown in Fig. P1.7(b).

Figure P1.7: (b) Plots of y 1 , y 2 , and y s versus x at ωt = π/4.

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

At ωt = π/2, 2πx , λ 2πx . y2 (x,t) = −Acos(π/2 + β x) = Asin β x = Asin λ

y1 (x,t) = Acos(π/2 − β x) = A sin β x = A sin

Plots of y1 , y2 , and y3 are shown in Fig. P1.7(c).

Figure P1.7: (c) Plots of y 1 , y 2 , and y s versus x at ωt = π/2.

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Problem 1.15 A laser beam traveling through fog was observed to have an intensity of 1 (µW/m2 ) at a distance of 2 m from the laser gun and an intensity of 0.2 (µW/m2 ) at a distance of 3 m. Given that the intensity of an electromagnetic wave is proportional to the square of its electric-field amplitude, find the attenuation constant α of fog. Solution: If the electric field is of the form E(x,t) = E0 e−α x cos(ωt − β x), then the intensity must have a form I(x,t) ≈ [E0 e−α x cos(ωt − β x)]2 ≈ E02e−2α x cos2 (ωt − β x)

or I(x,t) = I0 e−2α x cos2 (ωt − β x)

where we define I0 ≈ E02. We observe that the magnitude of the intensity varies as I0 e−2α x . Hence, at x = 2 m, at x = 3 m,

I0 e−4α = 1 × 10−6

(W/m2 ),

I0 e−6α = 0.2 × 10−6

(W/m2 ).

I0 e−4α 10−6 =5 = 0.2 × 10−6 I0 e−6α e−4α · e6α = e2α = 5 α = 0.8 (NP/m).

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Problem 1.18

Complex numbers z1 and z2 are given by z1 = −3 + j2

z2 = 1 − j2

Determine (a) z1 z2 , (b) z1 /z2∗, (c) z21 , and (d) z1 z1∗, all all in polar form. Solution: (a) We first convert z1 and z2 to polar form: p  −1 z1 = −(3 − j2) = − 32 + 22 e− j tan 2/3 √ ◦ = − 13 e− j33.7 √ ◦ ◦ = 13 e j(180 −33.7 ) √ ◦ = 13 e j146.3 . √ −1 1 + 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 j146.3 z1 13 j82.9◦ = e . √ = ◦ j63 . 4 5 z2∗ 5e

(c) √ ◦ ◦ z12 = ( 13)2 (e j146.3 )2 = 13e j292.6 ◦

= 13e− j360 e j292.6

◦

◦

= 13e− j67.4 . (c) z1 z1∗ =

√ √ ◦ ◦ 13 e j146.3 × 13 e− j146.3

= 13.

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Problem 1.20 Find 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 + j 1.5) + (−2.6 − j1.5) = 0, ◦

s = z1 − z2 = (2.6 + j1.5) − (−2.6 − j 1.5) = 5.2 + j 3 = 6e j30 .

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Problem 1.21

Complex numbers z1 and z2 are given by z1 = 5∠−60◦ , z2 = 4∠45◦ .

(a) (b) (c) (d) (e)

Determine the product z1 z2 in polar form. Determine the product z1 z2∗ in polar form. Determine the ratio z1 /z2 in polar form. Determine the ratio z1∗/z2∗ in polar form. √ Determine z1 in polar form.

Solution: ◦ ◦ z1 5e− j60 − j105 = (c) . ◦ = 1.25e j45 z2 4e

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Problem 1.24

If z = 3e jπ/6 , find the value of ez .

