Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machines September 5, 2005
Class Notes 1: Electromagnetic Forces c
2003 James L. Kirtley Jr.
1
Introduction
Stator
Bearings
Stator Conductors Rotor
Air Gap
Shaft
Rotor Conductors
End Windings
Figure 1: Form of Electric Machine This section of notes discusses some of the fundamental processes involved in electric machinery. In the section on energy conversion processes we examine the two major ways of estimating electromagnetic forces: those involving thermodynamic arguments (conservation of energy) and field methods (Maxwell’s Stress Tensor). But first it is appropriate to introduce the topic by describing a notional rotating electric machine. Electric machinery comes in many different types and a strikingly broad range of sizes, from those little machines that cause cell ’phones and pagers to vibrate (yes, those are rotating electric machines) to turbine generators with ratings upwards of a Gigawatt. Most of the machines with which we are familiar are rotating, but linear electric motors are widely used, from shuttle drives in weaving machines to equipment handling and amusement park rides. Currently under development are large linear induction machines to be used to launch aircraft. It is our purpose in this subject to develop an analytical basis for understanding how all of these different machines work. We start, however, with a picture of perhaps the most common of electric machines.
2
Electric Machine Description:
Figure 1 is a cartoon drawing of a conventional induction motor. This is a very common type of electric machine and will serve as a reference point. Most other electric machines operate in 1
a fashion which is the same as the induction machine or which differ in ways which are easy to reference to the induction machine. Most (but not all!) machines we will be studying have essentially this morphology. The rotor of the machine is mounted on a shaft which is supported on some sort of bearing(s). Usually, but not always, the rotor is inside. I have drawn a rotor which is round, but this does not need to be the case. I have also indicated rotor conductors, but sometimes the rotor has permanent magnets either fastened to it or inside, and sometimes (as in Variable Reluctance Machines) it is just an oddly shaped piece of steel. The stator is, in this drawing, on the outside and has windings. With most of the machines we will be dealing with, the stator winding is the armature, or electrical power input element. (In DC and Universal motors this is reversed, with the armature contained on the rotor: we will deal with these later). In most electrical machines the rotor and the stator are made of highly magnetically permeable materials: steel or magnetic iron. In many common machines such as induction motors the rotor and stator are both made up of thin sheets of silicon steel. Punched into those sheets are slots which contain the rotor and stator conductors. Figure 2 is a picture of part of an induction machine distorted so that the air-gap is straightened out (as if the machine had infinite radius). This is actually a convenient way of drawing the machine and, we will find, leads to useful methods of analysis.
Stator Core Stator Conductors In Slots
Air Gap
Rotor Conductors In Slots
Figure 2: Windings in Slots What is important to note for now is that the machine has an air gap g which is relatively small (that is, the gap dimension is much less than the machine radius r). The air-gap also has a physical length l. The electric machine works by producing a shear stress in the air-gap (with of course side effects such as production of “back voltage”). It is possible to define the average airgap shear stress, which we will refer to as τ . Total developed torque is force over the surface area times moment (which is rotor radius): T = 2πr 2 ℓ < τ > Power transferred by this device is just torque times speed, which is the same as force times
2
surface velocity, since surface velocity is u = rΩ:
Pm = ΩT = 2πrℓ < τ > u If we note that active rotor volume is , the ratio of torque to volume is just: T =2<τ > Vr Now, determining what can be done in a volume of machine involves two things. First, it is clear that the volume we have calculated here is not the whole machine volume, since it does not include the stator. The actual estimate of total machine volume from the rotor volume is actually quite complex and detailed and we will leave that one for later. Second, we need to estimate the value of the useful average shear stress. Suppose both the radial flux density Br and the stator surface current density Kz are sinusoidal flux waves of the form: √ Br = 2B0 cos (pθ − ωt) √ Kz = 2K0 cos (pθ − ωt) Note that this assumes these two quantities are exactly in phase, or oriented to ideally produce torque, so we are going to get an “optimistic” bound here. Then the average value of surface traction is: � 1 2π < τ >= Br Kz dθ = B0 K0 2π 0 The magnetic flux density that can be developed is limited by the characteristics of the magnetic materials (iron) used. Current densities are a function of technology and are typically limited by how much effort can be put into cooling and the temperature limits of insulating materials. In practice, the range of shear stress encountered in electric machinery technology is not terribly broad: ranging from a few kPa in smaller machines to about 100 kPa in very large, well cooled machines. It is usually said that electric machines are torque producing devices, meaning tht they are defined by this shear stress mechanism and by physical dimensions. Since power is torque times rotational speed, high power density machines necessarily will have high shaft speeds. Of course there are limits on rotational speed as well, arising from centrifugal forces which limit tip velocity. Our first step in understanding how electric machinery works is to understand the mechanisms which produce forces of electromagnetic origin.
3
Energy Conversion Process:
In a motor the energy conversion process can be thought of in simple terms. In “steady state”, electric power input to the machine is just the sum of electric power inputs to the different phase terminals: Pe =
�
vi ii
i
Mechanical power is torque times speed: Pm = T Ω 3
ElectroMechanical
Electric Power In
Mechanical Power Out
Converter
Losses: Heat, Noise, Windage,...
Figure 3: Energy Conversion Process And the sum of the losses is the difference: Pd = Pe − Pm It will sometimes be convenient to employ the fact that, in most machines, dissipation is small enough to approximate mechanical power with electrical power. In fact, there are many situations in which the loss mechanism is known well enough that it can be idealized away. The “thermodynamic” arguments for force density take advantage of this and employ a “conservative” or lossless energy conversion system.
3.1
Energy Approach to Electromagnetic Forces:
f + v -
Magnetic Field System
x
Figure 4: Conservative Magnetic Field System To start, consider some electromechanical system which has two sets of “terminals”, electrical and mechanical, as shown in Figure 4. If the system stores energy in magnetic fields, the energy stored depends on the state of the system, defined by (in this case) two of the identifiable variables: flux (λ), current (i) and mechanical position (x). In fact, with only a little reflection, you should be able to convince yourself that this state is a single-valued function of two variables and that the energy stored is independent of how the system was brought to this state. Now, all electromechanical converters have loss mechanisms and so are not themselves conservative. However, the magnetic field system that produces force is, in principle, conservative in the 4
sense that its state and stored energy can be described by only two variables. The “history” of the system is not important. It is possible to chose the variables in such a way that electrical power into this conservative system is: dλ P e = vi = i dt Similarly, mechanical power out of the system is: Pm = fe
dx dt
The difference between these two is the rate of change of energy stored in the system: dWm = Pe − Pm dt It is then possible to compute the change in energy required to take the system from one state to another by: � a
Wm (a) − Wm (b) =
b
idλ − f e dx
where the two states of the system are described by a = (λa , xa ) and b = (λb , xb ) If the energy stored in the system is described by two state variables, λ and x, the total differential of stored energy is: ∂Wm ∂Wm dλ + dx dWm = ∂λ ∂x and it is also: dWm = idλ − f e dx So that we can make a direct equivalence between the derivatives and: fe = −
∂Wm ∂x
In the case of rotary, as opposed to linear, motion, torque T e takes the place of force f e and angular displacement θ takes the place of linear displacement x. Note that the product of torque and angle has the same units as the product of force and distance (both have units of work, which in the International System of units is Newton-meters or Joules. In many cases we might consider a system which is electricaly linear, in which case inductance is a function only of the mechanical position x. λ(x) = L(x)i In this case, assuming that the energy integral is carried out from λ = 0 (so that the part of the integral carried out over x is zero), Wm = This makes
�
0
λ
1 1 λ2 λdλ = L(x) 2 L(x)
1 ∂ 1 f e = − λ2 2 ∂x L(x) 5
Note that this is numerically equivalent to 1 ∂ f e = − i2 L(x) 2 ∂x This is true only in the case of a linear system. Note that substituting L(x)i = λ too early in the derivation produces erroneous results: in the case of a linear system it produces a sign error, but in the case of a nonlinear system it is just wrong. 3.1.1
Example: simple solenoid
Consider the magnetic actuator shown in cartoon form in Figure 5. The actuator consists of a circular rod of ferromagnetic material (very highly permeable) that can move axially (the xdirection) inside of a stationary piece, also made of highly permeable material. A coil of N turns carries a current I. The rod has a radius R and spacing from the flat end of the stator is the variable dimension x. At the other end there is a radial clearance between the rod and the stator g. Assume g ≪ R. If the axial length of the radial gaps is ℓ = R/2, the area of the radial gaps is the same as the area of the gap between the rod and the stator at the variable gap. R/2
x
R µ
g CL
µ N turns
Figure 5: Solenoid Actuator
The permeances of the variable width gap is:
P1 =
µ0 πR2 x
and the permeance of the radial clearance gap is, if the gap dimension is small compared with the radius: 2µ0 πRℓ µ0 πR2 P2 = = g g The inductance of the coil system is: L=
P1 P2 µ0 πR2 N 2 N2 = N2 = R1 + R2 P2 + P2 x+g
Magnetic energy is: 6
Wm =
�
λ0
idλ =
0
λ2 x + g 1 λ2 = 0 2 L(x) 2 µ0 πR2 N 2
And then, of course, force of electric origin is: fe = −
∂Wm λ2 d 1 =− 0 ∂x 2 dx L(x)
Here that is easy to carry out: 1 d 1 = dx L µ0 πR2 N 2 So that the force is: f e (x) = −
1 λ20 2 µ0 πR2 N 2
Given that the system is to be excited by a current, we may at this point substitute for flux: λ = L(x)i =
µ0 πR2 N i x+g
and then total force may be seen to be: fe = −
µ0 πR2 N 2 i2 (x + g)2 2
The force is ‘negative’ in the sense that it tends to reduce x, or to close the gap. 3.1.2
Multiply Excited Systems
There may be (and in most electric machine applications there will be) more than one source of electrical excitation (more than one coil). In such systems we may write the conservation of energy expression as: dWm =
� k
ik dλk − f e dx
which simply suggests that electrical input to the magnetic field energy storage is the sum (in this case over the index k) of inputs from each of the coils. To find the total energy stored in the system it is necessary to integrate over all of the coils (which may and in general will have mutual inductance). Wm =
�
i · dλ
Of course, if the system is conservative, Wm (λ1 , λ2 , . . . , x) is uniquely specified and so the actual path taken in carrying out this integral will not affect the value of the resulting energy. 7
3.1.3
Coenergy
We often will describe systems in terms of inductance rather than its reciprocal, so that current, rather than flux, appears to be the relevant variable. It is convenient to derive a new energy variable, which we will call co-energy, by: ′ Wm =
� i
λi ii − Wm
and in this case it is quite easy to show that the energy differential is (for a single mechanical variable) simply: � ′ dWm = λk dik + f e dx k
so that force produced is:
′ ∂Wm ∂x
fe =
3.2
Example: Synchronous Machine
Stator
Rotor
A’
C
Gap
B
B’ F’
F
A
µ F
A
F’
B’
C’ µ
θ C’
B A’
C
Figure 6: Cartoon of Synchronous Machine Consider a simple electric machine as pictured in Figure 6 in which there is a single winding on a rotor (call it the field winding and a polyphase armature with three identical coils spaced at uniform locations about the periphery. We can describe the flux linkages as: λa = La ia + Lab ib + Lab ic + M cos(pθ)if 2π )if 3 2π = Lab ia + Lab ib + La ic + M cos(pθ + )if 3 2π 2π = M cos(pθ)ia + M cos(pθ − )ib + M cos(pθ + ) + Lf if 3 3
λb = Lab ia + La ib + Lab ic + M cos(pθ − λc λf
It is assumed that the flux linkages are sinusoidal functions of rotor position. As it turns out, many electrical machines work best (by many criteria such as smoothness of torque production) 8
if this is the case, so that techniques have been developed to make those flux linkages very nearly sinusoidal. We will see some of these techniques in later chapters of these notes. For the moment, we will simply assume these dependencies. In addition, we assume that the rotor is magnetically ’round’, which means the stator self inductances and the stator phase to phase mutual inductances are not functions of rotor position. Note that if the phase windings are identical (except for their angular position), they will have identical self inductances. If there are three uniformly spaced windings the phase-phase mutual inductances will all be the same. Now, this system can be simply described in terms of coenergy. With multiple excitation it is important to exercise some care in taking the coenergy integral (to ensure that it is taken over a valid path in the multi-dimensional space). In our case there are actually five dimensions, but only four are important since we can position the rotor with all currents at zero so there is no contribution to coenergy from setting rotor position. Suppose the rotor is at some angle θ and that the four currents have values ia0 , ib0 , ic0 and if 0 . One of many correct path integrals to take would be: ′ Wm =
+
�
ia0
La ia dia
0
�
ib0
0
+
�
ic0
(Lab ia0 + Lab ib0 + La ic ) dic
0
+
�
0
(Lab ia0 + La ib ) dib
if 0
2π 2π )ib0 + M cos(pθ + )ic0 + Lf if dif M cos(pθ)ia0 + M cos(pθ − 3 3 �
�
The result is: ′ Wm =
� 1 �2 La ia0 + i2b0 + i2co + Lab (iao ib0 + iao ic0 + ico ib0 ) 2 � � 2π 1 2π ) + ic0 cos(pθ + ) + Lf i2f 0 +M if 0 ia0 cos(pθ) + ib0 cos(pθ − 3 3 2
Since there are no variations of the stator inductances with rotor position θ, torque is easily given by: ′ ∂Wm 2π 2π Te = = −pM if 0 ia0 sin(pθ) + ib0 sin(pθ − ) + ico sin(pθ + ) ∂θ 3 3
�
3.2.1
�
Current Driven Synchronous Machine
Now assume that we can drive this thing with currents: ia0 = Ia cos ωt ib0 ic0 if 0
2π = Ia cos ω t − 3 � � 2π = Ia cos ω t + 3 = If �
9
�
and assume the rotor is turning at synchronous speed: pθ = ωt + δi Noting that cos x sin y =
1 2
sin(x − y) + 12 sin(x + y), we find the torque expression above to be:
Te = −pM Ia If
1 1 sin δi + sin (2ωt + δi ) 2 2 � � 1 1 sin δi + sin 2ωt + δi − + 2 2 � � 1 1 + sin δi + sin 2ωt + δi + 2 2 �
�
4π 3 �� 4π 3 ��
The sine functions on the left add and the ones on the right cancel, leaving: 3 Te = − pM Ia If sin δi 2 And this is indeed one way of looking at a synchronous machine, which produces steady torque if the rotor speed and currents all agree on frequency. Torque is related to the current torque angle δi . As it turns out such machines are not generally run against current sources, but we will take up actual operation of such machines later.
4
Field Descriptions: Continuous Media
While a basic understanding of electromechanical devices is possible using the lumped parameter approach and the principle of virtual work as described in the previous section, many phenomena in electric machines require a more detailed understanding which is afforded by a continuum approach. In this section we consider a fields-based approach to energy flow using Poynting’s Theorem and then a fields based description of forces using the Maxwell Stress Tensor. These techniques will both be useful in further analysis of what happens in electric machines.
4.1
Field Description of Energy Flow: Poyting’s Theorem
Start with Faraday’s Law: � � = − ∂B ∇×E ∂t and Ampere’s Law:
� = J� ∇×H
� and the second by E � and taking the difference: Multiplying the first of these by H � � � � ×H � = −H � · ∂B − E � ·∇×E � −E � ·∇×H � =∇· E � · J� H dt
On the left of this expression is the divergence of electromagnetic energy flow: �=E � ×H � S 10
� is the celebrated Poynting flow which describes power in an electromagnetic field Here, S sysstem. (The units of this quantity is watts per square meter in the International System). On � · ∂ B~ is rate of change of magnetic stored energy. The second the right hand side are two terms: H dt � · J� looks a lot like power dissipation. We will discuss each of these in more detail. For the term, E moment, however, note that the divergence theorem of vector calculus yields: �
��
� = S � · �nda ∇ · Sdv
volume
that is, the volume integral of the divergence of the Poynting energy flow is the same as the Poynting energy flow over the surface of the volume in question. This integral becomes: ��
� · �nda = −
S
� � · J� + H � · ∂B E ∂t volume �
�
�
dv
which is simply a realization that the total energy flow into a region of space is the same as the volume integral over that region of the rate of change of energy stored plus the term that looks like dissipation. Before we close this, note that, if there is motion of any material within the system, we can use the empirical expression for transformation of electric field between observers moving with respect to each other. Here the ’primed’ frame is moving with respeect to the ’unprimed’ frame with the velocity �v �′ = E � + �v × B � E This transformation describes, for example, the motion of a charged particle such as an electron under the influence of both electric and magnetic fields. Now, if we assume that there is material motion in the system we are observing and if we assign �v to be the velocity of that material, so that � ′ is measured in a frame in which thre is no material motion (that is the frame of the material E itself), the product of electric field and current density becomes: �
�
�
�
�
� � · J� = E � ′ · J� − �v × B � · J� = E � ′ · J� + �v · J� × B � · J� = E � ′ − �v × B E
�
In the last step we used the fact that in a scalar triple product the order of the scalar (dot) and vector (cross) products can be interchanged and that reversing the order of terms in a vector (cross) product simply changes the sign of that product. Now we have a ready interpretation for what we have calculated: If the ’primed’ coordinate system is actually the frame of material motion, � ′ · J� = 1 |J�|2 E σ which is easily seen to be dissipation and is positive definite if material conductivity σ is positive. The last term is obviously conversion of energy from electromagnetic to mechanical form: �
�
� = �v · F� �v · J� × B where we have now identified force density to be:
� = J� × B � F 11
This is the Lorentz Force Law, which describes the interaction of current with magnetic field to produce force. It is not, however, the complete story of force production in electromechanical systems. As we learned earlier, changes in geometry which affect magnetic stored energy can also produce force. Fortunately, a complete description of electromechanical force is possible using only magnetic fields and that is the topic of our next section.
4.2
Field Description of Forces: Maxwell Stress Tensor
Forces of electromagnetic origin, because they are transferred by electric and magnetic fields, are the result of those fields and may be calculated once the fields are known. In fact, if a surface can be established that fully encases a material body, the force on that body can be shown to be the integral of force density, or traction over that surface. The traction τ derived by taking the cross product of surface current density and flux density on the air-gap surface of a machine (above) actually makes sense in view of the empirically derived Lorentz Force Law: Given a (vector) current density and a (vector) flux density. This is actually enough to describe the forces we see in many machines, but since electric machines have permeable magnetic material and since magnetic fields produce forces on permeable material even in the absence of macroscopic currents it is necessary to observe how force appears on such material. A suitable empirical expression for force density is: � � � ·H � ∇µ �−1 H F� = J� × B 2
� is the magnetic field intensity and µ is the permeability.
where H Now, note that current density is the curl of magnetic field intensity, so that:
� F
And, since: �
force density is:
� � � � × µH � −1 H � ·H � ∇µ ∇×H 2 � � � � � ·H � ∇µ � ×H � −1 H = µ ∇×H 2 �
=
� � � � � � = H � ·H � � − 1∇ H � ·∇ H � ×H ∇×H 2
� � � � � � � − 1 µ∇ H � ·∇ H � ·H � −1 H � ·H � ∇µ = µ H 2� 2 � �� � � 1 � ·H � � ·∇ H � −∇ µ H = µ H 2
F�
This expression can be written by components: the component of force in the i’th dimension is: Fi = µ
∂ Hk ∂xk
�� k
�
∂ Hi − ∂xi
�
1 � 2 µ H 2 k k
�
The first term can be written as: µ
�� k
Hk
∂ ∂xk
�
Hi =
� ∂ k
∂xk 12
µHk Hi − Hi
� ∂ k
∂xk
µHk
The last term in this expression is easily shown to be divergence of magnetic flux density, which is zero: � ∂ � = µHk = 0 ∇·B ∂xk k Using this, we can write force density in a more compact form as: ∂ Fk = ∂xi
�
µ � 2 Hn µHi Hk − δik 2 n
�
where we have used the Kroneker delta δik = 1 if i = k, 0 otherwise. Note that this force density is in the form of the divergence of a tensor: ∂ Tik ∂xi
Fk = or
F� = ∇ · T
In this case, force on some object that can be surrounded by a closed surface can be found by using the divergence theorem: f� =
�
vol
� dv = F
or, if we note surface traction to be τi = total force in direction i is just: f� =
�
�
k
�
s
vol
��
∇ · T dv = T · �nda
Tik nk , where n is the surface normal vector, then the
τi da =
� �
Tik nk da
k
The interpretation of all of this is less difficult than the notation suggests. This field description of forces gives us a simple picture of surface traction, the force per unit area on a surface. If we just integrate this traction over the area of some body we get the whole force on the body. Note one more thing about this notation. Sometimes when subscripts are repeated as they are � here the summation symbol is omitted. Thus we would write τi = k Tik nk = Tik nk .
4.3
Example: Linear Induction Machine
Figure 7 shows a highly simplified picture of a single sided linear induction motor. This is not how most linear induction machines are actually built, but it is possible to show through symmetry arguments that the analysis we can carry out here is actually valid for other machines of this class. This machine consists of a stator (the upper surface) which is represented as a surface current on the surface of a highly permeable region. The moving element consists of a thin layer of conducting material on the surface of a highly permeable region. The moving element (or ’shuttle’) has a velocity u with respect to the stator and that motion is in the x direction. The stator surface current density is assumed to be: �
Kz = Re K z ej(ωt−kx) 13
�
g
K
z
111111111111111111111 000000000000000000000 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 µ µ
σ s
K
y u
x
s
Figure 7: Simple Model of single sided linear induction machine Note that we are ignoring some important effects, such as those arising from finite length of the stator and of the shuttle. Such effects can be quite important, but we will leave those until later, as they are what make linear motors interesting. Viewed from the shuttle for which the dimension in the direction of motion is x′ − x − ut′ , the relative frequency is: ωt − kx = (ω − ku) t − kx′ = ωs t − kx′
Now, since the shuttle surface can support a surface current and is excited by magnetic fields which are in turn excited by the stator currents, it is reasonable to assume that the form of rotor current is the same as that of the stator: n
′
Ks = Re K s ej(ωs t−kx ) Ampere’s Law is, in this situation: g
o
∂Hy = Kz + Ks ∂x
which is, in complex amplitudes: Hy =
Kz + Ks −jkg
The y- component of Faraday’s Law is, assuming the problem is uniform in the z- direction: −jωs B y = jkE ′z or
ωs µ0 H y k A bit of algebraic manipulation yields expressions for the complex amplitudes of rotor surface current and gap magnetic field: E ′z = −
Ks = Hy =
−j µ0kω2sgσs
1 + j muk02ωgs σs
Kz
Kz j kg 1 + j muk02ωgs σs 14
To find surface traction, the Maxwell Stress Tensor can be evaluated at a surface just below the stator (on this surface the x- directed magnetic field is simply H x = K z . Thus the traction is τx = Txy = µ0 Hx Hy and the average of this is: < τx >= This is: < τx >=
� µ0 � Re H x H y ∗ 2
2 µ0 ωs σs µ0 1 |K z | k2 g � � 2 kg 1 + µ0 ωs σs 2 k2 g
Now, if we consider electromagnetic power flow (Poynting’s Theorem): in the y- direction: Sy = Ez Hx And since in the frame of the shuttle E ′z = − ωks µ0 H y µ0 ωs σs
< Sy′ >= −
1 ω s µ0 ωs k2 g 2 < τx > � �2 |K z | = − k 2 k kg 1 + µ0 ωs σs k2 g
Similarly, evaluated in the frame of the stator: ω < Sy >= − < τ − x > k This shows what we already suspected: the electromagnetic power flow from the stator is the force density on the shuttle times the wave velocity. The electromagnetic ower flow into the shuttle is the same force density times the ’slip’ velocity. The difference between these two is the power converted to mechanical form and it is the force density times the shuttle velocity.
4.4
Rotating Machines
The use of this formulation in rotating machines is a bit tricky because, at lest formally, directional vectors must have constant identity if an integral of forces is to become a total force. In cylindrical coordinates, of course, the directional vectors are not of constant identity. However, with care and understanding of the direction of traction and how it is integrated we can make use of the MST approach in rotating electric machines. Now, if we go back to the case of a circular cylinder and are interested in torque, it is pretty clear that we can compute the circumferential force by noting that the normal vector to the cylinder is just the radial unit vector, and then the circumferential traction must simply be: τ θ = µ0 H r H θ Assuming that there are no fluxes inside the surface of the rotor, simply integrating this over the surface gives azimuthal force. In principal this is the same as surrounding the surface of the rotor by a continuum of infinitely small boxes, one surface just outside the rotor and with a normal facing outward, the other surface just inside with normal facing inward. (Of course the MST is zero on this inner surface). Then multiplying by radius (moment arm) gives torque. The last step is to note that, if the rotor is made of highly permeable material, the azimuthal magnetic field just outside the rotor is equal to surface current density. 15
5
Generalization to Continuous Media
Now, consider a system with not just a multiplicity of circuits but a continuum of current-carrying paths. In that case we could identify the co-energy as: ′ Wm
=
�
area
�
λ(�a)dJ� · d�a
where that area is chosen to cut all of the current carrying conductors. This area can be picked to be perpedicular to each of the current filaments since the divergence of current is zero. The flux λ is calculated over a path that coincides with each current filament (such paths exist since current has zero divergence). Then the flux is: �
λ(�a) =
� · d�n B
� for which the magnetic flux density is: Now, if we use the vector potential A � =∇×A � B the flux linked by any one of the current filaments is: λ(�a) =
�
� · d�ℓ A
where d�ℓ is the path around the current filament. This implies directly that the coenergy is: ′ Wm
=
�
area
� � J
� · d�ℓdJ� · d�a A
Now: it is possible to make d�ℓ coincide with d�a and be parallel to the current filaments, so that: ′ Wm =
5.1
�
vol
� · dJdv � A
Permanent Magnets
Permanent magnets are becoming an even more important element in electric machine systems. Often systems with permanent magnets are approached in a relatively ad-hoc way, made equivalent to a current that produces the same MMF as the magnet itself. � to The constitutive relationship for a permanent magnet relates the magnetic flux density B � and the property of the magnet itself, the magnetization M �. magnetic field H �
� � = µ0 H � +M B
�
Now, the effect of the magnetization is to act as if there were a current (called an amperian current) with density: � J�∗ = ∇ × M Note that this amperian current “acts” just like ordinary current in making magnetic flux density. Magnetic co-energy is: � ′ � · ∇ × dM � dv Wm = A vol 16
�
�
�
�
�
�
� ×D � =D � · ∇×C � −C � · ∇×D � Now, Next, note the vector identity ∇ · C ′ Wm =
�
vol
�
�
� × dM � dv + −∇ · A
�
vol
�
�
� · dM � dv ∇×A
� = ∇ × A: � Then, noting that B ��
′ � × dM � d�s + Wm =− A
�
vol
� · dM � dv B
The first of these integrals (closed surface) vanishes if it is taken over a surface just outside the � is zero. Thus the magnetic co-energy in a system with only a permanent magnet magnet, where M source is � ′ � · dM � dv Wm = B vol Adding current carrying coils to such a system is done in the obvious way.
17
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machines September 5, 2005
Class Notes 2 Magnetic Circuit Basics c
2003 James L. Kirtley Jr.
1
Introduction
Magnetic Circuits offer, as do electric circuits, a way of simplifying the analysis of magnetic field systems which can be represented as having a collection of discrete elements. In electric circuits the elements are sources, resistors and so forth which are represented as having discrete currents and voltages. These elements are connected together with ‘wires’ and their behavior is described by network constraints (Kirkhoff’s voltage and current laws) and by constitutive relationships such as Ohm’s Law. In magnetic circuits the lumped parameters are called ‘Reluctances’ (the inverse of ‘Reluctance’ is called ‘Permeance’). The analog to a ‘wire’ is referred to as a high permeance magnetic circuit element. Of course high permeability is the analog of high conductivity. By organizing magnetic field systems into lumped parameter elements and using network constraints and constitutive relationships we can simplify the analysis of such systems.
2
Electric Circuits
First, let us review how Electric Circuits are defined. We start with two conservation laws: conservation of charge and Faraday’s Law. From these we can, with appropriate simplifying assumptions, derive the two fundamental circiut constraints embodied in Kirkhoff’s laws.
2.1
KCL
Conservation of charge could be written in integral form as: dρf dv = 0 (1) volume dt This simply states that the sum of current out of some volume of space and rate of change of free charge in that space must be zero. Now, if we define a discrete current to be the integral of current density crossing through a part of the surface: ��
J~ · ~nda +
ik = −
��
�
J~ · ~nda
surfacek
(2)
and if we assume that there is no accumulation of charge within the volume (in ordinary circuit theory the nodes are small and do not accumulate charge), we have: ��
J~ · ~nda = −
1
� k
ik = 0
(3)
which holds if the sum over the index k includes all current paths into the node. This is, of course, KCL.
2.2
KVL
Faraday’s Law is, in integral form: ~ · d~ℓ = − d ~ · ~nda E B (4) dt where the closed loop in the left hand side of the equation is the edge of the surface of the integral on the right hand side. Now if we define voltage in the usual way, between points a and b for element k: �
��
vk =
�
bk
ak
~ · d~ℓ E
(5)
Then, if we assume that the right-hand side of Faraday’s Law (that is, magnetic induction) is zero, the loop equation becomes: �
vk = 0
(6)
k
This works for circuit analysis because most circuits do not involve magnetic induction in the loops. However, it does form the basis for much head scratching over voltages encountered by ‘ground loops’.
2.3
Constitutive Relationship: Ohm’s Law
Many of the materials used in electric circuits carry current through a linear conduction mechanism. That is, the relationship between electric field and electric current density is ~ J~ = σ E
(7)
Suppose, to start, we can identify a piece of stuff which has constant area and which is carrying current over some finite length, as shown in Figure 1. Assume this rod is carrying current density J~ (We won’t say anything about how this current density managed to get into the rod, but assume that it is connected to something that can carry current (perhaps a wire....). Total current carried by the rod is simply I = |J|A and then voltage across the element is: v=
�
~ · dℓ = ℓ I E σA
from which we conclude the resistance is R=
V ℓ = I σA
2
J
E
l
Figure 1: Simple Rod Shaped Resistor Of course we can still employ the lumped parameter picture even with elements that are more complex. Consider the annular resistor shown in Figure 2. This is an end-on view of something which is uniform in cross-section and has depth D in the direction you can’t see. Assume that the inner and outer elements are very good conductors, relative to the annular element in between. Assume further that this element has conductivity σ and inner and outer radii Ri and Ro , respectively.
Electrodes
+
v
−
Resistive Material
Figure 2: Annular Resistor Now, if the thing is carrying current from the inner to the outer electrode, current density would be: J~ = ~ir Jr (r) = Electric field is Er =
I 2πDr
Jr I = 2πDrσ σ
Then voltage is v=
�
Ro
Ri
Er (r) =
Ro I log 2πσD Ri
so that we conclude the resistance of this element is R=
log
Ro Ri
2πσD
3
3
Magnetic Circuit Analogs
In the electric circuit, elements for which voltage and current are defined are connected together by elements thought of as ‘wires’, or elements with zero or negligible voltage drop. The interconnection points are ‘nodes’. In magnetic circuits the analogous thing occurs: elements for which magnetomotive force and flux can be defined are connected together by high permeability magnetic circuit elements (usually iron) which are the analog of wires in electric circuits.
3.1
Analogy to KCL
Gauss’ Law is: ��
~ · ~nda = 0
B
(8)
which means that the total amount of flux coming out of a region of space is always zero. Now, we will define a quantity which is sometimes called simply ‘flux’ or a ‘flux tube’. This might be thought to be a collection of flux lines that can somehow be bundled together. Generally it is the flux that is identified with a magnetic circuit element. Mathematically it is: Φk =
��
~ · ~nda B
(9)
In most cases, flux as defined above is carried in magnetic circuit elements which are made of high permeability material, analogous to the ‘wires’ of high conductivity material which carry current in electric circuits. It is possible to show that flux is largely contained in such high permeability materials. If all of the flux tubes out of some region of space (’node’) are considered in the sum, they must add to zero: � Φk = 0 (10) k
3.2
Analogy to KVL: MMF
Ampere’s Law is
�
~ · d~ℓ = H
��
J~ · ~nda
(11)
Where, as for Faraday’s Law, the closed contour on the left is the periphery of the (open) surface on the right. Now we define what we call Magnetomotive Force, in direct analog to ‘Electromotive Force’, (voltage). Fk =
�
bk
ak
~ · d~ℓ H
(12)
Further, define the current enclosed by a loop to be: F0 = Then the analogy to KVL is:
��
�
J~ · ~nda
Fk = F0
k
4
(13)
Note that the analog is not exact as there is a source term on the right hand side whereas KVL has no source term. Note also that sign counts here. The closed integral is taken in such direction so that the positive sense of the surface enclosed is positive (upwards) when the surface is to the left of the contour. (This is another way of stating the celebrated ‘right hand rule’: if you wrap your right hand around the contour with your fingers pointing in the direction of the closed contour integration, your thumb is pointing in the positive direction for the surface).
3.3
Analog to Ohm’s Law: Reluctance
Consider a ‘gap’ between two high permeability pieces as shown in Figure 3. If we assume that their permeability is high enough, we can assume that there is no magnetic field H in them and so the MMF or ‘magnetic potential’ is essentially constant, just like in a wire. For the moment, assume that the gap dimension g is ‘small’ and uniform over the gap area A. Now, assume that some flux Φ is flowing from one of these to the other. That flux is Φ = BA where B is the flux density crossing the gap and A is the gap area. Note that we are ignoring ‘fringing’ fields in this simplified analysis. This neglect often requires correction in practice. Since the permeability of free space is µ0 , (assuming the gap is indeed filled with ’free space’), magnetic field intensity is B H= µ0 and gap MMF is just magnetic field intensity times gap dimension. This, of course, assumes that the gap is uniform and that so is the magnetic field intensity: F =
B g µ0
Which means that the reluctance of the gap is the ratio of MMF to flux: R=
F g = Φ µ0 A
Φ
y Area A µ g Figure 3: Air Gap
5
x
3.4
Simple Case
Consider the magnetic circuit situation shown in Figure 4. Here there is a piece of highly permeable material shaped to carry flux across a single air-gap. A coil is wound through the window in the magnetic material (this shape is usually referred to as a ‘core’). The equivalent circuit is shown in Figure 5.
Region 1
Region 2 I
Figure 4: Single air-capped Core Note that in Figure 4, if we take as the positive sense of the closed loop a direction which goes vertically upwards through the leg of the core through the coil and then downwards through the gap, the current crosses the surface surrounded by the contour in the positive sense direction.
+
F = N I
−
Φ
Figure 5: Equivalent Circuit
3.5
Flux Confinement
The gap in this case has the same reluctance as computed earlier, so that the flux in the gap is simply Φ = NRI . Now, by focusing on the two regions indicated we might make a few observations about magnetic circuits. First, consider ‘region 1’ as shown in Figure 6.
µ
Figure 6: Flux Confinement Boundary: This is ’Region 1’
6
~ parallel to the surface must be the same inside the In this picture, note that magnetic field H material as it is outside. Consider Ampere’s Law carried out about a very thin loop consisting of the two arrows drawn at the top boundary of the material in Figure 6 with very short vertical paths joining them. If there is no current singularity inside that loop, the integral around it must be zero which means the magnetic field just inside must be the same as the magnetic field outside. Since ~ = µH, ~ and ’highly permeable’ means µ is very large, the material is very highly permeable and B ~ must be quite small. Thus the magnetic circuit has small magnetic field unless B is really large, H H and therfore flux densities parallel to and just outside its boundaries aer also small.
µ
B is perpendicular
Figure 7: Gap Boundary At the surface of the magnetic material, since the magnetic field parallel to the surface must be very small, any flux lines that emerge from the core element must be perpendicular to the surface as shown for the gap region in Figure 7. This is true for region 1 as well as for region 2, but note that the total MMF available to drive fields across the gap is the same as would produce field lines from the area of region 1. Since any lines emerging from the magnetic material in region 1 would have very long magnetic paths, they must be very weak. Thus the magnetic circuit material largely confines flux, with only the relatively high permeance (low reluctance) gaps carrying any substantive amount of flux.
3.6
Example: C-Core
Consider a ‘gapped’ c-core as shown in Figure 8. This is two pieces of highly permeable material shaped generally like ‘C’s. They have uniform depth in the direction you cannot see. We will call that dimension D. Of course the area A = wD, where w is the width at the gap. We assume the two gaps have the same area. Each of the gaps will have a reluctance R=
g µ0 A
Suppose we wind a coil with N turns on this core as shown in Figure 9. Then we put a current I in that coil. The magnetic circuit equivalent is shown in Figure 10. The two gaps are in series and, of course, in series with the MMF source. Since the two fluxes are the same and the MMF’s add: F0 = N I = F1 + F2 = 2RΦ and then Φ=
NI µ0 AN I = 2R 2g 7
µ
g Area A
Figure 8: Gapped Core
N Turns
Φ
I
Figure 9: Wound, Gapped Core and corresponding flux density in the gaps would be: By =
3.7
µ0 N I 2g
Example: Core with Different Gaps
As a second example, consider the perhaps oddly shaped core shown in Figure 11. Suppose the gap on the right has twice the area as the gap on the left. We would have two gap reluctances: g g R2 = R1 = µ0 A 2µ0 A Since the two gaps are in series the flux is the same and the total reluctance is R=
3 g 2 µ0 A
Flux in the magnetic circuit loop is Φ=
F 2 µ0 AN I = R 3 g
and the flux density across, say, the left hand gap would be: By =
Φ 2 µ0 N I = A 3 g
8
−
F
+ Φ
Figure 10: Equivalent Magnetic Circuit
N Turns
Φ
I
Figure 11: Wound, Gapped Core: Different Gaps
9
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machines Class Notes 3: Eddy Currents, Surface Impedances and Loss Mechanisms September 5, 2005 c
2005 James L. Kirtley Jr.
1
Introduction
Losses in electric machines arise from conduction and magnetic hysteresis. Conduction losses are attributed to straightforward transport conduction and to eddy currents. Transport losses are relatively easy to calculate so we will not pay them much attention. Eddy currents are more interesting and result in frequency dependent conduction losses in machines. Eddy currents in linear materials can often be handled rigorously, but eddy currents in saturating material are more difficult and are often handled in a heuristic fashion. We present here both analytical and semi-emiprical ways of dealing with such losses. We start with surface impedance: the ratio of electric field to surface current. This is important not just in calculating machine losses, but also in describing how some machines operate.
2
Surface Impedance of Uniform Conductors
The objective of this section is to describe the calculation of the surface impedance presented by a layer of conductive material. Two problems are considered here. The first considers a layer of linear material backed up by an infinitely permeable surface. This is approximately the situation presented by, for example, surface mounted permanent magnets and is probably a decent approximation to the conduction mechanism that would be responsible for loss due to asynchronous harmonics in these machines. It is also appropriate for use in estimating losses in solid rotor induction machines and in the poles of turbogenerators. The second problem, which we do not work here but simply present the previously worked solution, concerns saturating ferromagnetic material.
2.1
Linear Case
The situation and coordinate system are shown in Figure 1. The conductive layer is of thicknes T and has conductivity σ and permeability µ0 . To keep the mathematical expressions within bounds, we assume rectilinear geometry. This assumption will present errors which are small to the extent that curvature of the problem is small compared with the wavenumbers encountered. We presume that the situation is excited, as it would be in an electric machine, by a current sheet of the form � � j(ωt−kx) Kz = Re Ke In the conducting material, we must satisfy the diffusion equation: ∇ 2 H = µ0 σ
1
∂H ∂t
Conductive Region
Kz
Hx
y x Permeable Surface Figure 1: Axial View of Magnetic Field Problem In view of the boundary condition at the back surface of the material, taking that point to be y = 0, a general solution for the magnetic field in the material is: �
Hx = Re A sinh αyej(ωt−kx) Hy
�
k = Re j A cosh αyej(ωt−kx) α �
�
where the coefficient α satisfies: α2 = jωµ0 σ + k2 and note that the coefficients above are chosen so that H has no divergence. Note that if k is small (that is, if the wavelength of the excitation is large), this spatial coefficient α becomes 1+j α= δ where the skin depth is: � 2 δ= ωµ0 σ Faraday’s law: ∇×E =− gives:
∂B ∂t
ω E z = −µ0 H y k
Now: the “surface current” is just K s = −H x so that the equivalent surface impedance is: Z=
Ez ω = jµ0 coth αT −H x α
A pair of limits are interesting here. Assuming that the wavelength is long so that k is negligible, then if αT is small (i.e. thin material), Z → jµ0
1 ω = 2 α T σT 2
On the other hand as αT → ∞,
1+j σδ Next it is necessary to transfer this surface impedance across the air-gap of a machine. So, assume a new coordinate system in which the surface of impedance Z s is located at y = 0, and we wish to determine the impedance Z = −E z /H x at y = g. In the gap there is no current, so magnetic field can be expressed as the gradient of a scalar potential which obeys Laplace’s equation: Z→
H = −∇ψ and ∇2 ψ = 0
Ignoring a common factor of ej(ωt−kx) , we can express H in the gap as: �
H x = jk ψ + eky + ψ − e−ky �
�
H y = −k ψ + eky − ψ − e−ky At the surface of the rotor,
�
E z = −H x Z s or
�
�
�
−ωµ0 ψ + − ψ − = jkZ s ψ + + ψ − and then, at the surface of the stator,
�
kg −kg E ω ψ+ e − ψ − e Z = − z = jµ0 k ψ + ekg + ψ − e−kg Hx
�
�
A bit of manipulation is required to obtain: ω Z = jµ0 k
�
ekg (ωµ0 − jkZ s ) − e−kg (ωµ0 + jkZ s ) ekg (ωµ0 − jkZ s ) + e−kg (ωµ0 + jkZ s )
�
It is useful to note that, in the limit of Z s → ∞, this expression approaches the gap impedance Zg = j
ωµ0 k2 g
and, if the gap is small enough that kg → 0, Z → Z g ||Z s
3
3
Iron
Electric machines employ ferromagnetic materials to carry magnetic flux from and to appropriate places within the machine. Such materials have properties which are interesting, useful and problematical, and the designers of electric machines must deal with this stuff. The purpose of this note is to introduce the most salient properties of the kinds of magnetic materials used in electric machines. We will be concerned here with materials which exhibit magnetization: flux density is something ~ = µ0 H. ~ Generally, we will speak of hard and soft magnetic materials. Hard materials other than B are those in which the magnetization tends to be permanent, while soft materials are used in magnetic circuits of electric machines and transformers. Since they are related we will find ourselves talking about them either at the same time or in close proximity, even though their uses are widely disparite.
3.1
Magnetization:
It is possible to relate, in all materials, magnetic flux density to magnetic field intensity with a consitutive relationship of the form: �
~ ~ = µ0 H ~ +M B
�
where magnetic field intensity H and magnetization M are the two important properties. Now, in linear magnetic material magnetization is a simple linear function of magnetic field: ~ = χm H ~ M so that the flux density is also a linear function: ~ = µ0 (1 + χm ) H ~ B Note that in the most general case the magnetic susceptibility χm might be a tensor, leading to flux density being non-colinear with magnetic field intensity. But such a relationship would still be linear. Generally this sort of complexity does not have a major effect on electric machines.
3.2
Saturation and Hysteresis
In useful magnetic materials this nice relationship is not correct and we need to take a more general view. We will not deal with the microscopic picture here, except to note that the magnetization is due to the alignment of groups of magnetic dipoles, the groups often called domaines. There are only so many magnetic dipoles available in any given material, so that once the flux density is high enough the material is said to saturate, and the relationship between magnetic flux density and magnetic field intensity is nonlinear. Shown in Figure 2, for example, is a “saturation curve” for a magnetic sheet steel that is sometimes used in electric machinery. Note the magnetic field intensity is on a logarithmic scale. If this were plotted on linear coordinates the saturation would appear to be quite abrupt. At this point it is appropriate to note that the units used in magnetic field analysis are not always the same nor even consistent. In almost all systems the unit of flux is the weber (Wb), 4
Figure 2: Saturation Curve: Commercial M-19 Silicon Iron
Courtesy of United States Steel Corporation. (U.S. Steel). U.S. Steel accepts no liability for reliance on any information contained in the graphs shown above.
5
Flux Density
Remanent Flux Density B r
Saturation Flux Density Bs Magnetic Field
Coercive Field Hc
Saturation Field H s
Figure 3: Hysteresis Curve Nomenclature which is the same as a volt-second. In SI the unit of flux density is the tesla (T), but many people refer to the gauss (G), which has its origin in CGS. 10,000 G = 1 T. Now it gets worse, because there is an English system measure of flux density generally called kilo-lines per square inch. This is because in the English system the unit of flux is the line. 108 lines is equal to a weber. Thus a Tesla is 64.5 kilolines per square inch. The SI and CGS units of flux density are easy to reconcile, but the units of magnetic field are a bit harder. In SI we generally measure H in amperes/meter (or ampere-turns per meter). Often, however, you will see magnetic field represented as Oersteds (Oe). One Oe is the same as the magnetic field required to produce one gauss in free space. So 79.577 A/m is one Oe. In most useful magnetic materials the magnetic domaines tend to be somewhat “sticky”, and a more-than-incremental magnetic field is required to get them to move. This leads to the property called “hysteresis”, both useful and problematical in many magnetic systems. Hysteresis loops take many forms; a generalized picture of one is shown in Figure 3. Salient features of the hysteresis curve are the remanent magnetization Br and the coercive field Hc . Note that the actual loop that will be traced out is a function of field amplitude and history. Thus there are many other “minor loops” that might be traced out by the B-H characteristic of a piece of material, depending on just what the fields and fluxes have done and are doing. Hysteresis is important for two reasons. First, it represents the mechanism for “trapping” magnetic flux in a piece of material to form a permanent magnet. We will have more to say about that anon. Second, hysteresis is a loss mechanism. To show this, consider some arbitrary chunk of material for which we can characterize an MMF and a flux: F
= NI =
Φ =
�
�
~ · d~ℓ H
V dt = N 6
��
Area
~ · dA ~ B
Energy input to the chunk of material over some period of time is
w=
�
V Idt =
�
F dΦ =
� � t
~ · d~ℓ H
��
~ · dA ~ dt dB
Now, imagine carrying out the second (double) integral over a continuous set of surfaces which are perpendicular to the magnetic field H. (This IS possible!). The energy becomes: w=
� ���
~ · dBdvol ~ dt H
t
and, done over a complete cycle of some input waveform, that is: w = Wm =
���
�
t
Wm dvol vol ~ · dB ~ H
That last expression simply expresses the area of the hysteresis loop for the particular cycle. Generally, for most electric machine applications we will use magnetic material characterized as “soft”, having as narrow a hysteresis loop (and therefore as low a hysteretic loss) as possible. At the other end of the spectrum are “hard” magnetic materials which are used to make permanent magnets. The terminology comes from steel, in which soft, annealed steel material tends to have narrow loops and hardened steel tends to have wider loops. However permanent magnet technology has advanced to the point where the coercive forces possible in even cheap ceramic magnets far exceed those of the hardest steels.
