Ulaby应用电磁学基础答案01

Chapter 1: Introduction: Waves and Phasors

Lesson #1

Chapter — Section: Chapter 1

Topics: EM history and how it relates to other fields

Highlights:

?EM in Classical era: 1000 BC to 1900

?Examples of Modern Era Technology timelines

?Concept of “fields” (gravitational, electric, magnetic)

?Static vs. dynamic fields

?The EM Spectrum

Special Illustrations:

?Timelines from CD-ROM

Timeline for Electromagnetics in the Classical Era

ca. 900 Legend has it that while walking

BC across a field in northern Greece, a

shepherd named Magnus experiences

a pull on the iron nails in his sandals

by the black rock he was standing on.

The region was later named Magnesia

and the rock became known as

magnetite [a form of iron with

permanent magnetism].

ca. 600 Greek philosopher Thales

BC describes how amber,

after being rubbed

with cat fur, can pick

up feathers [static

electricity].

ca. 1000 Magnetic compass used as

a navigational device. 1752 Benjamin Franklin

(American) invents the

lightning rod and

demonstrates that

lightning is electricity.

1785Charles-Augustin de

Coulomb (French) demonstrates that

the electrical force between charges is

proportional to the inverse of the

square of the distance between them. 1800 Alessandro Volta

(Italian) develops the

first electric battery.

1820 Hans Christian Oersted

(Danish) demonstrates the

interconnection between

electricity and magnetism

through his discovery that an electric

current in a wire causes a compass

needle to orient itself perpendicular to

the wire.

2

Lessons #2 and 3

Chapter — Sections: 1-1 to 1-6 Topics: Waves

Highlights:

?Wave properties

?Complex numbers

?Phasors

Special Illustrations:

?CD-ROM Modules 1.1-1.9

?CD-ROM Demos 1.1-1.3

CHAPTER13 Chapter1

Section1-3:Traveling Waves

Problem1.1A2-kHz sound wave traveling in the x-direction in air was observed to have a differential pressure p x t10N/m2at x0and t50μs.If the reference phase of p x t is36,?nd a complete expression for p x t.The velocity of sound in air is330m/s.

Solution:The general form is given by Eq.(1.17),

p x t A cos 2πt

λ

φ0

where it is given thatφ036.From Eq.(1.26),T1f1210305ms. From Eq.(1.27),

λu p

2103

0165m

Also,since

p x0t50μs10(N/m2)A cos 2π50106

180

A cos126rad031A

it follows that A100313236N/m2.So,with t in(s)and x in(m),

p x t3236cos2π106t

165

36(N/m2)

3236cos4π103t1212πx36(N/m2)

4

CHAPTER 1

0.000.250.500.751.001.251.501.752.002.252.502.753.00

-12.

-10.

-8.-6.-4.-2.0.2.4.6.

8.10.12.A m p l i t u d e (N /m 2)

Distance x (m)p(x,t=0)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

-12.

-10.-8.-6.-4.-2.0.2.4.6.

8.10.12.A m p l i t u d e (N /m 2)

Time t (ms)

p(x=0,t)

(a)(b)

Figure P1.2:(a)Pressure wave as a function of distance at t 0and (b)pressure

wave as a function of time at x 0.

603Hz u p

300cm

f

03

Problem 1.4Two waves,y 1t and y 2t ,have identical amplitudes and oscillate at the same frequency,but y 2t leads y 1t by a phase angle of 60.If

y 1t

4cos 2π

103t

write down the expression appropriate for y 2t and plot both functions over the time span from 0to 2ms.Solution:

y 2t

4cos 2π

103t

60

CHAPTER 1

5

z

Figure P1.4:Plots of y 1t and y 2t .

λ

06rad/m

u p

ω

06

83m/s

λ

2π

06

1047m

6CHAPTER1

At t2s,y x215sin106x(m),with the argument of the cosine function given in radians.Plot is shown in Fig.P1.5.