Solution: z = 3e jπ/6 = 3 cos π/6 + j3 sin π/6 = 2.6 + j1.5 ez = 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 Electromagnetics

c 2010 Prentice Hall

Problem 1.26 Find 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 Electromagnetics

c 2010 Prentice Hall

Problem 1.27 Find the instantaneous time sinusoidal functions corresponding to the following phasors: jπ/3 e V = −5e (V) (a) − jπ/4 e (V) (b) V = j6e (c) Ie= (6 + j8) (A) (d) I˜ = −3 + j2 (A) (e) I˜ = j (A) (f) I˜ = 2e jπ/6 (A) Solution: (d) ◦ e I = −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 Electromagnetics

c 2010 Prentice Hall

Problem 1.29

The voltage source of the circuit shown in Fig. P1.29 is given by vs (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

The voltage equation for the left-hand side loop is

e Vs = 25e− j45

◦

(V).

(1.9)

R1 i + R2 iR2 = vs

(1.10)

diL , dt

(1.11)

For the right-hand loop, R2 iR2 = L and at node A, i = iR2 + iL .

(1.12)

es IR2 = V I + R2 e R1 e R2IeR2 = jω LIeL e IL I = IeR2 + e

(1.13)

Next, we convert Eqs. (2)–(4) into phasor form:

e we have: Upon combining (6) and (7) to solve for IeR2 in terms of I, e IR2 =

jω L I. R2 + jω L

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

(1.14) (1.15)

(1.16)

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Substituting (8) in (5) and then solving for Ieleads to:

jR2 ω L e e R1 e I = Vs I+ R2 + jω L   jR2 ω L e = Ves I R1 + R2 + jω L   R1 R2 + jR1 ω L + jR2 ω L e = Ves I R2 + jω L   R + jω L 2 e Ve . I= R1 R2 + jω L(R1 + R2 ) s

I gives Combining (6) and (7) to solve for IeL in terms of e

e IL =

Combining (9) and (10) leads to

R2 e I. R2 + jω L

(1.17)

(1.18)



  R2 R2 + jω L Ve R2 + jω L R1 R2 + jω L(R1 + R2 ) s R2 = Ve. R1 R2 + + jω L(R1 + R2 ) s

e IL =

Using (1) for Ves and replacing R1 , R2 , L and ω with their numerical values, we have

30 ◦ 25e− j45 20 × 30 + j4 × 104 × 0.4 × 10−3 (20 + 30) 30 × 25 − j45◦ e = 600 + j800 ◦ 7.5 − j45◦ 7.5e− j45 − j98.1◦ (A). e = = ◦ = 0.75e 10e j53.1 6 + j8

e IL =

Finally, jωt ] iL (t) = Re[Ie Le

= 0.75 cos(4 × 104t − 98.1◦ ) (A).

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Chapter 2 Solved Problems Problem 2-5 Problem 2-16 Problem 2-34 Problem 2-45 Problem 2-48 Problem 2-64 Problem 2-75

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Problem 2.5 For the parallel-plate transmission line of Problem 2.4, the line parameters are given by: R′ = 1 Ω/m, ′ L = 167 nH/m, G′ = 0, and C′ = 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: p γ = (R′ + jω L′ )(G′ + jωC′ )

= [(1 + j2π × 109 × 167 × 10−9 )(0 + j 2π × 109 × 172 × 10−12)]1/2 = [(1 + j1049)( j1.1)]1/2 1/2 q j tan−1 1049 j90◦ 2 1 + (1049) e × 1.1e = ,

◦

( j = e j90 )

i h ◦ 1/2 ◦ = 1049e j89.95 × 1.1e j90 i h ◦ 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. ω 2π f 2π × 109 = = = 1.85 × 108 m/s. β β 34 1/2  ′ R + jω L′ Z0 = ′ G + jωC′  ◦ 1/2 1049e j89.95 = ◦ 1.1e j90 i h ◦ 1/2 = 954e− j0.05

up =

◦

= 31e− j0.025 ≃ (31 − j0.01) Ω.