3.3
Conduction, Eddy Currents and Laminations:
Steel, being a metal, is an electrical conductor. Thus when time varying magnetic fields pass through it they cause eddy currents to flow, and of course those produce dissipation. In fact, for almost all applications involving “soft” iron, eddy currents are the dominant source of loss. To reduce the eddy current loss, magnetic circuits of transformers and electric machines are almost invariably laminated, or made up of relatively thin sheets of steel. To further reduce losses the steel is alloyed with elements (often silicon) which poison the electrical conductivity. There are several approaches to estimating the loss due to eddy currents in steel sheets and in the surface of solid iron, and it is worthwhile to look at a few of them. It should be noted that this is a “hard” problem, since the behavior of the material itself is difficult to characterize.
3.4
Complete Penetration Case
Consider the problem of a stack of laminations. In particular, consider one sheet in the stack represented in Figure 4. It has thickness t and conductivity σ. Assume that the “skin depth” is much greater than the sheet thickness so that magnetic field penetrates the sheet completely. Further, assume that the applied magnetic flux density is parallel to the surface of the sheets: � �√ ~ = ~iz Re 2B0 ejωt B 7
y t
z
x
Figure 4: Lamination Section for Loss Calculation Now we can use Faraday’s law to determine the electric field and therefore current density in the sheet. If the problem is uniform in the x- and z- directions, ∂E x = −jω0 B0 ∂y Note also that, unless there is some net transport current in the x- direction, E must be antisymmetric about the center of the sheet. Thus if we take the origin of y to be in the center, electric field and current are: E x = −jωB0 y
J x = −jωB0 σy
Local power dissipated is
|J|2 σ To find average power dissipated we integrate over the thickness of the lamination: P (y) = ω 2 B02 σy 2 =
2 < P >= t
�
0
t 2
2 P (y)dy = ω 2 B02 σ t
�
t 2
y 2 dy =
0
1 2 2 2 ω B0 t σ 12
Pay attention to the orders of the various terms here: power is proportional to the square of flux density and to the square of frequency. It is also proportional to the square of the lamination thickness (this is average volume power dissipation). As an aside, consider a simple magnetic circuit made of this material, with some length ℓ and area A, so that volume of material is ℓA. Flux lined by a coil of N turns would be: Λ = N Φ = N AB0 and voltage is of course just V = jwL. Total power dissipated in this core would be: Pc = Aℓ
V2 1 2 2 2 ω B0 t σ = 12 Rc
where the equivalent core resistance is now
Rc =
A 12N2
ℓ σt2 8
B
B0
H
Figure 5: Idealized Saturating Characteristic
3.5
Eddy Currents in Saturating Iron
The same geometry holds for this pattern, although we consider only the one-dimensional problem (k → 0). The problem was worked by McLean and his graduate student Agarwal [2] [1]. They assumed that the magnetic field at the surface of the flat slab of material was sinusoidal in time and of high enough amplitude to saturate the material. This is true if the material has high permeability and the magnetic field is strong. What happens is that the impressed magnetic field saturates a region of material near the surface, leading to a magnetic flux density parallel to the surface. The depth of the region affected changes with time, and there is a separating surface (in the flat problem this is a plane) that moves away from the top surface in response to the change in the magnetic field. An electric field is developed to move the surface, and that magnetic field drives eddy currents in the material. Assume that the material has a perfectly rectangular magnetization curve as shown in Figure 5, so that flux density in the x- direction is: Bx = B0 sign(Hx ) The flux per unit width (in the z- direction) is: Φ=
� −∞ 0
Bx dy
and Faraday’s law becomes:
∂Φ ∂t while Ampere’s law in conjunction with Ohm’s law is: Ez =
∂Hx = σEz ∂y Now, McLean suggested a solution to this set in which there is a “separating surface” at depth ζ below the surface, as shown in Figure 6 . At any given time: �
Hx = Hs (t) 1 + Jz = σEz =
9
Hs ζ
y ζ
�
y x
Bs Separating Surface
Bs Penetration Depth
Figure 6: Separating Surface and Penetration Depth That is, in the region between the separating surface and the top of the material, electric field Ez is uniform and magnetic field Hx is a linear function of depth, falling from its impressed value at the surface to zero at the separating surface. Now: electric field is produced by the rate of change of flux which is: ∂Φ ∂ζ Ez = = 2Bx ∂t ∂t Eliminating E, we have: ∂ζ Hs 2ζ = ∂t σBx and then, if the impressed magnetic field is sinusoidal, this becomes: dζ 2 H0 = | sin ωt| dt σB0 This is easy to solve, assuming that ζ = 0 at t = 0, ζ=
�
ωt 2H0 sin ωσB0 2
Now: the surface always moves in the downward direction (as we have drawn it), so at each half cycle a new surface is created: the old one just stops moving at a maximum position, or penetration depth: � 2H0 δ= ωσB0 This penetration depth is analogous to the “skin depth” of the linear theory. However, it is an absolute penetration depth. The resulting electric field is: Ez =
ωt 2H0 cos σδ 2
0 < ωt < π
This may be Fourier analyzed: noting that if the impressed magnetic field is sinusoidal, only the time fundamental component of electric field is important, leading to: Ez =
8 H0 (cos ωt + 2 sin ωt + . . .) 3π σδ 10
Complex surface impedance is the ratio between the complex amplitude of electric and magnetic field, which becomes: E 8 1 (2 + j) Zs = z = Hx 3π σδ Thus, in practical applications, we can handle this surface much as we handle linear conductive surfaces, by establishing a skin depth and assuming that current flows within that skin depth of 16 and the “power factor” of this surface is the surface. The resistance is modified by the factor of 3π about 89 % (as opposed to a linear surface where the “power factor” is about 71 %. Agarwal suggests using a value for B0 of about 75 % of the saturation flux density of the steel.
4
Semi-Empirical Method of Handling Iron Loss
Neither of the models described so far are fully satisfactory in describing the behavior of laminated iron, because losses are a combination of eddy current and hysteresis losses. The rather simple model employed for eddy currents is precise because of its assumption of abrupt saturation. The hysteresis model, while precise, would require an empirical determination of the size of the hysteresis loops anyway. So we must often resort to empirical loss data. Manufacturers of lamination steel sheets will publish data, usually in the form of curves, for many of their products. Here are a few ways of looking at the data. A low frequency flux density vs. magnetic field (“saturation”) curve was shown in Figure 2. Included with that was a measure of the incremental permeability µ′ =
dB dH
In some machine applications either the “total” inductance (ratio of flux to MMF) or “incremental” inductance (slope of the flux to MMF curve) is required. In the limit of low frequency these numbers may be useful. For designing electric machines, however, a second way of looking at steel may be more useful. This is to measure the real and reactive power as a function of magnetic flux density and (sometimes) frequency. In principal, this data is immediately useful. In any well-designed electric machine the flux density in the core is distributed fairly uniformly and is not strongly affected by eddy currents, etc. in the core. Under such circumstances one can determine the flux density in each part of the core. With that information one can go to the published empirical data for real and reactive power and determine core loss and reactive power requirements. Figure 7 shows core loss and “apparent” power per unit mass as a function of (RMS) induction for 29 gage, fully processed M-19 steel. The two left-hand curves are the ones we will find most useful. “P ” denotes real power while “Pa ” denotes “apparent power”. The use of this data is quite straightforward. If the flux density in a machine is estimated for each part of the machine and the mass of steel calculated, then with the help of this chart a total core loss and apparent power can be estimated. Then the effect of the core may be approximated with a pair of elements in parallel with the terminals, with: Rc = Xc =
q|V |2 P q|V |2 Q 11
Figure 7: Real and Apparent Loss: M19, Fully Processed, 29 Ga
Courtesy of United States Steel Corporation. (U.S. Steel). U.S. Steel accepts no liability for reliance on any information contained in the graphs shown above.
12
M-19, 29 Ga, Fully Processed 100
Loss, W/Lb
10
Flux Density 1
0.1 T 0.3 T 0.5 T 0.7 T
0.1
1.0 T
0.01 10
100
1000
10000
Frequency, Hz
Figure 8: Steel Sheet Core Loss Fit vs. Flux Density and Frequency �
Q =
Pa2 − P 2
Where q is the number of machine phases and V is phase voltage. Note that this picture is, strictly speaking, only valid for the voltage and frequency for which the flux density was calculated. But it will be approximately true for small excursions in either voltage or frequency and therefore useful for estimating voltage drop due to exciting current and such matters. In design program applications these parameters can be re-calculated repeatedly if necessary. “Looking up” this data is a bit awkward for design studies, so it is often convenient to do a “curve fit” to the published data. There are a large number of possible ways of doing this. One method that has been found to work reasonably well for silicon iron is an “exponential fit”: P ≈ P0
�
B B0
�ǫB �
f f0
�ǫF
This fit is appropriate if the data appears on a log-log plot to lie in approximately straight lines. Figure 8 shows such a fit for the same steel sheet as the other figures. For “apparent power” the same sort of method can be used. It appears, however, that the simple exponential fit which works well for real power is inadequate, at least if relatively high inductions are to be used. This is because, as the steel saturates, the reactive component of exciting current rises rapidly. I have had some success with a “double exponential” fit: VA ≈ VA0
�
B B0
�ǫ0
13
+ VA1
�
B B0
�ǫ1
Table 1: Exponential Fit Parameters for Two Steel Sheets
Base Flux Density Base Frequency Base Power (w/lb) Flux Exponent Frequency Exponent Base Apparent Power 1 Base Apparent Power 2 Flux Exponent Flux Exponent
B0 f0 P0 ǫB ǫF V A0 V A1 ǫ0 ǫ1
29 Ga, Fully Processed M-19 M-36 1T 1T 60 Hz 60 Hz 0.59 0.67 1.88 1.86 1.53 1.48 1.08 1.33 .0144 .0119 1.70 2.01 16.1 17.2
To first order the reactive component of exciting current will be linear in frequency.
References [1] W. MacLean, “Theory of Strong Electromagnetic Waves in Massive Iron”, Journal of Applied Physics, V.25, No 10, October, 1954 [2] P.D. Agarwal, “Eddy-Current Losses in Solid and Laminated Iron”, Trans. AIEE, V. 78, pp 169-171, 1959
14
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machines Class Notes 4: Elementary Synchronous Machine Models c
2005 James L. Kirtley Jr.
1
September 14, 2005
Introduction
The objective here is to develop a simple but physically meaningful model of the synchronous machine, one of the major classes of electric machine. We can look at this model from several different directions. This will help develop an understanding of analysis of machines, particularly in cases where one or another analytical picture is more appropriate than others. Both operation and sizing will be of interest here. Along the way we will approach machine windings from two points of view. On the one hand, we will approximate windings as sinusoidal distributions of current and flux linkage. Then we will take a concentrated coil point of view and generalize that into a more realistic and useful winding model.
2
Physical Picture: Current Sheet Description
Consider this simple picture. The ‘machine’ consists of a cylindrical rotor and a cylindrical stator which are coaxial and which have sinusoidal current distributions on their surfaces: the outer surface of the rotor and the inner surface of the stator.
Stator
µ→∞
µ→∞
R
θ
K zs K zr
g
z
r
Rotor
Figure 1: Elementary Machine Model: Axial View The ‘rotor’ and ‘stator’ bodies are made of highly permeable material (we approximate this as being infinite for the time being, but this is something that needs to be looked at carefully later).
1
We also assume that the rotor and stator have current distributions that are axially (z) directed and sinusoidal: KzS
= KS cos pθ
KzR
= KR cos p (θ − φ)
Here, the angle φ is the physical angle of the rotor. The current distribution on the rotor is fixed with respect to the rotor. Now: assume that the air-gap dimension g is much less than the radius: g << R. It is not difficult to show that with this assumption the radial flux density Br is nearly uniform across the gap (i.e. not a function of radius) and obeys: K S + KzR ∂Br = −µ0 z ∂Rθ g Then the radial magnetic flux density for this case is simply: Br = −
µ0 R (KS sin pθ + KR sin p (θ − φ)) pg
Now it is possible to compute the traction on rotor and stator surfaces by recognizing that the surface current distributions are the azimuthal magnetic fields: at the surface of the stator, Hθ = −KzS , and at the surface of the rotor, Hθ = KzR . So at the surface of the rotor, traction is: τθ = Trθ = −
µ0 R (KS sin pθ + KR sin p (θ − φ)) KR cos p (θ − φ) pg
The average of that is simply: < τθ >= −
µ0 R KS KR sin pφ 2pg
The same exercise done at the surface of the stator yields the same results (with opposite sign). To find torque, use: µ0 πR3 ℓ T = 2πR2 ℓ < τθ >= KS KR sin pφ pg We pause here to make a few observations: 1. For a given value of surface currents Ks and Kr, torque goes as the third power of linear dimension. That implies that the the achieved shear stres is constant with machine size. And the ratio of machine torque density to machine volume is constant. 2. If, on the other hand, gap is held constant, torque goes as the fourth power of machine volume. Since the volume of the machine goes as the third power, this implies that torque capability goes as the 4/3 power of machine volume. 3. Actually, this understates the situation since the assumed surface current densities are the products of volume current densities and winding depth, which one would expect to increase with machine size. As machine radius grows one would expect both stator and rotor surface current densities to grow. Thus machine torque (and power) densities tend to increase somewhat faster than linearly with machine volume. 2
4. The current distributions want to align with each other. In actual practice what is done is to generate a stator current distribution which is not static as implied here but which rotates in space: KzS = KS cos (pθ − ωt) and this pulls the rotor along. 5. For a given pair of current distributions there is a maximum torque that can be sustained, but as long as the torque that is applied to the rotor is less than that value the rotor will adjust to the correct angle.
3
Continuous Approximation to Winding Patterns:
Now let’s try to produce those surface current distributions with physical windings. In fact we can’t do exactly that yet, but we can approximate a physical winding with a turns distribution that would look like: NS cos pθ 2R NR cos p (θ − φ) 2R
nS = nR =
Note that this implies that NS and NR are the total number of turns on the rotor and stator. i.e.: p
�
π 2
− π2
nS Rdθ = NS
Then the surface current densities are as we assumed above, with: KS =
N S IS 2R
KR =
NR IR 2R
So far nothing is different, but with an assumed number of turns we can proceed to computing inductances. It is important to remember what these assumed winding distributions mean: they are the density of wires along the surface of the rotor and stator. A positive value implies a wire with sense in the +z direction, a negative value implies a wire with sense in the -z direction. That is, if terminal current for a winding is positive, current is in the +z direction if n is positive, in the -z direction if n is negative. In fact, such a winding would be made of elementary coils with one half (the negatively going half) separated from the other half (the positively going half) by a physical angle of π/p. So the flux linked by that elemental coil would be: Φi (θ) =
�
θ
θ−π/p
µ0 Hr (θ ′ )ℓRdθ ′
So, if only the stator winding is excited, radial magnetic field is: Hr = −
N S IS sin pθ 2gp
3
and thus the elementary coil flux is: Φi (θ) =
µ0 NS IS ℓR cos pθ p2 g
Now, this is flux linked by an elementary coil. To get flux linked by a whole winding we must ‘add up’ the flux linkages of all of the elementary coils. In our continuous approximation to the real coil this is the same as integrating over the coil distribution: λS = p
�
π 2p
π − 2p
Φi (θ)nS (θ)Rdθ
This evaluates fairly easily to: λS = µ 0
π ℓRNS2 Is 4 gp2
which implies a self-inductance for the stator winding of: LS = µ 0
π ℓRNS2 4 gp2
The same process can be used to find self-inductance of the rotor winding (with appropriate changes of spatial variables), and the answer is: LR = µ 0
π ℓRNR2 4 gp2
To find the mutual inductance between the two windings, excite one and compute flux linked by the other. All of the expressions here can be used, and the answer is: M (φ) = µ0
π ℓRNS NR cos pφ 4 gp2
Now it is fairly easy to compute torque using conventional methods. Assuming both windings are excited, magnetic coenergy is: ′ Wm =
1 1 2 + M (φ)IS IR LS IS2 + LR IR 2 2
and then torque is:
T =
′ π ℓRNS NR ∂Wm = −µ0 IS IR sin pφ ∂φ 4 gp
and then substituting for NS IS and NR IR :
NS IS
= 2RKS
NR IR = 2RKR we get the same answer for torque as with the field approach: T = 2πR2 ℓ < τθ >= − 4
µ0 πR3 ℓ KS KR sin pφ pg
4
Classical, Lumped-Parameter Synchronous Machine:
Now we are in a position to examine the simplest model of a polyphase synchronous machine. Suppose we have a machine in which the rotor is the same as the one we were considering, but the stator has three separate windings, identical but with spatial orientation separated by an electrical angle of 120◦ = 2π/3. The three stator windings will have the same self- inductance (La ). With a little bit of examination it can be seen that the three stator windings will have mutual inductance, and that inductance will be characterized by the cosine of 120◦ . Since the physical angle between any pair of stator windings is the same, 1 Lab = Lac = Lbc = − La 2 There will also be a mutual inductance between the rotor and each phase of the stator. Using M to denote the magnitude of that inductance: M Maf Mbf Mcf
π ℓRNa Nf 4 gp2 = M cos (pφ)
= µ0
2π 3 � � 2π = M cos pφ + 3 �
�
= M cos pφ −
We show in Chapter 1 of these notes that torque for this system is: 2π T = −pM ia if sin (pφ) − pM ib if sin pφ − 3 �
5
�
2π − pM ic if sin pφ + 3 �
�
Balanced Operation:
Now, suppose the machine is operated in this fashion: the rotor turns at a constant velocity, the field current is held constant, and the three stator currents are sinusoids in time, with the same amplitude and with phases that differ by 120 degrees. pφ = ωt + δi if
= If
ia = I cos (ωt) 2π 3 � � 2π = I cos ωt + 3 �
ib = I cos ωt − ic
�
Straightforward (but tedious) manipulation yields an expression for torque: 3 T = − pM IIf sin δi 2 5
Operated in this way, with balanced currents and with the mechanical speed consistent with the electrical frequency (pΩ = ω), the machine exhibits a constant torque. The phase angle δi is called the torque angle, but it is important to use some caution, as there is more than one torque angle. Now, look at the machine from the electrical terminals. Flux linked by Phase A will be: λa = La ia + Lab ib + Lac ic + M If cos pφ Noting that the sum of phase currents is, under balanced conditions, zero and that the mutual phase-phase inductances are equal, this simplifies to: λa = (La − Lab ) ia + M If cos pφ = Ld ia + M If cos pφ where we use the notation Ld to denote synchronous inductance. Now, if the machine is turning at a speed consistent with the electrical frequency we say it is operating synchronously, and it is possible to employ complex notation in the sinusoidal steady state. Then, note: � � ia = I cos (ωt + θi ) = Re Iejωt+θi If , we can write an expression for the complex amplitude of flux as: �
λa = Re Λa ejωt where we have used this complex notation:
�
I = Iejθi = If ejθm
If
Now, if we look for terminal voltage of this system, it is: va =
� � dλa = Re jωΛa ejωt dt
This system is described by the equivalent circuit shown in Figure 2. jXd
I
+
+
E af
V
-
-
Figure 2: Round Rotor Synchronous Machine Equivalent Circuit where the internal voltage is: E af = jωM If ejθm 6
Now, if that is connected to a voltage source (i.e. if V is fixed), terminal current is: I=
V − Eaf ejδ jXd
where Xd = ωLd is the synchronous reactance. Then real and reactive power (in phase A) are: 1 ∗ VI 2 � �∗ V − Eaf ejδ 1 V 2 jXd
P + jQ = =
1 |V |2 1 V Eaf ejδ − 2 −jXd 2 −jXd
= This makes real and reactive power:
1 V Eaf sin δ 2 Xd 1V2 1 V Eaf − cos δ 2 Xd 2 Xd
Pa = − Qa =
If we consider all three phases, real power is P =−
3 V Eaf sin δ 2 Xd
Now, at last we need to look at actual operation of these machines, which can serve either as motors or as generators. Vector diagrams that describe operation as a motor and as a generator are shown in Figures 3 and 4, respectively.
Ia V V
δ
δ
Ia jXdIa
jXdIa Eaf Eaf
Figure 3: Motor Operation, Over- and Under- Excited
Operation as a generator is not much different from operation as a motor, but it is common to make notations with the terminal current given the opposite (“generator”) sign. 7
Eaf
Eaf jXdIg
jXdIg Ig Ig
V
V
Figure 4: Generator Operation, Over- and Under- Excited
6
Reconciliation of Models
We have determined that we can predict its power and/or torque characteristics from two points of view : first, by knowing currents in the rotor and stator we could derive an expression for torque vs. a power angle: 3 T = − pM IIf sin δi 2 From a circuit point of view, it is possible to derive an expression for power: P =−
3 V Eaf sin δ 2 Xd
and of course since power is torque times speed, this implies that: T =−
3 pV Eaf 3 V Eaf sin δ = − sin δ 2 ΩXd 2 ωXd
In this section of the notes we will, first of all, reconcile these notions, look a bit more at what they mean, and then generalize our simple theory to salient pole machines as an introduction to two-axis theory of electric machines.
6.1
Torque Angles:
Figure 5 shows a vector diagram that shows operation of a synchronous motor. It represents the MMF’s and fluxes from the rotor and stator in their respective positions in space during normal operation. Terminal flux is chosen to be ‘real’, or occupy the horizontal position. In motor operation the rotor lags by angle δ, so the rotor flux M If is shown in that position. Stator current is also shown, and the torque angle between it and the rotor, δi is also shown. Now, note that the dotted line OA, drawn perpendicular to a line drawn between the stator flux Ld I and terminal flux Λt , has length: |OA| = Ld I sin δi = Λt sin δ
Then, noting that terminal voltage V = ωΛt , Ea = ωM If and Xd = ωLd , straightforward substitution yields: 3 3 pV Eaf sin δ = pM IIf sin δi 2 ωXd 2
8
LdI
A
i
t
O
MIf
Figure 5: Synchronous Machine Phasor Addition So the current- and voltage- based pictures do give the same result for torque.
7
Per-Unit Systems:
Before going on, we should take a short detour to look into per-unit systems, a notational device that, in addition to being convenient, will sometimes be conceptually helpful. The basic notion is quite simple: for most variables we will note a base quantity and then, by dividing the variable by the base we have a per-unit version of that variable. Generally we will want to tie the base quantity to some aspect of normal operation. So, for example, we might make the base voltage and current correspond with machine rating. If that is the case, then power base becomes: PB = 3VB IB and we can define, in similar fashion, an impedance base: ZB =
VB IB
Now, a little caution is required here. We have defined voltage base as line-neutral and current base as line current (both RMS). That is not necessary. In a three phase system we could very well have defined base voltage to have been line-line and base current to be current in a delta connected element: √ IB IBΔ = √ VBΔ = 3VB 3 In that case the base power would be unchanged but base impedance would differ by a factor of three: PB = VBΔ IBΔ ZBΔ = 3ZB However, if we were consistent with actual impedances (note that a delta connection of elements of
impedance 3Z is equivalent to a wye connection of Z), the per-unit impedances of a given system
are not dependent on the particular connection. In fact one of the major advantages of using a
9
per-unit system is that per-unit values are uniquely determined, while ordinary variables can be line-line, line-neutral, RMS, peak, etc., for a large number of variations. Perhaps unfortunate is the fact that base quantities are usually given as line-line voltage and base power. So that: VB 1 VBΔ V2 PB ZB = = BΔ IB = √ = IB 3 IBΔ PB 3VBΔ Now, we will usually write per-unit variables as lower-case versions of the ordinary variables: v=
V VB
p=
P PB
etc.
Thus, written in per-unit notation, real and reactive power for a synchronous machine operating in steady state are: veaf veaf v2 sin δ q= − cos δ p=− xd xd xd These are, of course, in motor reference coordinates, and represent real and reactive power into the terminals of the machine.
8
Normal Operation:
The synchronous machine is used, essentially interchangeably, as a motor and as a generator. Note that, as a motor, this type of machine produces torque only when it is running at synchronous speed. This is not, of course, a problem for a turbogenerator which is started by its prime mover (e.g. a steam turbine). Many synchronous motors are started as induction machines on their damper cages (sometimes called starting cages). And of course with power electronic drives the machine can often be considered to be “in synchronism” even down to zero speed. As either a motor or as a generator, the synchronous machine can either produce or consume reactive power. In normal operation real power is dictated by the load (if a motor) or the prime mover (if a generator), and reactive power is determined by the real power and by field current. Figure 6 shows one way of representing the capability of a synchronous machine. This picture represents operation as a generator, so the signs of p and q are reversed, but all of the other elements of operation are as we ordinarily would expect. If we plot p and q (calculated in the normal way) against each other, we see the construction at the right. If we start at a location q = −v 2 /xd , (and remember that normally v = 1 per-unit) , then the locus of p and q is what would be obtained by swinging a vector of length veaf /xd over an angle δ. This is called a capability chart because it is an easy way of visualizing what the synchronous machine (in this case generator) can do. There are three easily noted limits to capability. The upper limit is a circle (the one traced out by that vector) which is referred to as field capability. The second limit is a circle that describes constant |p + jq |. This is, of course, related to the magnitude of armature current and so this limit is called armature capability. The final limit is related to machine stability, since the torque angle cannot go beyond 90 degrees. In actuality there are often other limits that can be represented on this type of a chart. For example, large synchronous generators typically have a problem with heating of the stator iron when they attempt to operate in highly underexcited conditions (q strongly negative), so that one will often see another limit that prevents the operation of the machine near its stability
10
Q
Field Limit
Stator Limit P 1 X
d
Stability Limit
Figure 6: Synchronous Generator Capability Diagram limit. In very large machines with more than one cooling state (e.g. different values of cooling hydrogen pressure) there may be multiple curves for some or all of the limits. Another way of describing the limitations of a synchronous machine is embodied in the Vee Curve. An example is shown in Figure 7 . This is a cross-plot of magnitude of armature current with field current. Note that the field and armature current limits are straightforward (and are the right-hand and upper boundaries, respectively, of the chart). The machine stability limit is what terminates each of the curves at the upper left-hand edge. Note that each curve has a minimum at unity power factor. In fact, there is yet another cross-plot possible, called a compounding curve, in which field current is plotted against real power for fixed power factor.
9
Salient Pole Machines: Two-Reaction Theory
So far, we have been describing what are referred to as “round rotor” machines, in which stator reactance is not dependent on rotor position. This is a pretty good approximation for large turbine generators and many smaller two-pole machines, but it is not a good approximation for many synchronous motors nor for slower speed generators. For many such applications it is more cost effective to wind the field conductors around steel bodies (called poles) which are then fastened onto the rotor body, with bolts or dovetail joints. These produce magnetic anisotropies into the machine which affect its operation. The theory which follows is an introduction to two-reaction theory and consequently for the rotating field transformations that form the basis for most modern dynamic analyses. Figure 8 shows a very schematic picture of the salient pole machine, intended primarily to show how to frame this analysis. As with the round rotor machine the stator winding is located in slots in the surface of a highly permeable stator core annulus. The field winding is wound around steel pole pieces. We separate the stator current sheet into two components: one aligned with and one in quadrature to the field. Remember that these two current components are themselves (linear) combinations of the stator phase currents. The transformation between phase currents and the d11
Vee Curves 1.2
Per-Unit Ia
1
0.8 0.6 0.4 0.2
0
0
0.5
1
1.5
2
2.5
3
Per-Unit Field Current Per-Unit Real Power:
0.00
0.32
0.64
0.96
Figure 7: Synchronous Machine Vee Curve and q- axis components is straightforward and will appear in Chapter 8 of these notes. The key here is to separate MMF and flux into two orthogonal components and to pretend that each can be treated as sinusoidal. The two components are aligned with the direct axis and with the quadrature axis of the machine. The direct axis is aligned with the field winding, while the quadrature axis leads the direct by 90 degrees. Then, if φ is the angle between the direct axis and the axis of phase a, we can write for flux linking phase a: λs = λd cos φ − λq sin φ Then, in steady state operation, if Va =
dλa dt
and φ = ωt + δ ,
Va = −ωλs sin φ − ωλq cos φ which allows us to define: Vd = −ωλq Vq = ωλd
one might think of the ‘voltage’ vector as leading the ‘flux’ vector by 90 degrees. Now, if the machine is linear, those fluxes are given by: λd = Ld Id + M If λ q = L q Iq Note that, in general, Ld 6= Lq . In wound-field synchronous machines, usually Ld > Lq . The reverse is true for most salient (buried magnet) permanent magnet machines. 12
d axis
Iq
q axis
I
Id
f
Figure 8: Cartoon of a Salient Pole Synchronous Machine
V
Vq
V d Figure 9: Resolution of Terminal Voltage
Referring to Figure 9, one can resolve terminal voltage into these components: Vd = V sin δ Vq = V cos δ or: Vd = −ωλq = −ωLq Iq = V sin δ
Vq = ωλd = ωLd Id + ωM If = V cos δ
which is easily inverted to produce: V cos δ − Eaf Xd V sin δ = − Xq
Id = Iq
13
where Xd = ωLd
Xq = ωLq
Eaf = ωM If
Now, we are working in ordinary variables (this discussion should help motivate the use of perunit!), and each of these variables is peak amplitude. Then, if we take up a complex frame of reference: = Vd + jVq
V
I = Id + jIq complex power is: 3 3 P + jQ = V I ∗ = {(Vd Id + Vq Iq ) + j (Vq Id − Vd Iq )} 2 2 or: P
�
3 = − 2
Q =
3 2
�
V2 V Eaf sin δ + Xd 2
V2 2
�
1 1 + Xd Xq
I
�
1 1 − Xd Xq
�
V2 − 2
d
ψ
�
�
sin 2δ
1 1 − Xd Xq
�
�
V Eaf cos δ cos 2δ − Xd
�
I V
I
q
δ
d axix
j X Id d jX I q q axis E
1
jX I q q E
af
Figure 10: Phasor Diagram: Salient Pole Machine A phasor diagram for a salient pole machine is shown in Figure 10. This is a little different from the equivalent picture for a round-rotor machine, in that stator current has been separated into its d- and q- axis components, and the voltage drops associated with those components have been drawn separately. It is interesting and helpful to recognize that the internal voltage Eaf can be expressed as: Eaf = E1 + (Xd − Xq ) Id 14
where the voltage E1 is on the quadrature axis. In fact, E1 would be the internal voltage of a round rotor machine with reactance Xq and the same stator current and terminal voltage. Then the operating point is found fairly easily: δ = − tan �
−1
�
Xq I cos ψ V + Xq I sin ψ
�
(V + Xq I sin ψ)2 + (Xq I cos ψ)2
E1 =
Power-Angle Curves 1.5
xd=2.2 1
xq = 1.6
Per-Unit
0.5
Round Rotor 0 -4
-3
-2
-1
Salient Rotor 0
1
2
3
4
-0.5
-1
-1.5 Torque Angle
Figure 11: Torque-Angle Curves: Round Rotor and Salient Pole Machines A comparison of torque-angle curves for a pair of machines, one with a round, one with a salient rotor is shown in Figure 11 . It is not too difficult to see why power systems analysts often neglect saliency in doing things like transient stability calculations.
10
Relating Rating to Size
It is possible, even with the simple model we have developed so far, to establish a quantitative relationship between machine size and rating, depending (of course) on elements such as useful flux and surface current density. To start, note that the rating of a machine (motor or generator) is: |P + jQ| = qV I where q is the number of phases, V is the RMS voltage in each phase and I is the RMS current. To establish machine rating we must establish voltage and current, and we do these separately. 15
10.1
Voltage
Assume that our sinusoidal approximation for turns density is valid: Na cos pθ 2R
na (θ) = And suppose that working flux density is:
Br (θ) = B0 sin p(θ − φ) Now, to compute flux linked by the winding (and consequently to compute voltage), we first compute flux linked by an incremental coil: λi (θ) =
�
θ+ π p θ
ℓBr (θ ′ )Rdθ ′
Then flux linked by the whole coil is: �
λa = p
π 2p
π − 2p
λi (θ)na (θ)Rdθ =
π 2ℓRNa B0 cos pφ 4 p
This is instantaneous flux linked when the rotor is at angle φ. If the machine is operating at some electrical frequency ω with a phase angle so that pφ = ωt + δ, the RMS magnitude of terminal voltage is: ωπ B0 Va = 2ℓRNa √ p4 2 Finally, note that the useful peak current density that can be used is limited by the fraction of machine periphery used for slots: B0 = Bs (1 − λs ) where Bs is the flux density in the teeth, limited by saturation of the magnetic material.
10.2
Current
The (RMS) magnitude of the current sheet produced by a current of (RMS) magnitude I is: Kz =
q Na I 2 2R
And then the current is, in terms of the current sheet magnitude: I = 2RKz
2 qNa
Note that the surface current density is, in terms of area current density Js , slot space factor λs and slot depth hs : Kz = λs Js hs This gives terminal current in terms of dimensions and useful current density:
I=
4R
λs hs Js qNa 16
10.3
Rating
Assembling these expressions, machine rating becomes: ω Bs 2πR2 ℓ √ λs (1 − λs ) hs Js p 2
|P + jQ| = qV I =
This expression is actually fairly easily interpreted. The product of slot factor times one minus slot factor optimizes rather quickly to 1/4 (when λs = 1). We could interpret this as: |P = jQ| = As us τ ∗ where the interaction area is: As = 2πRℓ The surface velocity of interaction is: us =
ω R = ΩR p
and the fragment of expression which “looks like” traction is: Bs τ ∗ = hs Js √ λs (1 − λs ) 2 Note that this is not quite traction since the current and magnetic flux may not be ideally aligned, and this is why the expression incorporates reactive as well as real power. This is not quite yet the whole story. The limit on Bs is easily understood to be caused by saturation of magnetic material. The other important element on shear stress density, hs Js is a little more involved.
10.4
Role of Reactance
The per-unit, or normalized synchronous reactance is: xd = Xd
I µ0 R λs √ hs Js 2 = V pg 1 − λs Bs
While this may be somewhat interesting by itself, it becomes useful if we solve it for hs Ja : hs Ja = xd g
p(1 − λs )Bs √ µ0 Rλs 2
That is, if xd is fixed, hs Ja (and so power) are directly related to air-gap g. Now, to get a limit on g, we must answer the question of how far the field winding can “throw” effective air-gap flux? To understand this question, we must calculate the field current to produce rated voltage, no-load, and then the excess of field current required to accommodate load current. Under rated operation, per- unit field voltage is: e2af = v 2 + (xd i)2 + 2xd i sin ψ Or, if at rated conditions v and i are both unity (one per- unit), then eaf =
�
1 + x2d + 2xd sin ψ 17
10.5
Field Winding
Thus, given a value for xd and ψ, per- unit internal voltage eaf is also fixed. Then field current required can be calculated by first estimating field winding current for “no-load operation”. Br =
µ0 Nf If nl 2gp
and rated field current is: If = If nl eaf or, required rated field current is: N f If =
2gp(1 − λs )Bs eaf µ0
Next, If can be related to a field current density: Nf If =
NRS ARS Jf 2
where NRS is the number of rotor slots and the rotor slot area ARS is ARS = wR hR where hR is rotor slot height and wR is rotor slot width: wR =
2πR λR NRS
Then: Nf If = πRλR hR Jf Now we have a value for air- gap g: g=
π µ0 RλR hR Jf 2 p(1 − λs )Bs eaf
This then gives us useful armature surface current density: π xd λR hs Js = √ hR Jf 2 2 eaf λs We will not have a lot more to say about this. Note that the ratio of xd /eaf can be quite small (if the per-unit reactance is small), will never be a very large number for any practical machine, and is generally less than one. As a practical matter it is unusual for the per-unit synchronous reatance of a machine to be larger than about 2 or 2.25 per-unit. What this tells us should be obvious: either the rotor or the stator of a machine can produce the dominant limitation on shear stress density (and so on rating). The best designs are “balanced”, with both limits being reached at the same time.
18
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machinery Class Notes 5: Winding Inductances c
2005 James L. Kirtley Jr.
1
September 5, 2005
Introduction
The purpose of this document is to show how the inductances of windings in round- rotor machines with narrow air gaps may be calculated. We deal only with the idealized air- gap magnetic fields, and do not consider slot, end winding, peripheral or skew reactances. We do, however, consider the space harmonics of winding magneto-motive force (MMF).
2
Description of Stators Back Iron
Slots
Slot Depression
Teeth
Figure 1: Stator Cross-Section Figure 1 shows a cartoon view of an axial cross-section of a twelve-slot stator. Actually, what is shown is the shape of a thin sheet of steel, or lamination that is used to make up the magnetic circuit. The iron is made of thin sheets to control eddy current losses. Thickness varies according to freuqency of operation, but in machines for 60 Hz (the vast bulk of machines made for industrial 1
use), lamination thickness is typically .014” (.355 mm). These are stacked to make the magnetic circuit of the appropriate length. Windings are carried in the slots of this structure. Figure 1 shows trapezoidal slots with teeth of approximately uniform cross-section over most of their length but wider extent near the air-gap. The tooth ends, in combination with the relatively narrow slot depression region, help control certain parasitic losses in the rotor of many machines by improving uniformity of the air-gap fields, increase the air-gap permeance and help hold the windings in the slots. It should be noted that large machines, with what are called “form wound” coils, have straight-sided rectangular slots and consequently teeth of non-uniform cross-section. The description that follows will hold for both types of machine.
A
A
C’
C’
B
B
A’
A’
C
C
B’
B’
1
2
3
4
5
6
7
8
9
10
11
12
Figure 2: Full-Pitched Winding To simplify the discussion, imagine the slot/tooth region to be “straightened out” as shown in Figure 2. This shows a three-phase, two-pole winding in the twelve slots. Such a winding would have two slots per pole per phase. One of the two coils of phase A would be wound in slots 1 and 7 (six slots apart).
A A
A
C’
C’
B
B
A’
A’
C
C
B’
C’
C’
B
B
A’
A’
C
C
B’
B’
B’ A
1
2
3
4
5
6
7
8
9
10
11
12
Figure 3: Five-Sixths-Pitched Winding Machines are seldom wound as shown in Figure 2 for a variety of reasons. It is usually advantageous in reducing the length of the end turns and to reducing space harmonic effects in the machine (usually bad effects!) to wind the machine with “short-pitched” windings as shown in Figure 3. Each phase in this case consists of four coils (two per slot). The four coils of Phase A would span between slots 1 and 6, slots 2 and 7, slots 7 and 12 and slots 8 and 1. Each of these coil spans is five slots, so this choice of winding pattern is referred to as “Five-Sixths” pitch. So this cartoon-figure machine stator (which could represent either a synchronous or induction motor or generator) has both breadth because there are more than one slots per pole per phase, and it may have the need for accounting for winding pitch. What follows in this note is a simple protocol for estimating the important air-gap fields and inductances.
2
3
Winding MMF
To start, consider the MMF of a full- pitch, concentrated winding as shown in schematic form in Figure 4. Assuming that the winding has a total of N turns over p pole- pairs, and is carrying current I the MMF is: ∞ � 4 NI sin npθ (1) F =− nπ 2p n=1 nodd This distribution is shown, as a function of angle θ in Figure 5. This leads directly to magnetic flux density in the air- gap: ∞ �
Br = −
n=1 nodd
µ0 4 N I sin npθ g nπ 2p
(2)
Note that a real winding, which will most likely not be full- pitched and concentrated, will have a winding factor which is the product of pitch and breadth factors, to be discussed later.
Magnetic Circuit: Stator Rotor NI p
µ
z
R
g
θ r
Air-Gap µ
Figure 4: Primitive Geometry Problem Now, suppose that there is a polyphase winding, consisting of more than one phase (we will use three phases), driven with one of two types of current. The first of these is balanced, current: Ia = I cos(ωt) 2π ) 3 2π = I cos(ωt + ) 3
Ib = I cos(ωt − Ic
(3)
Conversely, we might consider Zero Sequence currents: Ia = Ib = Ic = I cos ωt 3
(4)
F( θ ) NI p
π
π
2p
p
3π 2p
2
π
θ
p
Figure 5: Air-Gap MMF Then it is possible to express magnetic flux density for the two distinct cases. For the balanced case: ∞ Br =
�
Brn sin(npθ ∓ ωt)
n=1
where
(5)
• The upper sign holds for n = 1, 7, ... • The lower sign holds for n = 5, 11, ... • all other terms are zero and
3 µ0 4 N I 2 g nπ 2p The zero- sequence case is simpler: it is nonzero only for the triplen harmonics: Brn =
Br =
∞ �
µ0 4 N I 3 (sin(npθ − ωt) + sin(npθ + ωt)) g nπ 2p 2 n=3,9,...
(6)
(7)
Next, consider the flux from a winding on the rotor: that will have the same form as the flux produced by a single armature winding, but will be referred to the rotor position: ∞ �
Brf =
µ0 4 N I sin npθ ′ g nπ 2p
(8)
µ0 4 N I sin n(pθ − ωt) g nπ 2p
(9)
n=1 nodd which is, substituting θ ′ = θ −
ωt p ,
Brf =
∞ �
n=1 nodd
The next step here is to find the flux linked if we have some air- gap flux density of the form: Br =
∞ �
n=1
Brn sin(npθ ± ωt) 4
(10)
Now, it is possible to calculate flux linked by a single-turn, full-pitched winding by:
�
Φ=
π p
0
Br Rldθ
(11)
and, using (10), this is: Φ = 2Rl
∞ � Brn
n=1
np
cos(ωt)
(12)
This allows us to compute self- and mutual- inductances, since winding flux is: λ = NΦ
(13)
The end of this is a set of expressions for various inductances. It should be noted that, in the real world, most windings are not full-pitched nor concentrated. Fortunately, these shortcomings can be accommodated by the use of winding factors. The simplest and perhaps best definition of a winding factor is the ratio of flux linked by an actual winding to flux that would have been linked by a full- pitch, concentrated winding with the same number of turns. That is: λactual (14) kw = λf ull−pitch It is relatively easy to show, using reciprocity arguments, that the winding factors are also the ratio of effective MMF produced by an actual winding to the MMF that would have been produced by the same winding were it to be full- pitched and concentrated. The argument goes as follows: mutual inductance between any pair of windings is reciprocal. That is, if the windings are designated one and two, the mutual inductance is flux induced in winding one by current in winding two, and it is also flux induced in winding two by current in winding one. Since each winding has a winding factor that influences its linking flux, and since the mutual inductance must be reciprocal, the same winding factor must influence the MMF produced by the winding. The winding factors are often expressed for each space harmonic, although sometimes when a winding factor is referred to without reference to a harmonic number, what is meant is the space factor for the space fundamental. Two winding factors are commonly specified for ordinary, regular windings. These are usually called pitch and breadth factors, reflecting the fact that often windings are not full pitched, which means that individual turns do not span a full π electrical radians and that the windings occupy a range or breadth of slots within a phase belt. The breadth factors are ratios of flux linked by a given winding to the flux that would be linked by that winding were it full- pitched and concentrated. These two winding factors are discussed in a little more detail below. What is interesting to note, although we do not prove it here, is that the winding factor of any given winding is the product of the pitch and breadth factors: kw = kp kb (15) With winding factors as defined by (14) and the sections below, it is possible to define winding inductances. For example, the synchronous inductance of a winding will be the apparent inductance of one phase when the polyphase winding is driven by a balanced set of currents as in (3). This is, approximately: ∞ 2 � 3 4 µ0 N 2 Rlkwn (16) Ld = 2π p2 gn2 n=1,5,7,... 5
This expression is approximate because it ignores the asynchronous interactions between higher order harmonics and the rotor of the machine. These are beyond the scope of this note. Zero- sequence inductance is the ratio of flux to current if a winding is excited by zero sequence currents, as in (4): ∞ 2 � 4 µ0 N 2 Rlkwn (17) L0 = 3 π p2 gn2 n=3,9,... And then mutual inductance, as between a field winding (f ) and an armature winding (a), is: M (φ) =
∞ �
n=1 nodd
4
4 µ0 Nf Na kf n kan Rl cos(npφ) π p2 gn2
(18)
Winding Factors
Now we turn our attention to computing the winding factors for simple, regular winding patterns. We do not prove but only state that the winding factor can, for regular winding patterns, be expressed as the product of a pitch factor and a breadth factor, each of which can be estimated separately.
4.1
Pitch Factor
α θ z
r
Figure 6: Short-Pitched Coils Pitch factor is found by considering the flux linked by a less- than- full pitched winding. Consider the situation in which radial magnetic flux density is: Br = Bn sin(npθ − ωt)
(19)
A winding with pitch α will link flux (see Figure 6: λ = Nl
�
π + 2αp 2p π − 2αp 2p
Bn sin(npθ − ωt)Rdθ 6
(20)
Pitch α refers to the angular displacement between sides of the coil, expressed in electrical radians. For a full- pitch coil α = π. The flux linked is: nπ nα 2N lRBn λ= sin( ) sin( ) (21) np 2 2 Using the definition (14), the pitch factor is seen to be: kpn = sin
4.2
nα 2
(22)
Breadth Factor
Now for breadth factor. This describes the fact that a winding may consist of a number of coils, each linking flux slightly out of phase with the others. A regular winding will have a number (say m) coil elements, separated by electrical angle γ. (See Figure 7
θ
γ z
r
Figure 7: Distributed Coils A full- pitch coil with one side at angle ξ will, in the presence of magnetic flux density as described by (19), link flux: λ = Nl
�
π − pξ p ξ p
Bn sin(npθ − ωt)Rdθ
(23)
This is readily evaluated to be: λ=
� � 2N lRBn Re ej(ωt−nξ) np
(24)
where in (24), complex number notation has been used for convenience in carrying out the rest of this derivation. What happens here is that the coils link fluxes that differ in phase, so the addition of flux is as shown in vector form in Figure 8. 7
Individual Flux Linkages
Total Flux Linkage
Figure 8: Vector Flux Addition Now: if the winding is distributed into m sets of slots and the slots are evenly spaced, the angular position of each slot will be: ξi = iγ − and the number of turns in each slot will be λ=
N mp ,
m−1 γ 2
(25)
so that actual flux linked will be:
� � � 2N lRBn 1 m−1 Re ej(ωt−nξi ) np m i=0
(26)
� m−1 1 m−1 e−jn(iγ− 2 γ) m i=0
(27)
The breadth factor is then simply: kb = Note that (27) can be written as: m−1 m ejnγ 2 � kb = e−jniγ m i=0
(28)
Now, focus on that sum. We know that any coverging geometric sum has a simple sum: ∞ �
xi =
i=0
and that a truncated sum is:
m−1 � i=0
Then the sum in (28) can be written as: m−1 � i=0
�
=
1 1−x
∞ � i=0
e−jniγ = 1 − ejnmγ
−
(29)
∞ �
(30)
i=m
∞ �� i=0
e−jniγ =
1 − ejnmγ 1 − e−jnγ
(31)
Now, inserting the results of (31) into (28), and using the definitions for sine, the breadth factor is found: sin nmγ 2 (32) kbn = m sin nγ 2 8
4.3
Alternate Derivation of Breadth Factor
Most textbooks, if they bother to prove the Breadth Factor, use a geometric proof as shown in Figure 9.
