Fundamentals of Applied Electromagnetics 7e by Fawwaz T. Ulaby and Umberto Ravaioli Solutions PDF

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Fundamentals of Applied Electromagnetics 7e by Fawwaz T. Ulaby and Umberto Ravaioli Exercise Solutions...

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Fundamentals of Applied Electromagnetics 7e by Fawwaz T. Ulaby and Umberto Ravaioli

Exercise Solutions

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

c 2015 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 and Umberto Ravaioli, Fundamentals of Applied Electromagnetics

c 2015 Prentice Hall

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

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

c 2015 Prentice Hall

Exercise 1.1 Consider the red wave shown in Fig. E1.1. What is the wave’s (a) amplitude, (b) wavelength, and (c) frequency, given that its phase velocity is 6 m/s? υ (volts) 6 4 2 0 −2 −4 −6

x (cm) 1

2

3

4

5

6

7

8

9 10

Figure E1.1

Solution: (a) A = 6 V. (b) λ = 4 cm. (c) f =

6 up = = 150 Hz. λ 4 × 10−2

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

c 2015 Prentice Hall

Exercise 1.2

The wave shown in red in Fig. E1.2 is given by υ = 5cos2πt/8. Of the following four equations:

(1) υ = 5cos(2πt/8 − π/4), (2) υ = 5cos(2πt/8 + π/4), (3) υ = −5cos(2πt/8 − π/4), (4) υ = 5sin 2πt/8, (a) which equation applies to the green wave? (b) which equation applies to the blue wave? υ(volts) 5

t (s)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 −5 Figure E1.2

Solution: (a) The green wave has an amplitude of 5 V and a period T = 8 s. Its peak occurs earlier than that of the red wave; hence, its constant phase angle is positive relative to that of the red wave. A full cycle of 8 s corresponds to 2π in phase. The green wave crosses the time axis 1 s sooner than the red wave. Hence, its phase angle is φ0 =

π 1 × 2π = . 4 8

Consequently, υ = 5cos(2πt/T + φ0 ) = 5cos(2πt/7 + π/4), which is given by #2. (b) The blue wave’s period T = 8 s. Its phase angle is delayed relative to the red wave by 2 s. Hence, the phase angle is negative and given by 2 π φ0 = − × 2π = − , 2 8 and   2πt π − υ = 5cos 8 2 = 5sin 2πt/8, which is given by #4.

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

c 2015 Prentice Hall

Exercise 1.3

The electric field of a traveling electromagnetic wave is given by E(z,t) = 10cos(π × 107t + πz/15 + π/6) (V/m).

Determine (a) the direction of wave propagation, (b) the wave frequency f , (c) its wavelength λ , and (d) its phase velocity up . Solution: (a) −z-direction because the signs of the coefficients of t and z are both positive. (b) From the given expression, ω = π × 107

(rad/s).

Hence, f=

ω π × 107 = 5 × 106 Hz = 5 MHz. = 2π 2π

(c) From the given expression,

π 2π = . λ 15

Hence λ = 30 m. (d) up = f λ = 5 × 106 × 30 = 1.5 × 108 m/s.

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

c 2015 Prentice Hall

Exercise 1.4 Consider the red wave shown in Fig. E1.4. What is the wave’s (a) amplitude (at x = 0), (b) wavelength, and (c) attenuation constant? υ (volts) 5

(2.8, 4.23) (8.4, 3.02) x (cm)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 −5 Figure E1.4

Solution: The wave shown in the figure exhibits a sinusoidal variation in x and its amplitude decreases as a function of x. Hence, it can be described by the general expression   2πx −α x + φ0 . υ = Ae cos λ From the given coordinates of the first two peaks, we deduce that λ = 8.4 − 2.8 = 5.6 cm. At x = 0, υ = −5 V and it occurs exactly λ /2 before the first peak. Hence, the wave amplitude is 5 V, and from −5 = 5cos(0 + φ0 ), it follows that φ0 = π. Consequently, υ = 5e−α x cos



 2πx +π . 5.6

In view of the relation cos x = − cos(x ± π), υ can be expressed as υ = −5e−α x cos

2πx 5.6

(V).

