Electromagnetic waves
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ELECTROMAGNETIC WAVES
- Bu sahifa navigatsiya:
- Uniform Plane-wave Propagation
- Uniform Plane Wave Solutions in the Time Domain (Cont’d)
- Uniform Plane Wave Solutions in the Time Domain (Cont’d)
- The Wave Equations for a Conducting Medium
- The wave equation for a conducting medium
- Sinusoidal Time Variations
- Relationship between time-varying and phasor quantities
- Differential Integral
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ELECTROMAGNETIC WAVES ELECTROMAGNETIC WAVES • In the solution of any electromagnetic problem the fundamental relations that must be satisfied are the four field equations 𝛁𝛁 × 𝑯𝑯 = ̇𝑫𝑫 + 𝑱𝑱 𝛁𝛁 × 𝑬𝑬 = − ̇𝑩𝑩 𝛁𝛁 � 𝑫𝑫 = 𝝆𝝆 𝛁𝛁 � 𝑩𝑩 = 𝟎𝟎 𝛻𝛻 � 𝐽𝐽 = − ̇𝜌𝜌 • in which the dot superscript indicates partial differentiation with respect to time.
• Three constitutive relations that concern the characteristics of the medium in which the fields exist are 𝐷𝐷 = 𝜀𝜀𝜀𝜀 (1) 𝐵𝐵 = 𝜇𝜇𝜇𝜇 (2) 𝐽𝐽 = 𝜎𝜎𝜀𝜀 (3) • where 𝜀𝜀, μ, and σ are the permittivity, permeability, and conductivity of the medium, which is assumed to be homogeneous, isotropic, and source-free
• A homogeneous medium is one for which the quantities 𝜀𝜀, μ, and σ are constant throughout the medium. • The medium is isotropic if 𝜀𝜀 is a scalar constant, so that
everywhere the same direction. Solution for Free-space Conditions • The solution of electromagnetic phenomena in free space, or more generally, in a perfect dielectric containing no charges and no conduction currents. 𝛁𝛁 × 𝑯𝑯 = ̇𝑫𝑫 (𝟒𝟒) 𝛁𝛁 × 𝑬𝑬 = − ̇𝑩𝑩 (𝟓𝟓) 𝛁𝛁 � 𝑫𝑫 = 𝟎𝟎 𝟔𝟔 𝛁𝛁 � 𝑩𝑩 = 𝟎𝟎 (𝟕𝟕) 𝝏𝝏𝛁𝛁 × 𝑯𝑯 𝝏𝝏𝝏𝝏 = 𝛁𝛁 × ̇𝑯𝑯 • Also since 𝜀𝜀 and μ, are independent of time ̇𝐷𝐷 = 𝜖𝜖 ̇𝜀𝜀 (8)
̇𝐵𝐵 = 𝜇𝜇 ̇𝜇𝜇 (9)
so that there results 𝛻𝛻 × ̇𝜇𝜇 = 𝜀𝜀 ̈𝜀𝜀 (10) The symbol ̈𝜀𝜀 means ∂ 2 E/∂t 2 Take the curl of both sides of (5) and using (9), obtain 𝞩𝞩 × 𝞩𝞩 × 𝜀𝜀 = −𝜇𝜇𝞩𝞩 × ̇𝜇𝜇 (11) Substitute eq. (10) into (11) 𝞩𝞩 × 𝞩𝞩 × 𝜀𝜀 = −𝜇𝜇𝞩𝞩 × ̇𝜇𝜇 = −𝜇𝜇𝜀𝜀 ̈𝜀𝜀 (12)
It was shown in vector identity that 𝞩𝞩 × 𝞩𝞩 × 𝜀𝜀 = 𝞩𝞩𝞩𝞩. 𝜀𝜀 − 𝞩𝞩 2 𝜀𝜀
𝞩𝞩𝞩𝞩. 𝜀𝜀 − 𝞩𝞩 2 𝜀𝜀 = −𝜇𝜇𝜀𝜀 ̈𝜀𝜀 (13) 𝞩𝞩. 𝑬𝑬 =
1 𝜀𝜀 𝞩𝞩. 𝑫𝑫 = 𝟎𝟎 and therefore eq. (13) becomes 𝞩𝞩 2 𝜀𝜀 = 𝜇𝜇𝜀𝜀 ̈𝜀𝜀 (14)
This is the law that E must obey. • Differentiating (5) with respect to time and taking the curl of ( 4) it will be found on combining that
𝞩𝞩 2 𝜇𝜇 = 𝜇𝜇𝜀𝜀 ̈𝜇𝜇 15 • Equations (14) and (15) are known as the wave equations. • Thus the first condition on either E or H is that it must satisfy the wave equation.