C A
B
γ 2
mγ 2
O
Figure 9: Alternate Proof of Breadth Factor The short vectors (e.g. AC) represent the voltages induced in individual coils. In fact, what is shown in this figure is the same as is shown in Figure 8, but spread out to show the actual addition. Now, note that if each of the vectors is bisected by a line segment at right angles, all of those line segments meet at point O. The line segment that includes OB is one of these. Line segments that run from O to the ends of the vectors will have an angle γ2 from the bisectors of the vectors.
Similarly, the line segment OA has an angle of mγ 2 with respect to the bisector of the resultant voltage vector. Now, if we note F1 as the length of each of the individual coil voltage vectors and F as the length of the resultant sum, the length of half of the bisector is: AB = but then Then the resultant vector is:
mγ F = OA sin 2 2
(33)
1 γ 1 AC = F1 = OA sin 2 2 2
(34)
sin m γ2 F = 2AB = mF1 m sin γ2
(35)
9
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machines Class Notes 6: DC (Commutator) and Permanent Magnet Machines c
2005 James L. Kirtley Jr.
1
September 5, 2005
Introduction
Virtually all electric machines, and all practical electric machines employ some form of rotating or alternating field/current system to produce torque. While it is possible to produce a “true DC” machine (e.g. the “Faraday Disk”), for practical reasons such machines have not reached application and are not likely to. In the machines we have examined so far the machine is operated from an alternating voltage source. Indeed, this is one of the principal reasons for employing AC in power systems. The first electric machines employed a mechanical switch, in the form of a carbon brush/commutator system, to produce this rotating field. While the widespread use of power electronics is making “brushless” motors (which are really just synchronous machines) more popular and common, commutator machines are still economically very important. They are relatively cheap, particularly in small sizes, and they tend to be rugged and simple. You will find commutator machines in a very wide range of applications. The starting motor on all automobiles is a series-connected commutator machine. Many of the other electric motors in automobiles, from the little motors that drive the outside rear-view mirrors to the motors that drive the windshield wipers are permanent magnet commutator machines. The large traction motors that drive subway trains and diesel/electric locomotives are DC commutator machines (although induction machines are making some inroads here). And many common appliances use “universal” motors: series connected commutator motors adapted to AC.
1.1
Geometry:
Stator Yoke Field Poles
Rotor
Ω
Armature Winding Field Winding
Figure 1: Wound-Field DC Machine Geometry
1
A schematic picture (“cartoon”) of a commutator type machine is shown in 1. The armature of this machine is on the rotor (this is the part that handles the electric power), and current is fed to the armature through the brush/commutator system. The interaction magnetic field is provided (in this picture) by a field winding. A permanent magnet field is applicable here, and we will have quite a lot more to say about such arrangements below. Now, if we assume that the interaction magnetic flux density averages Br , and if there are Ca conductors underneath the poles at any one time, and if there are m parallel paths, then we may estimate torque produced by the machine by: Te =
Ca RℓBr Ia m
where R and ℓ are rotor radius and length, respectively and Ia is terminal current. Note that Ca is not necessarily the total number of conductors, but rather the total number of active conductors (that is, conductors underneath the pole and therefore subject to the interaction field). Now, if we note Nf as the number of field turns per pole, the interaction field is just: Br = µ 0
N f If g
leading to a simple expression for torque in terms of the two currents: Te = GIa If where G is now the motor coefficient (units of N-m/ampere squared): G = µ0
Ca Nf Rℓ m g
Now, let’s go back and look at this from the point of view of voltage. Start with Faraday’s Law: ~ ~ = − ∂B ∇×E ∂t Integrating both sides and noting that the area integral of a curl is the edge integral of the quantity, we find: �� � ~ ∂B ~ · dℓ = − E ∂t Now, that is a bit awkward to use, particularly in the case we have here in which the edge of the contour is moving (note we will be using this expression to find voltage). We can make this a bit more convenient to use if we note: d dt
��
~ · ~n da = B
��
~ ∂B · ~n da + ∂t
�
~ · d~ℓ ~v × B
where ~v is the velocity of the contour. This gives us a convenient way of noting the apparent electric field within a moving object (as in the conductors in a DC machine): ~′ = E ~ + ~v × B ~ E Now, note that the armature conductors are moving through the magnetic field produced by the stator (field) poles, and we can ascribe to them an axially directed electric field: Ez = −RΩBr 2
H
B
H
dl
H
v
Figure 2: Motion of a contour through a magnetic field produces flux change and electric field in the moving contour If the armature conductors are arranged as described above, with Ca conductors in m parallel paths underneath the poles and with a mean active radial magnetic field of Br , we can compute a voltage induced in the stator conductors: Ca RΩBr m Note that this is only the voltage induced by motion of the armature conductors through the field and does not include brush or conductor resistance. If we include the expression for effective magnetic field, we find that the back voltage is: Eb =
Eb = GΩIf which leads us to the conclusion that newton-meters per ampere squared equals volt seconds per ampere. This stands to reason if we examine electric power into the interaction and mechanical power out: Pem = Eb Ia = Te Ω Now, a more complete model of this machine would include the effects of armature, brush and lead resistance, so that in steady state operation: Va = Ra Ia + GΩIf Now, consider this machine with its armatucre connected to a voltage source and its field operating at steady current, so that: Ia =
Va − GΩIf Ra
Then torque, electric power in and mechanical power out are:
Va − GΩIf
Ra Va − GΩIf = Va Ra Va − GΩIf = GΩIf Ra
Te = GIf Pe Pm
Now, note that these expressions define three regimes defined by rotational speed. The two “break points” are at zero speed and at the “zero torque” speed: Ω0 = 3
Va GIf
Ra
+
+ Va
G ΩI f -
Figure 3: DC Machine Equivalent Circuit
Electrical
Mechanical
Figure 4: DC Machine Operating Regimes For 0 < Ω < Ω0 , the machine is a motor: electric power in and mechanical power out are both positive. For higher speeds: Ω0 < Ω , the machine is a generator, with electrical power in and mechanical power out being both negative. For speeds less than zero, electrical power in is positive and mechanical power out is negative. There are few needs to operate machines in this regime, short of some types of “plugging” or emergency braking in tractions systems.
1.2
Hookups:
We have just described a mode of operation of a commutator machine usually called “separately excited”, in which field and armature circuits are controlled separately. This mode of operation is used in some types of traction applications in which the flexibility it affords is useful. For example, some traction applications apply voltage control in the form of “choppers” to separately excited machines. Note that the “zero torque speed” is dependent on armature voltage and on field current. For high torque at low speed one would operate the machine with high field current and enough armature voltage to produce the requisite current. As speed increases so does back voltage, and field current may need to be reduced. At any steady operating speed there will be some optimum mix of field and armature currents to produced the required torque. For braking one could (and this is often done) re-connect the armature of the machine to a braking resistor and turn the machine into a generator. Braking torque is controlled by field current. A subset of the separately excited machine is the shunt connection in which armature and field are supplied by the same source, in parallel. This connection is not widely used any more: it does 4
Ra
+
GΩ I f -
Figure 5: Two-Chopper, separately excited machine hookup
+ V
Figure 6: Series Connection not yield any meaningful ability to control speed and the simple applications to which it used to be used are handled by induction machines. Another connection which is still widely used is the series connection, in which the field winding is sized so that its normal operating current level is the same as normal armature current and the two windings are connected in series. Then: Ia = If = And then torque is: Te =
V Ra + Rf + GΩ
GV 2 (Ra + Rf + GΩ)2
It is important to note that this machine has no “zero-torque” speed, leading to the possibility that an unloaded machine might accelerate to dangerous speeds. This is particularly true because the commutator, made of pieces of relatively heavy material tied together with non-conductors, is not very strong. Speed control of series connected machines can be achieved with voltage control and many appliances using this type of machine use choppers or phase control. An older form of control used in traction applications was the series dropping resistor: obviously not a very efficient way of controlling the machine and not widely used (except in old equipment, of course).
5
A variation on this class of machine is the very widely used “universal motor”, in which the stator and rotor (field and armature) of the machine are both constructed to operate with alternating current. This means that both the field and armature are made of laminated steel. Note that such a machine will operate just as it would have with direct current, with the only addition being the reactive impedance of the two windings. Working with RMS quantities: I = Te =
V Ra + Rf + GΩ + jω (La + Lf ) G|V |2 (Ra + Rf + GΩ)2 + (ωLa + ωLf )2
where ω is the electrical supply frequency. Note that, unlike other AC machines, the universal motor is not limited in speed to the supply frequency. Appliance motors typically turn substantially faster than the 3,600 RPM limit of AC motors, and this is one reason why they are so widely used: with the high rotational speeds it is possible to produce more power per unit mass (and more power per dollar).
1.3
Commutator:
The commutator is what makes this machine work. The brush and commutator system of this class of motor involves quite a lot of “black art”, and there are still aspects of how they work which are poorly understood. However, we can make some attempt to show a bit of what the brush/commutator system does. To start, take a look at the picture shown in Figure 7. Represented are a pair of poles (shaded) and a pair of brushes. Conductors make a group of closed paths. Current from one of the brushes takes two parallel paths. You can follow one of those paths around a closed loop, under each of the two poles (remember that the poles are of opposite polarity) to the opposite brush. Open commutator segments (most of them) do not carry current into or out of the machine.
Figure 7: Commutator and Current Paths A commutation interval occurs when the current in one coil must be reversed. (See Figure 8 In the simplest form this involves a brush bridging between two commutator segments, shorting out that coil. The resistance of the brush causes the current to decay. When the brush leaves the leading segment the current in the leading coil must reverse. We will not attempt to fully understand the commutation process in this type of machine, but we can note a few things. Resistive commutation is the process relied upon in small machines. 6
Figure 8: Commutator at Commutation When the current in one coil must be reversed (because it has left one pole and is approaching the other), that coil is shorted by one of the brushes. The brush resistance causes the current in the coil to decay. Then the leading commutator segment leaves the brush the current MUST reverse (the trailing coil has current in it), and there is often sparking.
1.4
Commutation Stator Yoke Field Poles
Rotor
Ω
Armature Winding Field Winding Commutation Interpoles
Figure 9: Commutation Interpoles In larger machines the commutation process would involve too much sparking, which causes brush wear, noxious gases (ozone) that promote corrosion, etc. In these cases it is common to use separate commutation interpoles. These are separate, usually narrow or seemingly vestigal pole pieces which carry armature current. They are arranged in such a way that the flux from the interpole drives current in the commutated coil in the proper direction. Remember that the coil being commutated is located physically between the active poles and the interpole is therefore in the right spot to influence commutation. The interpole is wound with armature current (it is in series with the main brushes). It is easy to see that the interpole must have a flux density proportional to the current to be commutated. Since the speed with which the coil must be commutated is proportional to rotational velocity and so is the voltage induced by the interpole, if the right 7
number of turns are put around the interpole, commutation can be made to be quite accurate.
1.5
Compensation:
Field Poles
Pole−Face Compensation Winding
Rotor
Ω
Armature Winding Field Winding Commutation Interpoles
Figure 10: Pole Face Compensation Winding The analysis of commutator machines often ignores armature reaction flux. Obviously these machines DO produce armature reaction flux, in quadrature with the main field. Normally, commutator machines are highly salient and the quadrature inductance is lower than direct-axis inductance, but there is still flux produced. This adds to the flux density on one side of the main poles (possibly leading to saturation). To make the flux distribution more uniform and therefore to avoid this saturation effect of quadrature axis flux, it is common in very highly rated machines to wind compensation coils: essentially mirror-images of the armature coils, but this time wound in slots in the surface of the field poles. Such coils will have the same number of ampere-turns as the armature. Normally they have the same number of turns and are connected directly in series with the armature brushes. What they do is to almost exactly cancel the flux produced by the armature coils, leaving only the main flux produced by the field winding. One might think of these coils as providing a reaction torque, produced in exactly the same way as main torque is produced by the armature. A cartoon view of this is shown in Figure 10.
2
Permanent Magnets in Electric Machines
Of all changes in materials technology over the last several years, advances in permanent magnets have had the largest impact on electric machines. Permanent magnets are often suitable as replacements for the field windings in machines: that is they can produce the fundamental interaction field. This does three things. First, since the permanent magnet is lossless it eliminates the energy required for excitation, usually improving the efficiency of the machine. Second, since eliminating the excitation loss reduces the heat load it is often possible to make PM machines more compact. Finally, and less appreciated, is the fact that modern permanent magnets have very large coercive
8
force densities which permit vastly larger air gaps than conventional field windings, and this in turn permits design flexibility which can result in even better electric machines. These advantages come not without cost. Permanent magnet materials have special characteristics which must be taken into account in machine design. The highest performance permanent magnets are brittle ceramics, some have chemical sensitivities, all are sensitive to high temperatures, most have sensitivity to demagnetizing fields, and proper machine design requires understanding the materials well. These notes will not make you into seasoned permanent magnet machine designers. They are, however, an attempt to get started, to develop some of the mathematical skills required and to point to some of the important issues involved.
2.1
Permanent Magnets:
Hysteresis Loop: Permanent Magnet
1 0.8 0.6 0.4 Tesla
0.2 -400
-300
-200
0 -100-0.2 0
100
200
300
400
-0.4 -0.6 -0.8 -1 Kilo Am peres/Meter
Figure 11: Hysteresis Loop Of Ceramic Permanent Magnet Permanent magnet materials are, at core, just materials with very wide hysteresis loops. Figure 11 is an example of something close to one of the more popular ceramic magnet materials. Note that this hysteresis loop is so wide that you can see the effect of the permeability of free space. It is usual to display only part of the magnetic characteristic of permanent magnet materials (see Figure 12), the third quadrant of this picture, because that is where the material is normally operated. Note a few important characteristics of what is called the “demagnetization curve”. The remanent flux density Br , is the value of flux density in the material with zero magnetic field H. The coercive field Hc is the magnetic field at which the flux density falls to zero. Shown also on the curve are loci of constant energy product. This quantity is unfortunately named, for although it has the same units as energy it represents real energy in only a fairly general sense. It is the product of flux density and field intensity. As you already know, there are three commonly used systems of units for magnetic field quantities, and these systems are often mixed up to form very confusing units. We will try to stay away from the English system of units in which field intensity H is measured in amperes per inchand flux density B in lines (actually, usually kilolines) per square inch. In CGS units flux density is measured in Gauss (or kilogauss) and magnetic field intensity in Oersteds. And in SI the unit of flux density is the Tesla, which is one Weber per square meter, and the unit of field intensity is the Ampere per meter . Of these, only the last one, A/m is obvious. A Weber is a volt-second. A Gauss is 10−4 Tesla. And, finally, an Oersted is that field 9
Demagnetization Curve 0.5 0.45 0.4
Energy Product Loci
Br
0.35 B, Tesla
0.3 0.25 0.2 0.15 0.1
Hc
0.05 0
-250
-200
-150
-100
-50
0
H, kA/m
Figure 12: Demagnetization Curve intensity required to produce one Gauss in the permeability of free space. Since the permeability of free space µ0 = 4π × 10−7 Hy/m, this means that one Oe is about 79.58 A/m. Commonly, the energy product is cited in MgOe (Mega-Gauss-Oersted)s. One MgOe is equal to 7.958kJ/m3 . A commonly used measure for the performance of a permanent magnet material is the maximum energy product, the largest value of this product along the demagnetization curve. To start to understand how these materials might be useful, consider the situation shown in Figure 13: A piece of permanent magnet material is wrapped in a magnetic circuit with effectively infinite permeability. Assume the thing has some (finite) depth in the direction you can’t see. Now, if we take Ampere’s law around the path described by the dotted line, I
~ · d~ℓ = 0 H
since there is no current anywhere in the problem. If magnetization is upwards, as indicated by the arrow, this would indicate that the flux density in the permanent magnet material is equal to the remanent flux density (also upward). A second problem is illustrated in Figure 14, in which the same magnet is embedded in a magnetic circuit with an air gap. Assume that the gap has width g and area Ag. The magnet has height hm and area Am. For convenience, we will take the positive reference direction to be up (as we see it here) in the magnet and down in the air-gap. Thus we are following the same reference direction as we go around the Ampere’s Law loop. That becomes: I ~ · d ~ℓ = Hm hm + Hg g H
Now, Gauss’ law could be written for either the upper or lower piece of the magnetic circuit. Assuming that the only substantive flux leaving or entering the magnetic circuit is either in the magnet or the gap: ZZ ~ · dA ~ = Bm Am − µ0 Hg Ag
B Solving this pair we have:
Bm = −µ0
Ag hm Hm = µ0 Pu Hm Am g 10
Permanent Magnet
Magnetic Circuit, µ→∞
Figure 13: Permanent Magnet in Magnetic Circuit
Permanent Magnet g hm
Magnetic Circuit, µ→∞
Figure 14: Permanent Magnet Driving an Air-Gap This defines the unit permeance, essentially the ratio of the permeance facing the permanent magnet to the internal permeance of the magnet. The problem can be, if necessary, solved graphically, since the relationship between Bm and Hm is inherently nonlinear, as shown in Figure 15 “load line” analysis of a nonlinear electronic circuit. Now, one more ‘cut’ at this problem. Note that, at least for fairly large unit permeances the slope of the magnet characteristic is fairly constant. In fact, for most of the permanent magnets used in machines (the one important exception is the now rarely used ALNICO alloy magnet), it is generally acceptable to approximate the demagnitization curve with:
~m + M ~0 ~ m = µm H B
Here, the magnetization M0 is fixed. Further, for almost all of the practical magnet materials the magnet permeability is nearly the same as that of free space (µm ≈ µ0 ). With that in mind, consider the problem shown in Figure 16, in which the magnet fills only part of a gap in a magnetic circuit. But here the magnet and gap areas are essentially the same. We could regard the magnet as simply a magnetization. 11
− µ 0℘u
Figure 15: Load Line, Unit Permeance Analysis
Permanent Magnet
Magnetic Circuit, µ→∞
Figure 16: Surface Magnet Primitive Problem In the region of the magnet and the air-gap, Ampere’s Law and Gauss’ law can be written:
∇ · µ0
~ = 0 ∇×H ~m + M ~0 = 0 H ~g = 0 ∇ · µ0 H
Now, if in the magnet the magnetization is constant, the divergence of H in the magnet is zero. Because there is no current here, H is curl free, so that everywhere: ~ = −∇ψ H
∇2 ψ = 0
That is, magnetic field can be expressed as the gradient of a scalar potential which satisfies Laplace’s equation. It is also pretty clear that, if we can assign the scalar potential to have a value of zero anywhere on the surface of the magnetic circuit it will be zero over all of the magnetic circuit (i.e. at both the top of the gap and the bottom of the magnet). Finally, note that we can’t actually assume that the scalar potential satisfies Laplace’s equation everywhere in the problem. In fact the divergence of M is zero everywhere except at the top surface of the magnet where it is singular! In fact, we can note that there is a (some would say fictitious) magnetic charge density: ~ ρm = −∇ · M 12
At the top of the magnet there is a discontinuous change in M and so the equivalent of a magnetic surface charge. Using Hg to note the magnetic field above the magnet and Hm to note the magnetic field in the magnet, µ0 Hg = µ0 (Hm + M0 ) σm = M0 = Hg − Hm and then to satisfy the potential condition, if hm is the height of the magnet and g is the gap: gHg = −hm Hm Solving,
hm hm + g Now, one more observation could be made. We would produce the same air-gap flux density if we regard the permanent magnet as having a surface current around the periphery equal to the magnetization intensity. That is, if the surface current runs around the magnet: Hg = M0
Kφ = M0 This would produce an MMF in the gap of: F = Kφ hm and then since the magnetic field is just the MMF divided by the total gap: Hg =
hm F = M0 hm + g hm + g
The real utility of permanent magnets comes about from the relatively large magnetizations: numbers of a few to several thousand amperes per meter are common, and these would translate into enormous current densities in magnets of ordinary size.
3 Simple Permanent Magnet Machine Structures: Machines
Commutator
Figure 17 is a cartoon picture of a cross section of the geometry of a two-pole commutator machine using permanent magnets. This is actually the most common geometry that is used. The rotor (armature) of the machine is a conventional, windings-in-slots type, just as we have already seen for commutator machines. The field magnets are fastened (often just bonded) to the inside of a steel tube that serves as the magnetic flux return path. Assume for the purpose of first-order analysis of this thing that the magnet is describable by its remanent flux density Br and has permeability of µ0 . First, we will estimate the useful magnetic flux density and then will deal with voltage generated in the armature.
3.1
Interaction Flux Density
Using the basics of the analysis presented above, we may estimate the radial magnetic flux density at the air-gap as being: Br Bd = 1 + P1c 13
Back Iron h
m
g R
Rotor
Permanent Magnets
Figure 17: PM Commutator Machine where the effective unit permeance is: Pc =
fl hm Ag ff g Am
A book on this topic by James Ireland suggests values for the two “fudge factors”: 1. The “leakage factor” fl is cited as being about 1.1. 2. The “reluctance factor” ff is cited as being about 1.2. We may further estimate the ratio of areas of the gap and magnet by: R + g2 Ag = Am R + g + h2m Now, there are a bunch of approximations and hand wavings in this expression, but it seems to work, at least for the kind of machines contemplated. A second correction is required to correct the effective length for electrical interaction. The reason for this is that the magnets produce fringing fields, as if they were longer than the actual ”stack length” of the rotor (sometimes they actually are). This is purely empirical, and Ireland gives a value for effective length for voltage generation of: ℓeff =
ℓ∗ fℓ
where ℓ∗ = ℓ + 2N R , and the empirical coefficient A hm N ≈ log 1 + B B R �
where B = 7.4 − 9.0 A = 0.9 14
hm R
�
3.1.1
Voltage:
It is, in this case, simplest to consider voltage generated in a single wire first. If the machine is running at angular velocity Ω, speed voltage is, while the wire is under a magnet, vs = ΩRℓBr Now, if the magnets have angular extent θm the voltage induced in a wire will have a waveform as shown in Figure 18: It is pulse-like and has the same shape as the magnetic field of the magnets.
vs
π θm
Ωt
Figure 18: Voltage Induced in One Conductor The voltage produced by a coil is actually made up of two waveforms of exactly this form, but separated in time by the ”coil throw” angle. Then the total voltage waveform produced will be the sum of the two waveforms. If the coil thrown angle is larger than the magnet angle, the two voltage waveforms add to look like this: There are actually two coil-side waveforms that add with a slight phase shift.
vc
0m 0m Figure 19: Voltage Induced in a Coil If, on the other hand, the coil thrown is smaller than the magnet angle, the picture is the same, only the width of the pulses is that of the coil rather than the magnet. In either case the average voltage generated by a coil is: θ∗ v = ΩRℓNs Bd π where θ ∗ is the lesser of the coil throw or magnet angles and Ns is the number of series turns in the coil. This gives us the opportunity to develop the number of “active” turns: Ca θ∗ θ∗ C = Ns = tot m π m π 15
Here, Ca is the number of active conductors, Ctot is the total number of conductors and m is the number of parallel paths. The motor coefficient is then: K=
3.2
Rℓeff Ctot Bd θ ∗ m π
Armature Resistance
The last element we need for first-order prediction of performance of the motor is the value of armature resistance. The armature resistance is simply determined by the length and area of the wire and by the number of parallel paths (generally equal to 2 for small commutator motors). If we note Nc as the number of coils and Na as the number of turns per coil, Ns =
Nc Na m
Total armature resistance is given by: Ra = 2ρw ℓt
Ns m
where ρw is the resistivity (per unit length) of the wire: ρw =
1 π 2 4 d w σw
(dw is wire diameter, σw is wire conductivity and ℓt is length of one half-turn). This length depends on how the machine is wound, but a good first-order guess might be something like this: ℓt ≈ ℓ + πR
16
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machines Class Notes 7: Permanent Magnet “Brushless DC” Motors c �2005 James L. Kirtley Jr.
1
September 5, 2005
Introduction
This document is a brief introduction to the design evaluation of permanent magnet motors, with an eye toward servo and drive applications. It is organized in the following manner: First, we describe three different geometrical arrangements for permanent magnet motors: 1. Surface Mounted Magnets, Conventional Stator, 2. Surface Mounted Magnets, Air-Gap Stator Winding, and 3. Internal Magnets (Flux Concentrating). After a qualitative discussion of these geometries, we will discuss the elementary rating parameters of the machine and show how to arrive at a rating and how to estimate the torque and power vs. speed capability of the motor. Then we will discuss how the machine geometry can be used to estimate both the elementary rating parameters and the parameters used to make more detailed estimates of the machine performance. Some of the more involved mathematical derivations are contained in appendices to this note.
2
Motor Morphologies
There are, of course, many ways of building permanent magnet motors, but we will consider only a few in this note. Actually, once these are understood, rating evaluations of most other geometrical arrangements should be fairly straightforward. It should be understood that the “rotor inside” vs. “rotor outside” distinction is in fact trivial, with very few exceptions, which we will note.
2.1
Surface Magnet Machines
Figure 1 shows the basic magnetic morphology of the motor with magnets mounted on the surface of the rotor and an otherwise conventional stator winding. This sketch does not show some of the important mechanical aspects of the machine, such as the means for fastening the permanent magnets to the rotor, so one should look at it with a bit of caution. In addition, this sketch and the other sketches to follow are not necessarily to a scale that would result in workable machines. This figure shows an axial section of a four-pole (p = 2) machine. The four magnets are mounted on a cylindrical rotor “core”, or shaft, made of ferromagnetic material. Typically this would simply be a steel shaft. In some applications the magnets may be simply bonded to the steel. For applications in which a glue joint is not satisfactory (e.g. for high speed machines) some sort of rotor banding or retaining ring structure is required.
1
Rotor Core (Shaft)
Stator Winding in Slots
Stator Core
Air−Gap Rotor Magnets
Figure 1: Axial View of a Surface Mount Motor The stator winding of this machine is “conventional”, very much like that of an induction motor, consisting of wires located in slots in the surface of the stator core. The stator core itself is made of laminated ferromagnetic material (probably silicon iron sheets), the character and thickness of the sheets determined by operating frequency and efficiency requirements. They are required to carry alternating magnetic fields, so must be laminated to reduce eddy current losses. This sort of machine is simple in construction. Note that the operating magnetic flux density in the air-gap is nearly the same as in the magnets, so that this sort of machine cannot have air-gap flux densities higher than that of the remanent flux density of the magnets. If low cost ferrite magnets are used, this means relatively low induction and consequently relatively low efficiency and power density. (Note the qualifier “relatively” here!). Note, however, that with modern, high performance permanent magnet materials in which remanent flux densities can be on the order of 1.2 T, air-gap working flux densities can be on the order of 1 T. With the requirement for slots to carry the armature current, this may be a practical limit for air-gap flux density anyway. It is also important to note that the magnets in this design are really in the “air gap” of the machine, and therefore are exposed to all of the time- and space- harmonics of the stator winding MMF. Because some permanent magnets have electrical conductivity (particularly the higher performance magnets), any asynchronous fields will tend to produce eddy currents and consequent losses in the magnets.
2.2
Interior Magnet or Flux Concentrating Machines
Interior magnet designs have been developed to counter several apparent or real shortcomings of surface mount motors: • Flux concentrating designs allow the flux density in the air-gap to be higher than the flux density in the magnets themselves. 2
• In interior magnet designs there is some degree of shielding of the magnets from high order space harmonic fields by the pole pieces. • There are control advantages to some types of interior magnet motors, as we will show anon. Essentially, they have relatively large negative saliency which enhances “flux weakening” for high speed operation, in rather direct analogy to what is done in DC machines. • Some types of internal magnet designs have (or claim) structural advantages over surface mount magnet designs. Rotor Pole Pieces
Armature in Slots
Rotor Magnets
Non−magnetic Rotor Core (shaft)
Stator Core
Figure 2: Axial View of a Flux Concentrating Motor The geometry of one type of internal magnet motor is shown (crudely) in Figure 2. The permanent magnets are oriented so that their magnetization is azimuthal. They are located between wedges of magnetic material (the pole pieces) in the rotor. Flux passes through these wedges, going radially at the air- gap, then azimuthally through the magnets. The central core of the rotor must be non-magnetic, to prevent “shorting out” the magnets. No structure is shown at all in this drawing, but quite obviously this sort of rotor is a structural challenge. Shown is a six-pole machine. Typically, one does not expect flux concentrating machines to have small pole numbers, because it is difficult to get more area inside the rotor than around the periphery. On the other hand, a machine built in this way but without substantial flux concentration will still have saliency and magnet shielding properties. A second morphology for an internal magnet motor is shown in Figure 3. This geometry has been proposed for highly salient synchronous machines without permanent magnets: such machines would run on the saliency torque and are called synchronous reluctance motors. however, the saliency slots may be filled with permanent magnet material, giving them some internally generated flux as well. The rotor iron tends to short out the magnets, so that the ’bridges’ around the ends of the permanent magnets must be relatively thin. They are normally saturated. 3
Stator Core Stator Slots
Air Gap
Rotor
Saliency Slots
Figure 3: Axial View of Internal Magnet Motor At first sight, these machines appear to be quite complicated to analyze, and that judgement seems to hold up.
2.3
Air Gap Armature Windings
Shown in Figure 4 is a surface-mounted magnet machine with an air-gap, or surface armature winding. Such machines take advantage of the fact that modern permanent magnet materials have very low permeabilities and that, therefore, the magnetic field produced is relatively insensitive to the size of the air-gap of the machine. It is possible to eliminate the stator teeth and use all of the periphery of the air-gap for windings. Not shown in this figure is the structure of the armature winding. This is not an issue in “conventional” stators, since the armature is contained in slots in the iron stator core. The use of an air-gap winding gives opportunities for economy of construction, new armature winding forms such as helical windings, elimination of “cogging” torques, and (possibly) higher power densities.
3
Zeroth Order Rating
In determining the rating of a machine, we may consider two separate sets of parameters. The first set, the elementary rating parameters, consist of the machine inductances, internal flux linkage and stator resistance. From these and a few assumptions about base and maximum speed it is possible to get a first estimate of the rating and performance of the motor. More detailed performance estimates, including efficiency in sustained operation, require estimation of other parameters. We will pay more attention to that first set of parameters, but will attempt to show how at least some of the more complete operating parameters can be estimated.
4
Rotor Core (Shaft)
Stator Winding
Stator Core
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Rotor Magnets
Figure 4: Axial View of a PM Motor With an Air-Gap Winding
3.1
Voltage and Current: Round Rotor
To get started, consider the equivalent circuit shown in Figure 5. This is actually the equivalent circuit which describes all round rotor synchronous machines. It is directly equivalent only to some of the machines we are dealing with here, but it will serve to illustrate one or two important points.
X ∩∩∩∩
�� + −
Ea
��
Ia � + Vt -
Figure 5: Synchronous Machine Equivalent Circuit What is shown here is the equivalent circuit of a single phase of the machine. Most motors are three-phase, but it is not difficult to carry out most of the analysis for an arbitrary number of phases. The circuit shows an internal voltage Ea and a reactance X which together with the terminal current I determine the terminal voltage V . In this picture armature resistance is ignored. If the machine is running in the sinusoidal steady state, the major quantities are of the form: Ea = ωλa cos (ωt + δ) Vt = V cos ωt 5
I
� �� �� � � ψ V �� � � �� �� � � �� � δ �� � ��� � �� � jXI �� � �� � � Ea ��
Figure 6: Phasor Diagram For A Synchronous Machine Ia = I cos (ωt − ψ) The machine is in synchronous operation if the internal and external voltages are at the same frequency and have a constant (or slowly changing) phase relationship (δ). The relationship between the major variables may be visualized by the phasor diagram shown in Figure 3.1. The internal voltage is just the time derivative of the internal flux from the permanent magnets, and the voltage drop in the machine reactance is also the time derivative of flux produced by armature current in the air-gap and in the “leakage” inductances of the machine. By convention, the angle ψ is positive when current I lags voltage V and the angle δ is positive then internal voltage Ea leads terminal voltage V . So both of these angles have negative sign in the situation shown in Figure 3.1. If there are q phases, the time average power produced by this machine is simply: q P = V I cos ψ 2 For most polyphase machines operating in what is called “balanced” operation (all phases doing the same thing with uniform phase differences between phases), torque (and consequently power) are approximately constant. Since we have ignored power dissipated in the machine armature, it must be true that power absorbed by the internal voltage source is the same as terminal power, or: q P = Ea I cos (ψ − δ) 2
Since in the steady state:
ω P = T p where T is torque and ω/p is mechanical rotational speed, torque can be derived from the terminal quantities by simply: q T = p λa I cos (ψ − δ) 2 In principal, then, to determine the torque and hence power rating of a machine it is only necessary to determine the internal flux, the terminal current capability, and the speed capability of the rotor. In fact it is almost that simple. Unfortunately, the model shown in Figure 5 is not quite complete for some of the motors we will be dealing with, and we must go one more level into machine theory. 6
3.2
A Little Two-Reaction Theory
The material in this subsection is framed in terms of three-phase (q = 3) machine theory, but it is actually generalizable to an arbitrary number of phases. Suppose we have a machine whose three-phase armature can be characterized by internal fluxes and inductance which may, in general, not be constant but is a function of rotor position. Note that the simple model we presented in the previous subsection does not conform to this picture, because it assumes a constant terminal inductance. In that case, we have: λph = Lph I ph + λR
(1)
where λR is the set of internally produced fluxes (from the permanent magnets) and the stator winding may have both self- and mutual- inductances. Now, we find it useful to do a transformation on these stator fluxes in the following way: each armature quantity, including flux, current and voltage, is projected into a coordinate system that is fixed to the rotor. This is often called the Park’s Transformation. For a three phase machine it is:
ua ud uq = udq = T uph = T ub uc u0
(2)
Where the transformation and its inverse are:
cos θ cos(θ − 23π ) cos(θ + 23π ) 2 2π T = − sin θ − sin(θ − 2π 3 ) − sin(θ + 3 ) 3 1 1 1
2
T −1
2
(3)
2
cos θ − sin θ 1 2π ) − sin(θ − ) 1 = cos(θ − 2π 3 3 2π 2π cos(θ + 3 ) − sin(θ + 3 ) 1
(4)
It is easy to show that balanced polyphase quantities in the stationary, or phase variable frame, translate into constant quantities in the so-called “d-q” frame. For example: Ia = I cos ωt 2π ) 3 2π Ic = I cos(ωt + ) 3 θ = ωt + θ0 Ib = I cos(ωt −
maps to: Id = I cos θ0 Iq = −I sin θ0 Now, if θ = ωt + θ0 , the transformation coordinate system is chosen correctly and the “d-” axis will correspond with the axis on which the rotor magnets are making positive flux. That happens 7
if, when θ = 0, phase A is linking maximum positive flux from the permanent magnets. If this is the case, the internal fluxes are: λaa = λf cos θ 2π ) 3 2π = λf cos(θ + ) 3
λab = λf cos(θ − λac
Now, if we compute the fluxes in the d-q frame, we have: λdq = Ldq I dq + λR = T Lph T −1 I dq + λR
(5)
Now: two things should be noted here. The first is that, if the coordinate system has been chosen as described above, the flux induced by the rotor is, in the d-q frame, simply:
λf λR = 0 0
(6)
That is, the magnets produce flux only on the d- axis. The second thing to note is that, under certain assumptions, the inductances in the d-q frame are independent of rotor position and have no mutual terms. That is: Ldq = T Lph T −1
Ld 0 0 Lq 0 = 0 0 0 L0
(7)
The assertion that inductances in the d-q frame are constant is actually questionable, but it is close enough to being true and analyses that use it have proven to be close enough to being correct that it (the assertion) has held up to the test of time. In fact the deviations from independence on rotor position are small. Independence of axes (that is, absence of mutual inductances in the d-q frame) is correct because the two axes are physically orthogonal. We tend to ignore the third, or “zero” axis in this analysis. It doesn’t couple to anything else and has neither flux nor current anyway. Note that the direct- and quadrature- axis inductances are in principle straightforward to compute. They are direct axis the inductance of one of the armature phases (corrected for the fact of multiple phases) with the rotor aligned with the axis of the phase, and quadrature axis the inductance of one of the phases with the rotor aligned 90 electrical degrees away from the axis of that phase. Next, armature voltage is, ignoring resistance, given by: d d λph = T −1 λdq dt dt and that the transformed armature voltage must be: V ph =
8
(8)
V dq = T V ph d = T (T −1 λdq ) dt d d (9) λdq + (T T −1 )λdq = dt dt The second term in this expresses “speed voltage”. A good deal of straightforward but tedious manipulation yields: T
d −1 T = dt
0 dθ dt
0
− dθ dt 0 0
0 0 0
The direct- and quadrature- axis voltage expressions are then:
dλd − ωλq Vd = dt dλq Vq = + ωλd dt where dθ ω= dt Instantaneous power is given by: P = Va Ia + Vb Ib + Vc Ic
(10)
(11) (12)
(13)
Using the transformations given above, this can be shown to be: 3 3 (14) P = Vd Id + Vq Iq + 3V0 I0 2 2 which, in turn, is: 3 dλd dλq dλ0 3 Id + Iq ) + 3 I0 (15) P = ω (λd Iq − λq Id ) + ( dt dt 2 dt 2 Then, noting that ω = pΩ and that (15) describes electrical terminal power as the sum of shaft power and rate of change of stored energy, we may deduce that torque is given by: q T = p(λd Iq − λq Id ) (16) 2 Note that we have stated a generalization to a q- phase machine even though the derivation given here was carried out for the q = 3 case. Of course three phase machines are by far the most common case. Machines with higher numbers of phases behave in the same way (and this generalization is valid for all purposes to which we put it), but there are more rotor variables analogous to “zero axis”. Now, noting that, in general, Ld and Lq are not necessarily equal,
then torque is given by:
λd = Ld Id + λf
(17)
λq = Lq Iq
(18)
q T = p (λf + (Ld − Lq ) Id ) Iq 2 9
(19)
3.3
Finding Torque Capability
For high performance drives, we will generally assume that the power supply, generally an inverter, can supply currents in the correct spatial relationship to the rotor to produce torque in some reasonably effective fashion. We will show in this section how to determine, given a required torque (or if the torque is limited by either voltage or current which we will discuss anon), what the values of Id and Iq must be. Then the power supply, given some means of determining where the rotor is (the instantaneous value of θ), will use the inverse Park’s transformation to determine the instantaneous valued required for phase currents. This is the essence of what is known as “field oriented control”, or putting stator currents in the correct location in space to produce the required torque. Our objective in this section is, given the elementary parameters of the motor, find the capability of the motor to produce torque. There are three things to consider here: • Armature current is limited, generally by heating, • A second limit is the voltage capability of the supply, particularly at high speed, and • If the machine is operating within these two limits, we should consider the optimal placement of currents (that is, how to get the most torque per unit of current to minimize losses). Often the discussion of current placement is carried out using, as a tool to visualize what is going on, the Id , Iq plane. Operation in the steady state implies a single point on this plane. A simple illustration is shown in Figure 7. The thermally limited armature current capability is represented as a circle around the origin, since the magnitude of armature current is just the length of a vector from the origin in this space. In general, for permanent magnet machines with buried magnets, Ld < Lq , so the optimal operation of the machine will be with negative Id . We will show how to determine this optimum operation anon, but it will in general follow a curve in the Id , Iq plane as shown. Finally, an ellipse describes the voltage limit. To start, consider what would happen if the terminals of the machine were to be short-circuited so that V = 0. If the machine is operating at sufficiently high speed so that armature resistance is negligible, armature current would be simply: Id = −
λf Ld
Iq = 0 Now, loci of constant flux turn out to be ellipses around this point on the plane. Since terminal flux is proportional to voltage and inversely proportional to frequency, if the machine is operating with a given terminal voltage, the ability of that voltage to command current in the Id , Iq plane is an ellipse whose size “shrinks” as speed increases. To simplify the mathematics involved in this estimation, we normalize reactances, fluxes, currents and torques. First, let us define the base flux to be simply λb = λf and the base current Ib to be the armature capability. Then we define two per-unit reactances: xd = xq = 10
L d Ib λb L q Ib λb
(20) (21)
Voltage Limit Loci
Optimal Torque Locus
Speed = Base Speed > Base
i
q Armature Current Limit
id Short Circuit Point
Figure 7: Limits to Operation Next, define the base torque to be:
q Tb = p λb Ib 2 and then, given per-unit currents id and iq , the per-unit torque is simply: te = (1 − (xq − xd ) id ) iq
(22)
It is fairly straightforward (but a bit tedious) to show that the locus of current-optimal operation (that is, the largest torque for a given current magnitude or the smallest current magnitude for a given torque) is along the curve:
id
iq
v u u 2 ui = −t a + 2
2
v u u 2 ui = −t a − 2
2
1 4 (xq − xd ) 1 4 (xq − xd )
!2 !2
1 − 2 (xq − xd ) 1 + 2 (xq − xd )
v u u t v u u t
1 4 (xq − xd ) 1 4 (xq − xd )
!2
+
i2a 2
(23)
!2
+
i2a 2
(24)
The “rating point” will be the point along this curve when ia = 1, or where this curve crosses the armature capability circle in the id , iq plane. It should be noted that this set of expressions only works for salient machines. For non-salient machines, of course, torque-optimal current is on the q-axis. In general, for machines with saliency, the “per-unit” torque will not be unity at the rating, so that the rated, or “Base Speed” torque is not the “Base” torque, but: Tr = Tb × te where te is calculated at the rating point (that is, ia = 1 and id and iq as per (23) and (24)).
11
(25)
For sufficiently low speeds, the power electronic drive can command the optimal current to produce torque up to rated. However, for speeds higher than the “Base Speed”, this is no longer true. Define a per-unit terminal flux: V ψ= ωλb Operation at a given flux magnitude implies: ψ 2 = (1 + xd id )2 + (xq iq )2 which is an ellipse in the id , iq plane. The Base Speed is that speed at which this ellipse crosses the point where the optimal current curve crosses the armature capability. Operation at the highest attainable torque (for a given speed) generally implies d-axis currents that are higher than those on the optimal current locus. What is happening here is the (negative) d-axis current serves to reduce effective machine flux and hence voltage which is limiting q-axis current. Thus operation above the base speed is often referred to as “flux weakening”. The strategy for picking the correct trajectory for current in the id , iq plane depends on the value of the per-unit reactance xd . For values of xd > 1, it is possible to produce some torque at any speed. For values of xd < 1, there is a speed for which no point in the armature current capability is within the voltage limiting ellipse, so that useful torque has gone to zero. Generally, the maximum torque operating point is the intersection of the armature current limit and the voltage limiting ellipse: id = iq =
v u
u xd xd t − 2 2 2 xq − x d x q − x 2d
!2
q
+
x2q − ψ 2 + 1 x2q − x 2d
1 − i2d
(26) (27)
It may be that there is no intersection between the armature capability and the voltage limiting ellipse. If this is the case and if xd < 1, torque capability at the given speed is zero. If, on the other hand, xd > 1, it may be that the intersection between the voltage limiting ellipse and the armature current limit is not the maximum torque point. To find out, we calculate the maximum torque point on the voltage limiting ellipse. This is done in the usual way by differentiating torque with respect to id while holding the relationship between id and iq to be on the ellipse. The algebra is a bit messy, and results in: id = − iq =
3xd (xq − xd ) − 4x2d (xq − xd )
1 xq
q
x2d
v ! u u 3x (xq − x ) − x2 2 (xq − xd ) (ψ 2 − 1) + xd d d t d − +
4x2d (xq − xd )
ψ 2 − (1 + xd id )2
2 (xq − xd ) x2d
(28) (29)
Ordinarily, it is probably easiest to compute (28) and (29) first, then test to see if the currents are outside the armature capability, and if they are, use (26) and (27). These expressions give us the capability to estimate the torque-speed curve for a machine. As an example, the machine described by the parameters cited in Table 1 is a (nominal) 3 HP, 4-pole, 3000 RPM machine. The rated operating point turns out to have the following attributes: 12
Table 1: Example Machine
D- Axis Inductance Q- Axis Inductance Internal Flux Armature Current
2.53 mHy 6.38 mHy 58.1 mWb 30 A
Table 2: Operating Characteristics of Example Machine
Per-Unit D-Axis Current At Rating Point Per-Unit Q-Axis Current At Rating Point Per-Unit D-Axis Reactance Per-Unit Q-Axis Reactance Rated Torque (Nm) Terminal Voltage at Base Point (V)
id iq xd xq Tr
-.5924 .8056 1.306 3.294 9.17 97
The loci of operation in the Id , Iq plane is shown in Figure 8. The armature current limit is shown only in the second and third quadrants, so shows up as a semicircle. The two ellipses correspond with the rated point (the larger ellipse) and with a speed that is three times rated (9000 RPM). The torque-optimal current locus can be seen running from the origin to the rating point, and the higher speed operating locus follows the armature current limit. Figure 9 shows the torque/speed and power/speed curves. Note that this sort of machine only approximates “constant power” operation at speeds above the “base” or rating point speed.
4
Parameter Estimation
We are now at the point of estimating the major parameters of the motors. Because we have a number of different motor geometries to consider, and because they share parameters in not too orderly a fashion, this section will have a number of sub-parts. First, we calculate flux linkage, then reactance.