We can describe the amplitude as 5 V for a wave with a constant phase angle of π, or as −5 V with a phase angle of zero. At x = 2.8 cm,   2π × 2.8 −2.8α υ(x = 2.8) = 4.23 = −5e cos 5.6 = 5e−2.8α .

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

c 2015 Prentice Hall

Hence, e−2.8α = and α =−

4.23 , 5

  1 4.23 ln = 0.06 Np/cm. 2.8 5

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

c 2015 Prentice Hall

Exercise 1.5 The red wave shown in Fig. E1.5 is given by υ = 5cos4πx (V). What expression is applicable to (a) the blue wave and (b) the green wave? υ (volts) 5V 3.52 V 1.01 V

5

x (m)

0 0.25

0.5

0.75

1.0

1.25

−5 Figure E1.5

Solution: At x = 0, all three waves start at their peak value of 5 V. Also, λ = 0.5 m for all three waves. Hence, they share the general form 2πx λ = 5e−α x cos4πx (V).

υ = Ae−α x cos

For the red wave, α = 0. For the blue wave, 3.52 = 5e−0.5α

α = 0.7 Np/m.

1.01 = 5e−0.5α

α = 3.2 Np/m.

For the green wave,

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

c 2015 Prentice Hall

Exercise 1.6 An electromagnetic wave is propagating in the z-direction in a lossy medium with attenuation constant α = 0.5 Np/m. If the wave’s electric-field amplitude is 100 V/m at z = 0, how far can the wave travel before its amplitude will have been reduced to (a) 10 V/m, (b) 1 V/m, (c) 1 µ V/m? Solution: (a) 100e−α z = 10 100e−0.5z = 10 e−0.5z = 0.1 −0.5z = ln0.1 = −2.3 z = 4.6 m. (b) 100e−0.5z = 1 ln0.01 = 9.2 m. z= −0.5 (c) 100e−0.5z = 10−6 z=

ln10−8 = 37 m. −0.5

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

c 2015 Prentice Hall

Exercise 1.7

Express the following complex functions in polar form: z1 = (4 − j3)2 , z2 = (4 − j3)1/2 .

Solution: z1 = (4 − j3)2 i2 h −1 = (42 + 32 )1/2 ∠− tan 3/4 = [5∠−36.87◦ ]2 = 25∠−73.7◦ .

z2 = (4 − j3)1/2 i1/2 h −1 = (42 + 32 )1/2 ∠− j tan 3/4 √ = [5∠−36.87◦ ]1/2 = ± 5∠−18.4◦ .

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

c 2015 Prentice Hall

Exercise 1.8

Show that

√ 2 j = ±(1 + j).

Solution: e jπ/2 = 0 + j sin(π/2) = j p

√ 2 j = [2e jπ/2 ]1/2 = ± 2 e jπ/4 √ = ± 2(cosπ/4 + j sin π/4)   √ 1 1 =± 2 √ + j√ 2 2 = ±(1 + j ).

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

c 2015 Prentice Hall

Exercise 1.9 A series RL circuit is connected to a voltage source given by vs (t) = 150cos ωt (V). Find (a) the phasor current I˜ and (b) the instantaneous current i(t) for R = 400 Ω, L = 3 mH, and ω = 105 rad/s. Solution: (a) From Example 1–4, I˜ =

es V R + jω L

150 400 + j105 × 3 × 10−3 150 ◦ = = 0.3∠−36.9 400 + j300 =

(A).

(b) e jωt ] i(t) = Re[Ie

◦

5

= Re[0.3e− j36.9 e j10 t ]

= 0.3cos(105t − 36.9◦ ) (A).

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

c 2015 Prentice Hall

Exercise 1.10 Solution:

e = j5 V. Find v(t). A phasor voltage is given by V Ve = j5 = 5e jπ/2

v(t) = Re[Vee jωt ]

= Re[5e jπ/2 e jωt ]  π = −5sin ω = 5cos ωt + 2

(V).