Uniform Plane-wave Propagation • The wave equation reduces to a very simple form in the special case where E and
independent of two dimensions, say y and z. Then 𝛻𝛻 2
𝜕𝜕 2 𝜀𝜀 𝜕𝜕𝑥𝑥 2 + 𝜕𝜕 2 𝜀𝜀 𝜕𝜕𝑦𝑦 2 + 𝜕𝜕 2 𝜀𝜀 𝜕𝜕𝑧𝑧 2 = 𝜕𝜕 2 𝜀𝜀 𝜕𝜕𝑥𝑥 2 𝜕𝜕 2 𝜀𝜀 𝜕𝜕𝑥𝑥
2 = 𝜕𝜕 2 𝜀𝜀 𝑥𝑥 𝜕𝜕𝑥𝑥 2 𝑎𝑎 𝑥𝑥 + 𝜕𝜕 2 𝜀𝜀 𝑦𝑦 𝜕𝜕𝑥𝑥 2 𝑎𝑎 𝑦𝑦 + 𝜕𝜕
2 𝜀𝜀 𝑧𝑧 𝜕𝜕𝑥𝑥 2 𝑎𝑎 𝑧𝑧 so that (14) becomes 𝜕𝜕 2
𝜕𝜕𝑥𝑥 2 = 𝜇𝜇𝜀𝜀 𝜕𝜕 2 𝜀𝜀 𝜕𝜕𝑡𝑡 2 = 𝜇𝜇𝜀𝜀 𝜕𝜕 2 𝜀𝜀 𝑥𝑥 𝜕𝜕𝑡𝑡
2 𝑎𝑎 𝑥𝑥 + 𝜕𝜕 2 𝜀𝜀 𝑦𝑦 𝜕𝜕𝑡𝑡 2 𝑎𝑎 𝑦𝑦 + 𝜕𝜕 2 𝜀𝜀 𝑧𝑧 𝜕𝜕𝑡𝑡
2 𝑎𝑎 𝑧𝑧 (16) • Vector eq. (16) is equivalent to three scalar equations, one for each of the scalar components of E. • In general, for uniform plane wave propagation in the x direction, E may have components E y and E z , but not E x . • Without loss of generality attention can be restricted to one of the components, say E y , knowing that results for E, will be similar to those obtained for E y . • Then the equation to be solved has the form 𝜕𝜕 2 𝜀𝜀 𝑦𝑦 𝜕𝜕𝑥𝑥 2 = 𝜇𝜇𝜀𝜀
𝜕𝜕 2 𝜀𝜀 𝑦𝑦 𝜕𝜕𝑡𝑡
2 (16𝑎𝑎)
• Equation (16a) is a second-order partial differential equation, which occurs frequently in mechanics and engineering. • For example it is the differential equation for the displacement from equilibrium along a uniform string.
• Electrical engineers will recognize it as the differential equation for voltage or current along a lossless transmission line.
Uniform Plane-wave Propagation • Its general solution is of the form • The functions
and
f 2 (x+v 0 t) represent uniform waves propagating in the +x and -x directions respectively. • Once the electric field has been determined from the wave equation, the magnetic field must follow from Maxwell’s equations. ( )
( ) ( ) ( )
1 0 2 0 , 17 x E x t f x v t f x v t = − + +
• The expression f 1 (x-v 0 t) means a function f of the variable (
• Examples are: A cosβ(x-v 0 t) 0r A sinβ(x-v 0 t) , 𝐶𝐶𝑒𝑒
𝑘𝑘 (x−v 0 t) , (x−v 0 t) etc.
• All of these expressions representwave motion. • If a physical phenomenon that occurs at one place at a given time is reproduced at other places at later times, the time delay being proportional to the space separation from the first location, then the group of phenomena constitute a wave
. Uniform Plane Wave • The functions f 1 (x-v 0 t) and
f 2 (x+v 0 t) describe
such a wave mathematically, the variation of the wave being confined to one dimension in space. • If a fixed time is taken, say t 1 . then the function
f 1 (x-v 0 t 1 ) becomes a function of x since
is a constant. • Such a function is represented by the first curve.
