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Introduction
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SI units: m, kg, s, A, V, , K, …
Conversion factors:
 1” = 2.54 cm
 1 lb. = 0.454 kg
 1 gallon = 3.785 liter
Prefixes

T
G
M
k
base m

n
p
f
1012
109
106
103
1
10-3 10-6 10-9 10-12 10-15
Notations:
 Scalars: a, A, …
 Vectors: a, A, …
 Unit vector: x̂ , ŷ , …
~ ~
 Phasors: E , H , …
Fundamental forces:
 Nuclear force (strongest)
 EM force (strong)***
 Weak-interaction force (weak)
 Gravitational force (weakest)
Electric field
 Fe = Electrical force: The source of electrical force is electric charge.
 Elementary charge e = 1.6  10-19 (C)
 Coulomb’s law:
The magnitude of the force (Fe21) on q2 due to q1 is given by:
q1q2
F
.
40 R122
The direction of the force points from q1 to q2.
  is called the permittivity and 0 = 8.854  10-12 F/m is for free space.
 If q1 and q2 are like charges, the resultant force will try to push q2 away from
q1. Otherwise, the resultant force will try to pull q2 to q1.
 If a system of electric charges is placed in space, it will exert a force to any
surrounding charges. Since this force depends on the magnitude and polarity
of the surrounding charges, the concept of E-field, which equals to the force
applied on a unit charge, is used to describe the electrical properties of the
system of charges.
 The E-field for a point charge in free space is given by:
q
E  R̂
(V/m)
40 R 2
 The direction of the E-field points away from the point charge (i.e. toward the
point charge if q is negative).
 Two important properties of electric charges: Conservation and superposition.
 The E-field in a material composed of atoms is smaller because a fraction of
the force is needed to align (polarize) the atoms.
 Permittivity  is used to describe the material effect on E-field.
The relative permittivity or dielectric constant r =/0 is often used: A
material with r = 10 reduces the E-field by 10 times.
 r for free space (vacuum) = 1.
 Electric flux density: D =E (C/m2)
 D is material independent.
Magnetic field
 Fm = magnetic force: The sources of magnetic force are electric current or
magnetic poles.
 Magnetic poles cannot be separated (not yet).
 Biot-Savart law:The magnetic flux density induced by a current I flowing in
the z-direction is given by:
 I
B  ˆ 0
(T)
2r
  is called the permeability and 0 = 4  10-7 H/m is for free space.
 Permittivity  is used to describe the material effect.
 The relative permeability: r =/0
 r for free space (vacuum) = 1.
 Magnetic field intensity: B =H
 Magnetic field is intensified in materials with high relative permeability.
Static fields
 Q  E and I  H
 Since I = dQ/dt, E and H are independent of each other.
 Electrostatics: q/t = 0
 Magnetostatics: I/t = 0
Dynamic fields
 Time-varying
 Both E- and H- fields are present and related to each other.
Traveling waves
 Carries energy
 Speed  c = 3  108 m/s for EM waves 330 m/s for sound waves
 Linear wave: EM and sound waves; nonlinear wave: fluid
 Transient and continuous harmonic waves (sinusoidal)
 1-D (transmission lines), 2-D, 3-D waves
 Plane waves, cylindrical waves, spherical waves
Waves in medium
 2t 2x


 0 
 Lossless medium: y ( x, t )  A cos

 T

A: amplitude of the wave
T: time period of the wave
: wavelength of the wave
0: reference of the wave
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Phase velocity: The speed of the wave measured at a fixed phase.
up =  / T = f 
Phase constant: The amount of phase shift (in radian) per meter. Hence:
 = 2/  = 2f/ up) =  / up
 

x 
 y ( x, t )  A cost  x   0   A cos   t     0 
  u p 

Direction of propagation: The coefficients of t and x have opposite signs
indicate that the wave is traveling in the + x direction.
The coefficients of t and x have the same signs indicate that the wave is
traveling in the - x direction.
 2t 2x