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

c 2015 Prentice Hall

Chapter 2 Exercise Solutions Exercise 2.1 Exercise 2.2 Exercise 2.3 Exercise 2.4 Exercise 2.5 Exercise 2.6 Exercise 2.7 Exercise 2.8 Exercise 2.9 Exercise 2.10 Exercise 2.11 Exercise 2.12 Exercise 2.13 Exercise 2.14 Exercise 2.15 Exercise 2.16 Exercise 2.17

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

c 2015 Prentice Hall

Exercise 2.1 Use Table 2-1 to compute the line parameters of a two-wire air line whose wires are separated by a distance of 2 cm, and each is 1 mm in radius. The wires may be treated as perfect conductors with σc = ∞. Solution:

Two-wire air line: Because medium between wires is air, ε = ε0 , µ = µ0 and σ = 0. d = 2 cm, 1/2 π f µc Rs = =0 σc

a = 1 mm,



σc = ∞

R′ = 0 

  µ0 d ′  + L = ln 2a π

s 



d 2a

2



− 1

   s 2 20 20 ln  + = − 1 2 2 √ = 4 × 10−7 ln[10 + 99] = 1.2 (µ H/m). 

4π × 10−7 π

G′ = 0 C′ = ln



because σ = 0

 d  2a

π × 8.85 × 10−12 πε0 = √ = 9.29 q  ln[10 + 99] d 2 + − 1 2a

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

(pF/m).

c 2015 Prentice Hall

Exercise 2.2 Calculate the transmission line parameters at 1 MHz for a rigid coaxial air line with an inner conductor diameter of 0.6 cm and an outer conductor diameter of 1.2 cm. The conductors are made of copper [see Appendix B for µc and σc of copper]. Solution:

Coaxial air line: Because medium between wires is air, ε = ε0 , µ = µ0 and σ = 0. a = 0.3 cm, Rs =

p

b = 0.6 cm,

µc = µ0 ,

σc = 5.8 × 107 S/m

π f µc /σc

= [π × 106 × 4π × 10−7 /(5.8 × 107 )]1/2 = 2.6 × 10−4 Ω. Rs R = 2π ′

L′ =

1 1 + a b



2.6 × 10−4 = 2π



1 1 + 3 × 10−3 6 × 10−3

  µ0 4π × 10−7 b ln2 = 0.14 ln = 2π 2π a

G′ = 0 C′ =





= 2.08 × 10−2

(Ω/m)

( µ H/m)

because σ = 0

2πε 2π × 8.85 × 10−12 = 80.3 = ln2 ln(b/a)

(pF/m).

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

c 2015 Prentice Hall

Exercise 2.3

Verify that Eq. (2.26a) is indeed a solution of the wave equation given by Eq. (2.21).

Solution: Ve(z) = V0+ e−γz +V0− eγz d 2 Ve(z) ? − γ 2 Ve(z) = 0 dz2

d 2 + −γz ? (V e +V0− eγz ) − γ 2 (V0+ e−γz +V0− eγz ) = 0 dz2 0 γ 2V0+e−γz + γ 2V0− eγz − γ 2V0+ e−γz − γ 2V0− eγz = 0.

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

c 2015 Prentice Hall

Exercise 2.4 A two-wire air line has the following line parameters: R′ = 0.404 (mΩ/m), L′ = 2.0 (µH/m), G′ = 0, and ′ C = 5.56 (pF/m). For operation at 5 kHz, determine (a) the attenuation constant α, (b) the phase constant β , (c) the phase velocity up , and (d) the characteristic impedance Z0 . Solution:

Given: R′ = 0.404 (mΩ/m),

G′ = 0,

L′ = 2.0 (µH/m),

C′ = 5.56 (pF/m).

(a) o n α = Re [(R′ + jω L′ )(G′ + jωC′ )]1/2 o n = Re [(0.404 × 10−3 + j2π × 5 × 103 × 2 × 10−6 )(0 + j2π × 5 × 103 × 5.56 × 10−12)]1/2 = Re[3.37 × 10−7 + j1.05 × 10−4 ]

α = 3.37 × 10−7

(Np/m).