Figure 5-1. A wave traveling in the positive x direction. • If another time, say t 2 , is taken, another function of x is obtained, exactly the same shape as the first except that the second curve is displaced to the right by a distance
- t 1 ).
travelled in the positive x direction with a velocity V 0 • On the other hand, the function f 2 (x+v 0 t) corresponds to a wave traveling in the negative x direction.
Uniform Plane Wave • Thus the general solution of the wave equation in this case is seen to consist of two waves, • One traveling to the right (away from the source), and the other traveling to the left (back toward the source). • If there is no reflecting surface present to reflect the wave back to the source, the second term of (17) is zero and the solution is given by 𝜀𝜀 = f 1 (x−v 0 t) (18) 18 Uniform Plane Wave Solutions in the Time Domain (Cont’d) • The
velocity of propagation is determined solely by the medium: • The functions f 1 and
f 2 are determined by the source and the other boundary conditions. µε 1
= v 19 Uniform Plane Wave Solutions in the Time Domain (Cont’d) • Here we must have where ( )
ˆ ,
z H a H x t = ( ) ( ) ( ) { } 1 0 2 0 1 , z H x t f x v t f x v t η = − − + 20 Uniform Plane Wave Solutions in the Time Domain (Cont’d) • In free space (vacuum): 8 3 10 m/s 120 377
p v c η π = ≈ × = ≈ Ω 21 Uniform Plane Wave Solutions in the Time Domain (Cont’d) • Strictly speaking, uniform plane waves can be produced only by sources of infinite extent.
• However, point sources create spherical waves. Locally, a spherical wave looks like a plane wave. • Thus, an understanding of plane waves is very important in the
study of electromagnetics. Uniform Plane Wave • Equation (18) is a solution of the wave equation for the particular case where the electric field is independent of y and z and is a function of x and t only. • Such a wave is called a uniform plane wave. • Although this is a special case of electromagnetic wave propagation, it is a very important one practically and will be considered further.
