Exercices Solutions Fundamentals of Applied Electromagnetics 7th edition By Fawwaz T.Ulaby,Umberto Ravaioli
Fundamentals of Applied Electromagnetics
7e By
Fawwaz T. Ulaby and Umberto Ravaioli
Exercise Solutions
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
,Exercices Solutions Fundamentals of Applied Electromagnetics 7th edition By Fawwaz T.Ulaby,Umberto Ravaioli
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 x (cm)
2 1 2 3 4 5 6 7 8 9 10
4
6
Figure E1.1
Solution:
(a) A = 6 V.
(b) l = 4 cm.
(c) f = = 2 = 150 Hz.
l 4 10
,Exercices Solutions Fundamentals of Applied Electromagnetics 7th edition By Fawwaz T.Ulaby,Umberto Ravaioli
Exercise 1.2 The wave shown in red in Fig. E1.2 is given by u = 5 cos 2pt=8. Of the following four equations:
(1) u = 5 cos(2pt=8 p=4),
(2) u = 5 cos(2pt=8 + p =4),
(3) u = 5 cos(2pt=8 p=4),
(4) u = 5 sin 2pt=8,
(a) which equation applies to the green wave? (b) which equation applies to the blue wave?
υ (volts)
5
0 t (s)
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 2p in phase. The green
wave crosses the time axis 1 s sooner than the red wave. Hence, its phase angle is
f0 = 2p = :
8 4
Consequently,
u = 5 cos(2pt=T + f 0 )
= 5 cos(2pt=7 + p =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
f0 = 2p = ;
8 2
and
2pt p
u = 5 cos
= 5 sin 2pt=8;
which is given by #4.
, Exercices Solutions Fundamentals of Applied Electromagnetics 7th edition By Fawwaz T.Ulaby,Umberto Ravaioli
Exercise 1.3 The electric field of a traveling electromagnetic wave is given by
E (z; t ) = 10 cos(p 107t + p z=15 + p =6) (V/m):
Determine (a) the direction of wave propagation, (b) the wave frequency f , (c) its wavelength l , 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,
w = p 107 (rad/s):
Hence,
w p 107
f= = = 5 106 Hz = 5 MHz:
(c) From the given expression,
2p p
=
l 15
Hence l = 30 m.
(d) up = f l = 5 106 30 = 1:5 108 m/s.
Fundamentals of Applied Electromagnetics
7e By
Fawwaz T. Ulaby and Umberto Ravaioli
Exercise Solutions
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
,Exercices Solutions Fundamentals of Applied Electromagnetics 7th edition By Fawwaz T.Ulaby,Umberto Ravaioli
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 x (cm)
2 1 2 3 4 5 6 7 8 9 10
4
6
Figure E1.1
Solution:
(a) A = 6 V.
(b) l = 4 cm.
(c) f = = 2 = 150 Hz.
l 4 10
,Exercices Solutions Fundamentals of Applied Electromagnetics 7th edition By Fawwaz T.Ulaby,Umberto Ravaioli
Exercise 1.2 The wave shown in red in Fig. E1.2 is given by u = 5 cos 2pt=8. Of the following four equations:
(1) u = 5 cos(2pt=8 p=4),
(2) u = 5 cos(2pt=8 + p =4),
(3) u = 5 cos(2pt=8 p=4),
(4) u = 5 sin 2pt=8,
(a) which equation applies to the green wave? (b) which equation applies to the blue wave?
υ (volts)
5
0 t (s)
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 2p in phase. The green
wave crosses the time axis 1 s sooner than the red wave. Hence, its phase angle is
f0 = 2p = :
8 4
Consequently,
u = 5 cos(2pt=T + f 0 )
= 5 cos(2pt=7 + p =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
f0 = 2p = ;
8 2
and
2pt p
u = 5 cos
= 5 sin 2pt=8;
which is given by #4.
, Exercices Solutions Fundamentals of Applied Electromagnetics 7th edition By Fawwaz T.Ulaby,Umberto Ravaioli
Exercise 1.3 The electric field of a traveling electromagnetic wave is given by
E (z; t ) = 10 cos(p 107t + p z=15 + p =6) (V/m):
Determine (a) the direction of wave propagation, (b) the wave frequency f , (c) its wavelength l , 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,
w = p 107 (rad/s):
Hence,
w p 107
f= = = 5 106 Hz = 5 MHz:
(c) From the given expression,
2p p
=
l 15
Hence l = 30 m.
(d) up = f l = 5 106 30 = 1:5 108 m/s.