CHAPTER 1

7

x

x = 0

y

Figure P1.6:Wave on a string tied to a wall at x

0(Problem 1.6).

the two waves must have the same phase constant β.Hence,with its direction being in the negative x -direction,y 2x t is given by the general form

y 2x t

B cos ωt

βx

φ0

(1)

where B and φ0are yet-to-be-determined constants.The total displacement is

y s x t

y 1x t

y 2x t

A cos ωt

βx

B cos ωt

βx

φ0

Since the string cannot move at x 0,the point at which it is attached to the wall,

y s 0t 0for all t .Thus,

y s 0t

A cos ωt

B cos ωt

φ0

(2)

(i)Easy Solution:The physics of the problem suggests that a possible solution for (2)is B A and φ00,in which case we have

y 2x t

A cos ωt

βx

(3)

(ii)Rigorous Solution:By expanding the second term in (2),we have

A cos ωt

B cos ωt cos φ0

sin ωt sin φ0

or

A

B cos φ0cos ωt

B sin φ0sin ωt

(4)

This equation has to be satis?ed for all values of t .At t

0,it gives

A

B cos φ0

(5)

8CHAPTER1 and atωtπ2,(4)gives

B sinφ00(6) Equations(5)and(6)can be satis?ed simultaneously only if

A B0(7) or

A B andφ00(8) Clearly(7)is not an acceptable solution because it means that y1x t0,which is contrary to the statement of the problem.The solution given by(8)leads to(3). (b)Atωtπ4,

y1x t A cosπ4βx A cos π

λ

y2x t A cosωtβx A cos π

λ

Plots of y1,y2,and y3are shown in Fig.P1.6(b).

ωt=π/4

Figure P1.6:(b)Plots of y1,y2,and y s versus x atωtπ4. Atωtπ2,

y1x t A cosπ2βx A sinβx A sin

2πx

CHAPTER19

2πx

y2x t A cosπ2βx A sinβx A sin

Problem1.7Two waves on a string are given by the following functions:

y1x t4cos20t30x(cm)

y2x t4cos20t30x(cm)

where x is in centimeters.The waves are said to interfere constructively when their superposition y s y1y2is a maximum and they interfere destructively when y s is a minimum.

(a)What are the directions of propagation of waves y1x t and y2x t?

(b)At tπ50s,at what location x do the two waves interfere constructively,

and what is the corresponding value of y s?

(c)At tπ50s,at what location x do the two waves interfere destructively,

and what is the corresponding value of y s?

Solution:

(a)y1x t is traveling in positive x-direction.y2x t is traveling in negative x-direction.

10CHAPTER1 (b)At tπ50s,y s y1y24cos04π30x cos04πhttps://www.360docs.net/doc/6f12607164.html,ing the formulas from Appendix C,

2sin x sin y cos x y cos x y

we have

y s8sin04πsin30x761sin30x

Hence,

y s max761

and it occurs when sin30x1,or30x π

60

2nπ

30

cm.

3

xφ0(cm)

with x in meters.

(a)y00040cosφ0.Hence,φ0π2,and

y x t40cos20πt 20π

2

40sin20πt20π

3

x(cm),ifφ0π2

(b)At x75cm=75102m,y040cosπ2φ0.Hence,φ00orπ, and

y x t40cos 20πt20π

3

x(cm),ifφ0π

CHAPTER111

2025s u p2

8

Problem1.10The vertical displacement of a string is given by the harmonic function:

y x t6cos16πt20πx(m)

where x is the horizontal distance along the string in meters.Suppose a tiny particle were to be attached to the string at x5cm,obtain an expression for the vertical velocity of the particle as a function of time.

Solution:

y x t6cos16πt20πx(m)

u005t

dy x t

Problem1.11Given two waves characterized by

y1t3cosωt

y2t3sinωt36

does y2t lead or lag y1t,and by what phase angle?

Solution:We need to express y2t in terms of a cosine function:

y2t3sinωt36

3cosπ

12CHAPTER1 Hence,y2t lags y1t by54.