• The plane-wave equation 𝜕𝜕 2 𝐸𝐸 𝜕𝜕𝑥𝑥
2 = 𝜇𝜇𝜀𝜀
𝜕𝜕 2 𝐸𝐸 𝜕𝜕𝑡𝑡 2 may be written in terms of the components of E as 𝜕𝜕 2 𝜀𝜀 𝑥𝑥 𝜕𝜕𝑥𝑥 2 = 𝜇𝜇𝜀𝜀
𝜕𝜕 2 𝜀𝜀 𝑥𝑥 𝜕𝜕𝑡𝑡
2 (19𝑎𝑎)
𝜕𝜕 2 𝐸𝐸 𝑦𝑦 𝜕𝜕𝑥𝑥
2 = 𝜇𝜇𝜀𝜀
𝜕𝜕 2 𝐸𝐸 𝑦𝑦 𝜕𝜕𝑡𝑡
2 (19𝑏𝑏)
𝜕𝜕 2 𝜀𝜀 𝑧𝑧 𝜕𝜕𝑥𝑥
2 = 𝜇𝜇𝜀𝜀
𝜕𝜕 2 𝜀𝜀 𝑧𝑧 𝜕𝜕𝑡𝑡
2 (19𝑐𝑐)
• In a region in which there is no charge density 𝛻𝛻 � 𝜀𝜀 = 1 𝜀𝜀 𝛻𝛻 � 𝐷𝐷 = 0 That is 𝜕𝜕𝐸𝐸
𝑥𝑥 𝜕𝜕𝑥𝑥
+ 𝜕𝜕𝐸𝐸
𝑦𝑦 𝜕𝜕𝑦𝑦
+ 𝜕𝜕𝐸𝐸
𝑧𝑧 𝜕𝜕𝑧𝑧
= 0 • For a uniform plane wave in which E is independent of y and z, the last two terms of this relation are equal to zero so that it reduces to 𝜕𝜕𝜀𝜀 𝑥𝑥
Therefore there is no variation of Ex in the x direction. • From eq. (19a) it is seen that the second derivative with respect to time of Ex must then be zero. • This requires that Ex be either zero, constant in time, or increasing uniformly with time. • A field satisfying either of the last two of these conditions would not be a part of the wave motion, and so Ex can be put equal to zero. • Therefore a uniform plane wave progressing in the x direction has no x component of E. • A similar analysis would show that there is no x component of
• It follows, therefore, that uniform plane electromagnetic waves are transverse and have components of
in directions perpendicular to the direction of propagation. Relation between E and H in a Uniform Plane Wave. • For a uniform plane wave traveling in the x direction E and H are both independent of y and z, and E and H have no x component. In this case • Then the first Maxwell Equation (I) can be written • −
𝜕𝜕𝐻𝐻 𝑧𝑧 𝜕𝜕𝑥𝑥 �𝑦𝑦 + 𝜕𝜕𝐻𝐻
𝑦𝑦 𝜕𝜕𝑥𝑥
̂𝑧𝑧 = 𝜀𝜀 𝜕𝜕𝐸𝐸
𝑦𝑦 𝜕𝜕𝑡𝑡
�𝑦𝑦 + 𝜀𝜀 𝜕𝜕𝐸𝐸
𝑧𝑧 𝜕𝜕𝑡𝑡
̂𝑧𝑧 z x H y x H H y z ˆ ˆ ∂ ∂ + ∂ ∂ − = × ∇ z x E y x E E y z ˆ ˆ ∂ ∂ + ∂ ∂ − = × ∇ • For three dimensional variation of H • 𝛻𝛻 × 𝜇𝜇 = 𝜀𝜀 𝜕𝜕𝐸𝐸 𝜕𝜕𝑡𝑡