 0 
Lossy medium: y ( x, t )  Ae x cos

 T

The coefficient  is called the attenuation factor with a unit of Np/m (Np is
dimensionless).
A more practical unit is dB/m = 8.686 
dB (power ratio)
3 dB loss = 50% left, 10 dB loss = 10 % left, …
–
1x
0.5 x
0.25 x
0.125 x
0.1 x
0.01 x
0.001 x


dBm (power unit)
+
1 mW
0.5 mW
0.25 mW
0.125 mW
0.1 mW
0.01 mW
0.001 mW
dB
0
3
6
9
10
20
30
+
1x
2x
4x
8x
10 x
100 x
1000 x
dBm
0
3
6
9
10
20
30
1 mW
2 mW
4 mW
8 mW
10 mW
100 mW
1000 mW
 Other dB units include dBW, dB, …
The EM spectrum
 Opacity: Atmosphere opaque and ionosphere opaque
 Windows: optical, IR, and RF.
 -ray, X-ray, UV, visible, IR, and RF.
 Radio bands
 -wave: 300 MHz to 300 GHz; mm-wave: 30 to 300 GHz
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Complex mathematics
 j =  -1
 z = x + jy = |z| ej = |z| 
 Euler’s identity: ej = cos + j sin  z = |z| ej = |z| cos + j |z| sin
 Re{z} = x = |z| cos , Im{z} = y = |z| sin ; |z| = x2 + y2 ,  = tan(y/x)
 Complex conjugate: z* = x – jy = |z| e-j = |z| - ; |z| = z* z
Operations
 Useful relations:
1  j 0  e j 0  e j 2  10  1360
j  1  e
j  e



j
j


4

2
 190
 (1  j )
2
j 2  1  e j  1180
 j  e
j

4

 (1  j )
2
Equality:
z1  z2  Re{z1}  Re{ z2 } and Im{z1}  Im{z2 } or z1  z2 and 1   2
Add/subtract:
z1  z2  (Re{ z1}  Re{z2 })  j (Im{z1}  Im{z2 })
Multiply/divide:
z1  z 2  z1  z 2 e j ( 1 2)  z1  z 2 cos( 1   2 )  j sin(  1   2 )

z
z
z1
 1 e j ( 1 2)  1 cos( 1   2 )  j sin(  1   2 )
z2 z2
z2
Powers and roots:
z n  z1 e jn  z1 cos(n )  j sin( n )
n

  
  
  z cos   j sin  
 2 
 2
Phasors: A shortcut for solving linear differential equations (DE) with sinusoidal
excitations.
 Because of the unique property of the exponential function, deax/dt= aeax, DEs
can be transformed into ordinary algebraic equations in the phasor-domain.
 Exponential (phasor) representation of a sinusoidal signal
z   ze

n
j
2
A cos t  Ae  A A sin t  Ae
j0
A cost  0   Ae
j0

2
  jA
A sin t  0   Ae
A cos t   x  0   Ae j (  x 0 )
dz
Zˆ
 jZˆ  z  dt 
dt
j
d
cos t  0   je j0
dt
j


 j  0  
2

Ae x cos t   x  0   Ae x e j (  x 0 )
 sin  t     dt 
0
e


j  0  
2

j

A more practical example: Use phasor method to find the total current of the
following circuit:
f  100KHz
v  10e
ZL  2f  L e
v
IR 
R
j 90 deg
IR  10 mA
IT  IR  IL  IC
ZT 
R
IR

1
ZL

L  1mH
V
ZL  0.628i k
v
IL 
ZL
ZC 
C  1000pF
1
2f  C
IL  15.915i mA
IT  10  9.632i mA
1
1
j 0deg
1
e
 j 90deg
IC 
R  1K
ZC  1.592i k
v
ZC
IC  6.283i mA
arg  IT   43.927 deg
ZT  0.519  0.5i k
IT 
v
ZT
IT  10  9.632i mA
ZC
+ IL
+ IC
= IT
IR
IC
IL
Itotal(t) = 13.885 cos(t + 0) mA
= 2f = 6.28310+5 rad/sec, 0 = -43.927
IT  13.885 mA
IT

Another practical example: Use phasor method to find the output voltage of the
following circuit:
f  100KHz
ZL  2f  L e
v  10e
j 90 deg
1
Zout 
1
ZL
Vout  v 

Zout
ZT
1
j 0deg
L  1mH
V
ZL  0.628i k
ZC 
Zout  1.038i k
C  1000pF
1
2f  C
e
R  1K
 j 90deg
ZC  1.592i k
ZT  R  Zout
ZT  1  1.038i K
ZC
Vout  5.187  4.996i V
arg  Vout  43.927 deg
Vout  7.202 V
Vout
Vout(t) = 7.202 cos(t + 0) V
V1
= 2f = 6.28310+5 rad, 0 = 43.927
6
1.2 10 s
1
 360deg  43.2 deg
vp
v p  14.1383V
v rms 
Vout_p  10.2760V
Vout_rms 
v rms  9.997 V
2
f
Vout_p
2
Vout_rms  7.266 V