(b) From part (a), o n β = Im [(R′ + jω L′ )(G′ + jωC′ )]1/2 = 1.05 × 10−4

(rad/m).

(c) up =

ω 2π × 5 × 103 = 3 × 108 = 1.05 × 10−4 β

(m/s).

(d) R′ + jω L′ α + jβ 0.404 × 10−3 + j5 × 103 × 2 × 10−6 = 3.37 × 10−7 + j 1.05 × 10−4

Z0 =

= (600 − j2) Ω.

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

c 2015 Prentice Hall

Exercise 2.5

For a lossless transmission line, λ = 20.7 cm at 1 GHz. Find εr of the insulating material.

Solution: λ0 λ=√ εr 2  2  2  c λ0 3 × 108 = 2.1. = εr = = λ fλ 1 × 108 × 20.7 × 10−2

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

c 2015 Prentice Hall

Exercise 2.6 A lossless transmission line uses a dielectric insulating material with εr = 4. If its line capacitance is C ′ = 10 (pF/m), find (a) the phase velocity up , (b) the line inductance L′ , and (c) the characteristic impedance Z0 . Solution: (a) 3 × 108 c up = √ = √ = 1.5 × 108 m/s. εr 4 (b) 1 up = √ , L′C′ L′ =

u2p =

1 . L′C′

1 1 = = 4.45 u2pC′ (1.5 × 108 )2 × 10 × 10−12

(c) Z0 =

r

L′ = C′



4.45 × 10−6 10 × 10−12

1/2

(µH/m).

= 667.1 Ω.

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

c 2015 Prentice Hall

Exercise 2.7 A 50-Ω lossless transmission line is terminated in a load impedance ZL = (30 − j200) Ω. Calculate the voltage reflection coefficient at the load. Solution: ZL − Z0 ZL + Z0 −20 − j 200 30 − j200 − 50 ◦ = = 0.93∠−27.5 . = (30 − j200) + 50 80 − j200

Γ=

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

c 2015 Prentice Hall

Exercise 2.8

A 150-Ω lossless line is terminated in a capacitor whose impedance is ZL = − j30 Ω. Calculate Γ.

Solution: ZL − Z0 ZL + Z0 − j30 − 150 ◦ = 1∠−157.4 . = − j30 + 150

Γ=

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

c 2015 Prentice Hall

Exercise 2.9 Use Module 2.4 to generate the voltage and current standing-wave patterns for a 50-Ω line of length 1.5λ , terminated in an inductance with ZL = j140 Ω. Solution:

Standing-wave patterns generated with the help of Module 2.4 are shown.

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

c 2015 Prentice Hall

Exercise 2.10 load.

◦

If Γ = 0.5∠−60 and λ = 24 cm, find the locations of the voltage maximum and minimum nearest to the

Solution: ◦

Γ = 0.5∠−60 ,

λ = 24 cm

θr λ λ + (because θr is negative) 4π 2   (−π/3) × 24 24 cm = [−2 + 12] cm = 10 cm. + = 2 4π

lmax =

λ (because lmax > λ /4) lmin = lmax − 4   24 cm = 4 cm. = 10 − 4

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

c 2015 Prentice Hall

Exercise 2.11 A 140-Ω lossless line is terminated in a load impedance ZL = (280 + j182) Ω. If λ = 72 cm, find (a) the reflection coefficient Γ, (b) the voltage standing-wave ratio S, (c) the locations of voltage maxima, and (d) the locations of voltage minima. Solution: Z0 = 140 Ω,

ZL = (280 + j182) Ω

(a) ZL − Z0 ZL + Z0 280 + j182 − 140 140 + j 182 ◦ = = = 0.5∠29 . 280 + j182 + 140 420 + j182

Γ=

(b) S= (c)

1 + |Γ| 1 + 0.5 1.5 = 3. = = 1 − |Γ| 1 − 0.5 0.5

θr λ nλ + , n = 0, 1, 2,. .. 4π 2 (29π/180) × 0.72 n × 0.72 + = 2 4π = (2.9 + 36n) (cm), n = 0, 1, 2,. ..

lmax =

(d) λ lmin = lmax + 4   72 = (2.9 + 36n) + cm 4 = (20.9 + 36n) cm, n = 0, 1, 2,. ..