= 𝜀𝜀 𝜕𝜕𝐸𝐸
𝑥𝑥 𝜕𝜕𝑡𝑡
�𝑥𝑥 + 𝜀𝜀 𝜕𝜕𝐸𝐸
𝑦𝑦 𝜕𝜕𝑡𝑡
�𝑦𝑦 + 𝜀𝜀 𝜕𝜕𝐸𝐸
𝑧𝑧 𝜕𝜕𝑡𝑡
̂𝑧𝑧 z y x H H H z y x z y x H ∂ ∂ ∂ ∂ ∂ ∂ = × ∇ ˆ ˆ ˆ z x H y x H y H x H z z H x H y z H y H x H y z x y x z y z ˆ ˆ ˆ ˆ ˆ ∂ ∂ + ∂ ∂ − = ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ = × ∇ • For three dimensional variation of E • 𝛻𝛻 × 𝜀𝜀 = −𝜇𝜇 𝜕𝜕𝐻𝐻 𝜕𝜕𝑡𝑡
= − 𝜇𝜇 𝜕𝜕𝐻𝐻
𝑥𝑥 𝜕𝜕𝑡𝑡
�𝑥𝑥 + 𝜇𝜇 𝜕𝜕𝐻𝐻
𝑦𝑦 𝜕𝜕𝑡𝑡
�𝑦𝑦 + 𝜇𝜇 𝜕𝜕𝐻𝐻
𝑧𝑧 𝜕𝜕𝑡𝑡
̂𝑧𝑧 z y x E E E z y x z y x E ∂ ∂ ∂ ∂ ∂ ∂ = × ∇ ˆ ˆ ˆ z x E y x E y E x E z z E x E y z E y E x E y z x y x z y z ˆ ˆ ˆ ˆ ˆ ∂ ∂ + ∂ ∂ − = ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ = × ∇ • and the second equation (II) becomes − 𝜕𝜕𝜀𝜀 𝑧𝑧 𝜕𝜕𝑥𝑥 �𝑦𝑦 + 𝜕𝜕𝜀𝜀 𝑦𝑦
𝜕𝜕𝜇𝜇 𝑦𝑦 𝜕𝜕𝑡𝑡 �𝑦𝑦 + 𝜕𝜕𝜇𝜇 𝑧𝑧 𝜕𝜕𝑡𝑡 ̂𝑧𝑧 • Equating the y terms and then the z terms yields the four relations 𝜕𝜕𝐸𝐸 𝑧𝑧
= 𝜇𝜇 𝜕𝜕𝐻𝐻
𝑦𝑦 𝜕𝜕𝑡𝑡
20(𝑎𝑎) 𝜕𝜕𝐸𝐸
𝑦𝑦 𝜕𝜕𝑥𝑥
= −𝜇𝜇 𝜕𝜕𝐻𝐻
𝑧𝑧 𝜕𝜕𝑡𝑡
20(𝑏𝑏) − 𝜕𝜕𝐻𝐻 𝑧𝑧 𝜕𝜕𝑥𝑥
= 𝜀𝜀 𝜕𝜕𝐸𝐸
𝑦𝑦 𝜕𝜕𝑡𝑡
20(c) 𝜕𝜕𝐻𝐻
𝑦𝑦 𝜕𝜕𝑥𝑥
= 𝜀𝜀 𝜕𝜕𝐸𝐸
𝑧𝑧 𝜕𝜕𝑡𝑡
20(𝑑𝑑) • Now if E y = f 1 (x - v o t), where 𝑣𝑣 0 = 1/√µ𝜀𝜀 then 𝜕𝜕𝜀𝜀 𝑦𝑦 𝜕𝜕𝑡𝑡 = 𝜕𝜕𝑓𝑓 1 𝜕𝜕 𝑥𝑥 − 𝑣𝑣 0 𝑡𝑡 𝜕𝜕 𝑥𝑥 − 𝑣𝑣 0 𝑡𝑡 𝜕𝜕𝑡𝑡 = −𝑣𝑣 0 𝜕𝜕𝑓𝑓 1 𝜕𝜕 𝑥𝑥 − 𝑣𝑣 0 𝑡𝑡
𝜕𝜕𝜀𝜀 𝑦𝑦 𝜕𝜕𝑡𝑡 = 𝑓𝑓 1 ′ 𝑥𝑥 − 𝑣𝑣 0 𝑡𝑡 𝜕𝜕 𝑥𝑥 − 𝑣𝑣 0 𝑡𝑡 𝜕𝜕𝑡𝑡 = −𝑣𝑣 0 𝑓𝑓 1 ′ 𝑥𝑥 − 𝑣𝑣 0 𝑡𝑡 • where f 1 ’ (x - v 0 t), or more simply f 1 ’ means
𝜕𝜕𝑓𝑓 1 𝑥𝑥 − 𝑣𝑣 0 𝑡𝑡 𝜕𝜕 𝑥𝑥 − 𝑣𝑣 0 𝑡𝑡
• Substituting for ∂E y /∂t in (20c) above gives 𝜕𝜕𝜇𝜇
𝑧𝑧 𝜕𝜕𝑥𝑥 = 𝑣𝑣 0 𝜀𝜀𝑓𝑓
′ 1 𝜇𝜇 𝑧𝑧 = 𝜀𝜀 𝜇𝜇 � 𝑓𝑓 ′ 1 𝑑𝑑𝑥𝑥 + 𝐶𝐶 𝜕𝜕𝑓𝑓
1 𝜕𝜕𝑥𝑥 =
𝜕𝜕𝑓𝑓 1 𝑥𝑥 − 𝑣𝑣 0 𝑡𝑡 𝜕𝜕 𝑥𝑥 − 𝑣𝑣 0 𝑡𝑡 𝜕𝜕 𝑥𝑥 − 𝑣𝑣 0 𝑡𝑡 𝜕𝜕𝑥𝑥 = 𝑓𝑓 ′ 1 Hence 𝜇𝜇 𝑧𝑧 = 𝜀𝜀 𝜇𝜇 ∫ 𝜕𝜕𝑓𝑓 1 𝜕𝜕𝑥𝑥 𝑑𝑑𝑥𝑥 + 𝐶𝐶 = 𝜀𝜀 𝜇𝜇 𝑓𝑓 1 + 𝐶𝐶 = 𝜀𝜀 𝜇𝜇 𝜀𝜀
𝑦𝑦 + 𝐶𝐶
• The constant of integration C that appears indicates that a field independent of x could be present . The relation between H
and E y becomes
𝜇𝜇 𝑧𝑧 = 𝜀𝜀 𝜇𝜇 𝜀𝜀
𝑦𝑦 𝜀𝜀 𝑦𝑦 𝜇𝜇 𝑧𝑧 = 𝜇𝜇 𝜀𝜀