2m ln

1

Problem1.13A certain electromagnetic wave traveling in sea water was observed to have an amplitude of98.02(V/m)at a depth of10m and an amplitude of81.87 (V/m)at a depth of100m.What is the attenuation constant of sea water? Solution:The amplitude has the form Aeαz.At z10m,

Ae10α9802

and at z100m,

Ae100α8187

The ratio gives

e10α

8187120

or

e10α12e100αTaking the natural log of both sides gives

ln e10αln12e100α

10αln12100α

90αln12018

Hence,

α

018

CHAPTER113

Section1-5:Complex Numbers

Problem1.14Evaluate each of the following complex numbers and express the result in rectangular form:

(a)z14e jπ3,

(b)z2

3e j3π4

4j sin

3π

2e jπ43

23cos3π4j sin3π4

2j221j

(g)

z71j12

Problem1.15Complex numbers z1and z2are given by

z13j2

z24j3

14CHAPTER1

(a)Express z1and z2in polar form.

(b)Find z1by applying Eq.(1.41)and again by applying Eq.(1.43).

(c)Determine the product z1z2in polar form.

(d)Determine the ratio z1z2in polar form.

(e)Determine z31in polar form.

Solution:

(a)Using Eq.(1.41),

z13j236e j337

z24j35e j1431

(b)By Eq.(1.41)and Eq.(1.43),respectively,

z13j213360

z113360

(c)By applying Eq.(1.47b)to the results of part(a),

z1z236e j3375e j143118e j1094

(d)By applying Eq.(1.48b)to the results of part(a),

z1

5e j1431

072e j1768

(e)By applying Eq.(1.49)to the results of part(a),

z3136e j3373

363e j33374666e j1011

CHAPTER115

(a)

1

2j41447e j116614471e j1166022e j1166 2j4

(b)z32j43447e j116634473e j35008944e j10.

(c)z2z z2j42j441620.

(d)z2j4 4.

(e)z2j444e jπ.

,z2330

,z23150

330

3150

Problem1.18Complex numbers z1and z2are given by

z1560

16CHAPTER1

(a)Determine the product z1z2in polar form.

(b)Determine the product z1z2in polar form.

(c)Determine the ratio z1z2in polar form.

(d)Determine the ratio z1z2in polar form.

(e)Determine

5e j60

z2

z1

z2

z15e j30.

5

3252583θtan1

176j103

180

18022918

π

Hence,e z2008cos22918j sin229181313j1520.

CHAPTER117

Section1-6:Phasors

Problem1.21A voltage source given by v s t25cos2π103t30(V)is connected to a series RC load as shown in Fig.1-19.If R1M?and C200pF, obtain an expression for v c t,the voltage across the capacitor.

Solution:In the phasor domain,the circuit is a voltage divider,and

V c V s 1jωC

1jωRC

Now V s25e j30V withω2π103rad/s,so

V c

25e j30V

1j2π5

1557e j815V.

Converting back to an instantaneous value,

v c t V c e jωt1557e jωt815V1557cos2π103t815V where t is expressed in seconds.

18CHAPTER1 (d)

i t2cosωt3π4

I2e j3π42e jπe j3π42e jπ4A

(e)

i t4sinωtπ33cosωtπ6

4cosπ2ωtπ33cosωtπ6

4cosωtπ63cosωtπ6

4cosωtπ63cosωtπ67cosωtπ6

I7e jπ6A

CHAPTER119 (e)

I j e jπ2

i t e jπ2e jωt cosωtπ2sinωt A

(f)

I2e jπ6

i t2e jπ6e jωt2cosωtπ6A

dt

1

jωC

(c)?I V s

R jωL1ωC

ωCV0e jπ3

Problem1.25A wave traveling along a string is given by

y x t2sin4πt10πx(cm)

20CHAPTER1 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 phaseφ0,(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 t2cos4πt10πx

Problem1.26A laser beam traveling through fog was observed to have an intensity of1(μW/m2)at a distance of2m from the laser gun and an intensity of0.2 (μW/m2)at a distance of3m.Given that the intensity of an electromagnetic wave is proportional to the square of its electric-?eld amplitude,?nd the attenuation constantαof fog.

Solution:If the electric?eld is of the form

E x t E0eαx cosωtβx

then the intensity must have a form

I x t E0eαx cosωtβx2

E20e2αx cos2ωtβx

or

I x t I0e2αx cos2ωtβx

where we de?ne I0E20.We observe that the magnitude of the intensity varies as I0e2αx.Hence,

at x2m I0e4α1106(W/m2)

at x3m I0e6α02106(W/m2)