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 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 the line parameters R′ , L′ , G′ , and C ′ . Solution: Given an arbitrary transmission line, f = 125 MHz, Z0 = 40 √Ω, α = 0.02 Np/m, p and β = 0.75 rad/m. Since Z0 is real and α 6= 0, the line is distortionless. From Problem 2.13, β = ω L′C ′ and Z0 = L′ /C ′ , therefore, L′ =

Then, from Z0 =

From α =

p

β Z0 0.75 × 40 = = 38.2 nH/m. ω 2π × 125 × 106

L′ /C ′ , C′ =

√ R′ G′ and R′C ′ = L′ G′ , √ R = R′ G′ ′

r

38.2 nH/m L′ = 23.9 pF/m. = 402 Z02

√ R′ = R′ G′ G′

and G′ =

r

L′ = αZ0 = 0.02 Np/m × 40 Ω = 0.6 Ω/m C′

α 2 (0.02 Np/m)2 = 0.5 mS/m. = 0.8 Ω/m R′

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Problem 2.34

A 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 to render Zi = Z0 , thereby eliminating reflection back to the end. Show that the proposed approach is valid and find the value of the shunt resistance. Solution: (a) ZL − Z0 30 − j20 − 50 −20 − j 20 −(20 + j20) ◦ = = 0.34e− j121 . = = 80 − j20 80 − j20 ZL + Z0 30 − j20 + 50 1 + |Γ| 1 + 0.34 = = 2. S= 1 − |Γ| 1 − 0.34

Γ=

(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 λ λ dmax = + = + = 0.33λ . 2 180◦ 4π 4π 2 Applying (2.79) at d = dmax = 0.33λ , for which β l = (2π/λ ) × 0.33λ = 2.07 radians, the value of Zin before adding the

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

shunt resistance is: 

 ZL + jZ0 tan β l Zin = Z0 Z + jZL tan β l  0  (30 − j20) + j 50 tan 2.07 = 50 = (102 + j0) Ω. 50 + j(30 − j20) tan 2.07 Thus, at the location A (at a distance dmax from the load), the input impedance is purely real. If we add a shunt resistor R in parallel 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 that 1 1 1 + = R 102 50 or R = 98 Ω.

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Problem 2.45 The circuit shown in Fig. P2.45 consists of a 100-Ω lossless transmission line terminated in a load with ZL = (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 + j 100 = 0.62e j82.9 . = = ZL + Z0 50 + j100 + 100 150 + j100

The time average power dissipated in the load is: 1 Pav = |IeL |2 RL 2  2 1 VeL  =   RL 2 ZL  =

(b)

1 |VeL |2 1 50 RL = × 122 × 2 = 0.29 W. 2 | Z | 2 2 L 50 + 1002 P av = Pavi (1 − |Γ|2 )

Hence, Piav = (c)

Pav 0.29 = 0.47 W. = 1 − |Γ|2 1 − 0.622

r i Pav = −|Γ|2 Pav = −(0.62)2 × 0.47 = −0.18 W.

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Problem 2.48

Repeat Problem 2.47 using CD Module 2.6.

Solution:

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2010 Prentice Hall

Problem 2.64 to a 50-Ω line.

Use CD Module 2.7 to design a quarter-wavelength transformer to match a load with ZL = (100 − j200) Ω

Solution: Figure P2.64(a) displays the first solution of Module 2.7 where a λ /4 section of Z02 = 15.5015 Ω is inserted at distance 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 /4 transformer.

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

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Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 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 Ω and 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 with Vg = 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 point midway along the length of the line from t = 0 to t = 25 ns. Solution: 50 1 Rg − Z0 100 − 50 = = , = 100 + 50 150 3 Rg + Z0 −25 −1 ZL − Z0 25 − 50 = = . ΓL = = 25 + 50 75 3 ZL + Z0 Γg =

From Eq. (2.149b), V1+ =

60 × 50 Vg Z0 = 20 V. = 100 + 50 Rg + Z0

Also, T=

l l 3 = 5 ns. = = up 2c/3 2 × 3 × 108

The 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 Electromagnetics

c 2010 Prentice Hall

Figure P2.75: (b) Time response of voltage.

Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Ele...