4.1
Flux Linkage
Given a machine which may be considered to be uniform in the axial direction, flux linked by a single, full-pitched coil which spans an angle from zero to π/p, is: φ=
Z
0
π p
Br Rldφ
where Br is the radial flux through the coil. And, if Br is sinusoidally distributed this will have a peak value of 2RlBr φp = p 13
PM Brushless Machine Current Loci 60
40
Q-Axis Current (A)
20
0
-20
-40
-60 -80
-60
-40 -20 0 D-Axis Current (A)
20
Figure 8: Operating Current Loci of Example Machine
PM Brushless Machine
10
Torque, N-m
8 6 4 2 0 0
1000
2000
3000
4000
5000
6000
7000
8000
9000
1000
2000
3000
4000 5000 Speed, RPM
6000
7000
8000
9000
Power, Watts
4000 3000 2000 1000 0 0
Figure 9: Torque- and Power-Speed Capability
14
Now, if the actual winding has Na turns, and using the pitch and breadth factors derived in Appendix 1, the total flux linked is simply: λf =
2RlB1 Na kw p
(30)
where kw = kp kb α kp = sin 2 sin m γ2 kb = m sin γ2 The angle α is the pitch angle, α = 2πp
Np Ns
where Np is the coil span (in slots) and Ns is the total number of slots in the stator. The angle γ is the slot electrical angle: 2πp γ= Ns Now, what remains to be found is the space fundamental magnetic flux density B1 . In Appendix 2 it is shown that, for magnets in a surface-mount geometry, the magnetic field at the surface of the magnetic gap is: B1 = µ0 M1 kg
(31)
where the space-fundamental magnetization is: M1 =
pθm Br 4 sin µ0 π 2
where Br is remanent flux density of the permanent magnets and θm is the magnet angle. and where the factor that describes the geometry of the magnetic gap depends on the case. For magnets inside and p = � 1, kg =
Rsp−1 Rs2p − Ri2p
p p p+1 Ri2p R11−p − R21−p R2 − R1p+1 + p+1 p−1
For magnets inside and p = 1, kg =
1 Rs2 − Ri2
1
2
R22 − R12 + Ri2 log
R2 R1
For the case of magnets outside and p �= 1: kg =
Rip−1 Rs2p − Ri2p
p p+1 p Rs2p R11−p − R21−p R2 − R1p+1 + p+1 p−1
and for magnets outside and p = 1, 15
1 kg = 2 Rs − Ri2
1
2
R22
−
R12
+
Rs2 log
R2 R1
Where Rs and Ri are the outer and inner magnetic boundaries, respectively, and R2 and R1 are the outer and inner boundaries of the magnets. Note that for the case of a small gap, in which both the physical gap g and the magnet thickness hm are both much less than rotor radius, it is straightforward to show that all of the above expressions approach what one would calculate using a simple, one-dimensional model for the permanent magnet: kg →
hm g + hm
This is the whole story for the winding-in-slot, narrow air-gap, surface magnet machine. For airgap armature windings, it is necessary to take into account the radial dependence of the magnetic field.
4.2
Air-Gap Armature Windings
With no windings in slots, the conventional definition of winding factor becomes difficult to apply. If, however, each of the phase belts of the winding occupies an angular extent θw , then the equivalent to (31) is: sin p θ2w kw = p θ2w Next, assume that the “density” of conductors within each of the phase belts of the armature winding is uniform, so that the density of turns as a function of radius is: N (r) =
2Na r 2 − Rwi
2 Rwo
This just expresses the fact that there is more azimuthal room at larger radii, so with uniform density the number of turns as a function of radius is linearly dependent on radius. Here, Rwo and Rwi are the outer and inner radii, respectively, of the winding. Now it is possible to compute the flux linked due to a magnetic field distribution: λf =
Z
Rwo
Rwi
2lNa kw r 2r 2 µ0 Hr (r)dr 2 p Rwo − Rwi
(32)
Note the form of the magnetic field as a function of radius expressed in 80 and 81 of the second appendix. For the “winding outside” case it is:
Hr = A r p−1 + Rs2p r −p−1
Then a winding with all its turns concentrated at the outer radius r = Rwo would link flux: λc =
2lRwo kw 2lRwo kw −p−1 p−1 + Rs2p Rwo µ0 Hr (Rwo ) = µ0 A Rwo p p
16
Carrying out (32), it is possible, then, to express the flux linked by a thick winding to the flux that would have been linked by a radially concentrated winding at its outer surface by: kt =
λf λc
where, for the winding outside, p �= 2 case: 1 − x2+p ξ 2p 1 − x2−p + 2+p 2−p
2 kt = 2 (1 − x ) (1 + ξ 2p )
!
(33)
where we have used the definitions ξ = Rwo /Rs and x = Rwi /Rwo . In the case of winding outside, p = 2, ! 1 − x4 ξ 4 2 − log x (34) kt = (1 − x2 ) (1 + ξ 2p ) 4 In a very similar way, we can define a winding factor for a thick winding in which the reference radius is at the inner surface. (Note: this is done because the inner surface of the inside winding is likely to be coincident with the inner ferromagnetic surface, as the outer surface of the outer winding ls likely to be coincident with the outer ferromagnetic surface). For p �= 2: 2x−p kt = (1 − x2 ) (1 + η 2p )
1 − x2+p 1 − x2−p + (ηx)2p 2+p 2−p
!
(35)
and for p = 2: 2x−2 kt = (1 − x2 ) (1 + η 2p )
1 − x4 − (ηx)4 log x 4
!
(36)
where η = Ri /Rwi So, in summary, the flux linked by an air-gap armature is given by: λf =
2RlB1 Na kw kt p
(37)
where B1 is the flux density at the outer radius of the physical winding (for outside winding machines) or at the inner radius of the physical winding (for inside winding machines). Note that the additional factor kt is a bit more than one (it approaches unity for thin windings), so that, for small pole numbers and windings that are not too thick, it is almost correct and in any case “conservative” to take it to be one.
4.3
Interior Magnet Motors:
For the flux concentrating machine, it is possible to estimate air-gap flux density using a simple reluctance model. The air- gap permeance of one pole piece is: ℘ag = µ0 l where θp is the angular width of the pole piece. 17
Rθp g
And the incremental permeance of a magnet is: ℘m = µ0
hm l wm
The magnet sees a unit permeance consisting of its own permeance in series with one half of each of two pole pieces (in series) : ℘u =
Rθp wm ℘ag = ℘m 4g hm
Magnetic flux density in the magnet is: Bm = B 0
℘u 1 + ℘u
And then flux density in the air gap is: Bg =
2hm 2hm wm Bm = B 0 Rθp 4ghm + Rθp wm
The space fundamental of that can be written as: B1 =
4 pθp wm sin B0 γm π 2 2g
where we have introduced the shorthand:
γm =
1
1+
w m θp R g 4 hm
The flux linkage is then computed as before: λf =
4.4
2RlB1 Na kw p
(38)
Winding Inductances
The next important set of parameters to compute are the d- and q- axis inductances of the machine. We will consider three separate cases, the winding-in-slot, surface magnet case, which is magnetically “round”, or non-salient, the air-gap winding case, and the flux concentrating case which is salient, or has different direct- and quadrature- axis inductances. 4.4.1
Surface Magnets, Windings in Slots
In this configuration there is no saliency, so that Ld = Lq . There are two principal parts to inductance, the air-gap inductance and slot leakage inductance. Other components, including end turn leakage, may be important in some configurations, and they would be computed in the same way as for an induction machine. If magnet thickness is not too great, we may make the narrow air-gap assumption, in which case the fundamental part of air-gap inductance is: Ld1 =
2 lR q 4 µ0 Na2 kw s 2 π p2 (g + hm )
18
(39)
Here, g is the magnetic gap, including the physical rotational gap and any magnet retaining means that might be used. hm is the magnet thickness. Since the magnet thickness is included in the air-gap, the air-gap permeance may not be very large, so that slot leakage inductance may be important. To estimate this, assume that the slot shape is rectangular, characterized by the following dimensions: hs height of the main portion of the slot ws width of the top of the main portion of the slot hd height of the slot depression wd slot depression opening Of course not all slots are rectangular: in fact in most machines the slots are trapezoidal in shape to maintain teeth cross-sections that are radially uniform. However, only a very small error (a few percent) is incurred in calculating slot permeance if the slot is assumed to be rectangular and the top width is used (that is the width closest to the air-gap). Then the slot permeance is, per unit length: hd 1 hs + P = µ0 3 ws wd Assume for the rest of this discussion a standard winding, with m slots in each phase belt (this assumes, then, that the total number of slots is Ns = 2pqm), and each slot holds two halfcoils. (A half-coil is one side of a coil which, of course, is wound in two slots). If each coil has Nc turns (meaning Na = 2pmNc ) , then the contribution to phase self-inductance of one slot is, if both half-coils are from the same phase, 4lPNc2 . If the half-coils are from different phases, then the contribution to self inductance is lPNc2 and the magnitude of the contribution to mutual inductance is lPNc2 . (Some caution is required here. For three phase windings the mutual inductance is negative, so are the senses of the currents in the two other phases, so the impact of “mutual leakage” is to increase the reactance. This will be true for other numbers of phases as well, even if the algebraic sign of the mutual leakage inductance is positive, in which case so will be the sense of the other- phase current.) We will make two other assumptions here. The standard one is that the winding “coil throw”, s or span between sides of a coil, is N 2p − Nsp . Nsp is the coil “short pitch”. The other is that each phase belt will overlap with, at most two other phases: the ones on either side in sequence. This last assumption is immediately true for three- phase windings (because there are only two other phases. It is also likely to be true for any reasonable number of phases. Noting that each phase occupies 2p(m − Nsp ) slots with both coil halves in the same slot and 2pNsp slots in which one coil half shares a slot with each of two different phases, we can write down the two components of slot leakage inductance, self- and mutual: h
Las = 2pl (m − Nsp ) (2Nc )2 + 2Nsp Nc2
Lam = −2plNsp Nc2
i
For a three- phase machine, then, the total slot leakage inductance is: La = Las − Lam = 2plPNc2 (4m − Nsp ) For a uniform, symmetric winding with an odd number of phases, it is possible to show that the effective slot leakage inductance is: La = Las − 2Lam cos 19
2π q
Total synchronous inductance is the sum of air-gap and leakage components: so far this is:
Ld = Ld1 + La 4.4.2
Air-Gap Armature Windings
It is shown in Appendix 1 that the inductance of a single-phase of an air-gap winding is: X
La =
Lnp
n
where the harmonic components are: Lk =
8 π + +
2 N2 µ0 lkwn a k(1 − x2 )2
1 − x2−k γ 2k
1 − x2+k
(4 − k2 ) (1 − γ 2k )
ξ 2k 1 − xk+2
2
(2 + k)2 (1 − γ 2k )
1 − γ −2k x2+k
+
ξ −2k 1 − x2−k
2
(2 − k)2 (γ −2k − 1)
1 − x2−k
(4 − k2 ) (γ −2k − 1)
where we have used the following shorthand coefficients:
k 1 − x2 − 4 − k2 2
Rwi Rwo Ri Rs Rwo Rs
x = γ = ξ =
This fits into the conventional inductance framework: Ln =
2 4 µ0 Na2 Rs Lkwn ka π N 2 p2 g
if we assign the “thick armature” coefficient to be: ka =
1 − x2−k γ 2k 1 − x2+k 1 2gk Rwo (1 − x2 )2 (4 − k2 ) (1 − γ 2k ) + +
ξ 2k 1 − xk+2
2
(2 + k)2 (1 − γ 2k )
1 − γ −2k x2+k
+
2
(2 − k)2 (γ −2k − 1)
1 − x2−k
(4 − k2 ) (γ −2k − 1)
20
ξ −2k 1 − x2−k
−
k 1− 4 − k2 2
x2
and k = np and g = Rs − Ri is the conventionally defined “air gap”. If the aspect ratio Ri /Rs is not too far from unity, neither is ka . In the case of p = 2, the fundamental component of ka is: "
2 #
2gk γ4 ξ 4 1 − x4 1 1 − x4 2γ 4 + x4 1 − γ 4 2 ka = − log x + (log x) + 2 Rwo (1 − x2 ) 8 4 (1 − γ 4 ) ξ 4 (1 − γ 4 ) 16 (1 − γ 4 )
For a q-phase winding, a good approximation to the inductance is given by just the first space harmonic term, or: 2 q 4 µ0 Na2 Rs Lkwn ka Ld = 2π n 2 p2 g 4.4.3
Internal Magnet Motor
The permanent magnets will have an effect on reactance because the magnets are in the main flux path of the armature. Further, they affect direct and quadrature reactances differently, so that the machine will be salient. Actually, the effect on the direct axis will likely be greater, so that this type of machine will exhibit “negative” saliency: the quadrature axis reactance will be larger than the direct- axis reactance. A full- pitch coil aligned with the direct axis of the machine would produce flux density: Br =
µ 0 Na I
2g 1 +
Rθp wm 4g hm
Note that only the pole area is carrying useful flux, so that the space fundamental of radial flux density is: B1 =
µ0 Na I 4 sin pθ2m 2g π 1 + wm Rθp hm 4g
Then, since the flux linked by the winding is: λa =
2RlNa kw B1 p
The d- axis inductance, including mutual phase coupling, is (for a q- phase machine): Ld =
2 pθp q 4 µ0 Na2 Rlkw γm sin 2 2π p g 2
The quadrature axis is quite different. On that axis, the armature does not tend to push flux through the magnets, so they have only a minor effect. What effect they do have is due to the fact that the magnets produce a space in the active air- gap. Thus, while a full- pitch coil aligned with the quadrature axis will produce an air- gap flux density: Br =
µ0 N I g
the space fundamental of that will be: 21
µ0 N I 4 pθt B1 = 1 − sin g π 2
where θt is the angular width taken out of the pole by the magnets. So that the expression for quadrature axis inductance is: 2 q 4 µ0 Na2 Rlkw Lq = 2π p2 g
5
pθt 1 − sin 2
Current Rating and Resistance
The last part of machine rating is its current capability. This is heavily influenced by cooling methods, for the principal limit on current is the heating produced by resistive dissipation. Generally, it is possible to do first-order design estimates by assuming a current density that can be handled by a particular cooling scheme. Then, in an air-gap winding:
2 2 Na Ia = Rwo − Rwi
θ we
Ja 2 and note that, usually, the armature fills the azimuthal space in the machine: 2qθwe = 2π For a winding in slots, nearly the same thing is true: if the rectangular slot model holds true: 2qNa Ia = Ns hs ws Js where we are using Js to note slot current density. Now, suppose we can characterize the total slot area by a “space factor” λs which is the ratio between total slot area and the annulus occupied by the slots: for the rectangular slot model: λs =
Ns hs ws 2 − R2 π Rwo wi
where Rwi = R+hd and Rwo = Rwi +hs in a normal, stator outside winding. In this case, Ja = Js λs and the two types of machines can be evaluated in the same way. It would seem apparent that one would want to make λs as large as possible, to permit high currents. The limit on this is that the magnetic teeth between the conductors must be able to carry the air-gap flux, and making them too narrow would cause them to saturate. The peak of the time fundamental magnetic field in the teeth is, for example, Bt = B 1
2πR Ns wt
where wt is the width of a stator tooth: wt =
2π(R + hd ) − ws Ns
so that Bt ≈
B1 1 − λs 22
5.1
Resistance
Winding resistance may be estimated as the length of the stator conductor divided by its area and its conductivity. The length of the stator conductor is: lc = 2lNa fe where the “end winding factor” fe is used to take into account the extra length of the end turns (which is usually not negligible). The area of each turn of wire is, for an air-gap winding : Aw =
2 − R2 θwe Rwo wi λw 2 Na
where λw , the “packing factor” relates the area of conductor to the total area of the winding. The resistance is then just: 4lNa2 Ra = 2 − R2 λ σ θwe Rwo w wi and, of course, σ is the conductivity of the conductor. For windings in slots the expression is almost the same, simply substituting the total slot area: Ra =
2qlNa2 Ns hs ws λw σ
The end turn allowance depends strongly on how the machine is made. One way of estimating what it might be is to assume that the end turns follow a roughly circular path from one side of the machine to the other. The radius of this circle would be, very roughly, Rw /p, where Rw is the average radius of the winding: Rw ≈ (Rwo + Rwi )/2 Then the end-turn allowance would be: fe = 1 +
6
πRw pl
Appendix 1: Air-Gap Winding Inductance
In this appendix we use a simple two-dimensional model to estimate the magnetic fields and then inductances of an air-gap winding. The principal limiting assumption here is that the winding is uniform in the zˆ direction, which means it is long in comparison with its radii. This is generally not true, nevertheless the answers we will get are not too far from being correct. The style of analysis used here can be carried into a three-dimensional, or quasi-three dimensional domain to get much more precise answers, at the expense of a very substantial increase in complexity. The coordinate system to be used is shown in Figure 10. To maintain generality we have four radii: Ri and Rs are ferromagnetic boundaries, and would of course correspond with the machine shaft and the stator core. The winding itself is carried between radii R1 and R2 , which correspond with radii Rwi and Rwo in the body of the text. It is assumed that the armature is carrying a current in the z- direction, and that this current is uniform in the radial dimension of the armature. If a single phase of the armature is carrying current, that current will be: Jz0 =
θwe 2
N a Ia R22 − R12
23
Winding
Outer Magnetic Boundary 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 R 11111111111111111111111111111 00000000000000000000000000000 s 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 R 11111111111111111111111111111 00000000000000000000000000000 i 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 R 11111111111111111111111111111 00000000000000000000000000000 111111111111 000000000000 1 11111111111111111111111111111 00000000000000000000000000000 R Inner Magnetic 11111111111111111111111111111 00000000000000000000000000000 2 Boundary 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 00000000000000000000000000000 11111111111111111111111111111
Figure 10: Coordinate System for Inductance Calculation over the annular wedge occupied by the phase. The resulting distribution can be fourier analyzed, and the n-th harmonic component of this will be (assuming the coordinate system has been chosen appropriately): 4 Na Ia 4 θwe = Jzn = Jz0 sin n kwn nπ 2 π R22 − R12 where the n-th harmonic winding factor is: kwn =
sin n θwe 2 n θwe 2
and note that θwe is the electrical winding angle: θwe = pθw Now, it is easiest to approach this problem using a vector potential. Since the divergence of flux density is zero, it is possible to let the magnetic flux density be represented by the curl of a vector potential: B =∇×A Taking the curl of that:
∇ × ∇ × A = µ0 J = ∇∇ · A − ∇2 A and using the Coulomb gage
∇·A =0
we have a reasonable tractable partial differential equation in the vector potential: ∇2 A = −µ0 J Now, since in our assumption there is only a z- directed component of J, we can use that one component, and in circular cylindrical coordinates that is: 1 ∂ ∂Az 1 ∂2 r + 2 2 Az = −µ0 Jz r ∂r ∂r r ∂θ 24
For this problem, all variables will be varying sinusoidally with angle, so we will assume that angular dependence ejkθ . Thus: k2 1 ∂ ∂Az r − 2 Az = −µ0 Jz r ∂r ∂r r
(40)
This is a three-region problem. Note the regions as: i w o
Ri < r < R1 R1 < r < R2 R2 < r < Rs
For i and o, the current density is zero and an appropriate solution to (40) is: Az = A+ r k + A− r −k In the region of the winding, w, a particular solution must be used in addition to the homogeneous solution, and Az = A+ r k + A− r −k + Ap where, for k �= 2, Ap = − or, if k = 2,
µ0 Jz r 2 4 − k2
1 µ0 Jz r 2 log r − Ap = − 4 4
And, of course, the two pertinent components of the magnetic flux density are: 1 ∂Az r ∂θ ∂Az = − ∂r
Br = Bθ
Next, it is necessary to match boundary conditions. There are six free variables and correspondingly there must be six of these boundary conditions. They are the following: • At the inner and outer magnetic boundaries, r = Ri and r = Rs , the azimuthal magnetic field must vanish. • At the inner and outer radii of the winding itself, r = R1 and r = R2 , both radial and azimuthal magnetic field must be continuous. These conditions may be summarized by: kAi+ Rik−1 − kAi− Ri−k−1 = 0
kAo+ Rsk−1 − kAo− Rs−k−1 = 0 µ0 Jz R2 −k−1 k−1 o k−1 − + Aw Aw = A+ R2 + Ao− R2−k−1 + R2 − R2 4 − k2 25
2µ0 Jz R2 4 − k2 µ0 Jz R1 −k−1 k−1 − + Aw Aw + R1 − R1 4 − k2 2µ0 Jz R1 k−1 −k−1 −kAw + kAw + + R1 − R1 4 − k2 k−1 w −k−1 R2 + + kA− −kAw + R2
= −kAo+ R2k−1 + kAo− R2−k−1 i = A+ R1k−1 + Ai− R1−k−1
= −kAi+ R1k−1 + kAi− R1−k−1
Note that we are carrying this out here only for the case of k �= 2. The k = 2 case may be obtained by substituting its particular solution in at the beginning or by using L’Hopital’s rule on the final solution. This set may be solved (it is a bit tedious but quite straightforward) to yield, for the winding region: Az =
µ0 Jz 2k
+
R22+k − R12+k
Rs2k R22−k − Ri2k R12−k + (2 + k) Rs2k − Ri2k (2 − k) Rs2k − Ri2k
"
R22−k − R12−k
(2 − k) Ri−2k − Rs−2k
+
!
r k
Rs−2k R22+k − Ri−2k R12+k
(2 + k) Ri−2k − Rs−2k
2k 2 r −k − r 4 − k2
Now, the inductance linked by any single, full-pitched loop of wire located with one side at azimuthal position θ and radius r is: λi = 2lAz (r, θ) To extend this to the whole winding, we integrate over the area of the winding the incremental flux linked by each element times the turns density. This is, for the n-th harmonic of flux linked: 4lkwn Na λn = 2 R2 − R12
Z
R2
R1
Az (r)rdr
Making the appropriate substitutions for current into the expression for vector potential, this becomes: λn =
2 N 2I 8 µ0 lkwn a a 2 π k R2 − R12 2
"
R22+k − R12+k Rs2k R22−k − Ri2k R12−k + (2 − k) Rs2k − Ri2k (2 + k) R22k − Ri2k
R22−k − R12−k
!
R2k+2 − R1k+2 k+2
R−2k R22+k − Ri−2k R12+k R22−k − R12−k 2k R24 − R14 + s + − 2−k 4 − k2 4 (2 − k) R−2k − Rs−2k (2 + k) R−2k − Rs−2k
7
i
i
Appendix 2: Permanent Magnet Field Analysis
This section is a field analysis of the kind of radially magnetized, permanent magnet structures commonly used in electric machinery. It is a fairly general analysis, which will be suitable for use with either surface or in-slot windings, and for the magnet inside or the magnet outside case. This is a two-dimensional layout suitable for situations in which field variation along the length of the structure is negligible.
26
8
Layout
The assumed geometry is shown in Figure 11. Assumed iron (highly permeable) boundaries are at radii Ri and Rs . The permanent magnets, assumed to be polarized radially and alternately (i.e. North-South ...), are located between radii R1 and R2 . We assume there are p pole pairs (2p magnets) and that each magnet subsumes an electrical angle of θme . The electrical angle is just p times the physical angle, so that if the magnet angle were θme = π, the magnets would be touching.
Magnets
Outer Magnetic Boundary 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 R 11111111111111111111111111111 00000000000000000000000000000 s 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 R 11111111111111111111111111111 00000000000000000000000000000 i 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 R 11111111111111111111111111111 00000000000000000000000000000 000000000000 111111111111 1 11111111111111111111111111111 00000000000000000000000000000 R Inner Magnetic 11111111111111111111111111111 00000000000000000000000000000 2 Boundary 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 00000000000000000000000000000 11111111111111111111111111111
Figure 11: Axial View of Magnetic Field Problem If the magnets are arranged so that the radially polarized magnets are located around the azimuthal origin (θ = 0), the space fundamental of magnetization is: M = ir M0 cos pθ
(41)
where the fundamental magnitude is: M0 =
4 θme Brem sin π 2 µ0
(42)
and Brem is the remanent magnetization of the permanent magnet. Since there is no current anywhere in this problem, it is convenient to treat magnetic field as the gradient of a scalar potential: H = −∇ψ
(43)
∇2 ψ = −∇ · H
(44)
The divergence of this is:
27
Since magnetic flux density is divergence-free, ∇·B =0
(45)
∇ · H = −∇ · M
(46)
we have:
or:
1 (47) ∇2 ψ = ∇ · M = M0 cos pθ r Now, if we let the magnetic scalar potential be the sum of particular and homogeneous parts: ψ = ψp + ψh
(48)
where ∇2 ψh = 0, then:
1 ∇2 ψp = M0 cos pθ (49) r We can find a suitable solution to the particular part of this in the region of magnetization by trying: (50) ψp = Cr γ cos pθ Carrying out the Laplacian on this: 1 ∇2 ψp = Cr γ−2 γ 2 − p2 cos pθ = M0 cos pθ r
(51)
which works if γ = 1, in which case:
ψp =
M0 r cos pθ 1 − p2
(52)
Of course this solution holds only for the region of the magnets: R1 < r < R2 , and is zero for the regions outside of the magnets. A suitable homogeneous solution satisfies Laplace’s equation, ∇2 ψh = 0, and is in general of the form: (53) ψh = Ar p cos pθ + Br −p cos pθ Then we may write a trial total solution for the flux density as: A1 r p + B1 r −p cos pθ M0 r cos pθ ψ= A2 r p + B2 r −p + 1 − p2 ψ = A3 r p + B3 r −p cos pθ
Ri < r < R1 ψ = R1 < r < R2 R2 < r < Rs
(54) (55) (56)
The boundary conditions at the inner and outer (assumed infinitely permeable) boundaries at r = Ri and r = Rs require that the azimuthal field vanish, or ∂ψ ∂θ = 0, leading to: B1 = −Ri2p A1
B3 =
−Rs2p A3
28
(57) (58)
At the magnet inner and outer radii, Hθ and Br must be continuous. These are: 1 ∂ψ r∂θ ∂ψ + Mr = µ0 − ∂r
Hθ = −
(59)
Br
(60)
These become, at r = R1 :
M0 1 − p2 M0 −pA1 R1p−1 + Ri2p R1−p−1 = −p A2 R1p−1 − B2 R1−p−1 − + M0 1 − p2
−pA1 R1p−1 − Ri2p R1−p−1
= −p A2 R1p−1 + B2 R1−p−1 − p
(61) (62)
and at r = R2 : M0 1 − p2 M0 −pA3 R2p−1 + Rs2p R2−p−1 = −p A2 R2p−1 − B2 R2−p−1 − + M0 1 − p2
−pA3 R2p−1 − Rs2p R2−p−1
= −p A2 R2p−1 + B2 R2−p−1 − p
(63) (64)
Some small-time manipulation of these yields: M0 1 − p2 M0 A1 R1p + Ri2p R1−p = A2 R1p − B2 R1−p + pR1 1 − p2 M0 A3 R2p − Rs2p R2−p = A2 R2p + B2 R2−p + R2 1 − p2 M0 A3 R2p + Rs2p R2−p = A2 R2p − B2 R2−p + pR2 1 − p2
A1 R1p − Ri2p R1−p
= A2 R1p + B2 R1−p + R1
(65) (66) (67) (68)
Taking sums and differences of the first and second and then third and fourth of these we obtain: 1+p 1 − p2 p−1 = −2B2 R1−p + R1 M0 1 − p2 1+p = 2A2 R2p + R2 M0 1 − p2 p−1 = −2B2 R2−p + R2 M0 1 − p2
2A1 R1p = 2A2 R1p + R1 M0 2A1 Ri2p R1−p 2A3 R2p 2A3 Rs2p R2−p
(69) (70) (71) (72)
and then multiplying through by appropriate factors (R2p and R1p ) and then taking sums and differences of these, M0 p + 1 2 1 − p2 M p−1 0 = R1 R2−p − R2 R1−p 2 1 − p2
(A1 − A3 ) R1p R2p = (R1 R2p − R2 R1p )
A1 Ri2p − A3 Rs2p R1−p R2−p
29
(73) (74)
Dividing through by the appropriate groups: A1 − A3 = A1 Ri2p − A3 Rs2p =
R1 R2p − R2 R1p M0 1 + p R1p R2p 2 1 − p2
(75)
R1 R2−p − R2 R1−p M0 p − 1 2 1 − p2 R1−p R2−p
(76)
and then, by multiplying the top equation by Rs2p and subtracting:
A1 Rs2p −
Ri2p
=
R1 R2p − R2 R1p M0 1 + p R1p R2p 2 1 − p2
!
Rs2p −
R1 R2−p − R2 R1−p M0 p − 1 2 1 − p2 R1−p R2−p
(77)
This is readily solved for the field coefficients A1 and A3 : M0
p + 1 1−p p − 1 1+p 1−p 1+p 2p R + R R − R − R s 1 2 2 1 p2 − 1 p2 − 1
(78)
M0
1 1−p 1 1+p R1 − R21−p Ri2p − R2 − R11+p 1−p 1+p
(79)
A1 = − 2p 2 Rs − Ri2p A3 = − 2p 2 Rs − Ri2p
Now, noting that the scalar potential is, in region 1 (radii less than the magnet), ψ = A1 (r p − Ri2p r −p ) cos pθ
r < R1
ψ = A3 (r p − Rs2p r −p ) cos pθ
r > R2
and noting that p(p + 1)/(p2 − 1) = p/(p − 1) and p(p − 1)/(p2 − 1) = p/(p + 1), magnetic field is:
Hr =
Hr =
r < R1 M0 2
Rs2p
−
(80) Ri2p
r > R2 M0 2
Rs2p
−
Ri2p
p p R11−p − R21−p Rs2p + R21+p − R11+p p−1 p+1
p p R21+p − R11+p R11−p − R21−p Ri2p + p−1 p+1
r p−1 + Ri2p r −p−1 cos pθ (81)
r p−1 + Rs2p r −p−1 cos pθ
The case of p = 1 appears to be a bit troublesome here, but is easily handled by noting that: p 1−p R2 R1 − R21−p = log p→1 p − 1 R1
lim
Now: there are a number of special cases to consider.
For the iron-free case, Ri → 0 and R2 → ∞, this becomes, simply, for r < R1 :
Hr =
M0 p 1−p R1 − R21−p r p−1 cos pθ 2 p−1
30
(82)
Note that for the case of p = 1, the limit of this is
Hr =
R2
M0 log cos θ 2 R1
and for r > R2 :
Hr =
M0 p p+1 R2 − R1p+1 r −(p+1) cos pθ 2 p+1
For the case of a machine with iron boundaries and windings in slots, we are interested in the fields at the boundaries. In such a case, usually, either Ri = R1 or Rs = R2 . The fields are: at the outer boundary: r = Rs : Hr = M0
Rsp−1 Rs2p − Ri2p
p p p+1 1−p Ri2p R1
− R21−p cos pθ R2 − R1p+1 + p+1 p−1
or at the inner boundary: r = Ri :
Hr = M0
Rip−1 Rs2p − Ri2p
p p p+1 Rs2p R11−p − R21−p cos pθ R2 − R1p+1 + p+1 p−1
31
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machinery Class Notes 8: Analytic Design Evaluation of Induction Machines c
2005 James L. Kirtley Jr.
1
January 12, 2006
Introduction
Induction machines are perhaps the most widely used of all electric motors. They are generally simple to build and rugged, offer reasonable asynchronous performance: a manageable torque-speed curve, stable operation under load, and generally satisfactory efficiency. Because they are so widely used, they are worth understanding. In addition to their current economic importance, induction motors and generators may find application in some new applications with designs that are not similar to motors currently in commerce. An example is very high speed motors for gas compressors, perhaps with squirrel cage rotors, perhaps with solid iron (or perhaps with both). Because it is possible that future, high performance induction machines will be required to have characteristics different from those of existing machines, it is necessary to understand them from first principles, and that is the objective of this document. It starts with a circuit theoretical view of the induction machine. This analysis is strictly appropriate only for wound-rotor machines, but leads to an understanding of more complex machines. This model will be used to explain the basic operation of induction machines. Then we will derive a model for squirrel-cage machines. Finally, we will show how models for solid rotor and mixed solid rotor/squirrel cage machines can be constructed. The view that we will take in this document is relentlessly classical. All of the elements that we will use are calculated from first principles, and we do not resort to numerical analysis or empirical methods unless we have no choice. While this may seem to be seriously limiting, it serves our basic objective here, which is to achieve an understanding of how these machines work. It is our feeling that once that understanding exists, it will be possible to employ more sophisticated methods of analysis to get more accurate results for those elements of the machines which do not lend themselves to simple analysis. An elementary picture of the induction machine is shown in Figure 1. The rotor and stator are coaxial. The stator has a polyphase winding in slots. The rotor has either a winding or a cage, also in slots. This picture will be modified slightly when we get to talking of “solid rotor” machines, anon. Generally, this analysis is carried out assuming three phases. As with many systems, this generalizes to different numbers of phases with little difficulty.
2
Induction Motor Transformer Model
The induction machine has two electrically active elements: a rotor and a stator. In normal operation, the stator is excited by alternating voltage. (We consider here only polyphase machines). The stator excitation creates a magnetic field in the form of a rotating, or traveling wave, which induces currents in the circuits of the rotor. Those currents, in turn, interact with the traveling 1
Stator Core
Stator Winding in Slots Rotor Winding or Cage in Slots
Rotor
Air−Gap
Figure 1: Axial View of an Induction Machine wave to produce torque. To start the analysis of this machine, assume that both the rotor and the stator can be described by balanced, three – phase windings. The two sets are, of course, coupled by mutual inductances which are dependent on rotor position. Stator fluxes are (λa , λb , λc ) and rotor fluxes are (λA , λB , λC ). The flux vs. current relationship is given by:
λa λb λc λA λB λC
=
L S M TSR
M SR
LR
ia ib ic iA iB iC
(1)
where the component matrices are:
La Lab Lab LS = Lab La Lab Lab Lab La
(2)
LA LAB LAB LAB LR = LAB LA LAB LAB LA
(3)
The mutual inductance part of (1) is a circulant matrix:
M SR
M cos(pθ) = M cos(pθ − M cos(pθ +
M cos(pθ + 2π ) M cos(pθ) 3 2π 3 ) M cos(pθ − 2
2π 3 )
M cos(pθ − M cos(pθ + 2π 3 ) M cos(pθ)
2π 3 ) 2π 3 )
(4)
To carry the analysis further, it is necessary to make some assumptions regarding operation. To start, assume balanced currents in both the stator and rotor: ia = IS cos(ωt)
ib = IS cos(ωt − ic = IS cos(ωt +
2π 3 ) 2π 3 )
iA = IR cos(ωR t + ξR )
iB = IR cos(ωR t + ξR − iC = IR cos(ωR t + ξR +
(5)
2π 3 ) 2π 3 )
(6)
The rotor position θ can be described by θ = ωm t + θ0
(7)
Under these assumptions, we may calculate the form of stator fluxes. As it turns out, we need only write out the expressions for λa and λA to see what is going on: λa = (La − Lab )Is cos(ωt) + M IR (cos(ωR t + ξR ) cos p(ωm + θ0 ) (8) 2π 2π 2π 2π + cos(ωR t + ξR + ) cos(p(ωm t + θ0 ) − ) + cos(ωR t + ξR − ) cos(p(ωm t + θ0 ) + ) 3 3 3
3 which, after reducing some of the trig expressions, becomes:
3 λa = (La − Lab )Is cos(ωt) + M IR cos((pωm + ωR )t + ξR + pθ0 ) 2
Doing the same thing for the rotor phase A yields:
λA = M Is (cos p(ωm t + θ0 ) cos(ωt)) + cos(p(ωm t + θ0 ) − + cos(p(ωm t + θ0 ) +
2π 2π ) cos(ωt − ) 3 3
(9)
(10)
2π 2π ) cos(ωt + ) + (LA − LAB )IR cos(ωR t + ξR ) 3 3
This last expression is, after manipulating:
3
λA = M Is cos((ω − pωm )t − pθ0 ) + (LA − LAB )IR cos(ωR t + ξR ) 2
(11)
These two expressions, 9 and 11 give expressions for fluxes in the armature and rotor windings in terms of currents in the same two windings, assuming that both current distributions are sinusoidal in time and space and represent balanced distributions. The next step is to make another assumption, that the stator and rotor frequencies match through rotor rotation. That is: ω − pωm = ωR It is important to keep straight the different frequencies here: ω ωR ωm
is stator electrical frequency is rotor electrical frequency is mechanical rotation speed
3
(12)
so that pωm is electrical rotation speed. To refer rotor quantities to the stator frame (i.e. non- rotating), and to work in complex amplitudes, the following definitions are made: λa = Re(Λa ejωt )
(13)
λA = Re(ΛA ejωR t )
(14)
ia = Re(I a ejωt )
(15)
iA = Re(I A ejωR t )
(16)
With these definitions, the complex amplitudes embodied in 56 and 64 become: 3 Λa = LS I a + M I A ej(ξR +pθ0 ) 2
(17)
3 ΛA = M I a e−jpθ0 + LR I A ejξR (18) 2 There are two phase angles embedded in these expressions: θ0 which describes the rotor physical phase angle with respect to stator current and ξR which describes phase angle of rotor currents with respect to stator currents. We hereby invent two new rotor variables: ΛAR = ΛA ejpθ)
(19)
I AR = I A ej(pθ0 +ξR )
(20)
These are rotor flux and current referred to armature phase angle. Note that ΛAR and I AR have the same phase relationship to each other as do ΛA and I A . Using 19 and 20 in 17 and 18, the basic flux/current relationship for the induction machine becomes: "
Λa ΛAR
#
=
"
LS 3 2M
3 2M LR
#"
Ia I AR
#
(21)
This is an equivalent single- phase statement, describing the flux/current relationship in phase a, assuming balanced operation. The same expression will describe phases b and c. Voltage at the terminals of the stator and rotor (possibly equivalent) windings is, then:
or:
V a = jωΛa + Ra I a
(22)
V AR = jωR ΛAR + RA I AR
(23)
3 V a = jωLS I a + jω M I AR + Ra I a 2
(24)
3 (25) V AR = jωR M I a + jωR LR I AR + RA I AR 2 To carry this further, it is necessary to go a little deeper into the machine’s parameters. Note that LS and LR are synchronous inductances for the stator and rotor. These may be separated into space fundamental and “leakage” components as follows: 4
LS = La − Lab =
3 4 µ0 RlNS2 kS2 + LSl 2π p2 g
LR = LA − LAB =
2 3 4 µ0 RlNR2 kR + LRl 2π p2 g
(26) (27)
Where the normal set of machine parameters holds: R l g p N k S R Ll
is rotor radius is active length
is the effective air- gap
is the number of pole- pairs
represents number of turns
represents the winding factor
as a subscript refers to the stator
as a subscript refers to the rotor
is “leakage” inductance
The two leakage terms LSl and LRl contain higher order harmonic stator and rotor inductances, slot inducances, end- winding inductances and, if necessary, a provision for rotor skew. Essentially, they are used to represent all flux in the rotor and stator that is not mutually coupled. In the same terms, the stator- to- rotor mutual inductance, which is taken to comprise only a space fundamental term, is: M=
4 µ0 RlNS NR kS kR π p2 g
(28)
Note that there are, of course, space harmonic mutual flux linkages. If they were to be included, they would hair up the analysis substantially. We ignore them here and note that they do have an effect on machine behavior, but that effect is second- order. Air- gap permeance is defined as: ℘ag =
4 µ0 Rl π p2 g
(29)
so that the inductances are: 3 LS = ℘ag kS2 NS2 + LSl 2
(30)
3 2 LR = ℘ag kR NR2 + LRl 2
(31)
M = ℘ag NS NR kS kR
(32)
ωR = sω
(33)
Here we define “slip” s by:
5
so that
s=1−
pωm ω
(34)
Then the voltage balance equations become: 3 3 V a = jω ℘ag kS2 NS2 + LSl I a + jω ℘ag NS NR kS kR I AR + Ra I a 2 2
V AR
3 3 2 ℘ag kR NR2 + LRl I AR + RA I AR = jsω ℘ag NS NR kS kR I a + jsω 2 2
(35) (36)
At this point, we are ready to define rotor current referred to the stator. This is done by assuming an effective turns ratio which, in turn, defines an equivalent stator current to produce the same fundamental MMF as a given rotor current: I2 =
NR kR I NS kS AR
(37)
Now, if we assume that the rotor of the machine is shorted so that V AR = 0 and do some manipulation we obtain: V a = j(XM + X1 )I a + jXM I 2 + Ra I a
(38)
R2 I s 2
(39)
0 = jXM I a + j(XM + X2 )I 2 + where the following definitions have been made: XM =
3 ω℘ag NS2 kS2 2
X1 = ωLSl X2 = ωLRl R2 = RA
(41)
NS kS NR kR
NS kS NR kR
(40)
2
2
(42) (43)
These expressions describe a simple equivalent circuit for the induction motor shown in Figure 2. We will amplify on this equivalent circuit anon.
6
I� a Ra
∧∧∧ ∨∨
X1 ∩∩∩∩
X2 I2 ∩∩∩∩ � < ⊃ ⊃ Xm <> R2 ⊃ > < s ⊃
Figure 2: Equivalent Circuit
3
Operation: Energy Balance
Now we are ready to see how the induction machine actually works. Assume for the moment that Figure 2 represents one phase of a polyphase system and that the machine is operated under balanced conditions and that speed is constant or varying only slowly. “Balanced conditions” means that each phase has the same terminal voltage magnitude and that the phase difference between phases is a uniform. Under those conditions, we may analyze each phase separately (as if it were a single phase system). Assume an RMS voltage magnitude of Vt across each phase. The “gap impedance”, or the impedance looking to the right from the right-most terminal of X1 is: R2 Zg = jXm ||(jX2 + ) (44) s A total, or terminal impedance is then Zt = jX1 + Ra + Zg and terminal current is It =
Vt Zt
(45) (46)
Rotor current is found by using a current divider: I2 = It
jXm jX2 + Rs2 + jXm
(47)
“Air-gap” power is then calculated (assuming a three-phase machine): Pag = 3|I2 |2
R2 s
(48)
This is real (time-average) power crossing the air-gap of the machine. Positive slip implies rotor speed less than synchronous and positive air-gap power (motor operation). Negative slip means rotor speed is higher than synchronous, negative air-gap power (from the rotor to the stator) and generator operation. Now, note that this equivalent circuit represents a real physical structure, so it should be possible to calculate power dissipated in the physical rotor resistance, and that is: Ps = Pag s 7
(49)
(Note that, since both Pag and s will always have the same sign, dissipated power is positive.) The rest of this discussion is framed in terms of motor operation, but the conversion to generator operation is simple. The difference between power crossing the air-gap and power dissipated in the rotor resistance must be converted from mechanical form: Pm = Pag − Ps
(50)
Pin = Pag + Pa
(51)
Pa = 3|It |2 Ra
(52)
and electrical inputpower is: where armature dissipation is: Output (mechanical) power is Pout = Pag − Pw Where Pw describes friction, windage and certain stray losses which we will discuss later. And, finally, efficiency and power factor are: η=
Pout Pin
cos ψ =
3.1
(53)
(54)
Pin 3Vt It
(55)
Example of Operation
The following MATLAB script generates a torque-speed and power-speed curve for the simple induction motor model described above. Note that, while the analysis does not require that any of the parameters, such as rotor resistance, be independent of rotor speed, this simple script does assume that all parameters are constant.
3.2
Example
That MATLAB script has been run for a standard motor with parameters given in Table 1. Torque vs. speed and power vs. speed are plotted for this motor in Figure 3. These curves were generated by the MATLAB script shown above.
4
Squirrel Cage Machine Model
Now we derive a circuit model for the squirrel-cage motor using field analytical techniques. The model consists of two major parts. The first of these is a description of stator flux in terms of stator and rotor currents. The second is a description of rotor current in terms of air- gap flux. The result of all of this is a set of expressions for the elements of the circuit model for the induction machine. To start, assume that the rotor is symmetrical enough to carry a surface current, the fundamental of which is:
′
K r = ız Re K r ej(sωt−pφ )
= ız Re K r ej(ωt−pφ) 8
(56)
% ------------------------------------------------------
% Torque-Speed Curve for an Induction Motor
% Assumes the classical model
% This is a single-circuit model
% Required parameters are R1, X1, X2, R2, Xm, Vt, Ns
% Assumed is a three-phase motor
% This thing does a motoring, full speed range curve
% Copyright 1994 James L. Kirtley Jr.
% -------------------------------------------------------
s = .002:.002:1; % vector of slip N = Ns .* (1 - s); % Speed, in RPM oms = 2*pi*Ns/60; % Synchronous speed Rr = R2 ./ s; % Rotor resistance Zr = j*X2 + Rr; % Total rotor impedance Za = par(j*Xm, Zr); % Air-gap impedance Zt = R1 + j*X1 +Za; % Terminal impedance Ia = Vt ./ Zt; % Terminal Current I2 = Ia .* cdiv (Zr, j*Xm); % Rotor Current Pag = 3 .* abs(I2) .^2 .* Rr; % Air-Gap Power Pm = Pag .* (1 - s); % Converted Power Trq = Pag ./ oms; % Developed Torque subplot(2,1,1) plot(N, Trq) title(’Induction Motor’); ylabel(’N-m’); subplot(2,1,2) plot(N, Pm); ylabel(’Watts’); xlabel(’RPM’);
Table 1: Example, Standard Motor
Rating Voltage Stator Resistance R1 Rotor Resistance R2 Stator Reactance X1 Rotor Reactance X2 Magnetizing Reactance Xm Synchronous Speed Ns
9
300 440 254 .73 .64 .06 .06 2.5 1200
kw VRMS, l-l VRMS, l-n Ω Ω Ω Ω Ω RPM
Induction Motor 300
250
N−m
200
150
100
50
0
0
200
400
600
800
1000
1200
200
400
600 RPM
800
1000
1200
4
3
x 10
2.5
Watts
2
1.5
1
0.5
0
0
Figure 3: Torque and Power vs. Speed for Example Motor Note that in 56 we have made use of the simple transformation between rotor and stator coordinates: φ′ = φ − ωm t (57) and that pωm = ω − ωr = ω(1 − s)
(58)
Here, we have used the following symbols: Kr s ω ωr ωm
is is is is is
complex amplitude of rotor surface current per- unit “slip” stator electrical frequency rotor electrical frequency rotational speed
The rotor current will produce an air- gap flux density of the form:
Br = Re B r ej(ωt−pφ) where
B r = −jµ0 10
R K pg r
(59)
(60)
Note that this describes only radial magnetic flux density produced by the space fundamental of rotor current. Flux linked by the armature winding due to this flux density is: λAR = lNS kS
Z
0
−π p
Br (φ)Rdφ
(61)
This yields a complex amplitude for λAR :
λAR = Re ΛAR ejωt where ΛAR =
2lµ0 R2 NS kS Kr p2 g
(62)
(63)
Adding this to flux produced by the stator currents, we have an expression for total stator flux: Λa =
!