Similarly it can be shown that 𝜀𝜀 𝑧𝑧 𝜇𝜇 𝑦𝑦 = − 𝜇𝜇 𝜀𝜀 Since 𝜀𝜀 = 𝜀𝜀 2 𝑦𝑦 + 𝜀𝜀 2 𝑧𝑧 H= 𝜇𝜇 2 𝑦𝑦 + 𝜇𝜇 2 𝑧𝑧 where E and H are the total electric and magnetic field strengths, there also results 𝜀𝜀 𝜇𝜇 = 𝜇𝜇 𝜀𝜀
• Above equation states that in a traveling plane electromagnetic wave there is a definite ratio between the amplitudes of E and H and that this ratio is equal to the square root of the ratio of permeability to the dielectric constant of the medium. • Since the units of E are volts per meter and the units of H are amperes per meter, the ratio 𝜀𝜀 𝜇𝜇 =
𝜇𝜇 𝜀𝜀 will have the dimensions of impedance or ohms. • For this reason it is customary to refer to the ratio µ/𝜀𝜀 as the characteristic impedance or intrinsic impedance of the (non conducting) medium. For free space 𝜇𝜇 = 𝜇𝜇 𝑣𝑣
−7 ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑦𝑦𝑒𝑒/𝑚𝑚 𝜀𝜀 = 𝜀𝜀 𝑣𝑣
1 36𝜋𝜋 × 10 9 𝑓𝑓/𝑚𝑚
𝜇𝜇 𝜀𝜀 =
𝜇𝜇 𝑣𝑣 𝜀𝜀 𝑣𝑣 ≈ 120𝜋𝜋 = 377 𝑜𝑜ℎ𝑚𝑚𝑒𝑒 • For any medium, whether conducting or not, the intrinsic impedance is designated by the symbol η. When the medium is free space or a vacuum, the subscript v is used. That is, the intrinsic impedance of free space is η 𝑣𝑣 = µ 𝑣𝑣 𝜀𝜀 𝑣𝑣 = 377 ohms • The relative orientation of E and H may be determined by taking their dot product. 𝜀𝜀. 𝜇𝜇 = 𝜀𝜀 𝑦𝑦 𝜇𝜇 𝑦𝑦 + 𝜀𝜀
𝑧𝑧 𝜇𝜇 𝑧𝑧 = η𝜇𝜇 𝑦𝑦 𝜇𝜇 𝑧𝑧 − η𝜇𝜇
𝑦𝑦 𝜇𝜇 𝑧𝑧 = 0 • Thus in a uniform plane wave, E and H are at right angles to each other. 𝜀𝜀𝑋𝑋𝜇𝜇 = �𝑥𝑥(𝜀𝜀 𝑦𝑦 𝜇𝜇 𝑧𝑧 − 𝜀𝜀
𝑧𝑧 𝜇𝜇 𝑦𝑦 ) = �𝑥𝑥(η𝜇𝜇 𝑧𝑧 2 + η𝜇𝜇 𝑦𝑦 2 ) = �𝑥𝑥η𝜇𝜇
2 • Thus the electric field vector crossed into the magnetic field vector gives the direction in which the wave travels. • Notice that the electric and magnetic fields are at right angles to one another. • They are also perpendicular to the direction of motion of the wave. • This picture defines the coordinate system we will use in our discussion. Wave propagates along the x-axis. The electric field varies in the y-direction and the magnetic field in the z- direction.