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

c 2015 Prentice Hall

Exercise 2.12 A 50-Ω lossless transmission line uses an insulating material with εr = 2.25. When terminated in an open circuit, how long should the line be for its input impedance to be equivalent to a 10-pF capacitor at 50 MHz? Solution:

For a 10-pF capacitor at 50 MHz, 1000 1 −j = −j Ω = 6 −12 2 π × 50 × 10 × 10 × 10 π jωC √ √ 2π 2π εr 2π f εr = β= = c λ0 λ √ 2π × 5 × 107 2.25 = 1.57 (rad/m). = 3 × 108 Zc =

For lossless lines with open-circuit termination, Zin = − jZ0 cotβ l = − j50 cot1.57l Hence, −j

1000 = − j50cot 1.57l π

or l = 9.92

(cm).

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

c 2015 Prentice Hall

Exercise 2.13 A 300-Ω feedline is to be connected to a 3-m long, 150-Ω line terminated in a 150-Ω resistor. Both lines are lossless and use air as the insulating material, and the operating frequency is 50 MHz. Determine (a) the input impedance of the 3-m long line, (b) the voltage standing-wave ratio on the feedline, and (c) the characteristic impedance of a quarter-wave transformer were it to be used between the two lines in order to achieve S = 1 on the feedline. Solution:

At 50 MHz, λ = λ0 =

(a)

c 3 × 108 = 6 m. = 5 × 107 f

3 l = = 0.5. λ 6 Hence, Zin = ZL = 150 Ω. (Zin = ZL if Z = nλ /2.) (b) Zin − Z0 150 − 300 −150 1 = = =− . 3 Zin + Z0 150 + 300 450 4/3 1 + 31 1 + |Γ| = 2. = S= = 2/3 1 − |Γ| 1 − 31

Γ=

(c) 2 = Z1 Z3 = 300 × 150 = 45, 000 Z02

Z02 = 212.1 Ω. where Z1 is the feedline and Z3 is Zin of part (a).

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

c 2015 Prentice Hall

Exercise 2.14 For a 50-Ω lossless transmission line terminated in a load impedance ZL = (100 + j50) Ω, determine the fraction of the average incident power reflected by the load. Solution: ZL − Z0 ZL + Z0 50 + j 50 100 + j50 − 50 = = 0.45∠26.6◦ . = 100 + j50 + 50 150 + j50

Γ=

Fraction of reflected power = |Γ|2 = (0.45)2 = 20%.

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

c 2015 Prentice Hall

Exercise 2.15

For the line of Exercise 2.14, what is the magnitude of the average reflected power if |V0 +| = 1 V?

Solution: r Pav = |Γ|2

|V0+ |2 0.2 × 1 =2 = 2Z0 2 × 50

(mW).

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

c 2015 Prentice Hall

Exercise 2.16 Use the Smith chart to find the values of Γ corresponding to the following normalized load impedances: (a) zL = 2 + j0, (b) zL = 1 − j1, (c) zL = 0.5 − j2, (d) zL = − j3, (e) zL = 0 (short circuit), (f) zL = ∞ (open circuit), (g) zL = 1 (matched load). Solution: 0.12

0.1 1

6

1.0

1 .4

1 .6

(+ jX

/Z

2 .0

0 .5

4 0.4

0

5

14

0 .0

5 0 .4

C OM PO NE NT

4

15

6

0

0 .3

1 .0

UCT IVE R

0 .2 8 0 2

5 .0

0 .2

IND

0.25 0.24 0.26 0.27 0.25 0.24 0.26 0.23 EC TION CO EFF IC IEN 0.27 T IN DEGR EE NG LE OF REFL S

0.23

A

0 .6

10 0.1

0.49

0 .4

17 0

0 .2 2

EA CT A N CE

4 .0

1 .0

0.8

0 .0

0.4

0 .8

9

0 .4

3 .0
...