3 4 µ0 NS2 RlkS2 2lµ0 R2 NS kS + + L Kr I Sl a 2π p2 g p2 g
(64)
Expression 64 motivates a definiton of an equivalent rotor current I2 in terms of the space fundamental of rotor surface current density: I2 =
π R K 3 NS kS z
(65)
Then we have the simple expression for stator flux: Λa = (Lad + LSl )I a + Lad I 2
(66)
where Lad is the fundamental space harmonic component of stator inductance: Lad =
4.1
3 4 µ0 NS2 kS2 Rl 2π p2 g
(67)
Effective Air-Gap: Carter’s Coefficient
In induction motors, where the air-gap is usually quite small, it is necessary to correct the air-gap permeance for the effect of slot openings. These make the permeance of the air-gap slightly smaller than calculated from the physical gap, effectively making the gap a bit bigger. The ratio of effective to physical gap is: t+s (68) geff = g t + s − gf (α) where
f (α) = f
s 2g
= α tan(α) − log sec α
11
(69)
4.2
Squirrel Cage Currents
The second part of this derivation is the equivalent of finding a relationship between rotor flux and I2 . However, since this machine has no discrete windings, we must focus on the individual rotor bars. Assume that there are NR slots in the rotor. Each of these slots is carrying some current. If the machine is symmetrical and operating with balanced currents, we may write an expression for current in the kth slot as:
ik = Re I k ejsωt where
(70)
−j 2πp N
I k = Ie
(71)
R
and I is the complex amplitude of current in slot number zero. Expression 71 shows a uniform progression of rotor current phase about the rotor. All rotor slots carry the same current, but that current is phase retarded (delayed) from slot to slot because of relative rotation of the current wave at slip frequency. The rotor current density can then be expressed as a sum of impulses:
Kz = Re
NX R −1 k=0
1 j(ωr t−k 2πp 2πk ) NR Ie ) δ(φ′ − R NR
(72)
The unit impulse function δ() is our way of approximating the rotor current as a series of impulsive currents around the rotor. This rotor surface current may be expressed as a fourier series of traveling waves: ∞ X
Kz = Re
j(ωr t−npφ′ )
K ne
n=−∞
!
(73)
Note that in 73, we are allowing for negative values of the space harmonic index n to allow for reverse- rotating waves. This is really part of an expansion in both time and space, although we are considering only the time fundamental part. We may recover the nth space harmonic component of 73 by employing the following formula: 1 K n =< π
Z
2π 0
Kr (φ, t)e−j(ωr t−npφ) dφ >
(74)
Here the brackets <> denote time average and are here beause of the two- dimensional nature of the expansion. To carry out 74 on 72, first expand 72 into its complex conjugate parts: NR −1 1 X I j(ωr t−k 2πp I ∗ −j(ωr t−k 2πp 2πk ) ) NR NR ) Kr = e + e δ(φ′ − 2 k=0 R R NR
(75)
If 75 is used in 74, the second half of 75 results in a sum of terms which time average to zero. The first half of the expression results in:
12
Kn =
I 2πR
Z
0
R −1 2π NX
−j 2πpk jnpφ NR
e
e
k=0
δ(φ −
2πk )dφ NR
(76)
The impulse function turns the integral into an evaluation of the rest of the integrand at the impulse. What remains is the sum: Kn =
NR −1 I X j(n−1) 2πkp NR e 2πR k=0
(77)
(
(78)
The sum in 77 is easily evaluated. It is: NX R −1
j
e
2πkp(n−1) NR
=
k=0
NR if (n − 1) NPR = integer 0 otherwise
The integer in 78 may be positive, negative or zero. As it turns out, only the first three of these (zero, plus and minus one) are important, because these produce the largest magnetic fields and therefore fluxes. These are: (n − 1)
p NR
= −1 or n = − NRp−p =0
or n = 1
=1
or n =
NR +p p
(79)
Note that 79 appears to produce space harmonic orders that may be of non- integer order. This is not really true: is is necessary that np be an integer, and 79 will always satisfy that condition. So, the harmonic orders of interest to us are one and NR +1 p NR = − −1 p
n+ =
(80)
n−
(81)
Each of the space harmonics of the squirrel- cage current will produce radial flux density. A surface current of the form: NR I j(ωr t−npφ′ ) e Kn = Re 2πR
produces radial magnetic flux density:
′
Brn = Re B rn ej(ωr t−npφ ) where
(82)
(83)
µ 0 NR I (84) 2πnpg In turn, each of the components of radial flux density will produce a component of induced voltage. To calculate that, we must invoke Faraday’s law: B rn = −j
13
∂B ∂t
The radial component of 85, assuming that the fields do not vary with z, is:
∇×E =−
∂Br 1 ∂ Ez = − R ∂φ ∂t
(85)
(86)
Or, assuming an electric field component of the form:
Ezn = Re E n ej(ωr t−npφ)
(87)
Using 84 and 87 in 86, we obtain an expression for electric field induced by components of airgap flux: ωr R En = (88) B np n E n = −j
µ 0 NR ω r R I 2πg(np)2
(89)
Now, the total voltage induced in a slot pushes current through the conductors in that slot. We may express this by: E 1 + E n− + E n+ = Z slot I
(90)
Now: in 90, there are three components of air- gap field. E1 is the space fundamental field, produced by the space fundamental of rotor current as well as by the space fundamental of stator current. The other two components on the left of 90 are produced only by rotor currents and actually represent additional reactive impedance to the rotor. This is often called zigzag leakage inductance. The parameter Zslot represents impedance of the slot itself: resistance and reactance associated with cross- slot magnetic fields. Then 90 can be re-written as: E 1 = Z slot I + j
µ 0 NR ω r R 2πg
1 1 + I 2 (n+ p) (n− p)2
(91)
To finish this model, it is necessary to translate 91 back to the stator. See that 65 and 77 make the link between I and I 2 : I2 =
NR I 6NS kS
(92)
Then the electric field at the surface of the rotor is: 6NS kS 3 µ0 NS kS R E1 = Z slot + jωr NR π g
1 1 + 2 (n+ p) (n− p)2
I2
(93)
This must be translated into an equivalent stator voltage. To do so, we use 88 to translate 93 into a statement of radial magnetic field, then find the flux liked and hence stator voltage from that. Magnetic flux density is:
14
pE 1 ω R r 6NS kS p Rslot 1 1 3 µ0 NS kS p + jLslot + j + = I2 NR R ωr π g (n+ p)2 (n− p)2
Br =
(94)
where the slot impedance has been expressed by its real and imaginary parts: Z slot = Rslot + jωr Lslot
(95)
Flux linking the armature winding is: λag = NS kS lR
Z
0
π − 2p
Which becomes:
Re B r ej(ωt−pφ) dφ
(96)
(97)
λag = Re Λag ejωt where: Λag = j
2NS kS lR B r p
(98)
Then “air- gap” voltage is: 2ωNS kS lR Br p " # 12lNS2 kS2 1 1 R2 6 µ0 RlNS2 kS2 = −I 2 + jωLslot + + jω π NR s g (n+ p)2 (n− p)2
V ag = jωΛag = −
(99)
Expression 99 describes the relationship between the space fundamental air- gap voltage V ag and rotor current I 2 . This expression fits the equivalent circuit of Figure 4 if the definitions made below hold: X2 I2 ∩∩∩∩ �
<> <> R2 < s
Figure 4: Rotor Equivalent Circuit
X2 R2
12lNS2 kS2 6 µ0 RlNS2 kS2 = ω Lslot + ω NR π g 12lNS2 kS2 = Rslot NR 15
1 1 + 2 (NR + p) (NR − p)2
(100) (101)
The first term in 100 expresses slot leakage inductance for the rotor. Similarly, 101 expresses rotor resistance in terms of slot resistance. Note that Lslot and Rslot are both expressed per unit length. The second term in 100 expresses the “zigzag” leakage inductance resulting from harmonics on the order of rotor slot pitch. Next, see that armature flux is just equal to air- gap flux plus armature leakage inductance. That is, 66 could be written as: Λa = Λag + Lal I a
4.3
(102)
Stator Leakage
There are a number of components of stator leakage Lal , each representing flux paths that do not directly involve the rotor. Each of the components adds to the leakage inductance. The most prominent components of stator leakage are referred to as slot, belt, zigzag, end winding, and .skew Each of these will be discussed in the following paragraphs. 4.3.1
Belt Leakage
Belt and zigzag leakage components are due to air- gap space harmonics. As it turns out, these are relatively complicated to estimate, but we may get some notion from our first- order view of the machine. The trouble with estimating these leakage components is that they are not really independent of the rotor, even though we call them “leakage”. Belt harmonics are of order n = 5 and n = 7. If there were no rotor coupling, the belt harmonic leakage terms would be: Xag5 =
3 4 µ0 NS2 k52 Rl ω 2π 52 p2 g
(103)
3 4 µ0 NS2 k72 Rl ω (104) 2π 72 p2 g The belt harmonics link to the rotor, however, and actually appear to be in parallel with components of rotor impedance appropriate to 5p and 7p pole- pair machines. At these harmonic orders we can usually ignore rotor resistance so that rotor impedance is purely inductive. Those components are: Xag7 =
X2,5
6 µ0 RlNS2 k52 12lNS2 k52 Lslot + ω =ω NR π g
1 1 + 2 (NR + 5p) (NR − 5p)2
(105)
X2,7
12lNS2 k72 6 µ0 RlNS2 k72 =ω Lslot + ω NR π g
1 1 + 2 (NR + 7p) (NR − 7p)2
(106)
In the simple model of the squirrel cage machine, because the rotor resistances are relatively small and slip high, the effect of rotor resistance is usually ignored. Then the fifth and seventh harmonic components of belt leakage are: X5 = Xag5 kX2,5
(107)
X7 = Xag7 kX2,7
(108)
16
4.3.2
Zigzag Leakage
Stator zigzag leakage is from those harmonics of the orders pns = Nslots ± p where Nslots . 3 4 µ0 NS2 Rl ω Xz = 2π g
kns + kns − + 2 (Nslots + p) (Nslots − p)2
(109)
Note that these harmonic orders do not tend to be shorted out by the rotor cage and so no direct interaction with the cage is ordinarily accounted for. 4.3.3
Skew Leakage
In order to reduce saliency effects that occur because the rotor teeth will tend to try to align with the stator teeth, induction motor designers always use a different number of slots in the rotor and stator. There still may be some tendency to align, and this produces “cogging” torques which in turn produce vibration and noise and, in severe cases, can retard or even prevent starting. To reduce this tendency to “cog”, rotors are often built with a little “skew”, or twist of the slots from one end to the other. Thus, when one tooth is aligned at one end of the machine, it is un-aligned at the other end. A side effect of this is to reduce the stator and rotor coupling by just a little, and this produces leakage reactance. This is fairly easy to estimate. Consider, for example, a space-fundamental flux density Br = B1 cos pθ, linking a (possibly) skewed full-pitch current path: λ=
Z
l 2
− 2l
Z
π + pς xl 2p
− 2πp + pς
x l
B1 cos pθRdθdx
Here, the skew in the rotor is ς electrical radians from one end of the machine to the other. Evaluation of this yields: 2B1 Rl sin 2ς λ= ς p 2 Now, the difference between what would have been linked by a non-skewed rotor and what is linked by the skewed rotor is the skew leakage flux, now expressible as:
Xk = Xag 1 − 4.3.4
sin 2ς ς 2
!2
Stator Slot Leakage
Currents in the stator slots produce fluxes that link the stator conductors but not the rotor. To estimate these fluxes, refer to the slot geometry shown in Figure 4.3.4. This shows a possibly unrealistic straight-sided stator slot. Typical in induction machines is for such slots to be trapezoidal in shape. A more careful field analysis than we will do here shows that this analysis will be no more than a few percent in error if the slot width used in the calculation is the slot top (the end of the slot closest to the air-gap). There are five important dimensions here: the slot height h, width w and the slot depression height d and width u, and (not shown) length ℓ. To estimate slot leakage inductance we assume some current in the slot, calculate the magnetic energy that results and then use the expression:
17
1 wm = Lℓ I 2 2 If there are N conductors in the slot, each carrying current I, the current density in the slot is: J=
NI hw
Using Ampere’s Law around a loop (shown dotted in the figure), magnetic field in the x direction at height y from the bottom of the slot is: Hx = −
NI y w h
In the slot depression that field is:
NI u Magnetic energy stored in the slot and slot depression are then conveniently calculated as: Hxd = −
1 wm = Lℓ I 2 = wℓ 2
Z
h 0
1 1h d 1 2 + = ℓµ0 µ0 Hx2 dy + udℓ Hxd N 2I 2 2 2 3w u
Noting the slot permeance as:
1h d + 3w u We have the total inductance of the slot to be: P = ℓµ0
Lℓ = PN 2 For the purpose of this estimate we will assume an ordinary winding consisting of coils of Nc turns each. For such a winding if there are m slots per pole per phase and p pole pairs and if the winding is short-pitched by Nsp slots, there will be 2p(m − Nsp ) slots per phase with two coils from the same phase and 2pNsp slots per phase sharing another phase. (We assume here a three phase machine). Then the ’self’ slot leakage inductance must be:
Lsℓ = P 4Nc2 2p(m − Nsp ) + Nc2 2pNsp
Since there are a total of pNsp ’mutual’ slots between each pair of phases, and the sense of the windings is opposite, the mutual component of slot leakage is: Lmℓ = −PpNsp Nc2 Total slot leakage is then: Lℓ = Lsℓ − Lmℓ = PpNc2 (8m − 5Nsp ) Expressed in terms of the total number of stator turns, Na = 2pmNc , N2 Lℓ = P a p
2 5 Nsp − m 4 m2
18
u
w
10111 1111 0000 00000 1011111 000 10111 0000 1111 10000 00000 10 1011111 000 000 111 10111 0 1 10 J 1010 1010 1010 10 1010 10 1010 1010 000000 111111 10111111 10 000000
d
y x h
Figure 5: Stator Slot Geometry for Leakage Calculation 4.3.5
End Winding Leakage
The final component of leakage reactance is due to the end windings. This is perhaps the most difficult of the machine parameters to estimate, being essentially three-dimensional in nature. There are a number of ways of estimating this parameter, but for our purposes we will use a simplified parameter from Alger[1]: Xe =
14 q µ0 RNa2 (p − 0.3) 4π 2 2 p2
As with all such formulae, extreme care is required here, since we can give little guidance as to when this expression is correct or even close. And we will admit that a more complete treatment of this element of machine parameter construction would be an improvement.
4.4
Stator Winding Resistance
Estimating stator winding resistance is fairly straightforward once end winding geometry is known. Total length of the armature winding is, per phase: ℓw = Na 2 (ℓ + ℓe ) Estimating ℓe , the length of one end winding, requires knowing how the winding is laid out and is beyond our scope here. (But once you see it you will know that length.) The area of the winding may be estimated by knowing wire diameter and how many strands are in parallel: Aw =
π 2 d Ninh 4 w
The area of the winding is related to slot area by a winding factor: λa =
2Nc Aw Aslot 19
Winding resistance, per phase, is simply Ra =
ℓw σAw
where σ is wire conductivity. Note that conductivity of the materials used in induction machines is a function of temperature and so will be winding resistance (and rotor resistance for that matter). The Fitzgerald, Kingsley and Umans textbook[2] gives the following correction for resistance of copper: RT = R t
T0 + T T0 + t
where RT and Rt are resistances at temperatures T and t. T0 = 234.5 for copper with basic conductivity of IACS (5.8 × 107 S/m)[3]. For aluminum with conductivity of 63% of IACS, T0 ≈ 212.9 Temperatures are given in Celcius.
4.5
Harmonic Order Rotor Resistance and Stray Load Losses
It is important to recognize that the machine rotor “sees” each of the stator harmonics in essentially the same way, and it is quite straightforward to estimate rotor parameters for the harmonic orders, as we have done just above. Now, particularly for the “belt” harmonic orders, there are rotor currents flowing in response to stator mmf’s at fifth and seventh space harmonic order. The resistances attributable to these harmonic orders are: R2,5 =
12lNs2 k52 Rslot,5 NR
(110)
R2,7 =
12lNs2 k72 Rslot,7 NR
(111)
The higher-order slot harmonics will have relative frequencies (slips) that are: sn = 1 ∓ (1 − s)n
(
n = 6k + 1 n = 6k − 1
)
k an integer
(112)
The induction motor electromagnetic interaction can now be described by an augmented magnetic circuit as shown in Figure 20. Note that the terminal flux of the machine is the sum of all of the harmonic fluxes, and each space harmonic is excited by the same current so the individual harmonic components are in series. Each of the space harmonics will have an electromagnetic interaction similar to the fundamental: power transferred across the air-gap is: 2 Pem,n = 3I2,n
R2,n sn
Of course dissipation in each circuit is: Pd,n = 3I22,n R2,n 20
leaving
R2,n (1 − sn ) sn Note that this equivalent circuit has provision for two sets of circuits which look like “cages”. In fact one of these sets is for the solid rotor body if that exists. We will discuss that anon. There is also a provision (rc ) for loss in the stator core iron. Power deposited in the rotor harmonic resistance elements is characterized as “stray load” loss because it is not easily computed from the simple machine equivalent circuit. 2 Pm,n = 3I2,n
4.6
Slot Models
Some of the more interesting things that can be done with induction motors have to do with the shaping of rotor slots to achieve particular frequency-dependent effects. We will consider here three cases, but there are many other possibilities. First, suppose the rotor slots are representable as being rectangular, as shown in Figure 6, and assume that the slot dimensions are such that diffusion effects are not important so that current in the slot conductor is approximately uniform. In that case, the slot resistance and inductance per unit length are: 1 ws hs σ hs Lslot = µ0 3ws
(113)
Rslot =
(114)
The slot resistance is obvious, the slot inductance may be estimated by recognizing that if the current in the slot is uniform, magnetic field crossing the slot must be: Hy =
I x ws hs
Then energy stored in the field in the slot is simply: 1 L I 2 = ws 2 slot
4.7
Z
0
hs
µ0 2
Ix ws hs
2
dx =
1 µ0 hs 2 I 6 ws
Deep Slots
Now, suppose the slot is not small enough that diffusion effects can be ignored. The slot becomes “deep” to the extent that its depth is less than (or even comparable to) the skin depthfor conduction at slip frequency. Conduction in this case may be represented by using the Diffusion Equation: ∇ 2 H = µ0 σ
∂H ∂t
In the steady state, and assuming that only cross-slot flux (in the y direction) is important, and the only variation that is important is in the radial (x) direction: ∂ 2 Hy = jωs µ0 σHy ∂x2 21
wd
hd
x
hs y
ws Figure 6: Single Slot This is solved by solutions of the form: x
Hy = H± e±(1+j) δ where the skin depth is δ=
s
2 ω s µ0 σ
Since Hy must vanish at the bottom of the slot, it must take the form: Hy = Htop
sinh(1 + j) xδ sinh(1 + j) hδs
Since current is the curl of magnetic field, Jz = σEz =
∂Hy 1 + j cosh(1 + j) hδs = Htop ∂x δ sinh(1 + j) hδs
Then slot impedance, per unit length, is: 1 1+j hs Zslot = coth(1 + j) ws σδ δ Of course the impedance (purely reactive) due to the slot depression must be added to this. It is possible to extract the real and imaginary parts of this impedance (the process is algebraically a bit messy) to yield: Rslot =
1 sinh 2 hδs + sin 2 hδs ws σδ cosh 2 hδs − cos 2 hδs
Lslot = µ0
1 1 sinh 2 hδs − sin 2 hδs hd + wd ωs ws σδ cosh 2 hδs − cos 2 hδs 22
4.8
Arbitrary Slot Shape Model
It is possible to obtain a better model of the behavior of rotor conductor slots by using simple numerical methods. In many cases rotor slots are shaped with the following objectives in mind: 1. A substantial part of the periphery of the rotor should be devoted to active conductor, for good running performance. 2. The magnetic iron of the rotor must occupy a certain fraction of the periphery, to avoid saturation. 3. For good starting performance, some means of forcing current to flow only in the top part of the rotor bar should be devised. Generally the rotor teeth, which make up part of the machine’s magnetic circuit, are of roughly constant width to avoid flux concentration. The rotor conductor bars are therefore tapered, with their narrow ends towards the center of the rotor. To provide for current concentration on starting they often have a ’starting bar’ at the outer periphery of the rotor with a much narrower region which has high inductance just below. The bulk of the rotor bar occupies the tapered region allowed between the teeth. This geometry is quite a bit more complicated than that described in the previous section. Note that, if we can describe the slot impedance per unit length as a function of frequency: Zs (ω) = Rs (ω) + jXs (ω), we can carry out the analysis of the machine as described previously. Thus our analysis is directed toward frequency response modeling of the rotor slot. Focusing then on a single slot, use the notation as described in Figure 7.
Ez [n]
w[n]
x = nΔ x
Δ x
Ez [n−1]
x y
z
Figure 7: Slot Geometry Notation The impedance per unit length is the ratio between slot current and axial electric field: Zs =
Ez I
For the purpose of this analysis we will use the symbol x as the radial distance from the bottom of the slot. Assume the slot can be divided radially into a number of regions or ’slices’, each with 23
radial height Δx. We further assume that currents are axially (z) directed and that magnetic field crosses the slot in the y direction. Under these assumptions the electric field at the top of one of the slices is related to the electric field at the bottom of the slice by magnetic field crossing through the slice. Using the trapezoidal rule for integration: E z (x) − E z (x − Δx) = jωµ0 The magnetic field is simply: H y (x) =
1 w(x)
Z
Δx H y (x) + H y (x − Δx) 2
x
σw(x)Ez (x)dx =
0
1 X nI wn i=1 n
where In is the total current flowing in one slice. Note that this can be reformulated into a ladder network by again using the trapezoidal rule for integration: current flowing in slice number n would be: Δx (wn E n + wn−1 E n−1 ) In = σ 2 Now the slot may be described as is shown in the ladder network of Figure 8. The incremental reactance of one slice is: Xn = ωµ0 Δx and the resistance of a slice is: Rn =
2 (wn + wn−1 )
1 2 σΔx wn + wn−1
L[n]
L[n]
2
2
L[n−1]
L[n−1]
2
2
R[n]
R[n−1]
Figure 8: Slot Impedance Ladder Network The procedure is to start at the bottom of the slot, corresponding to the right-hand end of the ladder (the inductance at the bottom of the slot is infinite so the first slice has only the resistance), and building toward the top of the slot.
4.9
Multiple Cages
In some larger induction motors the rotor cage is built in such a way as to separate the functions of ’starting’ and ’running’. The purpose of a “deep” slot is to improve starting performance of a motor. When the rotor is stationary, the frequency seen by rotor conductors is relatively high, and current crowding due to the skin effect makes rotor resistance appear to be high. As the rotor accelerates the frequency seen from the rotor drops, lessening the skin effect and making more use of the rotor conductor. This, then, gives the machine higher starting torque (requiring high resistance) without compromising running efficiency. 24
This effect can be carried even further by making use of multiple cages, such as is shown in Figure 9. Here there are two conductors in a fairly complex slot. Estimating the impedance of this slot is done in stages to build up an equivalent circuit. wd hd h2 hs w2 ws h1
w1
Figure 9: Double Slot Assume for the purposes of this derivation that each section of the multiple cage is small enough that currents can be considered to be uniform in each conductor. Then the bottom section may be represented as a resistance in series with an inductance: 1 σw1 h1 µ0 h1 3 w1
Ra = La =
The narrow slot section with no conductor between the top and bottom conductors will contribute an inductive impedance: hs Ls = µ 0 ws The top conductor will have a resistance: Rb =
1 σw2 h2
Now, in the equivalent circuit, current flowing in the lower conductor will produce a magnetic field across this section, yielding a series inductance of Lb = µ 0
h2 w2
By analogy with the bottom conductor, current in the top conductor flows through only one third of the inductance of the top section, leading to the equivalent circuit of Figure 10, once the inductance of the slot depression is added on: hd Lt = µ 0 wd Now, this rotor bar circuit fits right into the framework of the induction motor equivalent circuit, shown for the double cage case in Figure 11, with R2a =
12lNS2 kS2 Ra NR 25
1 3 Lb
Lt ∩∩∩∩
∩∩∩∩
2 3 Lb
Ls ∩∩∩∩
∩∩∩∩
< <> Rb <>
La ∩∩∩∩
< <> Ra <>
Figure 10: Equivalent Circuit: Double Bar 12lNS2 kS2 Rb NR 12lNS2 kS2 2 ( Lb + Ls + La ) = ω NR 3 12lNS2 kS2 1 = ω (Lt + Lb ) NR 3
R2b = X2a X2a
I� a Ra
∧∧∧ ∨∨
X1 ∩∩∩∩
X2b I2 X2a ∩∩∩∩ � ∩∩∩∩ < < ⊃ ⊃ Xm <> R2b <> R2a ⊃ > s > s < < ⊃
Figure 11: Equivalent Circuit: Double Cage Rotor
4.10
Rotor End Ring Effects
It is necessary to correct for “end ring” resistance in the rotor. To do this, we note that the magnitude of surface current density in the rotor is related to the magnitude of individual bar current by: 2πR Iz = Kz (115) NR Current in the end ring is: IR = Kz
R p
(116)
Then it is straightforward to calculate the ratio between power dissipated in the end rings to power dissipated in the conductor bars themselves, considering the ratio of current densities and volumes. Assuming that the bars and end rings have the same radial extent, the ratio of current densities is: JR NR wr = Jz 2πp lr 26
(117)
where wr is the average width of a conductor bar and lr is the axial end ring length. Now, the ratio of losses (and hence the ratio of resistances) is found by multiplying the square of current density ratio by the ratio of volumes. This is approximately: Rend = Rslot
4.11
NR wr 2πp lr
2
2
2πR lr NR Rwr = NR l wr πllr p2
(118)
Windage
Bearing friction, windage loss and fan input power are often regarded as elements of a “black art”. We approach them with some level of trepidation, for motor manufacturers seem to take a highly empirical view of these elements. What follows is an attempt to build reasonable but simple models for two effects: loss in the air gap due to windage and input power to the fan for cooling. Some caution is required here, for these elements of calculation have not been properly tested, although they seem to give reasonable numbers The first element is gap windage loss. This is produced by shearing of the air in the relative rotation gap. It is likely to be a signifigant element only in machines with very narrow air gaps or very high surface speeds. But these include, of course, the high performance machines with which we are most interested. We approach this with a simple “couette flow” model. Air-gap shear loss is approximately: Pw = 2πR4 Ω3 lρa f (119) where ρa is the density of the air-gap medium (possibly air) and f is the friction factor, estimated by: .0076 (120)
f= 1 Rn4 and the Reynold’s NumberRn is ΩRg (121) Rn = νair and νair is the kinematic viscosity of the air-gap medium. The second element is fan input power. We base an estimate of this on two hypotheses. The first of these is that the mass flow of air circulated by the fan can be calculated by the loss in the motor and an average temperature rise in the cooling air. The second hypothesis is the the pressure rise of the fan is established by the centrifugal pressure rise associated with the surface speed at the outside of the rotor. Taking these one at a time: If there is to be a temperature rise ΔT in the cooling air, then the mass flow volume is: Pd Cp ΔT
m ˙ = and then volume flow is just v˙ =
m ˙ ρair
Pressure rise is estimated by centrifugal force: ΔP = ρair
27
ω r p fan
2
then power is given by: Pfan = ΔP v˙ For reference, the properties of air are: Density Kinematic Viscosity Heat Capacity
4.12
ρair νair Cp
1.18 1.56 × 10−5 1005.7
kg/m2 m2 /sec J/kg
Magnetic Circuit Loss and Excitation
There will be some loss in the stator magnetic circuit due to eddy current and hysteresis effects in the core iron. In addition, particularly if the rotor and stator teeth are saturated there will be MMF expended to push flux through those regions. These effects are very difficult to estimate from first principles, so we resort to a simple model. Assume that the loss in saturated steel follows a law such as: Pd = PB
ωe ωB
ǫf
B BB
ǫb
(122)
This is not too bad an estimate for the behavior of core iron. Typically, ǫf is a bit less than two (between about 1.3 and 1.6) and ǫb is a bit more than two (between about 2.1 and 2.4). Of course this model is good only for a fairly restricted range of flux density. Base dissipation is usually expressed in “watts per kilogram”, so we first compute flux density and then mass of the two principal components of the stator iron, the teeth and the back iron. In a similar way we can model the exciting volt-amperes consumed by core iron by something like:
Qc = V a1
B BB
ǫv1
+ V a2
B BB
ǫv2
ω ωB
(123)
This, too, is a form that appears to be valid for some steels. Quite obviously it may be necessary to develop different forms of curve ’fits’ for different materials. Flux density (RMS) in the air-gap is: Br =
pVa 2RlNa k1 ωs
(124)
wt + w1 wt
(125)
Then flux density in the stator teeth is: Bt = B r
where wt is tooth width and w1 is slot top width. Flux in the back-iron of the core is Bc = B r where dc is the radial depth of the core.
28
R pdc
(126)
One way of handling this loss is to assume that the core handles flux corresponding to terminal voltage, add up the losses and then compute an equivalent resistance and reactance: rc =
3|Va |2 Pcore
xc =
3|Va |2 Qcore
then put this equivalent resistance in parallel with the air-gap reactance element in the equivalent circuit.
5
Solid Iron Rotor Bodies
Solid steel rotor electric machines (SSRM) can be made to operate with very high surface speeds and are thus suitable for use in high RPM situations. They resemble, in form and function, hysteresis machines. However, asynchronous operation will produce higher power output because it takes advantage of higher flux density. We consider here the interactions to be expected from solid iron rotor bodies. The equivalent circuits can be placed in parallel (harmonic-by-harmonic) with the equivalent circuits for the squirrel cage, if there is also a cage in the machine. To estimate the rotor parameters R2s and X2s , we assume that important field quantities in the machine are sinusoidally distributed in time and space, so that radial flux density is:
(127)
(128)
Br = Re B r ej(ωt−pφ) and, similarly, axially directed rotor surface current is:
Kz = Re K z ej(ωt−pφ) Now, since by Faraday’s law:
∇×E =− we have, in this machine geometry:
∂B ∂t
(129)
1 ∂ ∂Br Ez = − R ∂φ ∂t
(130)
The transformation between rotor and stator coordinates is: φ′ = φ − ωm t
(131)
pωm = ω − ωr = ω(1 − s)
(132)
where ωm is rotor speed. Then: and Now, axial electric field is, in the frame of the rotor, just:
Ez = Re E z ej(ωt−pφ)
= Re E z ej(ωr 29
t−pφ′ )
(133)
(134)
and
ωr R Br p
Ez =
(135)
Of course electric field in the rotor frame is related to rotor surface current by: Ez = Z sK z
(136)
Now these quantities can be related to the stator by noting that air-gap voltage is related to radial flux density by: p (137) Br = V 2lNa k1 Rω ag The stator-equivalent rotor current is: I2 =
π R K 3 Na ka z
(138)
Then we can find stator referred, rotor equivalent impedance to be: Z2 =
V ag 3 4 l 2 2 ω Ez N k = I2 2 π R a a ωr K z
(139)
Now, if rotor surface impedance can be expressed as: Z s = Rs + jωr Ls then Z2 =
R2 + jX2 s
(140)
(141)
where R2 = X2 =
34 l 2 2 N k Rs 2πR a 1 34 l 2 2 N k Xs 2πR a 1
(142) (143)
Now, to find the rotor surface impedance, we make use of a nonlinear eddy-current model proposed by Agarwal. First we define an equivalent penetration depth (similar to a skin depth): δ=
s
2Hm ωr σB0
(144)
where σ is rotor surface material volume conductivity, B0 , ”saturation flux density” is taken to be 75 % of actual saturation flux density and Hm = |K z | =
3 Na ka |I 2 | π R
(145)
Then rotor surface resistivity and surface reactance are: 16 1 3π σδ = .5Rs
Rs =
(146)
Xs
(147)
30
Note that the rotor elements X2 and R2 depend on rotor current I2 , so the problem is nonlinear. We find, however, that a simple iterative solution can be used. First we make a guess for R2 and find currents. Then we use those currents to calculate R2 and solve again for current. This procedure is repeated until convergence, and the problem seems to converge within just a few steps. Aside from the necessity to iterate to find rotor elements, standard network techniques can be used to find currents, power input to the motor and power output from the motor, torque, etc.
5.1
Solution
Not all of the equivalent circuit elements are known as we start the solution. To start, we assume a value for R2 , possibly some fraction of Xm , but the value chosen doesn not seem to matter much. The rotor reactance X2 is just a fraction of R2 . Then, we proceed to compute an “air-gap” impedance, just the impedance looking into the parallel combination of magnetizing and rotor branches: R2 (148) Zg = jXm ||(jX2 + ) s (Note that, for a generator, slip s is negative). A total impedance is then Zt = jX1 + R1 + Zg (149) and terminal current is It = Rotor current is just: I2 = It
Vt Zt
jXm jX2 + Rs2
(150)
(151)
Now it is necessary to iteratively correct rotor impedance. This is done by estimating flux density at the surface of the rotor using (145), then getting a rotor surface impedance using (146) and using that and (143 to estimate a new value for R2 . Then we start again with (148). The process “drops through” this point when the new and old estimates for R2 agree to some criterion.
5.2
Harmonic Losses in Solid Steel
If the rotor of the machine is constructed of solid steel, there will be eddy currents induced on the rotor surface by the higher-order space harmonics of stator current. These will produce magnetic fields and losses. This calculation assumes the rotor surface is linear and smooth and can be characterized by a conductivity and relative permeability. In this discussion we include two space harmonics (positive and negative going). In practice it may be necessary to carry four (or even more) harmonics, including both ‘belt’ and ‘zigzag’ order harmonics. Terminal current produces magnetic field in the air-gap for each of the space harmonic orders, and each of these magnetic fields induces rotor currents of the same harmonic order. The “magnetizing” reactances for the two harmonic orders, really the two components of the zigzag leakage, are: Xzp = Xm
31
kp2 Np2 k12
(152)
Xzn = Xm
kn2 Nn2 k12
(153)
where Np and Nn are the positive and negative going harmonic orders: For ‘belt’ harmonics these orders are 7 and 5. For ‘zigzag’ they are: Np =
Ns + p p
(154)
Nn =
Ns − p p
(155)
Now, there will be a current on the surface of the rotor at each harmonic order, and following 65, the equivalent rotor element current is: I 2p =
π R K 3 Na kp p
(156)
I 2n =
π R K 3 Na kn n
(157)
These currents flow in response to the magnetic field in the air-gap which in turn produces an axial electric field. Viewed from the rotor this electric field is: E p = sp ωRB p
(158)
E n = sn ωRB n
(159)
where the slip for each of the harmonic orders is: sp = 1 − Np (1 − s)
(160)
sn = 1 + Np (1 − s)
(161)
and then the surface currents that flow in the surface of the rotor are: Kp =
Ep Zsp
(162)
Kn =
En Zsn
(163)
where Zsp and Zsn are the surface impedances at positive and negative harmonic slip frequencies, respectively. Assuming a linear surface, these are, approximately: Zs =
1+j σδ
(164)
where σ is material restivity and the skin depth is δ=
s
2 ωs µσ
32
(165)
and ωs is the frequency of the given harmonic from the rotor surface. We can postulate that the appropriate value of µ to use is the same as that estimated in the nonlinear calculation of the space fundamental, but this requires empirical confirmation. The voltage induced in the stator by each of these space harmonic magnetic fluxes is: Vp =
2Na kp lRω Bp Np p
(166)
Vn =
2Na kn lRω Bn Nn p
(167)
Then the equivalent circuit impedance of the rotor is just: Z2p =
3 4 Na2 kp2 l Zsp Vp = Ip 2 π Np R s p
(168)
Z2n =
Vn 3 4 Na2 kn2 l Zsn = In 2 π Nn R s n
(169)
The equivalent rotor circuit elements are now: R2p =
3 4 Na2 kp2 l 1 2 π Np R σδp
3 4 Na2 kn2 l 1 2 π Nn R σδn 1 X2p = R2p 2 1 X2n = R2n 2
R2n =
5.3
(170) (171) (172) (173)
Stray Losses
So far in this document, we have outlined the major elements of torque production and consequently of machine performance. We have also discussed, in some cases, briefly, the major sources of loss in induction machines. Using what has been outlined in this document will give a reasonable impression of how an induction machine works. We have also discussed some of the stray load losses: those which can be (relatively) easily accounted for in an equivalent circuit description of the machine. But there are other losses which will occur and which are harder to estimate. We do not claim to do a particularly accurate job of estimating these losses, and fortunately they do not normally turn out to be very large. To be accounted for here are: 1. No-load losses in rotor teeth because of stator slot opening modulation of fundamental flux density, 2. Load losses in the rotor teeth because of stator zigzag mmf, and 3. No-load losses in the solid rotor body (if it exists) due to stator slot opening modulation of fundamental flux density. 33
Note that these losses have a somewhat different character from the other miscellaneous losses we compute. They show up as drag on the rotor, so we subtract their power from the mechanical output of the machine. The first and third of these are, of course, very closely related so we take them first. The stator slot openings ‘modulate’ the space fundamental magnetic flux density. We may estimate a slot opening angle (relative to the slot pitch): θD =
wd Ns 2πwd Ns = 2πr r
Then the amplitude of the magnetic field disturbance is: BH = Br1
2 θD sin π 2
In fact, this flux disturbance is really in the form of two traveling waves, one going forward and one backward with respect to the stator at a velocity of ω/Ns . Since operating slip is relatively small, the two variations will have just about the same frequency as viewed from the rotor, so it seems reasonable to lump them together. The frequency is: ωH = ω
Ns p
Now, for laminated rotors this magnetic field modulation will affect the tips of rotor teeth. We assume (perhaps arbitrarily) that the loss due to this magnetic field modulation can be estimated from ordinary steel data (as we estimated core loss above) and that only the rotor teeth, not any of the rotor body, are affected. The method to be used is straightforward and follows almost exactly what was done for core loss, with modification only of the frequency and field amplitude. For solid steel rotors the story is only a little different. The magnetic field will produce an axial electric field: ω E z = R BH p and that, in turn, will drive a surface current Kz =
Ez Zs
√ Now, what is important is the magnitude of the surface current, and since |Z s | = 1 + .52 Rs ≈ 1.118Rs , we can simply use rotor resistance. The nonlinear surface penetration depth is: δ=
s
2B0 ωH σ |K z |
A brief iterative substitution, re-calculating δ and then |K z | quickly yields consistent values for δ and Rs . Then the full-voltage dissipation is: Prs = 2πRl and an equivalent resistance is: Rrs =
|K z |2 σδ
3|Va |2 Prs
34
Finally, the zigzag order current harmonics in the stator will produce magnetic fields in the air gap which will drive magnetic losses in the teeth of the rotor. Note that this is a bit different from the modulation of the space fundamental produced by the stator slot openings (although the harmonic order will be the same, the spatial orientation will be different and will vary with load current). The magnetic flux in the air-gap is most easily related to the equivalent circuit voltage on the nth harmonic: Bn =
npvn 2lRNa kn ω
This magnetic field variation will be substantial only for the zigzag order harmonics: the belt harmonics will be essentially shorted out by the rotor cage and those losses calculated within the equivalent circuit. The frequency seen by the rotor is that of the space harmonics, already calculated, and the loss can be estimated in the same way as core loss, although as we have pointed out it appears as a ‘drag’ on the rotor.
6 6.1
Induction Motor Speed Control Introduction
The inherent attributes of induction machines make them very attractive for drive applications. They are rugged, economical to build and have no sliding contacts to wear. The difficulty with using induction machines in servomechanisms and variable speed drives is that they are “hard to control”, since their torque-speed relationship is complex and nonlinear. With, however, modern power electronics to serve as frequency changers and digital electronics to do the required arithmetic, induction machines are seeing increasing use in drive applications. In this chapter we develop models for control of induction motors. The derivation is quite brief for it relies on what we have already done for synchronous machines. In this chapter, however, we will stay in “ordinary” variables, skipping the per-unit normalization.
6.2
Volts/Hz Control
Remembering that induction machines generally tend to operate at relatively low per unitslip, we might conclude that one way of building an adjustable speed drive would be to supply an induction motor with adjustable stator frequency. And this is, indeed, possible. One thing to remember is that flux is inversely proportional to frequency, so that to maintain constant flux one must make stator voltage proportional to frequency (hence the name “constant volts/Hz”). However, voltage supplies are always limited, so that at some frequency it is necessary to switch to constant voltage control. The analogy to DC machines is fairly direct here: below some “base” speed, the machine is controlled in constant flux (“volts/Hz”) mode, while above the base speed, flux is inversely proportional to speed. It is easy to see that the maximum torque is then inversely to the square of flux, or therefore to the square of frequency. To get a first-order picture of how an induction machine works at adjustable speed, start with the simplified equivalent network that describes the machine, as shown in Figure 12 Earlier in this chapter, it was shown that torque can be calculated by finding the power dissipated in the virtual resistance R2 /s and dividing by electrical speed. For a three phase machine, and assuming we are dealing with RMS magnitudes: 35
I� a Ra
∧∧∧ ∨∨
X1 ∩∩∩∩
X2 I2 ∩∩∩∩ � < ⊃ ⊃ Xm <> R2 ⊃ > < s ⊃
Figure 12: Equivalent Circuit
p R2 Te = 3 |I2 |2 s ω where ω is the electrical frequency and p is the number of pole pairs. It is straightforward to find I2 using network techniques. As an example, Figure 13 shows a series of torque/speed curves for an induction machine operated with a wide range of input frequencies, both below and above its “base” frequency. The parameters of this machine are: Number of Phases Number of Pole Pairs RMS Terminal Voltage (line-line) Frequency (Hz) Stator Resistance R1 Rotor Resistance R2 Stator Leakage X1 Rotor Leakage X2 Magnetizing Reactance Xm
3 3 230 60 .06 Ω .055 Ω .34 Ω .33 Ω 10.6 Ω
Strategy for operating the machine is to make terminal voltage magnitude proportional to frequency for input frequencies less than the “Base Frequency”, in this case 60 Hz, and to hold voltage constant for frequencies above the “Base Frequency”. For high frequencies the torque production falls fairly rapidly with frequency (as it turns out, it is roughly proportional to the inverse of the square of frequency). It also falls with very low frequency because of the effects of terminal resistance. We will look at this next.
6.3
Idealized Model: No Stator Resistance
Ignore, for the moment, R1 . An equivalent circuit is shown in Figure 14. It is fairly easy to show that, from the rotor, the combination of source, armature leakage and magnetizing branch can be replaced by its equivalent circuit, as shown in in Figure 15. In the circuit of Figure 15, the parameters are: Xm Xm + X1 = Xm ||X1
V′ = V X′
36
Induction Motor Torque
250
N−m
200
150
100
50
0 0
50
100
150 Speed, RPM
200
250
Figure 13: Induction Machine Torque-Speed Curves
I� a X1 ∩∩∩∩
X2 I2 ∩∩∩∩ � �� < ⊃ + ⊃ Xm <> R2 ⊃ − V <> s �� ⊃
Figure 14: Idealized Circuit: Ignore Armature Resistance If the machine is operated at variable frequency ω, but the reactance is established at frequency ωB , current is: I=
V′ j(X1′ + X2 ) ωωB +
R2 s
and then torque is Te = 3|I2 |2
|V ′ |2 Rs2 3p R2 = s ω (X1′ + X2 )2 + ( Rs2 )2
Now, if we note that what counts is the absolute slip of the rotor, we might define a slip with respect to base frequency: ωr ωr ωB ωB s= = = sB ω ωB ω ω
37
X1′ I� a ∩∩∩∩
�� + −
X2 I2 ∩∩∩∩ �
<> <> R2 < s
V′
��
Figure 15: Idealized Equivalent Then, if we assume that voltage is applied proportional to frequency: V ′ = V0′
ω ωB
and with a little manipulation, we get: 2 |V0′ |2 R 3p sB Te = 2 2 ωB (X1′ + X2 )2 + ( R s ) B
This would imply that torque is, if voltage is proportional to frequency, meaning constant applied flux, dependent only on absolute slip. The torque-speed curve is a constant, dependent only on the difference between synchronous and actual rotor speed. This is fine, but eventually, the notion of “volts per Hz” runs out because at some number of Hz, there are no more volts to be had. This is generally taken to be the “base” speed for the drive. Above that speed, voltage is held constant, and torque is given by: Te =
2 |V ′ |2 R 3p sB 2 2 ωB (X1′ + X2 )2 + ( R sB )
The peak of this torque has a square-inverse dependence on frequency, as can be seen from Figure 16.
6.4
Peak Torque Capability
Assuming we have a smart controller, we are interested in the actual capability of the machine. At some voltage and frequency, torque is given by: Te = 3|I2 |2
3 ωp |V ′ |2 Rs2 R2 = s ((X1′ + X2 )( ωωB ))2 + (R1′ +
R2 2 s )
Now, we are interested in finding the peak value of that, which is given by the value of R2 s which maximizes power transfer to the virtual resistance. This is given by the matching condition: R2 = s
r
R1′2 + ((X1′ + X2 )( 38
ω 2 )) ωB
Induction Motor Torque
250
N−m
200
150
100
50
0 0
1500 1000 Speed, RPM
500
2000
Figure 16: Idealized Torque-Speed Curves: Zero Stator Resistance Then maximum (breakdown) torque is given by: Tmax =
3p ′ 2 ω |V |
q
R1′2 + ((X1′ + X2 )( ωωB ))2
((X1′ + X2 )( ωωB ))2 + (R1′ +
q
R1′2 + ((X1′ + X2 )( ωωB ))2 )2
This is plotted in Figure 17. Just as a check, this was calculated assuming R1 = 0, and the results are plotted in figure 18. This plot shows, as one would expect, a constant torque limit region to zero speed.