The Wave Equations for a Conducting Medium • For regions in which the conductivity is not zero and conduction currents may exist, the more general solution must be obtained. • Maxwell's equation 𝞩𝞩𝞩𝞩𝑯𝑯 = 𝜀𝜀 ̇𝑬𝑬 + 𝑱𝑱 (1) 𝞩𝞩𝞩𝞩𝑬𝑬 = −𝜇𝜇 ̇𝑯𝑯 (2) • If the medium has a conductivity σ (mhos/m), the conduction current density will be given by Ohm's law: 𝑱𝑱 = 𝝈𝝈𝑬𝑬
40 The wave equation for a conducting medium The derivation of the wave equation with J≠0 dt B E ∂ − = × ∇ dt D J H ∂ + = × ∇ H B B E × ∇ ∂ ∂ − = × ∇ ∂ ∂ − = ∂ ∂ × −∇ = × ∇ × ∇ t t t o µ 2 2 2 2 t t t t t t o o o o o ∂ ∂ − ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ − = × ∇ × ∇
E D J D J E ε µ σ µ µ µ µ
D E J ε σ = =
• so that eq. (I) becomes 𝞩𝞩𝞩𝞩𝑯𝑯 = 𝜀𝜀 ̇𝑬𝑬 + 𝜎𝜎𝑬𝑬 (3) • Take the curl of both sides of eq. (2) and then substitute into it the time derivative of (3). 𝞩𝞩𝑋𝑋𝞩𝞩𝑋𝑋𝜀𝜀 = −𝜇𝜇𝞩𝞩𝑋𝑋 ̇𝜇𝜇 = −𝜇𝜇𝜀𝜀 ̈𝜀𝜀 − 𝜎𝜎𝜇𝜇 ̇𝜀𝜀 Recall that 𝞩𝞩𝑋𝑋𝞩𝞩𝑋𝑋𝜀𝜀 = 𝞩𝞩𝞩𝞩. 𝜀𝜀 − 𝞩𝞩 2 𝜀𝜀
results 𝞩𝞩 2 𝜀𝜀 − 𝜇𝜇𝜀𝜀 ̈𝜀𝜀 − 𝜎𝜎𝜇𝜇 ̇𝜀𝜀 = 𝞩𝞩𝞩𝞩. 𝜀𝜀 (4)
• Now for any homogeneous medium in which 𝜀𝜀 is constant 𝞩𝞩. 𝑬𝑬 = 1 𝜀𝜀 𝞩𝞩. 𝑫𝑫 • But 𝛻𝛻·D =ρ , and since there is no net charge within a conductor (although there may be a charge on the surface), the charge density ρ equals zero and therefore 𝛻𝛻·D=O
𝞩𝞩 2 𝜀𝜀 − 𝜇𝜇𝜀𝜀 ̈𝜀𝜀 − 𝜎𝜎𝜇𝜇 ̇𝜀𝜀 = 0 (4) This is the wave equation for E. The wave equation for H is obtained in a similar manner. 𝞩𝞩𝑋𝑋𝞩𝞩𝑋𝑋𝜇𝜇 = 𝜀𝜀𝛻𝛻𝑋𝑋 ̇𝜀𝜀 + 𝜎𝜎𝛻𝛻𝑋𝑋𝜀𝜀 𝞩𝞩𝞩𝞩𝑬𝑬 = −𝜇𝜇 ̇𝑯𝑯 𝛻𝛻𝛻𝛻. 𝜇𝜇 − 𝞩𝞩 2 𝜇𝜇 = −𝜇𝜇𝜀𝜀 ̈𝜇𝜇 − 𝜎𝜎𝜇𝜇 ̇𝜇𝜇 𝞩𝞩. 𝜇𝜇 = 1 µ 𝞩𝞩. 𝐵𝐵 = 0 𝞩𝞩 2 𝜇𝜇 − 𝜇𝜇𝜀𝜀 ̈𝜇𝜇 − 𝜎𝜎𝜇𝜇 ̇𝜇𝜇 = 0 This is the wave equation for H. Sinusoidal Time Variations In practice most generators produce voltages and currents, and hence electric and magnetic fields, which vary sinusoidally with time. E = E 0 cos ωt E = E 0 sin ωt where
The above expressions suggest that a sinusoidal time factor would have to be attached to every term in any equation • Fortunately this is not necessary since the time factor may be suppressed through the use of phasor notation. • The time varying field E(r, t) may be expressed in terms of the corresponding phasor quantity E(r) as
�𝜀𝜀 𝑒𝑒, 𝑡𝑡 = 𝑅𝑅𝑒𝑒{𝜀𝜀 𝑒𝑒 𝑒𝑒 𝑗𝑗ω𝑡𝑡
} • in which the symbol (˜) has been placed over the time-varying quantity to distinguish it from the phasor quantity. • Consider one component at a time, say the x component. The phasor E x is defined by the relation 𝜀𝜀 𝑥𝑥 𝑒𝑒, 𝑡𝑡 = 𝑅𝑅𝑒𝑒{𝜀𝜀 𝑥𝑥 𝑒𝑒 𝑒𝑒 𝑗𝑗ω𝑡𝑡 } • Ex(r) is a complex number and thus at some fixed point r in space Ex may be represented as a point in the complex plane shown in Fig. 5 • Multiplication by e