6.5
Field Oriented Control
One of the more useful impacts of modern power electronics and control technology has enabled us to turn induction machines into high performance servomotors. In this note we will develop a picture of how this is done. Quite obviously there are many details which we will not touch here. The objective is to emulate the performance of a DC machine, in which (as you will recall), torque is a simple function of applied current. For a machine with one field winding, this is simply: T = GIf Ia This makes control of such a machine quite easy, for once the desired torque is known it is easy to translate that torque command into a current and the motor does the rest. Of course DC (commutator) machines are, at least in large sizes, expensive, not particularly efficient, have relatively high maintenance requirements because of the sliding brush/commutator interface, provide environmental problems because of sparking and carbon dust and are environmentally sensitive. The induction motor is simpler and more rugged. Until fairly recently the induction motor has not been widely used in servo applications because it was thought to be ”hard to control”. As we will show, it does take a little effort and even some computation to do the controls right, but this is becoming increasingly affordable. 39
Breakdown Torque 300
250
Newton−Meters
200
150
100
50
0 0
20
40
60 80 Drive Frequency, Hz
100
120
Figure 17: Torque-Capability Curve For An Induction Motor
6.6
Elementary Model:
We return to the elementary model of the induction motor. In ordinary variables, referred to the stator, the machine is described by flux-current relationships (in the d-q reference frame): "
λdS λdR
#
"
λqS λqR
#
=
"
LS M
M LR
#"
idS idR
#
=
"
LS M
M LR
#"
iqS iqR
#
Note the machine is symmetric (there is no saliency), and since we are referred to the stator, the stator and rotor self-inductances include leakage terms: LS
= M + LSℓ
LR = M + LRℓ The voltage equations are:
vdS
=
vqS
=
0 = 0 =
dλdS − ωλqS + rS idS dt dλqS + ωλdS + rS iqS dt dλdR − ωs λqR + rR idR dt dλqR + ωs λdR + rR iqR dt 40
Breakdown Torque 300
Newton−Meters
250
200
150
100
50 0
20
40
60 80 Drive Frequency, Hz
100
120
Figure 18: Idealized Torque Capability Curve: Zero Stator Resistance Note that both rotor and stator have “speed” voltage terms since they are both rotating with respect to the rotating coordinate system. The speed of the rotating coordinate system is w with respect to the stator. With respect to the rotor that speed is , where wm is the rotor mechanical speed. Note that this analysis does not require that the reference frame coordinate system speed w be constant. Torque is given by:
3
T e = p (λdS iqS − λqS idS ) 2
6.7
Simulation Model
As a first step in developing a simulation model, see that the inversion of the flux-current relationship is (we use the d- axis since the q- axis is identical): idS
=
idR =
LR M λdS − λdR LS LR − M 2 LS LR − M 2 LS M λdS − λdR LS LR − M 2 LS LR − M 2
Now, if we make the following definitions (the motivation for this should by now be obvious): Xd = ω0 LS Xkd = ω0 LR Xad = ω0 M Xd′
= ω0
41
M2 LS − LR
!
the currents become: idS
ω0 Xad ω0 λdS − λdR ′ Xd Xkd Xd′ Xd ω0 Xad ω0 λdR λdS − ′ Xd Xkd Xkd Xd′
=
idR =
The q- axis is the same. Torque may be, with these calculations for current, written as: 3 ω0 Xad 3 (λdS λqR − λqS λdR ) Te = p (λdS iqS − λqS idS ) = − p 2 2 Xkd Xd′ Note that the usual problems with ordinary variables hold here: the foregoing expression was written assuming the variables are expressed as peak quantities. If RMS is used we must replace 3/2 by 3! With these, the simulation model is quite straightforward. The state equations are: dλdS dt dλqS dt dλdR dt dλqR dt dΩm dt
= VdS + ωλqS − RS idS = VqS − ωλdS − RS iqS = ωs λqR − RR idR = −ωs λdR − RS iqR =
1 (Te + Tm ) J
where the rotor frequency (slip frequency) is: ωs = ω − pΩm For simple simulations and constant excitaion frequency, the choice of coordinate systems is arbitrary, so we can choose something convenient. For example, we might choose to fix the coordinate system to a synchronously rotating frame, so that stator frequency ω = ω0 . In this case, we could pick the stator voltage to lie on one axis or another. A common choice is Vd = 0 and Vq = V .
6.8
Control Model
If we are going to turn the machine into a servomotor, we will want to be a bit more sophisticated about our coordinate system. In general, the principle of field-oriented control is much like emulating the function of a DC (commutator) machine. We figure out where the flux is, then inject current to interact most directly with the flux. As a first step, note that because the two stator flux linkages are the sum of air-gap and leakage flux, λdS
= λagd + LSℓ idS
λqS
= λagq + LSℓ iqS
42
This means that we can re-write torque as: 3 T e = p (λagd iqS − λagq idS ) 2 Next, note that the rotor flux is, similarly, related to air-gap flux: λagd = λdR − LRℓ idR λagq = λqR − LRℓ iqR
Torque now becomes: 3 3 T e = p (λdR iqS − λqR idS ) − pLRℓ (idR iqS − iqR idS ) 2 2 Now, since the rotor currents could be written as: M λdR − idS LR LR M λqR − iqS LR LR
idR = iqR = That second term can be written as: idR iqS − iqR idS =
1 (λdR iqS − λqR idS ) LR
So that torque is now: 3 LRℓ T = p 1− 2 LR e
6.9
3 M (λdR iqS − λqR idS ) = p (λdR iqS − λqR idS ) 2 LR
Field-Oriented Strategy:
What is done in field-oriented control is to establish a rotor flux in a known position (usually this position is the d- axis of the transformation) and then put a current on the orthogonal axis (where it will be most effective in producing torque). That is, we will attempt to set λdR = Λ0 λqR = 0 Then torque is produced by applying quadrature-axis current: Te =
3 M p Λ0 iqS 2 LR
The process is almost that simple. There are a few details involved in figuring out where the quadrature axis is and how hard to drive the direct axis (magnetizing) current.
43
Now, suppose we can succeed in putting flux on the right axis, so that λqR = 0, then the two rotor voltage equations are: 0 = 0 =
dλdR − ωs λqR + rR IdR dt dλqR + ωs λdR + rR IqR dt
Now, since the rotor currents are: idR = iqR =
λdR M − idS LR LR M λqR − iqS LR LR
The voltage expressions become, accounting for the fact that there is no rotor quadrature axis flux: M dλdR λdR 0 = + rR − idS LR dt LR M iqS 0 = ωs λdR − rR LR
Noting that the rotor time constant is TR =
LR rR
we find: TR
dλdR + λdR = M idS dt M iqS ωs = TR λdR
The first of these two expressions describes the behavior of the direct-axis flux: as one would think, it has a simple first-order relationship with direct-axis stator current. The second expression, which describes slip as a function of quadrature axis current and direct axis flux, actually describes how fast to turn the rotating coordinate system to hold flux on the direct axis. Now, a real machine application involves phase currents ia , ib and ic , and these must be derived from the model currents idS and iqs . This is done with, of course, a mathematical operation which uses a transformation angle θ. And that angle is derived from the rotor mechanical speed and computed slip: Z θ=
(pωm + ωs ) dt
A generally good strategy to make this sort of system work is to measure the three phase currents and derive the direct- and quadrature-axis currents from them. A good estimate of direct-axis flux is made by running direct-axis flux through a first-order filter. The tricky operation involves dividing quadrature axis current by direct axis flux to get slip, but this is now easily done numerically (as 44
θ
D
ω
S
λ
dR
N÷ ÷D
M 1 + STa
T
M TR
N
i
i a*
* d
ib*
-1
T i q*
i
ia o
Amp
o o
* c
ic
Motor Load
ω
∫
θ
ib
m
Σ
Figure 19: Field Oriented Controller are the trigonometric operations required for the rotating coordinate system transformation). An elmentary block diagram of a (possbly) plausible scheme for this is shown in Figure 19. In this picture we start with commanded values of direct- and quadrature- axis currents, corresponding to flux and torque, respectively. These are translated by a rotating coordinate transformation into commanded phase currents. That transformation (simply the inverse Park’s transform) uses the angle q derived as part of the scheme. In some (cheap) implementations of this scheme the commanded currents are used rather than the measured currents to establish the flux and slip. We have shown the commanded currents i∗a , etc. as inputs to an “Amplifier”. This might be implemented as a PWM current-source, for example, and a tight loop here results in a rather high performance servo system.
References [1] P.L. Alger, “Induction Machines”, Gordon and Breach, 1969 [2] A.E. Fitzgerals, C. Kingsley Jr., S.D. Umans, ”Electric Machinery”, Sixth Edition, McGraw Hill, 2003
45
[3] D. Fink, H. W. Beaty, ”Standard Handbook for Electrical Engineers, Thirteenth Edition,
McGraw-Hill, 1993
46
r1 ∧∧∧ ∨∨
x1 ∩∩∩∩ ⊃ ⊃ jxag ⊃ ⊃
<> < rc <>
⊃ ⊃ xc ⊃ ⊃
⊃ ⊃ jx2s ⊃ ⊃ <> r < s2s <>
⊃ ⊃ jx2s5 ⊃ ⊃
⊃ ⊃ jxa5 ⊃ ⊃
< r 2s5 <> s <> 5
⊃ ⊃ jx2s7 ⊃ ⊃
⊃ ⊃ jxa7 ⊃ ⊃
<> r < s2s7 <> 7
⊃ ⊃ jx2sm ⊃ ⊃
⊃ ⊃ jxam ⊃ ⊃
<> r <> s2sm m <
⊃ ⊃ jx2sp ⊃ ⊃
⊃ ⊃ jxap ⊃ ⊃
<> r2sp <> s p <
Figure 20: Extended Equivalent Circuit
47
⊃ ⊃ jx2c ⊃ ⊃ <> r < s2c <>
⊃ ⊃ jx2c5 ⊃ ⊃ < r 2c5 <> s <> 5
⊃ ⊃ jx2c7 ⊃ ⊃ <> r < s2c7 <> 7
⊃ ⊃ jx2cm ⊃ ⊃ <> r <> s2cm m <
⊃ ⊃ jx2cp ⊃ ⊃ <> r2cp <> s p <
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machinery Class Notes 9: Synchronous Machine Simulation Models c �2005 James L. Kirtley Jr.
1
September 5, 2005
Introduction
In this document we develop models useful for calculating the dynamic behavior of synchronous machines. We start with a commonly accepted picture of the synchronous machine, assuming that the rotor can be fairly represented by three equivalent windings: one being the field and the other two, the d- and q- axis “damper” windings, representing the effects of rotor body, wedge chain, amortisseur and other current carrying paths. While a synchronous machine is assumed here, the results are fairly directly applicable to induction machines. Also, extension to situations in which the rotor representation must have more than one extra equivalent winding per axis should be straightforward.
2
Phase Variable Model
To begin, assume that the synchronous machine can be properly represented by six equivalent windings. Four of these, the three armature phase windings and the field winding, really are windings. The other two, representing the effects of distributed currents on the rotor, are referred to as the “damper” windings. Fluxes are, in terms of currents: "
λph λR
#
=
"
Lph MT
M LR
#"
Iph IR
#
(1)
where phase and rotor fluxes (and, similarly, currents) are:
λph
(2)
(3)
λa = λb λc
λf λR = λkd λkq
There are three inductance sub- matrices. The tances: La Lph = Lab Lac
first of these describes armature winding induc-
1
Lab Lac Lb Lbc Lbc Lc
(4)
where, for a machine that may have some saliency: La = La0 + L2 cos 2θ
(5)
2π ) (6) 3 2π Lc = La0 + L2 cos 2(θ + ) (7) 3 π (8) Lab = Lab0 + L2 cos 2(θ − ) 3 Lbc = Lab0 + L2 cos 2θ (9) π Lac = Lab0 + L2 cos 2(θ + ) (10) 3 Note that, in this last set of expressions, we have assumed a particular form for the mutual inductances. This is seemingly restrictive, because it constrains the form of phase- to- phase mutual inductance variations with rotor position. The coefficient L2 is actually the same in all six of these last expressions. As it turns out, this assumption does not really restrict the accuracy of the model very much. We will have more to say about this a bit later. The rotor inductances are relatively simply stated: Lb = La0 + L2 cos 2(θ −
Lf Lf kd 0 LR = Lf kd Lkd 0 0 0 Lkq
(11)
And the stator- to- rotor mutual inductances are:
M cos θ M = M cos(θ − M cos(θ +
3
Lakd cos θ 2π 3 ) Lakd cos(θ − 2π 3 ) Lakd cos(θ +
−Lakq sin θ 2π 3 ) −Lakq sin(θ − 2π 3 ) −Lakq sin(θ +
2π 3 ) 2π 3 )
(12)
Park’s Equations
The first step in the development of a suitable model is to transform the armature winding variables to a coordinate system in which the rotor is stationary. We identify equivalent armature windings in the direct and quadrature axes. The direct axis armature winding is the equivalent of one of the phase windings, but aligned directly with the field. The quadrature winding is situated so that its axis leads the field winding by 90 electrical degrees. The transformation used to map the armature currents, fluxes and so forth onto the direct and quadrature axes is the celebrated Park’s Transformation, named after Robert H. Park, an early investigator into transient behavior in synchronous machines. The mapping takes the form:
ua ud u = T u = T = u ub q dq ph uc u0
(13)
Where the transformation and its inverse are:
cos θ cos(θ − 23π ) cos(θ + 23π ) 2 2π T = − sin θ − sin(θ − 3 ) − sin(θ + 23π ) 3 1 1 1
2
2
2
2
(14)
T −1
cos θ − sin θ 1 2π = cos(θ − 2π ) − sin(θ − ) 1 3 3 2π 2π cos(θ + 3 ) − sin(θ + 3 ) 1
(15)
This transformation maps balanced sets of phase currents into constant currents in the d-q frame. That is, if rotor angle is θ = ωt + θ0 , and phase currents are: Ia = I cos ωt 2π ) 3 2π = I cos(ωt + ) 3
Ib = I cos(ωt − Ic Then the transformed set of currents is: Id = I cos θ0
Iq = −I sin θ0 Now, we apply this transformation to (1) to express fluxes and currents in the armature in the d-q reference frame. To do this, extract the top line in (1): λph = Lph I ph + M I R
(16)
The transformed flux is obtained by premultiplying this whole expression by the transformation matrix. Phase current may be obtained from d-q current by multiplying by the inverse of the transformation matrix. Thus: (17) λdq = T Lph T −1 I dq + T M I R The same process carried out for the lower line of (1) yields: λR = M T T −1 I dq + LR I R
(18)
Thus the fully transformed version of (1) is: "
λdq λR
#
=
"
Ldq 3 T 2 LC
LC LR
#"
Idq IR
#
(19)
If the conditions of (5) through (10) are satisfied, the inductance submatrices of (19) wind up being of particularly simple form. (Please note that a substantial amount of algebra has been left out here!) Ld 0 0 Lq 0 (20) Ldq = 0 0 0 L0
M LC = 0 0
Lakd 0 0 Lakq 0 0
3
(21)
Note that (19) through (21) express three separate sets of apparently independent flux/current relationships. These may be re-cast into the following form:
λd λkd = λf "
λq λkq
#
Ld Lakd M Id 3 L L L Ikd akd kd f kd 2 3 If Lf kd Lf 2M
=
"
Lq Lakq 3 L Lkq 2 akq
#"
Iq Ikq
#
λ0 = L0 I0
(22)
(23) (24)
Where the component inductances are: 3 Ld = La0 − Lab0 + L2 2 3 Lq = La0 − Lab0 − L2 2 L0 = La0 + 2Lab0
(25) (26) (27)
Note that the apparently restrictive assumptions embedded in (5) through (10) have resulted in the very simple form of (21) through (24). In particular, we have three mutually independent sets of fluxes and currents. While we may be concerned about the restrictiveness of these expressions, note that the orthogonality between the d- and q- axes is not unreasonable. In fact, because these axes are orthogonal in space, it seems reasonable that they should not have mutual flux linkages. The principal consequence of these assumptions is the de-coupling of the zero-sequence component of flux from the d- and q- axis components. We are not in a position at this time to determine the reasonableness of this. However, it should be noted that departures from this form (that is, coupling between the “direct” and “zero” axes) must be through higher harmonic fields that will not couple well to the armature, so that any such coupling will be weak. Next, armature voltage is, ignoring resistance, given by: d d λ = T −1 λdq dt ph dt and that the transformed armature voltage must be: V ph =
V dq = T V ph d = T (T −1 λdq ) dt d d = λdq + (T T −1 )λdq dt dt
(28)
(29)
A good deal of manupulation goes into reducing the second term of this, resulting in: T
d −1 T = dt
0 dθ dt
0
4
− dθ dt 0 0
0 0 0
(30)
This expresses the speed voltagethat arises from a coordinate transformation. The two voltage/flux relationships that are affected are: Vd = Vq =
dλd − ωλq dt dλq + ωλd dt
where we have used ω=
4
dθ dt
(31) (32)
(33)
Power and Torque
Instantaneous power is given by: P = V a Ia + V b Ib + V c Ic
(34)
Using the transformations given above, this can be shown to be: P =
3 3 Vd Id + Vq Iq + 3V0 I0 2 2
(35)
which, in turn, is: 3 3 dλd dλq dλ0 P = ω (λd Iq − λq Id ) + ( Id + Iq ) + 3 I0 (36) 2 dt dt 2 dt Then, noting that electrical speed ω and shaft speed Ω are related by ω = pΩ and that (36) describes electrical terminal power as the sum of shaft power and rate of change of stored energy, we may deduce that torque is given by: 3 T = p(λd Iq − λq Id ) 2
5
(37)
Per-Unit Normalization
The next thing for us to do is to investigate the way in which electric machine system are normalized, or put into what is called a per-unit system. The reason for this step is that, when the voltage, current, power and impedance are referred to normal operating parameters, the behavior characteristics of all types of machines become quite similar, giving us a better way of relating how a particular machine works to some reasonable standard. There are also numerical reasons for normalizing performance parameters to some standard. The first step in normalization is to establish a set of base quantities. We will be normalizing voltage, current, flux, power, impedance and torque, so we will need base quantities for each of these. Note, however, that the base quantities are not independent. In fact, for the armature, we need only specify three quantities: voltage (VB ), current (IB ) and frequency (ω0 ). Note that we do not normalize time nor frequency. Having done this for the armature circuits, we can derive each of the other base quantities:
5
• Base Power
3 PB = VB IB 2
• Base Impedance
• Base Flux • Base Torque
ZB =
VB IB
λB =
VB ω0
TB =
p PB ω0
Note that, for our purposes, base voltage and current are expressed as peak quantities. Base voltage is taken on a phase basis (line to neutral for a “wye” connected machine), and base current is similarly taken on a phase basis, (line current for a “wye” connected machine). Normalized, or per-unit quantities are derived by dividing the ordinary variable (with units) by the corresponding base. For example, per-unit flux is: ψ=
λ ω0 λ = VB λB
(38)
In this derivation, per- unit quantities will usually be designated by lower case letters. Two notable exceptions are flux, where we use the letter ψ, and torque, where we will still use the upper case T and risk confusion. Now, we note that there will be base quantities for voltage, current and frequency for each of the different coils represented in our model. While it is reasonable to expect that the frequency base will be the same for all coils in a problem, the voltage and current bases may be different. We might write (22) as:
ψd ψkd = ψf
ω0 IdB Vdb Ld ω0 IdB 3 Vkb 2 Lakd ω0 IdB 3 Vf b 2 M
ω0 IkB Vdb Lakd ω0 IkB Vkb Lkd ω0 IkB Vf b Lf kd
ω0 I f B Vdb M ω0 I f B Vkdb Lf kd ω0 I f B Vf b Lf
id ikd
(39)
if
where i = I/IB denotes per-unit, or normalized current. Note that (39) may be written in simple form:
ψd xd xakd xad id xf kd ikd ψkd = xakd xkd xad xf kd xf ψf if
(40)
It is important to note that (40) assumes reciprocity in the normalized system. To wit, the following expressions are implied: xd = ω 0
6
IdB Ld VdB
(41)
IkB Lkd VkB If B ω0 Lf Vf B IkB ω0 Lakd VdB 3 IdB ω0 Lakd 2 VkB If B M ω0 VdB 3 IdB ω0 M 2 Vf B IkB ω0 Lf kd Vf b If B Lf kd ω0 Vkb
xkd = ω0 xf
=
xakd = = xad = = xf kd = =
(42) (43)
(44)
(45)
(46)
These in turn imply: 3 VdB IdB = Vf B If B 2 3 VdB IdB = VkB IkB 2 Vf B If B = VkB IkB
(47) (48) (49)
These expressions imply the same power base on all of the windings of the machine. This is so because the armature base quantities Vdb and Idb are stated as peak values, while the rotor base quantities are stated as DC values. Thus power base for the three- phase armature is 23 times the product of peak quantities, while the power base for the rotor is simply the product of those quantities. The quadrature axis, which may have fewer equivalent elements than the direct axis and which may have different numerical values, still yields a similar structure. Without going through the details, we can see that the per-unit flux/current relationship for the q- axis is: "
ψq ψkq
#
=
"
xq xakq xakq xkq
#"
iq ikq
#
(50)
The voltage equations, including speed voltage terms, (31) and (32), may be augmented to reflect armature resistance: dλd (51) − ωλq + Ra Id Vd = dt dλq Vq = ωλd + + Ra Iq (52) dt The per-unit equivalents of these are: vd =
1 dψd ω − ψq + ra id ω0 dt ω0 7
(53)
ω 1 dψq + ra iq ψd + ω0 ω0 dt
vq =
(54)
Where the per-unit armature resistance is just ra = ZRBa Note that none of the other circuits in this model have speed voltage terms, so their voltage expressions are exactly what we might expect: vf
=
vkd = vkq = v0 =
1 dψf + rf if ω0 dt 1 dψkd + rkd ikd ω0 dt 1 dψkq + rkq ikq ω0 dt 1 dψ0 + ra i0 ω0 dt
(55) (56) (57) (58)
It should be noted that the damper winding circuits represent closed conducting paths on the rotor, so the two voltages vkd and vkq are always zero. Per-unit torque is simply: Te = ψd iq − ψq id (59) Often, we need to represent the dynamic behavior of the machine, including electromechanical dynamics involving rotor inertia. If we note J as the rotational inertia constant of the machine system, the rotor dynamics are described by the two ordinary differential equations: 1 dω J p dt dδ dt
= T e + T m
(60)
= ω − ω0
(61)
where T e and T m represent electrical and mechanical torques in “ordinary” variables. The angle δ represents rotor phase angle with respect to some synchronous reference. It is customary to define an “inertia constant” which is not dimensionless but which nevertheless fits into the per-unit system of analysis. This is: H ≡
Rotational kinetic energy at rated speed Base Power
Or: H=
1 2J
ω0 p
PB
2
=
J ω0 2pTB
(62)
(63)
Then the per-unit equivalent to (60) is: 2H dω = Te + Tm ω0 dt where now we use Te and Tm to represent per-unit torques.
8
(64)
6
Equal Mutual’s Base
In normalizing the differential equations that make up our model, we have used a number of base quantities. For example, in deriving (40), the per-unit flux- current relationship for the direct axis, we used six base quantities: VB , IB , Vf B , If B , VkB and IkB . Imposing reciprocity on (40) results in two constraints on these six variables, expressed in (47) through (49). Presumably the two armature base quantities will be fixed by machine rating. That leaves two more “degrees of freedom” in selection of base quantities. Note that the selection of base quantities will affect the reactance matrix in (40). While there are different schools of thought on just how to handle these degrees of freedom, a commonly used convention is to employ what is called the equal mutuals base system. The two degrees of freedom are used to set the field and damper base impedances so that all three mutual inductances of (40) are equal: xakd = xf kd = xad (65) The direct- axis flux- current relationship becomes:
id xd xad xad ψd ψkd = xad xkd xad ikd if xad xad xf ψf
7
(66)
Equivalent Circuit
i� d ra ∧∧∧ ∨∨
+
+
(ω0 vd + ωψq )
ψd
-
-
xal ∩∩∩∩ ⊃ ⊃ xad ⊃ ⊃
xf l ∩∩∩∩ ⊃ ⊃ xkdl ⊃ ⊃
if rf � ∧∧∧ ∨∨
<> <> rkd <
+ vf -
Figure 1: D- Axis Equivalent Circuit The flux- current relationship of (66) is represented by the equivalent circuit of Figure 1, if the “leakage” inductances are defined to be: xal = xd − xad
(67)
xf l = xf − xad
(69)
xkdl = xkd − xad
9
(68)
Many of the interesting features of the electrical dynamics of the synchronous machine may be discerned from this circuit. While a complete explication of this thing is beyond the scope of this note, it is possible to make a few observations. The apparent inductance measured from the terminals of this equivalent circuit (ignoring resistance ra ) will, in the frequency domain, be of the form: Pn (s) ψd (s) = xd id (s) Pd (s)
x(s) =
(70)
Both the numerator and denominator polynomials in s will be second order. (You may convince yourself of this by writing an expression for terminal impedance). Since this is a “diffusion” type circuit, having only resistances and inductances, all poles and zeros must be on the negative real axis of the “s-plane”. The per-unit inductance is, then: x(s) = xd
(1 + Td′ s)(1 + Td′′ s) ′ s)(1 + T ′′ s) (1 + Tdo do
(71)
The two time constants Td′ and Td′′ are the reciprocals of the zeros of the impedance, which are the poles of the admittance. These are called the short circuittime constants. ′ and T ′′ are the reciprocals of the poles of the impedance, and The other two time constants Tdo do so are called the open circuittime constants. We have cast this thing as if there are two sets of well- defined time constants. These are the ′ , and the subtransient time constants T ′′ and T ′′ . In many transient time constants Td′ and Tdo d do cases, these are indeed well separated, meaning that: Td′ ≫ Td′′
′ Tdo
≫
′′ Tdo
(72) (73)
If this is true, then the reactance is described by the pole-zero diagram shown in Figure 2. Under this circumstance, the apparent terminal inductance has three distinct values, depending on frequency. These are the synchronous inductance, the transient inductance, and the subtransient inductance, given by: Td′ ′ Tdo T ′′ = xd′ d′′ Tdo T ′ T ′′ = xd ′d d′′ Tdo Tdo
x′d = xd x′′d
(74)
(75)
A Bode Plotof the terminal reactance is shown in Figure 3.
If the time constants are spread widely apart, they are given, approximately, by:
′ Tdo = ′′ = Tdo
xf ω0 rf xkdl + xf l ||xad ω0 rkd 10
(76) (77)
�
1 Td ”
1 Tdo ”
1 ′
Tdo
�
×
×
1 Td′
Figure 2: Pole-Zero Diagram For Terminal Inductance
log |x(jω)|
� � �
1 ′ Tdo
� � �
1 T ′ d
1 1 Tdo ” Td ”
log ω
Figure 3: Frequency Response of Terminal Inductance Finally, note that the three reactances are found simply from the model:
8
xd = xal + xad
(78)
x′d x′′d
(79)
= xal + xad ||xf l
= xal + xad ||xf l ||xkdl
(80)
Statement of Simulation Model
Now we can write down the simulation model. Actually, we will derive more than one of these, since the machine can be driven by either voltages or currents. Further, the expressions for permanent magnet machines are a bit different. So the first model is one in which the terminals are all constrained by voltage. The state variables are the two stator fluxes ψd , ψq , two “damper” fluxes ψkd , ψkq , field flux ψf , and rotor speed ω and torque angle δ. The most straightforward way of stating the model employs currents as auxiliary variables, and these are:
−1
id xd xad xad ikd = xad xkd xad if xad xad xf "
iq ikq
#
=
"
xq xaq xaq xkq
11
#−1 "
ψd ψkd ψf
ψq ψkq
#
(81)
(82)
Then the state equations are: dψd dt dψq dt dψkd dt dψkq dt dψf dt dω dt dδ dt
= ω0 vd + ωψq − ω0 ra id
(83)
= ω0 vq − ωψd − ω0 ra iq
(84)
= −ω0 rkd ikd
(85)
= −ω0 rkq ikq
(86)
= ω0 vf − ω0 rf if
(87)
ω0 (Te + Tm ) 2H
(88)
=
= ω − ω0
(89)
and, of course, Te = ψd iq − ψq id
8.1
Statement of Parameters:
′ and Note that often data for a machine may be given in terms of the reactances xd , x′d , xd′′ , Tdo ′′ , rather than the elements of the equivalent circuit model. Note that there are four inductances Tdo in the equivalent circuit so we have to assume one. There is no loss in generality in doing so. Usually one assumes a value for the stator leakage inductance, and if this is done the translation is straightforward:
xad = xd − xal xad (x′d − xal ) xf l = xad − x′d + xal 1 xkdl = 1 1 x′′ −xal − xad − d
rf
=
rkd =
8.2
1 xf
l
xf l + xad ′ ω0 Tdo xkdl + xad ||xf l ′′ ω0 Tdo
Linearized Model
Often it becomes desirable to carry out a linearized analysis of machine operation to, for example, examine the damping of the swing mode at a particular operating point. What is done, then, is to assume a steady state operating point and examine the dynamics for deviations from that operating point that are “small”. The definition of “small” is really “small enough” that everything important appears in the first-order term of a Taylor series about the steady operating point. Note that the expressions in the machine model are, for the most part, linear. There are, however, a few cases in which products of state variables cause us to do the expansion of the 12
Taylor series. Assuming a steady state operating point [ψd0 ψkd0 ψf 0 ψq0 ψkq0 ω0 δ0 ], the firstorder (small-signal) variations are described by the following set of equations. First, since the flux-current relationship is linear:
−1
id1 xd xad xad ikd1 = xad xkd xad if 1 xad xad xf "
iq1 ikq1
#
=
"
xq xaq xaq xkq
#−1 "
ψd1 ψkd1 ψf 1
ψq1 ψkq1
#
(90)
(91)
Terminal voltage will be, for operation against a voltage source: Vd = V sin δ
Vq = V cos δ
Then the differential equations governing the first-order variations are: dψd1 dt
dψq1
dt
dψkd1
dt
dψkq1
dt
dψf 1
dt dω1 dt
= ω0 V cos δ0 δ1 + ω0 ψq1 + ω1 ψq0 − ω0 ra id1
(92)
= −ω0 V sin δ0 δ1 − ω0 ψd1 − ω1 ψd0 − ω0 ra iq1
(93)
= −ω0 rkd ikd1
(94)
= −ω0 rkq ikq1
(95)
= −ω0 rf if 1
(96)
=
ω0 (Te1 + Tm1 ) 2H
dδ1
= ω1 dt
(97) (98)
Te = ψd0 iq1 + ψd1 iq0 − ψq0 id1 − ψq1 id0
8.3
Reduced Order Model for Electromechanical Transients
In many situations the two armature variables contribute little to the dynamic response of the machine. Typically the armature resistance is small enough that there is very little voltage drop across it and transients in the difference between armature flux and the flux that would exist in the “steady state” decay rapidly (or are not even excited). Further, the relatively short armature time constant makes for very short time steps. For this reason it is often convenient, particularly when studying the relatively slow electromechanical transients, to omit the first two differential equations and set: ψd = vq = V cos δ ψq = −vd = −V sin δ
(99) (100)
The set of differential equations changes only a little when this approximation is made. Note, however, that it can be simulated with far fewer “cycles” if the armature time constant is short. 13
9
Current Driven Model: Connection to a System
The simulation expressions developed so far are useful in a variety of circumstances. They are, however, difficult to tie to network simulation programs because they use terminal voltage as an input. Generally, it is more convenient to use current as the input to the machine simulation and accept voltage as the output. Further, it is difficult to handle unbalanced situations with this set of equations. An alternative to this set would be to employ the phase currents as state variables. Effectively, this replaces ψd , ψq and ψ0 with ia , ib , and ic . The resulting model will, as we will show, interface nicely with network simulations. To start, note that we could write an expression for terminal flux, on the d- axis: ψd = x′′d id + ψf
xad ||xf l xad ||xkdl + ψkd xad ||xf l + xkdl xad ||xkdl + xf l
(101)
and here, of course, x′′d = xal + xad ||xkdl ||xf l This leads us to define a “flux behind subtransient reactance”: ψd′′ =
xad xkdl ψf + xad xf l ψkd xad xkdl + xad xf l + xkdl xf l
(102)
So that ψd = ψd′′ + xd′′ id On the quadrature axis the situation is essentially the same, but one step easier if there is only one quadrature axis rotor winding: ψq = x′′q iq + ψkq
xaq xaq + xkql
(103)
where x′′q = xal + xaq ||xkql
Very often these fluxes are referred to as “voltage behind subtransient reactance, with ψd′′ = eq′′ and ψq′′ = −ed′′ . Then: ψd = x′′d id + e′′q x′′q iq
ψq =
−
(104)
ed′′
(105)
Now, if id and iq are determined, it is a bit easier to find the other currents required in the simulation. Note we can write: "
ψkd ψf
ikd if
#
#
=
"
xkd xad xad xf
#"
ikd if
#
+
"
xad xad
#
id
(106)
and this inverts easily: "
=
"
xkd xad xad xf
#−1
14
"
ψkd ψf
#
−
"
xad xad
#
id
!
(107)
The quadrature axis rotor current is simply: ikq =
xaq 1 ψkq − iq xkq xkq
(108)
The torque equation is the same, but since it is usually convenient to assemble the fluxes behind subtransient reactance, it is possible to use: Te = e′′q iq + ed′′ id + (xd′′ − xq′′ )id iq
(109)
Now it is necessary to consider terminal voltage. This is most conveniently cast in matrix notation. The vector of phase voltages is: v ph
va = vb vc
(110)
Then, with similar notation for phase flux, terminal voltage is, ignoring armature resistance: 1 dψ ph ω0 dt o 1 d n −1 T ψ dq ω0 dt
v ph = =
(111)
Note that we may define the transformed vector of fluxes to be: ψ dq = x′′ idq + e′′
(112)
where the matrix of reactances shows orthogonality:
x′′ 0 0 d ′′ x = 0 x′′q 0 0 0 x0
(113)
and the vector of internal fluxes is:
e′′q ′′ e = −e ′′d 0
(114)
Now, of course, idq = T iph , so that we may re-cast (111) as: v ph =
o 1 d n −1 ′′ T x T iph + T −1 e ′′ ω0 dt
(115)
Now it is necessary to make one assumption and one definition. The assumption, which is only moderately restrictive, is that subtransient saliency may be ignored. That is, we assume that x′′d = x′′q . The definition separates the “zero sequence” impedance into phase and neutral components:
15
x0 = x′′d + 3xg
(116)
Note that according to this definition the reactance xg accounts for any impedance in the neutral of the synchronous machine as well as mutual coupling between phases. Then, the impedance matrix becomes:
x′′ 0 0 0 0 0 d ′′ x = 0 x′′d 0 + 0 0 0 0 0 3xg 0 0 x′′d
(117)
In compact notation, this is:
x′′ = x′′d I + xg
(118)
where I is the identity matrix. Now the vector of phase voltages is: v ph =
o 1 d n ′′ xd iph + T −1 xg T iph + T −1 e′′ ω0 dt
(119)
Note that in (119), we have already factored out the multiplication by the identity matrix. The next step is to carry out the matrix multiplication in the third term of (119). This operation turns out to produce a remarkably simple result:
1 1 1 −1 T xg T = xg 1 1 1 1 1 1
(120)
The impact of this is that each of the three phase voltages has the same term, and that is related to the time derivative of the sum of the three currents, multiplied by xg . The third and final term in (119) describes voltages induced by rotor fluxes. It can be written as: 1 1 d n −1 o ′′ de′′ 1 d n −1 ′′ o e + T −1 T e = T ω0 dt ω0 dt ω0 dt Now, the time derivative of the inverse transform is: 1 d −1 T ω0 dt
(121)
− sin(θ) − cos(θ) 0 ω 2π 2π − sin(θ − = ) − cos(θ − ) 0 3 3 ω0 − sin(θ + 23π ) − cos(θ + 2π 3 ) 0
(122)
Now the three phase voltages can be extracted from all of this matrix algebra: va = vb = vc =
xg d x′′d dia + (ia + ib + ic ) + e′′a ω0 dt ω0 dt x′′d dib xg d + (ia + ib + ic ) + eb′′ ω0 dt ω0 dt x′′d dic xg d + (ia + ib + ic ) + ec′′ ω0 dt ω0 dt 16
(123) (124) (125)
Where the internal voltages are:
ω ′′ (e sin(θ) − ed′′ cos(θ)) ω0 q de′′q 1 de′′ 1 + sin(θ) d + cos(θ) ω0 dt ω0 dt 2π 2π ω ′′ ) − e′′d cos(θ − )) = − (eq sin(θ − ω0 3 3 2π de′′q 1 2π de′′d 1 ) + sin(θ − ) + cos(θ − ω0 3 dt ω0 3 dt 2π 2π ω ) − e′′d cos(θ + )) = − (e′′q sin(θ + ω0 3 3 1 2π de′′q 1 2π de′′d + cos(θ + ) + sin(θ + ) ω0 3 dt ω0 3 dt
e′′a = −
e′′b
e′′c
(126)
(127)
(128)
This set of expressions describes the equivalent circuit shown in Figure 4.
′′
ea ��
′′ i� a xd va ∩∩∩∩
+ −
�� e′′b �� xg
ib x′′d �
vb ∩∩∩∩
+ −
∩∩∩∩
�� e′′c ��
ic x′′d � vc ∩∩∩∩
+ −
��
Figure 4: Equivalent Network Model
10
Restatement Of The Model
The synchronous machine model which uses the three phase currents as state variables may now be stated in the form of a set of differential and algebraic equations: dψkd dt dψkq dt dψf dt dδ dt
= −ω0 rkd ikd
(129)
= −ω0 rkq ikq
(130)
= −ω0 rf if
(131)
= ω − ω0
(132)
17
dω dt
=
ω0 Tm + e′′q iq + ed′′ id 2H
where: "
ikd if
#
=
"
xkd xad xad xf
and ikq =
#−1
"
ψkd ψf
#
−
"
(133)
xad xad
#
id
!
xaq 1 ψkq − iq xkq xkq
(It is assumed here that the difference between subtransient reactances is small enough to be neglected.) The network interface equations are, from the network to the machine: 2π ) + ic cos(θ + 3 2π iq = −ia sin(θ) − ib sin(θ − ) − ic sin(θ + 3 id = ia cos(θ) + ib cos(θ −
2π ) 3 2π ) 3
(134) (135)
and, in the reverse direction, from the machine to the network: ω ′′ (e sin(θ) − ed′′ cos(θ)) ω0 q de′′q 1 de′′ 1 + sin(θ) d + cos(θ) ω0 dt ω0 dt 2π 2π ω ′′ )) ) − e′′d cos(θ − = − (eq sin(θ − ω0 3 3 1 2π de′′q 2π de′′d 1 ) + ) sin(θ − + cos(θ − ω0 3 dt ω0 3 dt 2π 2π ω ) − e′′d cos(θ + )) = − (e′′q sin(θ + ω0 3 3 1 1 2π de′′q 2π de′′d + cos(θ + ) + ) sin(θ + ω0 3 dt ω0 3 dt
e′′a = −
e′′b
e′′c
(136)
(137)
(138)
And, of course, θ = ω0 t + δ e′′q e′′d
(139)
=
ψd′′
(140)
=
−ψq′′
(141)
ψd′′ = ψq′′ =
xad xkdl ψf + xad xf l ψkd xad xkdl + xad xf l + xkdl xf l xaq ψkq xaq + xkql
18
(142) (143)
11
Network Constraints
This model may be embedded in a number of networks. Different configurations will result in different constraints on currents. Consider, for example, the situation in which all of the terminal voltages are constrained, but perhaps by unbalanced (not entirely positive sequence) sources. In that case, the differential equations for the three phase currents would be: x′′d dia ω0 dt x′′d dib ω0 dt x′′d dic ω0 dt
12
x′′d + 2xg xg − (vb − eb′′ ) + (vc − ec′′ ) ′′ ′′ xd + 3xg xd + 3xg ′′ x + 2xg xg = (vb − e′′b ) d′′ − (va − ea′′ ) + (vc − ec′′ ) ′′ xd + 3xg xd + 3xg ′′ xg x + 2xg = (vc − e′′c ) ′′d − (vb − eb′′ ) + (va − ea′′ ) ′′ xd + 3xg xd + 3xg
= (va − e′′a )
(144) (145) (146)
Example: Line-Line Fault
We are not, however, constrained to situations defined in this way. This model is suitable for embedding into network analysis routines. It is also possible to handle many different situations directly. Consider, for example, the unbalanced fault represented by the network shown in Figure 5. This shows a line-line fault situation, with one phase still connected to the network.
i� a ra
va
∧∧∧ ∨∨
ib ra � ∧∧∧ ∨∨
ra ∧∧∧ ∨∨
′′
ea ��
x′′d ∩∩∩∩
+ −
�� e′′b �� xg
x′′d ∩∩∩∩
+ −
∩∩∩∩
�� e′′c ��
x′′d ∩∩∩∩
+ −
��
Figure 5: Line-Line Fault Network Model In this situation, we have only two currents to worry about, and their differential equations would be: dib dt dia dt
= =
ω0 ′′ (e − e′′b − 2ra ib ) 2x′′d c ω0 (va − e′′a − ra ia ) x′′d + xg
(147) (148)
and, of course, ic = −ib . Note that here we have included the effects of armature resistance, ignored in the previous section but obviously important if the results are to be believed. 19
13
Permanent Magnet Machines
Permanent Magnet machines are one state variable simpler than their wound-field counterparts. They may be accurately viewed as having constant field current. Assuming that we can define the internal (field) flux as: ψ0 = xad if 0 (149)
13.1
Model: Voltage Driven Machine
We have a reasonably simple expression for the rotor currents, in the case of a voltage driven machine: "
id ikd
#
"
iq ikq
#
=
"
xd xad xad xkd
#−1 "
=
"
xq xaq xaq xkq
#−1 "
ψd − ψ0 ψkd − ψ0 ψq ψkq
#
#
(150) (151)
The simulation model then has six states: dψd dt dψq dt dψkd dt dψkq dt dω dt dδ dt
13.2
= ω0 vd + ωψq − ω0 ra id
(152)
= ω0 vq − ωψd − ω0 ra iq
(153)
= −ω0 rkd ikd
(154)
= −ω0 rkq ikq
(155)
ω0 (ψd iq − ψq id + Tm ) 2H
=
= ω − ω0
(156) (157)
Curent-Driven Machine Model
In the case of a current-driven machine, rotor currents required in the simulation are: ikd = ikq =
1 (ψkd − xad id − ψ0 ) xkd 1 (ψkq − xaq iq ) xkq
(158) (159)
Here, the “flux behind subtransient reactance” is, on the direct axis: ψd′′ =
xkdl ψ0 + xad ψkd xad + xkdl
(160)
and the subtransient reactance is: x′′d = xal + xad ||xkdl 20
(161)
On the quadrature axis, ψq′′ =
xad ψkq xad + xkql
(162)
and x′′q = xal + xaq ||xkql
(163)
In this case there are only four state equations: dψkd dt dψkq dt dω dt dδ dt
= −ω0 rkd ikd
(164)
= −ω0 rkq ikq
(165)
=
ω0 ′′ eq iq + ed′′ id + Tm 2H
(166)
= ω − ω0
(167)
The interconnections to and from the network are the same as in the case of a wound-field machine: in the “forward” direction, from network to machine: 2π ) + ic cos(θ + 3 2π iq = −ia sin(θ) − ib sin(θ − ) − ic sin(θ + 3 id = ia cos(θ) + ib cos(θ −
2π ) 3 2π ) 3
(168) (169)
and, in the reverse direction, from the machine to the network: ω ′′ (e sin(θ) − ed′′ cos(θ)) ω0 q de′′q 1 de′′ 1 + sin(θ) d + cos(θ) ω0 dt ω0 dt ω 2π 2π = − (e′′q sin(θ − ) − e′′d cos(θ − )) ω0 3 3 2π de′′q 1 2π de′′d 1 ) + sin(θ − ) + cos(θ − ω0 3 dt ω0 3 dt 2π 2π ω )) ) − e′′d cos(θ + = − (e′′q sin(θ + ω0 3 3 2π de′′q 1 2π de′′d 1 ) + sin(θ + ) + cos(θ + ω0 3 dt ω0 3 dt
e′′a = −
e′′b
e′′c
13.3
(170)
(171)
(172)
PM Machines with no damper
PM machines without much rotor conductivity may often behave as if they have no damper winding at all. In this case the model simplifies even further. Armature currents are: id = iq =
1 (ψd − ψ0 ) xd 1 ψq xq 21
(173) (174)
The state equations are: dψd dt dψq dt dω dt dδ dt
13.4
= ω0 vd + ωψq − ω0 ra id
(175)
= ω0 vq − ωψd − ω0 ra iq
(176)
=
ω0 (ψd iq − ψq id + Tm ) 2H
(177)
= ω − ω0
(178)
Current Driven PM Machines with no damper
In the case of no damper the machine becomes quite simple. There is no “internal flux” on the quadrature axis. Further, there are no time derivatives of the internal flux on the d- axis. The only machine state equations are mechanical: dω dt dδ dt
=
ω0 (ψ0 iq + Tm ) 2H
(179)
= ω − ω0
(180)
The “forward” network interface is as before: 2π ) + ic cos(θ + 3 2π ) − ic sin(θ + iq = −ia sin(θ) − ib sin(θ − 3 id = ia cos(θ) + ib cos(θ −
2π ) 3 2π ) 3
(181) (182)
and, in the reverse direction, from the machine to the network, things are a bit simpler than before: ω ψ0 sin(θ) ω0 ω = − ψ0 sin(θ − ω0 ω = − ψ0 sin(θ + ω0
e′′a = − e′′b e′′c
(183) 2π ) 3 2π ) 3
(184) (185) (186)
22
Characterization of Left-Handed Materials Massachusetts Institute of Technology 6.635 lecture notes
1
Introduction 1. How are they realized? 2. Why the denomination “Left-Handed”? 3. What are their properties? 4. Does it really work?
It has already been shown (see previous classes) that rings, or split-rings, can realize a negative permeability (µ < 0) over a certain frequency band. In addition to this, we need to realize a negative permittivity (² < 0). It has also been shown (see previous classes) that: • lossless: ²metal = 1 −
ωp2 , ω2
where ωp =
ne2 ²0 me
(n: electron density, e: electron charge, me : effective mass of electrons). • lossy: ²metal
ωp2 =1− . ω(ω + iγ)
A typical transmission curve looks like shown in Fig. 1. T 1
PSfrag replacements
no transmission (² < 0)
transmission (² > 0)
ωp
ω
Figure 1: Transmission curve for a plasma-like medium.
.
1
2
Section 2. Why “left-handed”?
With these characteristics, ² < 0 has been realized already at infrared frequencies (where metals behave like plasmas). Problem: how to realize it at GHz frequencies? Solution: by reducing n, the electron density. One way of doing this is to confine the electrons in space. This can be achieved by an array of rods for example, as shown in Fig. 2.
a
r
PSfrag replacements
Figure 2: Array of rods confining the electrons in space.
Note: it is important that the wires are thin, so as to reduce the radiation interaction and allow penetration into the structure. Effect of the wires: to reduce n to neff = n
πr2 . a2
(1)
Finally, note also that the rods have to be parallel to the electric field. This, plus the (known already) fact that rings have to be perpendicular to the magnetic field, gives an idea on how to realize physically LH metamaterials (see Fig. 3).
2
Why “left-handed”?
At this point, we have a metamaterial which can realize ² < 0, µ < 0 .