results in a rotation through the angle ωt measured from the angle Φ. Relationship between time-varying and phasor quantities • As time progresses, the point e jωt traces out a circle with centre at the origin. • Taking the real part is the same as taking the projection on the real axis and this projection varies sinusoidally with time. • The phase of the sinusoid is determined by the argument Φ of the complex number Ex and thus the time-varying quantity may be expressed as � 𝜀𝜀
= 𝑅𝑅𝑒𝑒 |𝜀𝜀 𝑥𝑥 |𝑒𝑒 𝑗𝑗Φ 𝑒𝑒 𝑗𝑗ω𝑡𝑡 = |𝜀𝜀 𝑥𝑥 |cos(ω𝑡𝑡 + Φ) • Another possibility is to introduce a √2 factor on the right-hand side of the phasor definition, thus making |Ex| the rms value of the sinusoid rather than the peak value. Maxwell's Equations Using Phasor Notation. • Corresponding set of phasor equations applicable for sinusoidal time variations. • Consider for instance the equation given in time-varying form by 𝛻𝛻 × �
𝜇𝜇 = 𝜕𝜕�𝐷𝐷
𝜕𝜕𝑡𝑡 + ̃𝐽𝐽
• For the sinusoidal steady state we may substitute the phasor relations as follows: • 𝛻𝛻 × Re 𝑯𝑯𝑒𝑒 𝑗𝑗𝑗𝑗𝑡𝑡
= 𝜕𝜕 𝜕𝜕𝑡𝑡 𝑅𝑅𝑒𝑒 𝑫𝑫𝑒𝑒 𝑗𝑗𝑗𝑗𝑡𝑡
+ 𝑅𝑅𝑒𝑒 𝑱𝑱𝑒𝑒 𝑗𝑗𝑗𝑗𝑡𝑡
• The operations of taking the real part and differentiation may be interchanged with the result that Re 𝛻𝛻 × 𝜇𝜇 − 𝑗𝑗𝑗𝑗𝐷𝐷 − 𝐽𝐽 𝑒𝑒 𝑗𝑗𝑗𝑗𝑡𝑡 = 0
If the above relation is to be valid for all t, then
𝛻𝛻 × 𝜇𝜇 = 𝑗𝑗𝑗𝑗𝐷𝐷 + 𝐽𝐽 • which is the required differential equation in phasor form. • Evidently the phasor equation may be derived from the time-varying equation by replacing each time-varying quantity with a phasor quantity and each time derivative with a jω factor. • For sinusoidal time variations Maxwell's equations may be expressed in phasor form as Differential Integral 𝛁𝛁 × 𝑬𝑬 = −jω𝐵𝐵 ∮ 𝑬𝑬 � 𝒅𝒅𝒅𝒅 = − ∫ jω𝐁𝐁 � 𝒅𝒅𝒅𝒅 𝛁𝛁 � 𝑫𝑫 = 𝝆𝝆 ∮ 𝑫𝑫 � 𝒅𝒅𝒅𝒅 = ∫ 𝝆𝝆𝒅𝒅𝝆𝝆 𝛁𝛁 × 𝑯𝑯 = jω𝐷𝐷 + 𝑱𝑱 ∮ 𝑯𝑯 � 𝒅𝒅𝒅𝒅 = ∫ jω𝐃𝐃 + 𝑱𝑱 � 𝒅𝒅𝒅𝒅 𝛁𝛁 � 𝑩𝑩 = 𝟎𝟎 ∮ 𝑩𝑩 � 𝒅𝒅𝒅𝒅 = 𝟎𝟎 𝛻𝛻 � 𝐽𝐽 = −jω𝜌𝜌 ∮ 𝑱𝑱 � 𝒅𝒅𝒅𝒅 = − ∫ jω𝜌𝜌𝒅𝒅𝝆𝝆
• The constitutive relations retain their forms, being D = 𝜀𝜀
• For sinusoidal time variations the wave equation for the electric field in a lossless medium becomes 𝞩𝞩 2
2 𝜇𝜇𝜀𝜀𝜀𝜀
which is the vector Helmholtz equation. • In a conducting medium, the wave equation becomes 𝞩𝞩
𝜀𝜀 + (ω 2 𝜇𝜇𝜀𝜀𝜀𝜀 − 𝑗𝑗ω𝜎𝜎𝜇𝜇𝜀𝜀) = 0 • Equations of the same form may be written for
H, of course. Document Outline
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