(2)
We shall now see what does it imply on the electromagnetic fields. Let us write Maxwell’s curl equations for plane wave solutions and time harmonic notations:
¯ r) = ωµH(¯ ¯ r) , k¯ × E(¯ ¯ r) = − ω²E(¯ ¯ r) . k¯ × H(¯
(3a) (3b)
3 0.5 mm 5 mm
z 1.3 cm
x
εs FRONT
BACK
(a) 1 mm
y εs
x
(b)
Figure 3: A realization for LH material.
¯ forms a right-handed system. ¯ H, ¯ k) In standard materials, Eq. (3) implies that the tryad (E, However, under Eq. (2), we will have: ¯ form a left-handed (LH) tryad. ¯ H, ¯ k) (E, However, the time average Poynting power is still ¯ r) >= < S(¯
1 ¯ r) × H ¯ ? (¯ <{E(¯ r)} 2
and remains in the same direction so that we have the set up shown in Fig. 4. Characteristics: • k¯ is in the phase velocity direction. • phase velocity and energy flux are in opposite directions.
3
Properties of LH media
Some know characteristics are: • Reversed Doppler effect (track the phase), ˇ • Reversed Cerenkov radiation (cf. 6.632),
(4)
4
3.1
Reversed Snell’s law
¯ E
k¯
S¯
PSfrag replacements
¯ H ¯ and Poynting power ¯ magnetic field (H), ¯ wave-vector (k) Figure 4: Electric field (E), ¯ (S) in an LH medium.
• Negative index of refraction. This last item is very significant, and we shall spend some time discussing it. The index of refraction of a medium is defined as n=
√ ² r µr ,
(5)
or, writing explicitly the frequency dependence (cf. later), n(ω) =
p
²r (ω) µr (ω) .
(6)
For those frequencies inside the left-handed band (i.e. in the band where ² < 0 and µ < 0), we can write: ²(ω) < 0
⇒
µ(ω) < 0
⇒
²(ω) = |²(ω)| eiπ ,
µ(ω) = |µ(ω)| eiπ ,
(7a) (7b)
Eventually, we write n from Eq. (6): n=
3.1
p
|²(ω)µ(ω)| eiπ = −
p |²(ω)µ(ω)| .
Reversed Snell’s law
An important consequence of this fact is the reversal of Snell’s law. • Ray diagram:
(8)
5
n>0 PSfrag replacements
θr −θr
θi
n<0
• k¯ diagram for an LH medium:
PSfrag replacements kx
k¯2
k¯1
kx S¯
k1z
3.2
k2z
Energy
Traditionally, the energy is given by W = ²E 2 + µH 2 .
(9)
What happens if ² < 0 and µ < 0? Is W < 0? Actually no, but this direct conclusion from Eq. (9) shows that this equation is not valid as is. In fact, these materials have to be modeled by frequency dispersive permittivity and permeability. In that case, the relation of Eq. (9) becomes (from Poynting’s theorem): W =
∂(²ω) 2 ∂(µω) 2 E + H ∂ω ∂ω
(10)
and we must have: ∂(²ω) > 0, ∂ω ∂(µω) > 0. ∂ω
(11a) (11b)
6
3.3
Properties of an LHM slab
When LH materials are studies as bulk materials, two models are commonly used for the permittivity/permeability: 1. Drude model:
²r = 1 −
2 ωep , ω(ω + iγe )
(12a)
µr = 1 −
2 ωmp , ω(ω + iγm )
(12b)
which is schematically represented in Fig. 5. Real(eps/eps0) Imag(eps/eps0)
10 0.02
0.015
0
0.01
−10 0.005
−20
0
−0.005
−30 −0.01
−40 wp = 100 GHz wp = 266.5 GHz wp = 500 GHz wp = 1000 GHz −50
0
0.2
0.4
0.6
0.8
1 w/w0
1.2
1.4
1.6
1.8
wp = 100 GHz wp = 266.5 GHz wp = 500 GHz wp = 1000 GHz
−0.015
2
(a) <(²).
−0.02
0
0.2
0.4
0.6
0.8
1 w/w0
1.2
1.4
1.6
1.8
2
(b) =(²).
Figure 5: Permittivity for various values of ωep in the Drude model (f0 = 30 GHz, ωep = ωp ).
2. Resonant model:
²r = 1 − µr = 1 −
2 − ω2 ωep eo , 2 + iγ ω ω 2 − ωeo e
(13a)
2 − ω2 ωmp mo , 2 2 ω − ωmo + iγm ω
(13b)
where ω(em)o are the electric/magnetic resonant frequencies and ω(em)p are the electric/magnetic plasma frequencies. An illustration of this model is given in Fig. 6.
3.3
Properties of an LHM slab
Let us consider the situation depicted in Fig. 7.
7
Imag(eps/eps0)
Real(eps/eps0) 100
1 wp = 266.5 GHz wp = 500 GHz
80
60
0.6
40
0.4
20
0.2
0
0
−20
−0.2
−40
−0.4
−60
−0.6
−80
−0.8
−100
wp = 266.5 GHz wp = 500 GHz
0.8
−1 0
0.2
0.4
0.6
0.8
1 w/w0
1.2
1.4
1.6
1.8
2
0
0.2
0.4
0.6
(a) <(²).
0.8
1 w/w0
1.2
1.4
1.6
1.8
(b) =(²).
Figure 6: Permittivity for various values of ωep in the resonant model of Eq. (13a) (f0 = 30 GHz, ωep = ωp , ω0 = 100GHz).
x
y
z
PSfrag replacements
Figure 7: LH slab in free-space.
2
8
3.3
Properties of an LHM slab
Let us consider the case for which ²r = µr = −1 (working at the right frequency). 1. From simple ray diagrams, using the reversed Snell’s law: PSfrag replacements x #0 #1 #2
I2
S
z
I1
d We see that is the source is close enough to the slab (distance< d), the slab will produce two images, one inside the slab and one outside. The distance from source to the second image is S − I2 = 2d . 2. Rigorous calculation: Let us consider a TE wave impinging on this slab. We write, for a single interface: ¯0 =ˆ E y E0 eikx x−iωt [eikz z + re−ikz z ] , 0 ¯1 =ˆ E y E0 eikx x−iωt teikz z .
(14a) (14b)
We need to match the boundary conditions and, for simplicity, we set the boundary to be at z = 0. We get: ¯ field: • Tangential E ¯ field (H ¯ = • Tangential H
0
eikz z + re−ikz z = teikz z . 1 iωµ ∇
¯ × E):
eikz z − re−ikz z =
µ0 kz0 ikz0 z te . µ1 k z
Upon solving, we get the reflection/transmission coefficient from free-space to the medium: 2µ1 kz , µ1 kz + µ0 kz0 µ1 kz − µ0 kz0 r= . µ1 kz + µ0 kz0 t=
(15a) (15b)
9
In a similar way, the reflection/transmission coefficients from the medium to free-space are: 2µ0 kz0 , µ1 kz + µ0 kz0 µ0 kz0 − µ1 kz r0 = . µ1 kz + µ0 kz0 t0 =
(16a) (16b)
In order to obtain the field inside the slab, we shall compute the transmission coefficient as:
0
0
0
2
0
0
T = teikz d t0 + teikz d r0 eikz d r0 eikz d t0 + tt0 (r0 )2 e5ikz d + · · · 0
= tt0 eikz d
∞ X
0
2
0
(r0 )n e2inkz d =
n=0
tt0 eikz d . 1 − r 0 2 e2ikz0 d
(17)
Using the previous expressions, we obtain 0
T =
4µµ1 kz kz0 eikz d . (µ1 kz + µ0 kz0 )2 − (µ0 kz0 − µ1 kz )e2ikz0 d
(18)
For the specific case of ²r = µr = −1:
0
T =
−4kz kz0 eikz d . (−kz + kz0 )2 − (kz0 + kz )e2ikz0 d
(19)
At this point, we need to define kz and kz0 :
Propagating waves (k > kρ ) q k 2 − kρ2
kz
kz =
kz0
kz0 = −
Relation
q k 2 − kρ2
Evanescent waves (k < kρ ) kz = i
q
kρ2 − k 2
kz0 = i
q
kρ2 − k 2
kz0 = −kz
Performing the calculation, we get: • Propagating waves (kz0 = −kz ): T =
4kz2 e−ikz d = e−ikz d . 4kz2
T =
−4kz2 eikz d = e−ikz d . −4kz2 e2ikz d
• Evanescent waves (kz0 = kz ):
kz0 = kz
10
Section 4. Does it really work?
Therefore, for all waves, we get: T = e−ikz d .
(20)
Conclusions: • Evanescent waves are amplified by the medium. • Propagating waves are “backward waves”. • Taking an infinite amount of those creates a source. • Two images can be formed, like shown in the ray diagram of the previous subsection.
4
Does it really work?
The theoretical predictions have been verified in experiments. To date, essentially two experiments have been done: 1. Prism, 2. Gaussian beam.
4.1
Prism experiment
First experiment (2001). Later reproduced.
LHM
RHM PSfrag replacements
4.2
Gaussian beam experiment
First publication was theoretical. Later experiments followed, but still need to be improved.
11
RHM
LHM
PSfrag replacements
Dispertion relations in Left-Handed Materials Massachusetts Institute of Technology 6.635 lecture notes
1
Introduction
We know already the following properties of LH media: 1. ²r and µr are frequency dispersive. 2. ²r and µr are negative over a similar frequency band. ¯ is left-handed. ¯ H, ¯ k) 3. The tryad (E, 4. The index of refraction is negative. From the past lectures, we know that these materials can be realized by a succession of wires and rods: • Periodic arrangement of rods: realizes a plasma medium with negative ² r over a certain frequency band. The model for the permittivity is: ²r = 1 −
2 ωep . ω 2 + iγe ω
(1)
• Periodic arrangement of rings (split-rings) realizes a resonant µ r modeled as µr = 1 −
2 F ωmp , 2 + iγ ω ω 2 − ωmo m
(2)
where F is the fractional area of the unit cell occupied by the interior of the split-ring (F < 1). In the lossless case (γe = γm = 0), we can rewrite these two relations as: 2 ω 2 − ωep , ω2 ω 2 − ωb2 ω 2 − ω02 − F ω 2 µr = = (1 − F ) , ω 2 − ω02 ω 2 − ω02
²r =
(3a) (3b)
√ where ωb = ω0 / 1 − F > ω0 . Therefore:
2 )(ω 2 − ω 2 ) (ω 2 − ωep 1 b k 2 = ω 2 ²µ = (1 − F ) . 2 2 c ω − ω0
(4)
Upon identifying the regions where ²r and µr change signs, we can immediately get the relation for k:
1
2
Section 2. Argument on n < 0
ω
ωb
ω0
ωp
²r
−
−
−
+
µr
+
−
+
+
k2
−
+
−
+
The region ω ∈ [ω0 , ωb ], which also corresponds to ²r < 0 and µr < 0, corresponds to positive, which means k real. Therefore, there is propagation in this band, but not in the adjacent ones. k2
It may still not be clear that k is negative, even if we write q √ √ k = ω 2 ²µ = ω 2 ²20 µ20 ²r µr = k0 n .
(5)
A demonstration of the fact that n is negative follows.
2 2.1
Argument on n < 0 Complex Poynting theorem
We shall first recall the derivation of the complex Poynting theorem and the signification of the various terms. We start from Maxwell’s curl equation ¯ = iω B ¯, ∇×E ¯ = − iω D ¯ + J¯ . ∇×H
(6a) (6b)
¯ ? and substracting the complex conjugate of Eq. (6b) mulUpon multiplying Eq. (6a) by H ¯ we get: tuplied by E ¯ ? · ∇ × E− ¯ E ¯·∇×H ¯ ? = ∇ · (E ¯×H ¯ ?) H ¯ ·H ¯ ? − iω D ¯? · E ¯ − J¯? · E ¯ = iω B
¯ ·H ¯? − E ¯·D ¯ ?] − E ¯ · J¯? . = iω[B
(7)
Upon rewriting, we get: ¯ · J¯? = ∇ · (E ¯×H ¯ ? ) + iω[E ¯·D ¯? − B ¯ ·H ¯ ?] . −E
(8)
On the right-hand side of the equation, the first terms corresponds to the divergence of Poyting power, which is therefore positive. The second term relates to the complex EM energy, and is therefore also positive. Consequently, the left-hand side term must also be positive, and actually corresponds to the power supplied by J¯ to the volume. We shall use this result hereafter.
3
2.2
1D wave equation
For the sake of simplification, let us work with a 1D problem. The wave equation ¯ r) + k 2 E(¯ ¯ r) = −iωµJ(¯ ¯ r) , ∇2 E(¯
(9)
is rewritten with ¯ r) = zˆ E(x) , E(¯ ¯ r) = zˆ j0 δ(x − x0 ) , J(¯
(10a) (10b)
to yield
∂2 E(x) + k 2 E(x) = −iωµj0 δ(x − x0 ) . ∂x2 The solution to this equation is E(x) = α eik|x−x0 | ,
(11)
(12)
where α needs to be determined. From Eq. (12), we write: 1. First derivative:
∂E(x) ∂ = αik |x − x0 | eik|x−x0 | . ∂x ∂x2
(13)
2. Second derivative: ∂ 2 E(x) ∂2 ∂ =αik |x − x0 | eik|x−x0 | + α(−k 2 )( 2 |x − x0 |)2 eik|x−x0 | 2 2 ∂x ∂x ∂x 2 ik|x−x0 | = −αk e + 2iαkδ(x − x0 ) .
(14)
Therefore: ∂ 2 E(x) + k 2 E(x) =2iαkδ(x − x0 ) = 2iα k0 n δ(x − x0 ) . ∂x2
(15)
Comparing Eq. (11) to Eq. (15), we get α=−
j 0 η 0 µr ωµj0 =− , 2k0 n 2 n
(16)
so that finally the solution is: E(x) = −
j0 η0 µr ik|x−x0 | e . 2 n
(17)
If we now compute the power supplied by the current J¯ to the volume: 1 P =− 2
Z
V
2 ¯ · J¯? dV = η0 j0 µr > 0 . E 4 n
(18)
The source must, on average, do positive work on the field. Yet, in LH regime, we have µr < 0 so that we must have n < 0 as well.
4
Section 3. Dispersion relations
¯ field as: Finally, we can also write the E η E(x, t) = − j0 ei(k0 n|x−x0 |−ωt) . 2
(19)
Thus, plane waves appear to propagate from −∞ and +∞ to the source, seemingly running backward in time. Yet, the work done on the field is positive so clearly the energy propagates outward from the source.
3
Dispersion relations
At this point, we know that n < 0 and k < 0. The difference between phase and group velocity can be directly seen on the dispersion relation diagram. ω , kz µ ¶ ∂kz −1 vg = . ∂ω
vφ =
(20a) (20b)
√ • Free-space: k = ω ²µ where ² = cte and µ = cte. • Metamaterial: Let us take the following models ωp2 , ω 2 + iγe ω 2 − ω2 ωmp mo , µr =1 − 2 2 + iγ ω ω − ωmo m ²r =1 −
3.1
(21a) (21b)
Lossless case (γe = γm = 0), ωmp = ωp
We rewrite ω 2 − ωp2 , ω2 2 ω 2 − ωmp , µr = 2 2 ω − ωmo ²r =
and plot the relations with
(22a) (22b) (22c)
5
ωp ωmp ωmo γe = γ m
= = = =
20e9 rad/s 20e9 rad/s 5e9 rad/s 0 −6
k surface 10
3
x 10
Dispersion relation
2.5
4
2 2
k
ω [rad/s]
x 10
1.5 1
0 2
2
0
−6
x 10
k
0.5
0 −2
−2
z
k
−6
x 10
0
x
0
Relative permittivity 50
1
2
3
2
3
4
10 ω [rad/s] x 10 Relative permeability
300 200
r
r
100
µ
ε
0
0 −100 −50
0
1
2
ω [rad/s]
3
10
4
−200
4
0
1
10
x 10
ω [rad/s]
4 10
x 10
k surface (Gamma= 0)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0 −3
−2
−1
0 kz
1
2
3 −6
x 10
6
3.1
Lossless case (γe = γm = 0), ωmp = ωp
Other cases follow. = = = =
ωp ωmp ωmo γe = γ m
30e9 rad/s 20e9 rad/s 5e9 rad/s 0 rad/s −6
k surface 10
4
x 10
x 10
Dispersion relation
4
2
k
ω [rad/s]
3
0
2
1 2 0
−6
x 10
k
−2
−2
z
2
0 k
−6
x 10
0
x
0
1
Relative permittivity 50
2
3
2
3
4
10 ω [rad/s] x 10 Relative permeability
300 200
µr
εr
100 0
0 −100 −50
0
1
2
ω [rad/s]
3
0
1
x 10
10
4
−200
4 10
ω [rad/s]
4 10
x 10
k surface (Gamma= 0)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0 −4
−3
−2
−1
0 k
z
1
2
3
4 −6
x 10
7
= = = =
ωp ωmp ωmo γe = γ m
30e9 rad/s 20e9 rad/s 5e9 rad/s 10e7 rad/s Dispersion relation
−6
k surface 10
4
x 10
x 10
4
2
k
ω [rad/s]
3
0
2
1 2 0
−6
x 10
kz
−2
−2
2
0 k
−6
x 10
0
x
0
1
Relative permittivity 50
2
3
2
3
4
10 ω [rad/s] x 10 Relative permeability
300 200
µr
εr
100 0
0 −100 −50
0
1
2
ω [rad/s]
10
4
3
−200
4
0
1
10
x 10
ω [rad/s]
4 10
x 10
k surface (Gamma= 100000000)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0 −4
−3
−2
−1
0 k
z
1
2
3
4 −6
x 10
8
3.1
= = = =
ωp ωmp ωmo γe = γ m
30e9 rad/s 20e9 rad/s 5e9 rad/s 10e8 rad/s Dispersion relation
−6
k surface 10
2.5
x 10 4
x 10
2 1.5
2
k
ω [rad/s]
Lossless case (γe = γm = 0), ωmp = ωp
1 0 2
0.5
2
0
−6
x 10
0 −2
kz
−2
k
−6
x 10
0
x
0
Relative permittivity 50
1
2
3
2
3
4
10 ω [rad/s] x 10 Relative permeability
60 40
µr
εr
20 0
0 −20 −50
0
1
2
ω [rad/s]
10
4
3
−40
4
0
1
10
x 10
ω [rad/s]
4 10
x 10
k surface (Gamma= 1000000000)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0 −2.5
−2
−1.5
−1
−0.5
0 k
z
0.5
1
1.5
2
2.5 −6
x 10
9
= = = =
ωp ωmp ωmo γe = γ m
30e9 rad/s 20e9 rad/s 5e9 rad/s 10e9 rad/s −7
k surface 10
3
x 10
Dispersion relation
2.5
4
2 2
k
ω [rad/s]
x 10
1.5 1
0 2
0.5
2
0
−7
x 10
0
−2
kz
−2
−7
x 10
0
kx
0
1
Relative permittivity 20
0
15
−2
10
3
2
3
4
µ
εr
r
2
2
10 ω [rad/s] x 10 Relative permeability
−4
5
−6
0
−8
0
1
2
ω [rad/s]
10
4
3
−5
4
0
1
10
x 10
ω [rad/s]
4 10
x 10
k surface (Gamma= 1.000000e+10)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0 −4
−3
−2
−1
0 k
z
1
2
3
4 −7
x 10
10
3.1
= = = =
ωp ωmp ωmo γe = γ m
30e9 rad/s 20e9 rad/s 5e9 rad/s 10e10 rad/s −7
k surface 10
5
x 10
x 10
Dispersion relation
4 3
2
k
ω [rad/s]
4
2 0 4
1
2
0
−7
x 10
−2
kz
−4
−4
−2
0
4
2
−7
x 10
0
kx
0
Relative permittivity 0.966
0.964
0.964
εr
µr
0.966
0.962
0.96
1
2
3
2
3
4
10 ω [rad/s] x 10 Relative permeability
0.962
0
1
2
ω [rad/s]
10
4
Lossless case (γe = γm = 0), ωmp = ωp
3
0.96
4
0
1
10
x 10
ω [rad/s]
4 10
x 10
k surface (Gamma= 1.000000e+11)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0 −5
−4
−3
−2
−1
0 k
z
1
2
3
4
5 −7
x 10
11
Plotting all the 3D curves on the same scale:
10
x 10 4
3.5
3.5
3
3
2.5
2.5 ω [rad/s]
ω [rad/s]
k surface (Gamma= 10000000)
10
k surface (Gamma= 0)
x 10 4
2
2
1.5
1.5
1
1
0.5
0.5
0 −5
0 −5
0
0 5
−6
x 10
5
0
−5
x
5
−6
x 10
−6
3.5
3
3
2.5
2.5 ω [rad/s]
ω [rad/s]
3.5
2
!#"$&%'#(*)+,
k surface (Gamma= 1000000000)
10
x 10 4
1.5
2
1.5
1
1
0.5
0.5
0 −5
0 −5 0
0 5
−6
x 10
5
0
−5
!#"#1&%2'(*)+,
.34
k surface (Gamma= 1.000000e+10)
10
x 10
3.5
3.5
3
2.5
2.5 ω [rad/s]
3
2
1.5
!#"5&%'#(*)+,
k surface (Gamma= 1.000000e+11)
10
x 10 4
x 10
kz
kx
4
5
0
−5
−6
x 10
kz
.-/0
5
−6
x 10
−6
kx
ω [rad/s]
z
k surface (Gamma= 100000000)
10
x 10
x 10
k
kx
4
5
0
−5
−6
x 10
kz
k
2
1.5
1
1
0.5
0.5
0 −5
0 −5 0
0 5
−6
x 10
5
0
−5
kx
.6*7
5
−6
−6
kz
!#"8&%2'(*)+,
x 10
x 10
5
0
−5
−6
x 10
kz
kx
90
!#": !&%'()+,
;=< >@?4ACBED@F&G&H IJ@KLI2H M/NOLK#P Q:RSH M/N T LKQCP7UWVYX M/L Z:QCLH M/[@I\Z:QCP [WKI=MCX P MIIK#I\]7^=_\`WKba0MWc*K#P I&QCLKCd T \ V T lYm&kElYm eSfhgji&kElYm T lYm=prq ] l VS^ts fgui&k l m kElYm nWo v n v w o v w Q c@:I^ gC}/~ LSQ/c@:Ix l v=w gC~ LS/
LSQ/c@:Ix l vn
p q ] l
Vxzy=`KLSK
l n
g|{/}/~
12
3.1
Lossless case (γe = γm = 0), ωmp = ωp
For simplicity, we can study the lossy case for γe = γm amd ωmo = 0 (although we don’t really simulate the same medium, the fundamental behavior is similar, and simpler to carry out mathematically). The model therefore reads: ω 2 − ωp2 + iγω . ²r = µ r = ω 2 + iγω
(23)
We compute: ω 2 − ωp2 + iγω √ ² r µr = ω 2 + iγω [ω 2 − ωp2 + iγω] [ω 2 − iγω] = ω4 + γ 2 ω2 2 2 ω (ω − ωp2 ) + γ 2 ω 2 + iγωωp2 = . ω4 + γ 2 ω2 (24a) The real part is given by: ω 2 [ω 2 − (ωp2 − γ 2 )] √ <{ ²r µr } = . ω4 + γ 2 ω2 Losses have the effect to lower the plasma frequency to q ωp0 = ωp2 − γ 2 .
(25)
(26)
In addition, we also see that if γ is very large, the plasma effect will completey dissapear (cf. dispersion relation for γ = 10e10 rad/s).
Green’s functions for planarly layered media Massachusetts Institute of Technology 6.635 lecture notes
1
Introduction: Green’s functions
The Green’s functions is the solution of the wave equation for a point source (dipole). For scalar √ problems, the wave equation is written as (k0 = ω ²µ): (∇2 + k02 ) g(¯ r, r¯0 ) = −δ(¯ r − r¯0 ) ,
(1)
and the solution for an unbounded medium is: 0
eik0 |¯r−¯r | g(¯ r, r¯ ) = . 4π|¯ r − r¯0 | 0
(2)
From Maxwell’s equations in frequency domain with an eiωt dependency, the wave equation ¯ r) is: for electric field E(¯
¯ r) − k02 E(¯ ¯ r) = iω J(¯ ¯ r) , ∇ × ∇ × E(¯ =0
for source free case .
(3)
Therefore, the free-space dyadic Green’s function satisfies ∇ × ∇ × G(¯ r, r¯0 ) − k02 G(¯ r, r¯0 ) = I δ(¯ r − r¯0 ) ,
(4)
µ ¶ 1 G(¯ r, r¯ ) = I + 2 ∇∇ g(¯ r, r¯0 ) , k0
(5)
which solution is 0
(check using ∇ × ∇ × (Ig) = ∇∇g − ∇ · (∇g)I and Eq. (1)). For the use of Green’s functions in scattering problems, it is useful to express the Green’s function in the same coordinates as the problem, which can be rectangular, cylindrical, spherical, etc. Here we shall concentrate on the rectangular representation (Cartesian).
1
2
2 2.1
Section 2. Cartesian coordinates
Cartesian coordinates Scalar Green’s function
The formulae are derived from Eq. (1) and the Fourier transform of the quantities: ZZZ +∞ 1 0 ¯ ¯ , dk¯ eik·(¯r−¯r ) g(k) g(¯ r, r¯ ) = 3 (2π) −∞ ZZZ +∞ 1 0 ¯ δ(¯ r − r¯0 ) = dk¯ eik·(¯r−¯r ) , 3 (2π) −∞ 0
(6a) (6b)
where k¯ = kx x ˆ + ky yˆ + kz zˆ and dk¯ = dkx dky dkz . Upon using Eq. (1), we write: (∇2 + k02 )
ZZZ
+∞ −∞
¯ =− dk¯ eik·(¯r−¯r ) g(k)
ZZZ
∂2 ∂y 2
+
¯
0
Introducing the differential operator (∇2 =
2
(∇ +
k02 )
ZZZ
+∞
0 ¯ ¯ = dk¯ eik·(¯r−¯r ) g(k)
ZZZ
=
ZZZ
−∞
=−
∂2 ∂x2
+∞ −∞ +∞ −∞
ZZZ
+
+∞
¯
dk¯ eik·(¯r−¯r ) , 0
(7)
−∞ ∂2 ) ∂z 2
we write:
0 ¯ ¯ dk¯ (∇2 + k02 )eik·(¯r−¯r ) g(k) 0 ¯ ¯ dk¯ (−kx2 − ky2 − kz2 + k02 )eik·(¯r−¯r ) g(k)
+∞
¯
dk¯ eik·(¯r−¯r ) , 0
(8)
−∞
from which we conclude that ¯ = g(k)
kx2
+
ky2
1 . + kz2 − k02
(9)
Using Eq. (6a), we therefore need to evaluate the following integral: g(¯ r, r¯0 ) =
1 (2π)3
ZZZ
+∞
dkx dky dkz −∞
1 0 ¯ eik·(¯r−¯r ) . kx2 + ky2 + kz2 − k02
(10)
Note that Eq. (10) can be integrated along one of the three axis. In remote sensing application, the vertical axis is usually taken to be the z axis, (xy) being the transverse plane (planar components). We therefore choose to evaluate Eq. (10) along kz , and we split:
k¯ = kx x ˆ + ky yˆ + kz zˆ = k¯⊥ + kz zˆ ,
(11a)
r¯ = r¯⊥ + z zˆ ,
(11b)
0 r¯0 = r¯⊥ + z 0 zˆ .
(11c)
3
We will perform the integral of Eq. (10) in the complex plane, using Cauchy’s theorem and the Residue theorem. Before doing this, we have to be careful not to have divergent integrals. Since we integrate in kz , the condition is: lim eikz z < +∞ .
(12)
kz →∞
If we write kz as kz = kz0 + ikz00 (kz0 ∈ R, kz00 ∈ R), we see that • if z > 0, we have to choose kz00 > 0, which means that for complex plane integration, we need to deform the contour into the upper plane. • if z < 0, we have to choose kz00 < 0, which corresponds to a deformation into the lower plane. In addition, we see from Eq. (10) that the integrand has a pole at 2 k02z = k02 − kx2 − ky2 = k02 − k⊥ .
(13)
We therefore need to evaluate Eq. (10) via the Residue theorem. Calculus: let us just write the integral in dkz for z > 0: ¸ · Z +∞ 1 1 ¯ r−¯ ¯ r−¯ ik·(¯ r0 ) ik·(¯ r0 ) = 2iπ Res 2 e 2 e 2 kz − k02z −∞ kz − k0z kz − k 0z 0 0 ¯ = 2iπ lim eik⊥ ·(¯r⊥ −¯r⊥ ) eikz (z−z ) kz →k0z (kz − k0z )(kz + k0z ) 1 ik¯⊥ ·(¯r⊥ −¯r0 ) ik0z (z−z 0 ) ⊥ e = 2iπ , e 2k0z
(14)
so that i g(¯ r, r¯ ) = (2π)2 0
ZZ
∞
dk¯⊥ −∞
1 ik¯⊥ ·(¯r⊥ −¯r0 ) ik0z (z−z 0 ) ⊥ e , e 2k0z
for z − z 0 > 0.
(15)
The treatment for z < 0 follows the same reasoning so that we write for all (z − z 0 ): 0
∀(z − z ) ∈ R,
2.2
i g(¯ r, r¯ ) = (2π)2 0
ZZ
∞
dk¯⊥ −∞
1 ik¯⊥ ·(¯r⊥ −¯r0 ) ik0z |z−z 0 | ⊥ e . e 2k0z
(16)
Dyadic Green’s function
From Eq. (16) and Eq. (26), we can get the dyadic Green’s functions. Note that ∇∇ is a dyadic operator (give a dyad when applied to a scalar) and can be exchanged with the integral sign. In addition, it only applies to the exponential terms so that we actually need to evaluate: ·
∇∇ e
0 ) ik |z−z 0 | ¯⊥ ·(¯ ik r⊥ −¯ r⊥ 0z
e
¸
(17)
4
2.2
Dyadic Green’s function
or, by a simple change of variables: ·
∇∇ e
¯⊥ ·¯ ik r⊥ ik0z |z|
e
¸
(18)
Calculus: Let us first consider z > 0 and ¯
∇∇(eik⊥ ·¯r⊥ eik0z |z| ) = ∇∇ f (x, y, z) .
(19)
Various derivatives will be: •
∂ ∂ ∂x ∂x f (x, y, z)
= −kx2 f (x, y, z) ,
•
∂ ∂ ∂x ∂y f (x, y, z)
= −kx ky f (x, y, z) ,
• ... and identically for z < 0. At z = 0 however, ½ ¾ ∂ 2 ik0z |z| ∂ ik0z |z| ∂ = e ik |z| e 0z ∂z 2 ∂z ∂z ¶2 ¾ ½ µ ∂2 ∂ = ik0z ik0z eik0z |z| |z| + ik0z eik0z |z| 2 |z| ∂z ∂z = 2ik0z δ(z) − k02z eik0z |z| .
(20)
Using these results, we write: · ¸ ZZ ∞ ∂2 1 ik¯⊥ ·¯r⊥ i 2 ik0z |z| ¯ dk ⊥ g(¯ r) = e 2ik0z δ(z) − k0z e ∂z 2 (2π)2 −∞ 2k0z ZZ ∞ ZZ ∞ i δ(z) ¯ ¯⊥ ·¯ ik r⊥ ¯ − 2 dk ⊥ e dk¯⊥ k0z eik⊥ ·¯r⊥ eik0z |z| =− 2 (2π) 8π −∞ −∞ ZZ ∞ i ¯ i k ·¯ r = −δ(¯ r) − 2 dk¯⊥ k0z e ⊥ ⊥ eik0z |z| . 8π −∞
(21)
Again, all the other terms of ∇∇ applied to the integrand give −k¯k¯ so that the Green’s function becomes:
G(¯ r, r¯0 ) = −ˆ z zˆ where
i δ(¯ r) + 2 8π k02
ZZ ∞ ¸ · 1 k¯k¯ ik·¯ ¯ ¯ d k⊥ I− 2 e r for z > 0 , k0z · k0 ¸ ZZ−∞ ∞ ¯K ¯ K 1 ¯ dk¯⊥ I − 2 eiK·¯r for z < 0 , k0z k0 −∞
k¯ = kx x ˆ + ky yˆ + k0z zˆ , ¯ = kx x K ˆ + ky yˆ − k0z zˆ . Some notes:
(22)
(23a) (23b)
5
1. The Dirac delta function is known as the singularity of the Green’s function and is important in calculating the fields in the source region. 2. The different signs ensure that the integral converges for evanescent waves, i.e. when kx2 + ky2 > k02 . 3. The square bracket in the expression of the Green’s functions can be expressed in terms of superposition of TE and TM waves, as we shall see.
2.3
Superposition of TE and TM waves
¯ we can form an orthonormal system for TE/TM polarized waves: Based on k, 1 k¯ × zˆ 1 TE: eˆ(k0z ) = ¯ =q xky − yˆkx ) , [ˆ xky − yˆkx ] = (ˆ kρ |k × zˆ| kx2 + ky2 TM:
kρ ˆ 0 ) = 1 eˆ(k0 ) × k¯ = − k0z (ˆ h(k xky + yˆkx ) + zˆ . z z k0 ko kρ ko
(24a) (24b)
ˆ h ˆ and eˆ form an orthonormal system, in which: The three vectors k, ˆ 0 ). ˆ 0 )h(k e(k0z ) + h(k I = kˆkˆ + eˆ(k0z )ˆ z z
(25)
After translating to the origin, we get for the dyadic Green’s functions:
δ(¯ r) i G(¯ r, r¯ ) = −ˆ z zˆ 2 + 2 k0 8π 0
ZZ ∞ ¸ · 1 0 ¯ ˆ ˆ ¯ e(k0z ) + h(k0z )h(k0z ) eik·(¯r−¯r ) eˆ(k0z )ˆ dk ⊥ k0z · ¸ ZZ−∞ ∞ 1 0 ¯ ˆ ˆ ¯ ) eiK·(¯r−¯r ) ) h(−k ) + h(−k )ˆ e (−k e ˆ (−k d k 0z 0z 0z 0z ⊥ k 0z −∞
for z > z 0 , for z < z 0 , (26)
where eˆ(−k0z ) = eˆ(k0z ) ,
(27a)
¯ eˆ(−k0z ) × K ˆ h(−k . 0z ) = k0
(27b)
ˆ ¯ ¯ eˆ(−k0z ) and h(−k (Note that K, 0z ) form another orthonormal set of vectors about K).
2.4
Treatment of layered media
Depending upon the medium under study and the location of the source, the kernel of Eq. (26) will have to be modified. To make it more clear, we can gather the terms in the Green’s function relative to the source (primed coordinates) and those relative to the source.
6
2.4
G(¯ r,¯ r 0 )=−ˆ z zˆ
δ(¯ r) i 2 + 8π 2 k0
RR ¯⊥ ∞ dk −∞
1 k 0z
¯ 0
¯
Treatment of layered media
¯ 0
¯
r } r ] h(k r +[h(k r] e ˆ 0 ) e−ik¯ ˆ 0 ) eik·¯ ˆ(k0z ) e−ik¯ {[ˆ e(k0z ) eik·¯ z z
for z > z 0 ,
¯ 0 ¯ r ˆ ¯ 0 ¯ r iK·¯ ˆ ] h(−k0z ) e−iK r¯ } ] eˆ(−k0z ) e−iK r¯ +[h(−k {[ˆ e(−k0z ) eiK·¯ 0z ) e
for z < z 0 , (28) If we now consider a layered medium problem, with an arbitrary number of layers and a source in the top region (incident wave), we write: RR ∞ ¯ −∞ dk⊥
i G(¯ r, r¯ )i0 = 2 8π 0
1 k 0z
ZZ
∞ −∞
½ ¾ 1 ¯ r0 ¯ r0 −iK·¯ −iK·¯ ˆ ¯ ¯ Ke eˆ(−k0z ) e + Kh h(−k0z ) e dkx dky k0z
(29)
with 1. For z < z 0 , i = 0:
¯ ¯ ¯ e =ˆ K e(−k0z ) eiK·¯r + RT E eˆ(k0z ) ek0z eik·¯r
(30a)
¯r ¯ r iK·¯ ˆ ˆ 0 ) ek0z eik·¯ ¯ h =h(−k K + RT M h(k 0z ) e z
(30b)
¯ e =A` eˆ(k` ) eik¯` ·¯r + B` eˆ(−k` ) eiK¯ ` ·¯r K z z
(31a)
2. For region `, i = `:
ˆ ` )e ¯ h =C` h(k K z
¯` ·¯ ik r
ˆ + D` h(−k `z ) e
¯ ` ·¯ iK r
(31b) (31c)
3. For region t, i = t:
¯ e =T T E eˆ(−ktz ) eiK¯ t ·¯r K ¯ h =T K
TM
ˆ h(−k tz ) e
¯ t ·¯ iK r
(32a) (32b) (32c)
where q k`z = k`2 − kx2 − ky2 ,
k¯` =ˆ xkx + yˆky + zˆk`z , ¯ ` =ˆ K xkx + yˆky − zˆk`z ,
(33a) (33b) (33c)
and the coefficients A` , B` , C` and D` are determined from the boundary conditions. The boundary conditions apply to the tangential electric and magnetic fields. Thus, in r, r¯0 ) and zˆ × ∇ × terms of Green’s functions, we need to satisfy the continuity of zˆ × G(¯
7
G(¯ r, r¯0 ). Let us write this at the interface between media (`) and (` + 1), by separating the TE and TM components:
A` eik`z z + B` e−ik`z z = A`+1 eik`z z + B`+1 e−ik`z z ¸ ¸ · −ik`z z −ik`z z ik`z z ik`z z = kz `+1 A`+1 e − B`+1 e − B` e k`z A ` e ¸ ¸ · · k`z kz A` eik`z z − B` e−ik`z z = `+1 A`+1 eik`z z − B`+1 e−ik`z z k` k`+1 · ¸ · ¸ −ik`z z ik`z z −ik`z z ik`z z k`z C` e + D` e = kz `+1 C`+1 e + D`+1 e ·
(34a) (34b) (34c) (34d)
With the conditions in the first and last layer as: A0 = R T E , At = 0 ,
B0 = 1 ,
Bt = T
TE
,
C0 = R T M , Ct = 0 ,
D0 = 1 ,
Dt = T
TM
.
(35a) (35b)
Before evaluating these coefficients, we can build a recursive scheme to calculate the amplitudes from region ` to region ` + 1. For example, it is straightforward to build a propagation matrix for TE modes from Eq. (34a) and (34b): ! Ã ! Ã TE A` eikz ` d` A`+1 eikz `+1 d`+1 (36) =V B` e−ikz ` d` B`+1 e−ikz `+1 d`+1 A similar procedure of course applied to the TM modes: ! ! Ã Ã TM C`+1 eikz `+1 d`+1 C` eikz ` d` =V D`+1 e−ikz `+1 d`+1 D` e−ikz ` d`
(37)
In order to end up the recursive method, we have to express the reflection and transmission coefficient in the first and last regions, respectively. We shall only illustrated this point here, as it has been developed in previous classes. Let us consider a plane wave incident from region 0, with its plane of incidence parallel to ∂ the (xy) plane. All fields vectors are independent on y, so that ∂y = 0 in Maxwell’s equations. Thus, we can decompose the fields into their TE and TM components. We get in region `: • TE modes: 1 ∂ E`y , iωµ` ∂z 1 ∂ H`z = E`y , iωµ` ∂x µ 2 ¶ ∂ ∂ 2 + + ω ² µ ` ` E`y = 0 . ∂x2 ∂y H`x = −
(38a) (38b) (38c)
8
2.4
Treatment of layered media
• TM modes: 1 ∂ H`y , iω²` ∂z 1 ∂ H`y , E`z = − iω²` ∂x µ 2 ¶ ∂ ∂ 2 + + ω ² µ ` ` H`y = 0 . ∂x2 ∂y E`x =
(39a) (39b) (39c)
For a TE wave inside the stratified medium: E`y =(A` eik`z z + B` e−ik`z z ) eikx x , k` H`x = − z (A` eik`z z − B` e−ik`z z ) eikx x , ωµ` kx H`z = (A` eik`z z + B` e−ik`z z ) eikx x . ωµ`
(40a) (40b) (40c)
By matching the boundary conditions, and upon using the already known notation µ` kz (`+1) , µ`+1 k`z 1 − p`(`+1) = . 1 + p`(`+1)
p`(`+1) =
(41a)
R`(`+1)
(41b)
we get the recursive relation: 2 [1 − 1/R`(`+1) ] e2i[k(`+1) z +k`z ]d` A` e2ik`z d` , = + A B` R`(`+1) 1/R`(`+1) e2ik(`+1) z d` + B`+1 `+1
(42)
with the limiting condition: At = 0, , Bt
A0 = R. B0
(43)
Example: for a two-layer medium (t = 2): R=
R01 + R12 e2ik1z (d1 −d0 ) 2ikz d0 e . 1 + R01 R12 e2ik1z (d1 −d0 )
(44)
Fundamentals of Model Theory William Weiss and Cherie D'Mello Department of Mathematics University of Toronto
c 1997 W.Weiss and C. D'Mello
1
Introduction Model Theory is the part of mathematics which shows how to apply logic to the study of structures in pure mathematics. On the one hand it is the ultimate abstraction on the other, it has immediate applications to every-day mathematics. The fundamental tenet of Model Theory is that mathematical truth, like all truth, is relative. A statement may be true or false, depending on how and where it is interpreted. This isn't necessarily due to mathematics itself, but is a consequence of the language that we use to express mathematical ideas. What at rst seems like a deciency in our language, can actually be shaped into a powerful tool for understanding mathematics. This book provides an introduction to Model Theory which can be used as a text for a reading course or a summer project at the senior undergraduate or graduate level. It is also a primer which will give someone a self contained overview of the subject, before diving into one of the more encyclopedic standard graduate texts. Any reader who is familiar with the cardinality of a set and the algebraic closure of a eld can proceed without worry. Many readers will have some acquaintance with elementary logic, but this is not absolutely required, since all necessary concepts from logic are reviewed in Chapter 0. Chapter 1 gives the motivating examples and we recommend that you read it rst, before diving into the more technical aspects of Chapter 0. Chapters 2 and 3 are selections of some of the most important techniques in Model Theory. The remaining chapters investigate the relationship between Model Theory and the algebra of the real and complex numbers. Thirty exercises develop familiarity with the denitions and consolidate understanding of the main proof techniques. Throughout the book we present applications which cannot easily be found elsewhere in such detail. Some are chosen for their value in other areas of mathematics: Ramsey's Theorem, the Tarski-Seidenberg Theorem. Some are chosen for their immediate appeal to every mathematician: existence of innitesimals for calculus, graph colouring on the plane. And some, like Hilbert's Seventeenth Problem, are chosen because of how amazing it is that logic can play an important role in the solution of a problem from high school algebra. In each case, the derivation is shorter than any which tries to avoid logic. More importantly, the methods of Model Theory display clearly the structure of the main ideas of the proofs, showing how theorems of logic combine with theorems from other areas of mathematics to produce stunning results. The theorems here are all are more than thirty years old and due in great part to the cofounders of the subject, Abraham Robinson and Alfred Tarski. However, we have not attempted to give a history. When we attach a name to a theorem, it is simply because that is what mathematical logicians popularly call it. The bibliography contains a number of texts that were helpful in the preparation of this manuscript. They could serve as avenues of further study and in addition, they contain many other references and historical notes. The more recent titles were added to show the reader where the subject is moving today. All are worth a look. This book began life as notes for William Weiss's graduate course at the University of Toronto. The notes were revised and expanded by Cherie D'Mello and
2
William Weiss, based upon suggestions from several graduate students. The electronic version of this book may be downloaded and further modied by anyone for the purpose of learning, provided this paragraph is included in its entirety and so long as no part of this book is sold for prot.
Contents Chapter 0. Models, Truth and Satisfaction Formulas, Sentences, Theories and Axioms Prenex Normal Form Chapter 1. Notation and Examples Chapter 2. Compactness and Elementary Submodels Compactness Theorem Isomorphisms, elementary equivalence and complete theories Elementary Chain Theorem Lowenheim-Skolem Theorems The L os-Vaught Test Every complex one-to-one polynomial map is onto Chapter 3. Diagrams and Embeddings Diagram Lemmas Every planar graph can be four coloured Ramsey's Theorem The Leibniz Principle and innitesimals Robinson Consistency Theorem Craig Interpolation Theorem Chapter 4. Model Completeness Robinson's Theorem on existentially complete theories Lindstrom's Test Hilbert's Nullstellensatz Chapter 5. The Seventeenth Problem Positive denite rational functions are the sums of squares Chapter 6. Submodel Completeness Elimination of quantiers The Tarski-Seidenberg Theorem Chapter 7. Model Completions Almost universal theories Saturated models Blum's Test Bibliography Index 3
4 4 9 11 14 14 15 16 19 21 23 24 25 25 26 26 27 31 32 32 35 38 39 39 45 45 49 50 52 54 55 61 62
CHAPTER 0
Models, Truth and Satisfaction We will use the following symbols: logical symbols: { the connectives ^ ,_ , : , ! , $ called \and", \or", \not", \implies" and \i" respectively { the quantiers 8 , 9 called \for all" and \there exists" { an innite collection of variables indexed by the natural numbers N v0 ,v1 , v2 , : : : { the two parentheses ), ( { the symbol = which is the usual \equal sign" constant symbols : often denoted by the letter c with subscripts function symbols : often denoted by the letter F with subscripts each function symbol is an m-placed function symbol for some natural number m 1 relation symbols : often denoted by the letter R with subscripts each relational symbol is an n-placed relation symbol for some natural number n 1. We now dene terms and formulas. Definition 1. A term is dened as follows: (1) a variable is a term (2) a constant symbol is a term (3) if F is an m-placed function symbol and t1 : : : tm are terms, then F (t1 : : : tm ) is a term. (4) a string of symbols is a term if and only if it can be shown to be a term by a nite number of applications of (1), (2) and (3). Remark. This is a recursive denition. Definition 2. A formula is dened as follows : (1) if t1 and t2 are terms, then t1 = t2 is a formula. (2) if R is an n-placed relation symbol and t1 : : : tn are terms, then R(t1 : : : tn ) is a formula. (3) if ' is a formula, then (:') is a formula (4) if ' and are formulas then so are (' ^ ), (' _ ), (' ! ) and (' $ ) (5) if vi is a variable and ' is a formula, then (9vi )' and (8vi )' are formulas (6) a string of symbols is a formula if and only if it can be shown to be a formula by a nite number of applications of (1), (2), (3), (4) and (5). Remark. This is another recursive denition. :' is called the negation of '. ' ^ is called the conjunction of ' and . ' _ is called the disjunction of ' and . Definition 3. A subformula of a formula ' is dened as follows: 4
0. MODELS, TRUTH AND SATISFACTION
5
(1) ' is a subformula of ' (2) if (:) is a subformula of ' then so is (3) if any one of ( ^ ), ( _ ), ( ! ) or ( $ ) is a subformula of ' , then so are both and (4) if either (9vi ) or (8vi ) is a subformula of ' for some natural number i , then is also a subformula of ' (5) A string of symbols is a subformula of ', if and only if it can be shown to be such by a nite number of applications of (1), (2), (3) and (4). Definition 4. A variable vi is said to occur bound in a formula ' i for some subformula of ' either (9vi ) or (8vi ) is a subformula of '. In this case each occurance of vi in is said to be a bound occurance of vi . Other occurances of vi which do not occur bound in ' are said to be free. Exercise 1. Using the previous denitions as a guide, dene the substitution of a term t for a variable vi in a formula '. Example 1. F2 (v3 v4 ) is a term. ((8v3 )(v3 = v3 ^ v0 = v1 ) _ (9v0 )v0 = v0 ) is a formula. In this formula the variable v3 occurs bound, the variable v1 occurs free, but the variable v0 occurs both bound and free. Exercise 2. Reconsider Exercise 1 in light of substituting the above term for v0 in the formula. Definition 5. A language L is a set consisting of all the logical symbols with perhaps some constant, function and/or relational symbols included. It is understood that the formulas of L are made up from this set in the manner prescribed above. Note that all the formulas of L are uniquely described by listing only the constant, function and relation symbols of L. We use t(v0 : : : vk ) to denote a term t all of whose variables occur among v0 : : : vk . We use '(v0 : : : vk ) to denote a formula ' all of whose free variables occur among v0 : : : vk . Example 2. These would be formulas of any language : For any variable vi : vi = vi for any term t(v0 : : : vk ) and other terms t1 and t2 : t1 = t2 ! t(v0 : : : vi;1 t1 vi+1 : : : vk ) = t(v0 : : : vi;1 t2 vi+1 : : : vk ) for any formula '(v0 : : : vk ) and terms t1 and t2 : t1 = t2 ! '(v0 : : : vi;1 t1 vi+1 : : : vk ) $ '(v0 : : : vi;1 t2 vi+1 : : : vk ) Note the simple way we denote the substitution of t1 for vi . Definition 6. A model (or structure) A for a language L is an ordered pair hA Ii where A is a set and I is an interpretation function with domain the set of all constant, function and relation symbols of L such that: 1. if c is a constant symbol, then I (c) 2 A I (c) is called a constant
0. MODELS, TRUTH AND SATISFACTION
6
2. if F is an m-placed function symbol, then I (F ) is an m-placed function on
A
3. if R is an n-placed relation symbol, then I (R) is an n-placed relation on A. A is called the universe of the model A. We generally denote models with Gothic letters and their universes with the corresponding Latin letters in boldface. One set may be involved as a universe with many dierent interpretation functions of the language L. The model is both the universe and the interpretation function. Remark. The importance of Model Theory lies in the observation that mathematical objects can be cast as models for a language. For instance, the real numbers with the usual ordering < and the usual arithmetic operations, addition + and multiplication along with the special numbers 0 and 1 can be described as a model. Let L contain one two-placed (i.e. binary) relation symbol R0 , two two-placed function symbols F1 and F2 and two constant symbols c0 and c1 . We build a model by letting the universe A be the set of real numbers. The interpretation function I will map R0 to < , i.e. R0 will be interpreted as < . Similarly, I (F1 ) will be + , I (F2 ) will be , I (c0 ) will be 0 and I (c1 ) will be 1. So hA Ii is an example of a model. We now wish to show how to use formulas to express mathematical statements about elements of a model. We rst need to see how to interpret a term in a model. Definition 7. The value t x0 : : : xq ] of a term t(v0 : : : vq ) at x0 : : : xq in the model A is dened as follows: 1. if t is vi then t x0 xq ] is xi , 2. if t is the constant symbol c, then t x0 : : : xq ] is I (c), the interpretation of c in A, 3. if t is F (t1 : : : tm ) where F is an m-placed function symbol and t1 : : : tm are terms, then t x0 : : : xq ] is G(t1 x0 : : : xq ] : : : tm x0 : : : xq ]) where G is the m-placed function I (F ), the interpretation of F in A. Definition 8. Suppose A is a model for a language L. The sequence x0 : : : xq satises the formula '(v0 : : : vq ) all of whose free and bound variables are among v0 : : : vq , in the model A, written A j= ' x0 : : : xq ] provided we have: 1. if '(v0 : : : vq ) is the formula t1 = t2 , then A j= t1 = t2 x0 : : : xq ] means that t1 x0 : : : xq ] equals t2 x0 : : : xq ], 2. if '(v0 : : : vq ) is the formula R(t1 : : : tn ) where R is an n-placed relation symbol, then A j= R(t1 : : : tn ) x0 : : : xq ] means S (t1 x0 : : : xq ] : : : tn x0 : : : xq ]) where S is the n-placed relation I (R), the interpretation of R in A, 3. if ' is (:), then A j= ' x0 : : : xq ] means not A j= x0 : : : xq ], 4. if ' is ( ^ ), then A j= ' x0 : : : xq ] means both A j= x0 : : : xq ] and A j= x0 : : : xq ], 5. if ' is ( _ ) then A j= ' x0 : : : xq ] means either A j= x0 : : : xq ] or A j= x0 : : : xq ],
0. MODELS, TRUTH AND SATISFACTION
7
6. if ' is ( ! ) then A j= ' x0 : : : xq ] means that A j= x0 : : : xq ] implies A j= x0 : : : xq ], 7. if ' is ( $ ) then A j= ' x0 : : : xq ] means that A j= x0 : : : xq ] i A j= x0 : : : xq ], 8. if ' is 8vi , then A j= ' x0 : : : xq ] means for every x 2 A A j= x0 : : : xi;1 x xi+1 : : : xq ], 9. if ' is 9vi , then A j= ' x0 : : : xq ] means for some x 2 A A j= x0 : : : xi;1 x xi+1 : : : xq ]: Exercise 3. Each of the formulas of Example 2 is satised in any model A for any language L by any (long enough) sequence x0 x1 : : : xq of A. This is where you test your solution to Exercise 2. We now prove two lemmas which show that the preceeding concepts are welldened. In the rst one, we see that the value of a term only depends upon the values of the variables which actually occur in the term. In this lemma the equal sign = is used, not as a logical symbol in the formal sense, but in its usual sense to denote equality of mathematical objects | in this case, the values of terms, which are elements of the universe of a model. Lemma 1. Let A be a model for L and let t(v0 : : : vp ) be a term of L. Let x0 : : : xq and y0 : : : yr be sequences from A such that p q and p r, and let xi = yi whenever vi actually occurs in t(v0 : : : vp ). Then t x0 : : : xq ] = t y0 : : : yr ] . Proof. We use induction on the complexity of the term t. 1. If t is vi then xi = yi and so we have t x0 : : : xq ] = xi = yi = t y0 : : : yr ] since p q and p r: 2. If t is the constant symbol c, then t x0 : : : xq ] = I (c) = t y0 : : : yr ] where I (c) is the interpretation of c in A. 3. If t is F (t1 : : : tm) where F is an m-placed function symbol, t1 : : : tm are terms and I (F ) = G, then t x0 : : : xq ] = G(t1 x0 : : : xq ] : : : tm x0 : : : xq ]) and t y0 : : : yr ] = G(t1 y0 : : : yr ] : : : tm y0 : : : yr ]). By the induction hypothesis we have that ti x0 : : : xq ] = ti y0 : : : yr ] for 1 i m since t1 : : : tm have all their variables among fv0 : : : vp g. So we have t x0 : : : xq ] = t y0 : : : yr ]. In the next lemma the equal sign = is used in both senses | as a formal logical symbol in the formal language L and also to denote the usual equality of mathematical objects. This is common practice where the context allows the reader to distinguish the two usages of the same symbol. The lemma conrms that satisfaction of a formula depends only upon the values of its free variables.
0. MODELS, TRUTH AND SATISFACTION
8
Lemma 2. Let A be a model for L and ' a formula of L, all of whose free and bound variables occur among v0 : : : vp . Let x0 : : : xq and y0 : : : yr (q r p) be two sequences such that xi and yi are equal for all i such that vi occurs free in '. Then A j= ' x0 : : : xq ] i A j= ' y0 : : : yr ] Proof. Let A and L be as above. We prove the lemma by induction on the complexity of '. 1. If '(v0 : : : vp ) is the formula t1 = t2 , then we use Lemma 1 to get: A j= (t1 = t2 ) x0 : : : xq ] i t1 x0 : : : xq ] = t2 x0 : : : xq ] i t1 y0 : : : yr ] = t2 y0 : : : yr ] i A j= (t1 = t2 ) y0 : : : yr ]: 2. If '(v0 : : : vp ) is the formula R(t1 : : : tn ) where R is an n-placed relation symbol with interpretation S , then again by Lemma 1, we get: A j= R(t1 : : : tn ) x0 : : : xq ] i S (t1 x0 : : : xq ] : : : tn x0 : : : xq ]) i S (t1 y0 : : : yr ] : : : tn y0 : : : yr ]) i A j= R(t1 : : : tn ) y0 : : : yr ]: 3. If ' is (:), the inductive hypothesis gives that the lemma is true for . So, A j= ' x0 : : : xq ] i not A j= x0 : : : xq ] i not A j= y0 : : : yr ] i A j= ' y0 : : : yr ]: 4. If ' is ( ^ ), then using the inductive hypothesis on and we get A j= ' x0 : : : xq ] i both A j= x0 : : : xq ] and A j= x0 : : : xq ] i both A j= y0 : : : yr ] and A j= y0 : : : yr ] i A j= ' y0 : : : yr ]: 5. If ' is ( _ ) then A j= ' x0 : : : xq ] i either A j= x0 : : : xq ] or A j= x0 : : : xq ] i either A j= y0 : : : yr ] or A j= y0 : : : yr ] i A j= ' y0 : : : yr ]: 6. If ' is ( ! ) then A j= ' x0 : : : xq ] i A j= x0 : : : xq ] implies A j= x0 : : : xq ] i A j= y0 : : : yr ] implies A j= y0 : : : yr ] i A j= ' y0 : : : yr ]: 7. If ' is ( $ ) then A j= ' x0 : : : xq ] i we have A j= x0 : : : xq ] i A j= x0 : : : xq ] i we have A j= y0 : : : yr ] i A j= y0 : : : yr ] i A j= ' y0 : : : yr ]: 8. If ' is 8vi , then A j= ' x0 : : : xq ] i for every z 2 A A j= x0 : : : xi;1 z xi+1 : : : xq ] i for every z 2 A A j= y0 : : : yi;1 z yi+1 : : : yr ] i A j= ' y0 : : : yr ]: The inductive hypothesis uses the sequences x0 : : : xi;1 z xi+1 : : : xq and y0 : : : yi;1 z yi+1 : : : yr with the formula .
0. MODELS, TRUTH AND SATISFACTION
9
9. If ' is 9vi , then A j= ' x0 : : : xq ] i for some z 2 A A j= x0 : : : xi;1 z xi+1 : : : xq ] i for some z 2 A A j= y0 : : : yi;1 z yi+1 : : : yr ] i A j= ' y0 : : : yr ]: The inductive hypothesis uses the sequences x0 : : : xi;1 z xi+1 : : : xq and y0 : : : yi;1 z yi+1 : : : yr with the formula . Definition 9. A sentence is a formula with no free variables.
If ' is a sentence, we can write A j= ' without any mention of a sequence from
A since by the previous lemma, it doesn't matter which sequence from A we use.
In this case we say: A satises ' or A is a model of ' or ' holds in A or ' is true in A If ' is a sentence of L, we write j= ' to mean that A j= ' for every model A for L. Intuitively then, j= ' means that ' is true under any relevant interpretation (model for L). Alternatively, no relevant example (model for L) is a counterexample to ' | so ' is true. Lemma 3. Let '(v0 : : : vq ) be a formula of the language L. There is another formula '0 (v0 : : : vq ) of L such that 1. '0 has exactly the same free and bound occurances of variables as '. 2. '0 can possibly contain :, ^ and 9 but no other connective or quantier. 3. j= (8v0 ) : : : (8vq )(' $ '0 ) Exercise 4. Prove the above lemma by induction on the complexity of '. Definition 10. A formula ' is said to be in prenex normal form whenever
(1) no variable occuring in ' occurs both free and bound, (2) no bound variable occuring in ' is bound by more than one quantier, and (3) in the written order, all of the quantiers preceed all of the connectives. Remark. (3) is equivalent to saying that in the constructed order, all of the connective steps preceed all of the quantier steps. Lemma 4. Let '(v0 : : : vp ) be any formula of a language L. There is a formula ' of L which has the following properties: 1. ' is in prenex normal form 2. ' and ' have the same free occurances of variables, and 3. j= (8v0 ) : : : (8vp )(' $ ' ) Exercise 5. Prove this lemma by induction on the complexity of '. There is a notion of rank on prenex formulas | the number of alternations of quantiers. The usual formulas of elementary mathematics have prenex rank 0, i.e. no alternations of quantiers. For example: (8x)(8y)(2xy x2 + y2 ):
0. MODELS, TRUTH AND SATISFACTION
10
However, the ; denition of a limit of a function has prenex rank 2 and is much more di!cult for students to comprehend at rst sight: 89 8x((0 < ^ 0 < jx ; aj < ) ! jF (x) ; Lj < ): A formula of prenex rank 4 would make any mathematician look twice.
CHAPTER 1
Notation and Examples Although the formal notation for formulas is precise, it can become cumbersome and di!cult to read. Condent that the reader would be able, if necessary, to put formulas into their formal form, we will relax our formal behaviour. In particular, we will write formulas any way we want using appropriate symbols for variables, constant symbols, function and relation symbols. We will omit parentheses or add them for clarity. We will use binary function and relation symbols between the arguments rather than in front as is the usual case for \plus", \times" and \less than". Whenever a language L has only nitely many relation, function and constant symbols we often write, for example: L = f< R0 + F1 c0 c1 g omitting explicit mention of the logical symbols (including the innitely many variables) which are always in L. Correspondingly we may denote a model A for L as: < S0 + G1 a0 a1 i A = hA< where the interpretations of the symbols in the language L are given by I (<) = <, I (R0 ) = S0 , I (+) = + , I (F1 ) = G1 , I (c0 ) = a0 and I (c1 ) = a1 . < + 0 1i and Q = hQ < < + 0 1i, where R is the Example 3. R = hR< reals, Q the rationals , are models for the language L = f< + 0 1g. Here < is a binary relation symbol, + and are binary function symbols, 0 and 1 are constant symbols whereas <, + , , 0, 1 are the well known relations, arithmetic functions and constants. Similarly, C = hC + 0 1i, where C is the complex numbers, is a model for the language L = f+ 0 1g. Note the exceptions to the boldface convention for these popular sets. Example 4. Here L = f< + 0 1g, where < is a binary relation symbol, + and are binary function symbols and 0 and 1 are constant symbols. The following formulas are sentences. 1. (8x):(x < x) 2. (8x)(8y):(x < y ^ y < x) 3. (8x)(8y)(8z )(x < y ^ y < z ! x < z ) 4. (8x)(8y)(x < y _ y < x _ x = y) 5. (8x)(8y)(x < y ! (9z )(x < z ^ z < y)) 6. (8x)(9y)(x < y) 7. (8x)(9y)(y < x) 8. (8x)(8y)(8z )(x + (y + z ) = (x + y) + z ) 9. (8x)(x + 0 = x) 11
1. NOTATION AND EXAMPLES
12
(8x)(9y)(x + y = 0) (8x)(8y)(x + y = y + x) (8x)(8y)(8z )(x (y z ) = (x y) z ) (8x)(x 1 = x) (8x)(x = 0 _ (9y)(y x = 1) (8x)(8y)(x y = y x) (8x)(8y)(8z )(x (y + z ) = (x y) + (y z )) 0 6= 1 (8x)(8y)(8z )(x < y ! x + z < y + z ) (8x)(8y)(8z )(x < y ^ 0 < z ! x z < y z ) for each n 1 we have the formula (8x0 )(8x1 ) (8xn )(9y)(xn yn + xn;1 yn;1 + + x1 y + x0 = 0 _ xn = 0) 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
z
k
}|
{
where, as usual, yk abbreviates y y y The latter formulas express that each polynomial of degree n has a root. The following formulas express the intermediate value property for polynomials of degree n: if the polynomial changes sign from w to z , then it is zero at some y between w and z . 21. for each n 1 we have (8x0 ) : : : (8xn )(8w)(8z ) (xn wn + xn;1 wn;1 + + x1 w + x0 ) (xn z n + xn;1 z n;1 + + x1 z + x0 ) < 0 ! (9y )(((w < y ^ y < z ) _ (z < y ^ y < w)) ^ (xn y n + xn;1 y n;1 + + x1 y + x0 = 0))] The most fundamental concept is that of a sentence being true when interpreted in a model A. We write this as A j= , and we extend this concept in the following denitions. Definition 11. If " is a set of sentences, A is said to be a model of ", written A j= ", whenever A j= for each 2 ". " is said to be satisable i there is some A such that A j= ". Definition 12. A theory T is a set of sentences. If T is a theory and is a sentence, we write T j= whenever we have that for all A if A j= T then A j= . We say that is a consequence of T . A theory is said to be closed whenever it contains all of its consequences. Definition 13. If A is a model for the language L, the theory of A, denoted by ThA, is dened to be the set of all sentences of L which are true in A, f of L : A j= g This is one way that a theory can arise. Another way is through axioms. Definition 14. " T is said to be a set of axioms for T whenever " j= for every in T in this case we write " j= T . Remark. We will generally assume our theories are closed and we will often describe theories by specifying a set of axioms ". The theory will then be all consequences of ". Example 5. We will consider the following theories and their axioms:
1. NOTATION AND EXAMPLES
13
1. The theory of Linear Orderings (LOR) which has as axioms sentences 1-4 from Example 4. 2. The theory of Dense Linear Orders (DLO) which has as axioms all the axioms of LOR, and sentence 5, 6 and 7 of Example 4. 3. The theory of Fields (FEI) which has as axioms sentences 8-17 from Example 4. 4. The theory of Ordered Fields (ORF) which has as axioms all the axioms of FEI, LOR and sentences 18 and 19 from Example 4. 5. The theory of Algebraically Closed Fields (ACF) which has as axioms all the axioms of FEI and all sentences from 20 of Example 4, i.e. innitely many sentences, one for each n 1. 6. The theory of Real Closed Ordered Fields (RCF) which has as axioms all the axioms of ORF, and all sentences from 21 of Example 4, i.e. innitely many sentences, one for each n 1. Exercise 6. Show that : 1. Q j= DLO 2. R j= RCF using the Intermediate Value theorem 3. C j= ACF using the Fundamental Theorem of Algebra where Q, R and C are as in Example 3. Remark. The theory of Real Closed Ordered Fields is sometimes axiomatized dierently. All the axioms of ORF are retained, but the sentences from 21 of Example 4, which amount to an Intermediate Value Property, are replaced by the sentences from 20 for odd n and the sentence (8x)(0 < x ! (9y)y2 = x) which states that every positive element has a square root. A signicant amount of algebra would then be used to verify the Intermediate Value Property from these axioms.
CHAPTER 2
Compactness and Elementary Submodels Theorem 1. The Compactness Theorem (Malcev) A set of sentences is satisable i every nite subset is satisable. Proof. There are several proofs. We only point out here that it is an easy consequence of the following, a theorem which appears in all elementary logic texts: Proposition. The Completeness Theorem (G odel, Malcev) A set of sentences is consistent i it is satisable. Although we do not here formally dene \consistent", it does mean what you think it does. In particular, a set of sentences is consistent if and only if each nite subset is consistent. Remark. The Compactness Theorem is the only one for which we do not give a complete proof. If the reader has not previously seen the Completeness Theorem, there are other proofs of the Compactness Theorem which may be more easily absorbed: set theoretic (using ultraproducts), topological (using compact spaces, hence the name) or Boolean algebraic. However these topics are too far aeld to enter into the proofs here. We will use the Compactness Theorem as a starting point | in fact, all that follows can be seen as its corollaries. Exercise 7. Suppose T is a theory for the language L and is a sentence of L such that T j= . Prove that there is some nite T 0 T such that T 0 j= . Recall that T j= i T f: g is not satisable. Definition 15. If L, and L0 are two languages such that L L0 we say that 0 L is an expansion of L and L is a reduction of L0 Definition 16. Given a model A for the language L, we can expand it to a model A0 of L0 by giving appropriate interpretations to the symbols in L0 n L. We say that A0 is an expansion of A to L0 and that A is a reduct of A0 to L. We also use the notation A0 jL for the reduct of A0 to L. Theorem 2. If a theory T has arbitrarily large nite models, then it has an innite model. Proof. Consider new constant symbols ci for i 2 N , the usual natural numbers, and expand from L, the language of T , to L0 = L fci : i 2 N g. Let " = T f:ci = cj : i 6= j i j 2 N g: We rst show that every nite subset of " has a model by interpreting the nitely many relevant constant symbols as dierent elements in an expansion of some nite model of T . Then we use compactness to get a model A0 of ". 14
2. COMPACTNESS AND ELEMENTARY SUBMODELS
15
The model that we require is for the language L, so we take A to be the reduct of A0 to L.
< 0 1i, the set of all sentences of Example 6. Number theory is ThhN + <
< 0 1i, the standard model which we which are true in hN + < all learned in school. Theorem 3. (T. Skolem) There exist non-standard models of number theory. Proof. Add a new constant symbol c to L. Consider
L = f+ < 0 1g
z
n
}|
{
Th(hN + < 0 1i ) f1 + 1 + + 1 < c : n 2 N g and use the Compactness Theorem. The interpretation of the constant symbol c will not be a natural number. Definition 17. Two models A and A0 for L are said to be isomorphic whenever there is a bijection f : A ! A0 such that
1. for each n-placed relation symbol R of L and corresponding interpretations S of A and S 0 of A0 we have S (x1 : : : xn ) i S 0 (f (x1 ) : : : f (xn )) for all x1 : : : xn in A 2. for each n-placed function symbol F of L and corresponding interpretations G of A and G0 of A0 we have f (G(x1 : : : xn )) = G0 (f (x1 ) : : : f (xn )) for all x1 : : : xn in A 3. for each constant symbol c of L and corresponding constant elements a of A and a0 of A0 we have f (a) = a0 . We write A = A0 . This is an equivalence relation. Definition 18. Two models A and A0 for L are said to be elementarily equivalent whenever we have that for each sentence of L A j= i A0 j=
We write A A0 . This is another equivalence relation. Exercise 8. Suppose f : A ! A0 is an isomorphism and ' is a formula such that A j= ' a0 : : : ak ] for some a0 : : : ak from A then A0 j= ' f (a0 ) : : : f (ak )]. Use this to show that A = A0 implies A A0 . Definition 19. A model A0 is called a submodel of A i 6= A0 A and 1. each n-placed relation S 0 of A0 is the restriction to A0 of the corresponding relation S of A, i. e. S 0 = S \ (A0 )n 2. each m-placed function G0 of A0 is the restriction to A0 of the corresponding function G of A, i. e. G0 = G (A0 )m 3. each constant of A0 is the corresponding constant of A. We write A0 A. Definition 20. Let A and B be two models for L. We say A is an elementary submodel of B and B is an elementary extension of A and we write A B whenever 1. A B and
2. COMPACTNESS AND ELEMENTARY SUBMODELS
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2. for all formulas '(v0 : : : vk ) of L and all a0 : : : ak 2 A A j= ' a0 : : : ak ] i B j= ' a0 : : : ak ]: Exercise 9. Prove that if A B then A B and A B. Example 7. Let N be the usual natural numbers with < as the usual ordering. < i and A = hN n f0g<
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1. If t is the variable vi then both values are just ai . 2. If t is the constant symbol c then the values are equal because c has the same interpretation in A and in Ak . 3. If t is F (t1 : : : tm ) where F is a function symbol and t1 : : : tm are terms such that each value ti a0 : : : ap ] is the same in both A and Ak , then the value F (t1 : : : tm ) a0 : : : ap ] in A is G(t1 a0 : : : ap ] : : : tm a0 : : : ap ]) where G is the interpretation of F in A and the value of F (t1 : : : tm ) a0 : : : ap ] in Ak is Gk (t1 a0 : : : ap ] : : : tm a0 : : : ap ]) where Gk is the interpretation of F in Ak . But Gk is the restriction of G to Ak so these values are equal. In order to show that each Ak A it will su!ce to prove the following statement for each formula '(v0 : : : vp ) of L. \ For all k 2 N and all a0 : : : ap in Ak : A j= ' a0 : : : ap ] i Ak j= ' a0 : : : ap ]:" Claim. The statement is true whenever ' is t1 = t2 where t1 and t2 are terms. Proof of Claim. Fix k 2 N and a0 : : : ap in Ak . A j= (t1 = t2 ) a0 : : : ap ] i t1 a0 : : : ap ] = t2 a0 : : : ap ] in A i t1 a0 : : : ap ] = t2 a0 : : : ap ] in Ak i Ak j= (t1 = t2 ) a0 : : : ap ]: Claim. The statement is true whenever ' is R(t1 : : : tn ) where R is a relation symbol and t1 : : : tn are terms. Proof of Claim. Fix k 2 N and a0 : : : ap in Ak . Let S be the interpretation of R in A and Sk be the interpretation in Ak Sk is the restriction of S to Ak . A j= R(t1 : : : tn ) a0 : : : ap ] i S (t1 a0 : : : ap ] : : : tn a0 : : : ap ]) i Sk (t1 a0 : : : ap ] : : : tn a0 : : : ap ]) i Ak j= R(t1 : : : tn ) a0 : : : ap ] Claim. If the statement is true when ' is , then the statement is true when
' is :.
Proof of Claim. Fix k 2 N and a0 : : : ap in Ak . A j= (:) a0 : : : ap ] i not A j= a0 : : : ap ] i not Ak j= a0 : : : ap ] i Ak j= (:) a0 : : : ap ]:
2. COMPACTNESS AND ELEMENTARY SUBMODELS
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Claim. If the statement is true when ' is 1 and when ' is 2 then the statement is true when ' is 1 ^ 2 . Proof of Claim. Fix k 2 N and a0 : : : ap in Ak . A j= (1 ^ 2 ) a0 : : : ap ] i A j= 1 a0 : : : ap ] and A j= 2 a0 : : : ap ] i Ak j= 1 a0 : : : ap ] and Ak j= 2 a0 : : : ap ] i Ak j= (1 ^ 2 ) a0 : : : ap ]: Claim. If the statement is true when ' is then the statement is true when '
is 9vi .
Proof of Claim. Fix k 2 N and a0 : : : ap in Ak . Note that
A = fAj : j 2 Ng.
A j= 9vi a0 : : : ap ] i A j= 9vi a0 : : : aq ] where q is the maximum of i and p (by Lemma 2),
i A j= a0 : : : ai;1 a ai+1 : : : aq ] for some a 2 A i A j= a0 : : : ai;1 a ai+1 : : : aq ] for some a 2 Al for some l k i Al j= a0 : : : ai;1 a ai+1 : : : aq ] since the statement is true for , i Al j= 9vi a0 : : : aq ] i Ak j= 9vi a0 : : : aq ] since Ak Al i Ak j= 9vi a0 : : : ap ] (by Lemma 2): By induction on the complexity of ', we have proven the statement for all formulas ' which do not contain the connectives _, ! and $ or the quantier 8. To verify the statement for all ' we use Lemma 3. Let ' be any formula of L. By Lemma 3 there is a formula which does not use _, !, $ nor 8 such that j= (8v0 ) : : : (8vp )(' $ ): Now x k 2 N and a0 : : : ap in Ak . We have A j= (' $ ) a0 : : : ap ] and Ak j= (' $ ) a0 : : : ap ]: A j= ' a0 : : : ap ] i A j= a0 : : : ap ] i Ak j= a0 : : : ap ] i Ak j= ' a0 : : : ap ] which completes the proof of the theorem. Lemma 5. (The Tarski-Vaught Condition) Let A and B be models for L with A B. The following are equivalent: (1) A B
2. COMPACTNESS AND ELEMENTARY SUBMODELS
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(2) for any formula (v0 : : : vq ) and any i q and any a0 : : : aq from A: if there is some b 2 B such that B j= a0 : : : ai;1 b ai+1 : : : aq ] then we have some a 2 A such that B j= a0 : : : ai;1 a ai+1 : : : aq ]: Proof. Only the implication (2) ) (1) requires a lot of proof. We will prove that for each formula '(v0 : : : vp ) and all a0 : : : ap from A we will have: A j= ' a0 : : : ap ] i B j= ' a0 : : : ap ] by induction on the complexity of ' using only the negation symbol :, the connective ^ and the quantier 9 (recall Lemma 3). 1. The cases of formulas of the form t1 = t2 and R(t1 : : : tn ) come immediately from the fact that A B. 2. For negation: suppose ' is : and we have it for , then A j= ' a0 : : : ap ] i not A j= a0 : : : ap ] i not B j= a0 : : : ap ] i B j= ' a0 : : : ap ]: 3. The ^ case proceeds similarly. 4. For the 9 case we consider ' as 9vi . If A j= 9vi a0 : : : ap ], then the inductive hypothesis for and the fact that A B ensure that B j= 9vi a0 : : : ap ]. It remains to show that if B j= ' a0 : : : ap ] then A j= ' a0 : : : ap ]. Assume B j= 9vi a0 : : : ap ]. By Lemma 2, B j= 9vi a0 : : : aq ] where q is the maximum of i and p. By the denition of satisfaction, there is some b 2 B such that B j= a0 : : : ai;1 b ai+1 : : : aq ]: By (2), there is some a 2 A such that B j= a0 : : : ai;1 a ai+1 : : : aq ]: By the inductive hypothesis on , for that same a 2 A, A j= a0 : : : ai;1 a ai+1 : : : aq ]: By the denition of satisfaction, A j= 9vi a0 : : : aq ]: Finally, by Lemma 2, A j= a0 : : : ap ]. Recall that jBj is used to represent the cardinality, or size, of the set B. Note that since any language L contains innitely many variables, jLj is always innite, but may be countable or uncountable depending on the number of other symbols. We often denote an arbitrary innite cardinal by the lower case Greek letter . Theorem 5. (Downward L owenheim-Skolem Theorem) Let B be a model for L and let be any cardinal such that jLj < jBj. Then B has an elementary submodel A of cardinality . Furthermore if X B and jX j , then we can also have X A.
2. COMPACTNESS AND ELEMENTARY SUBMODELS
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Proof. Without loss of generosity assume jX j = . We recursively dene sets Xn for n 2 N such that X = X0 X1 Xn and such that for each formula '(v0 : : : vp ) of L and each i p and each a0 : : : ap from Xn such that B j= 9vi ' a0 : : : ap ] we have x 2 Xn+1 such that B j= ' a0 : : : ai;1 x ai+1 : : : ap ]: Since jLj and each formula of L is a nite string of symbols from L, there are at most many formulas of L. So there are at most elements of B that need to be added to each Xn and so, without loss of generosity each jXn j = . Let A = fXn : n 2 Ng then jAj = . Since A is closed under functions from B and contains all constants from B, A gives rise to a submodel A B. The Tarski-Vaught Condition is used to show that A B. Theorem 6. (The Upward L owenheim-Skolem Theorem) Let A be an innite model for L and be any cardinal such that jLj and jAj < . Then A has an elementary extension of cardinality . Proof. For each a 2 A, let ca be a new constant symbol let L0 = L fca : a 2 Ag: Note that sentences of L0 are just formulas of L with all free variables replaced by constant symbols. In addition, add many new constant symbols to L0 to make L00 . Dene " to be the following set of sentences of L00 : f:d = d0 : d and d0 are distinct new constant symbols of L00 n L0 g f : is a sentence of L0 obtained from the formula '(v0 : : : vp ) of L by replacing each free variable vi by the constant symbol cai and A j= ' a0 : : : ap ]g By interpreting ca as a and the d's as distinct elements of A we can transform A into a model of any nite subset of ". Using the Compactness Theorem, we obtain a model D00 for L00 such that D00 j= ". Note that D00 has size at least and that for any sentence of L0 , D00 j= i
2 ". Obtain a model C00 for L00 from D00 by simply switching elements of the universe of D00 with A to ensure that for each a 2 A the interpretation of ca in C00 is a. Hence the universe of C00 contains A and C00 j= ". Let C be the reduct of C00 to L. The following argument will show that A C. Let ' be any formula of L and a0 : : : ap any elements from A. Let be the sentence of L0 formed by replacing free occurances of vi with cai . We have A j= ' a0 : : : ap ] i 2 " i D00 j=
i C00 j=
i C j= ' a0 : : : ap ]: However, C may have size strictly larger than . In this case we obtain our nal B by using the previous theorem to get B C with A B. It is now straightforward to conclude that A B.
2. COMPACTNESS AND ELEMENTARY SUBMODELS
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Definition 23. A theory T for a language L is said to be complete whenever for each sentence of L either T j= or T j= : . Lemma 6. A theory T for L is complete i any two models of T are elementarily equivalent. Proof. ()) easy. (() easy. Definition 24. A theory T is said to be categorical in cardinality whenever any two models of T of cardinality are isomorphic. We also say that T is categorical. The most interesting cardinalities in the context of categorical theories are @0 , the cardinality of countably innite sets, and @1, the rst uncountable cardinal. Exercise 11. Show that DLO is @0 -categorical. There are two well-known proofs. One uses a back-and-forth construction of an isomorphism. The other constructs, by recursion, an isomorphism from the set of dyadic rational numbers between 0 and 1: n f m : m is a positive integer and n is an integer 0 < n < 2m g 2 onto a countable dense linear order without endpoints. Now use the following theorem to show that DLO is complete. Theorem 7. (The L o s-Vaught Test) Suppose that a theory T has only innite models for a language L and that T is -categorical for some cardinal jLj. Then T is complete. Proof. We will show that any two models of T are elementarily equivalent. Let A of cardinality 1 , and B of cardinality 2 , be two models of T . If 1 < use the Upward Lowenheim-Skolem Theorem to get A0 such that 0 jA j = and A A0 . If 1 > use the Downward Lowenheim-Skolem Theorem to get A0 such that 0 jA j = and A0 A. Either way, we can get A0 such that jA0 j = and A0 A. Similarly, we can get B0 such that jB0 j = and B0 B. Since T is -categorical, A0 = B0 . Hence A B. Recall that the characteristic of a eld is the prime number p such that z
p
}|
{
1+1+ +1 = 0 provided that such a p exists, and, if no such p exists the eld has characteristic 0. All of our best-loved elds: Q , R and C have characteristic 0. On the other hand, elds of characteristic p include the nite eld of size p (the prime Galois eld). Theorem 8. The theory of algebraically closed elds of characteristic 0 is complete. Proof. We use the L os-Vaught Test and the following Lemma. Lemma 7. Any two algebraically closed elds of characteristic 0 and cardinality @1 are isomorphic.
2. COMPACTNESS AND ELEMENTARY SUBMODELS
22
Proof. Let A be such a eld containing the rationals Q = hQ + 0 1i as a prime subeld. In a manner completely analogous to nding a basis for a vector space, we can nd a transcendence basis for A, that is, an indexed subset fa : 2 I g A such that A is the algebraic closure of the subeld A0 generated by fa : 2 I g but no a is in the algebraic closure of the subeld generated by the rest: fa : 2 I and 6= g. Since the subeld generated by a countable subset would be countable and the algebraic closure of a countable subeld would also be countable, we must have that the transcendence base is uncountable. Since jAj = @1 , the least uncountable cardinal, we must have in fact that jI j = @1 . Now let B be any other algebraically closed eld of characteristic 0 and size @1 . As above, obtain a transcendence basis fb : 2 J g with jJ j = @1 and its generated subeld B0 . Since jI j = jJ j, there is a bijection g : I ! J which we can use to build an isomorphism from A to B. Since B has characteristic 0, a standard theorem of algebra gives that the rationals are isomorphically embedded into B. Let this embedding be: f : Q ,! B: We extend f as follows: for each 2 I , let f (a ) = bg() , which maps the transcendence basis of A into the transcendence basis of B. We now extend f to map A0 onto B0 as follows: Each element of A0 is given by p(a1 : : : am ) q(a1 : : : am ) where p and q are polynomials with rational coe!cients and the a's come, of course, from the transcendence basis. Let f map such an element to p#(bg(1 ) : : : bg(m ) ) q#(bg(1 ) : : : bg(m ) ) where p# and q# are polynomials whose coe!cients are the images under f of the rational coe!cients of p and q. The nal extension of f to all of A and B comes from the uniqueness of algebraic closures. Remark. Lemma 7 is also true when 0 is replaced by any xed characteristic and @1 by any uncountable cardinal. Theorem 9. Let H be a set of sentences in the language of eld theory which are true in algebraically closed elds of arbitrarily high characteristic. Then H holds in some algebraically closed eld of characteristic 0. Proof. A eld is a model in the language f+ 0 1g of the axioms of eld theory. Let ACF be the set of axioms for the theory of algebraically closed elds see Example 5. For each n 2, let n denote the sentence z
n
}|
{
:(1 + 1 + + 1) = 0
Let " = ACF H fn : n 2g Let "0 be any nite subset of " and let m be the largest natural number such that m 2 "0 or let m = 1 by default.
2. COMPACTNESS AND ELEMENTARY SUBMODELS
23
Let A be an algebraically closed eld of characteristic p > m such that A j= H then in fact A j= "0 . So by compactness there is B such that B j= ". B is the required eld. Corollary 1. Let C denote, as usual, the complex numbers. Every one-toone polynomial map f : C m ! C m is onto. Proof. A polynomial map is a function of the form f (x1 : : : xm ) = hp1 (x1 : : : xm ) : : : pm(x1 : : : xm )i where each pi is a polynomial in the variables x1 : : : xm . We call max f degree of pi : i mg the degree of f . Let L be the language of eld theory and let mn be the sentence of L which expresses that \each polynomial map of m variables of degree < n which is one-toone is also onto". We wish to show that there are algebraically closed elds of arbitrarily high characteristic which satisfy H = fmn : m n 2 N g. We will then apply Theorem 9, Theorem 8, Lemma 6 and Exercise 6 and be nished. Let p be any prime and let Fp be the prime Galois eld of size p. The algebraic closure F~p is the countable union of a chain of nite elds Fp = A0 A1 A2 Ak Ak+1 obtained by recursively adding roots of polynomials. We nish the proof by showing thatmeach hF~pm + 0 1i satises H. Given any polynomial map f : (F~p ) ! (F~p ) which is one-to-one, we show that f is also onto. Given any elements b1 : : : bm 2 F~p , there is some Ak containing b1 : : : bm as well as all the coe!cients of f. Since f is one-to-one, f Am Amk ! Amk is a one-to-one polynomial map. k :m m Hence, since Ak is nite, f Ak is onto and so there are a1 : : : am 2 Ak such that f (a1 : : : am ) = hb1 : : : bmi. Therefore f is onto. Thus, for each prime number p and each m n 2 N , mn holds in a eld of characteristic p, i.e. hF~p + 0 1i satises H.
It is a signicant problem to replace \one-to-one" with \locally one-to-one".
CHAPTER 3
Diagrams and Embeddings Let A = hA Ii be a model for a language L and X A. Expand L to = L fca : a 2 X g by adding new constant symbols to L. We can expand A to a model AX = hA I 0 i for LX by choosing I 0 extending I such that I 0 (ca ) = a for each a 2 X . We sometimes write this as hA cx ix2X . We often deal with the case X = A, to obtain AA . Exercise 12. Let A and B be models for L with X A B. Prove: (i) if A B then AX BX . (ii) if A = B then AX = BX . (iii) if A B then AX BX . Hint: A j= ' a1 : : : ap ] i AA j= ' where ' is the sentence of LA formed by replacing each free occurance of vi with cai . Definition 25. Let A be a model for L. 1. The elementary diagram of A is ThAA , the set of all sentences of LA which hold in AA . 2. The diagram of A, denoted by 4A, is the set of all those sentences in ThAA without quantiers. Remark. There is a notion of atomic formula, which is a formula of the form t1 = t2 or R(t1 : : : tn ) where t1 : : : tn are terms. Sometimes 4A is dened to be the set of all atomic formulas and negations of atomic formulas which occur in ThAA . However this is not substantially dierent from Denition 25, since the reader can quickly show that for any model B, B j= 4A in one sense i B j= 4A in the other sense. Definition 26. A is said to be isomorphically embedded in B whenever 1. there is a model C and an isomorphism f such that f : A ! C and C B or 2. there is a model D and an isomorphism g such that A D and g : D ! B Exercise 13. Prove that, in fact, 1 and 2 are equivalent conditions. Definition 27. A is said to be elementarily embedded in B whenever 1. there is a model C and an isomorphism f such that f : A ! C and C B LX
24
3. DIAGRAMS AND EMBEDDINGS
25
2. there is a model D and an isomorphism g such that A D and g : D ! B Exercise 14. Again, prove that, in fact, 1 and 2 are equivalent. Theorem 10. (The diagram lemmas) Let A and B be models for L.
1. A is isomorphically embedded in B i B can be expanded to a model of 4A . 2. A is elementarily embedded in B i B can be expanded to a model of Th(AA ). Proof. We sketch the proof of 1. ()) If f is as in 1 of Denition 26 above, then hB f (a)ia2A j= 4A. (() If hB ba ia2A j= 4A , then let f (a) = ba . Exercise 15. Give a careful proof of part 2 of the theorem.
We now apply these notions to graph theory and to calculus. The natural language for graph theory has one binary relation symbol which we call E (to suggest the word \edge"), and the following two axioms: (8x)(8y)E (x y) $ E (y x) (8x):E (x x). A graph is, of course, a model of graph theory. Corollary 2. Every planar graph can be four coloured. Proof. We will have to use the famous result of Appel and Haken that every nite planar graph can be four coloured. Model Theory will take us from the nite to the innite. We recall that a planar graph is one that can be embedded, or drawn, in the usual Euclidean plane and to be four coloured means that each vertex of the graph can be assigned one of four colours in such a way that no edge has the same colour for both endpoints. Let A be an innite planar graph. Introduce four new unary relation symbols: R G B Y (for red, green, blue and yellow). We wish to prove that there is some expansion A0 of A such that A0 j= where is the sentence in the expanded language: (8x) R(x) _ G(x) _ B (x) _ Y (x)] ^ (8x) R(x) ! :(G(x) _ B (x) _ Y (x))] ^ : : : ^ (8x)(8y ):(R(x) ^ R(y ) ^ E (x y )) ^ which will ensure that the interpretations of R G B and Y will four colour the graph. Let " = 4A f g. Any nite subset of " has a model, based upon the appropriate nite subset of A. By the compactness theorem, we get B j= ". Since B j= , the interpretations of R G B and Y four colour it. By the diagram lemma A is isomorphically embedded in the reduct of B, and this isomorphism delivers the four-colouring of A. A graph with the property that every pair of vertices is connected with an edge is called complete. At the other extreme, a graph with no edges is called discrete. A very important theorem in nite combinatorics says that most graphs contain an
3. DIAGRAMS AND EMBEDDINGS
26
example of one or the other as a subgraph. A subgraph of a graph is, of course, a submodel of a model of graph theory. Corollary 3. (Ramsey's Theorem) For each n 2 N there is an r 2 N such that if G is any graph with r vertices, then either G contains a complete subgraph with n vertices or a discrete subgraph with n vertices. Proof. We follow F. Ramsey who began by proving an innite version of the theorem (also called Ramsey's Theorem). Claim. Each innite graph G contains either an innite complete subgraph or an innite discrete subgraph. Proof of Claim. By force of logical necessity, there are two possiblities: (1) there is an innite X G such that for all x 2 X there is a nite Fx X such that E (x y) for all y 2 X n Fx , (2) for all innite X G there is a x 2 X and an innite Y X such that :E (x y ) for all y 2 Y . If (1) occurs, we recursively pick x1 2 X , x2 2 X n Fx1 , x3 2 X n (Fx1 Fx2 ), etc, to obtain an innite complete subgraph. If (2) occurs we pick x0 2 G and Y0 G with the property and then recursively choose x1 2 Y0 and Y1 Y0 , x2 2 Y1 and Y2 Y1 and so on, to obtain an innite discrete subgraph.
We now use Model Theory to go from the innite to the nite. Let be the sentence, of the language of graph theory, asserting that there is no complete subgraph of size n. (8x1 : : : 8xn ) :E (x1 x2 ) _ :E (x1 x3 ) _ _ :E (xn;1 xn )]: Let be the sentence asserting that there is no discrete subgraph of size n. (8x1 : : : 8xn ) E (x1 x2 ) _ E (x1 x3 ) _ _ E (xn;1 xn )]: Let T be the set consisting of , and the axioms of graph theory. If there is no r as Ramsey's Theorem states, then T has arbitrarily large nite models. By Theorem 2, T has an innite model, contradicting the claim. The following theorem of A. Robinson nally solved the centuries old problem of innitesimals in the foundations of calculus. Theorem 11. (The Leibniz Principle) There is an ordered eld R called the hyperreals, containing the reals R and an innitesimal number such that any statement about the reals which holds in R also holds in R. < 0 1 i. We will make the statement of the theorem Proof. Let R be hR + < precise by proving that there is some model B, in the same language L as R and with the universe called R , such that R B and there is b 2 R such that 0 < b < a for each positive a 2 R.
3. DIAGRAMS AND EMBEDDINGS
27
For each real number a, we introduce a new constant symbol ca , and in addition another new constant symbol d. Let " be the set of sentences in the expanded language given by: ThRR f0 < d < ca : a is a positive real g We can obtain a model C j= " by the compactness theorem. Let C0 be the reduct of C to L. By the elementary diagram lemma R is elementarily embedded in C0 , and so there is a model B for L such that C0 = B and R B. Remark. This idea is extremely useful in understanding calculus. An element
x 2 R is said to be innitesimal whenever ;r < x < r for each positive r 2 R. 0 is innitesimal. Two elements x y 2 R are said to be innitely close, written x y whenever x ; y is innitesimal. Note: x is innitesimal i x 0. An element x 2 R is said to be nite whenever ;r < x < r for some positive r 2 R. Else it is
innite. Each nite x 2 R is innitely close to some real number, called the standard part of x, written st(x). To dierentiate f , for each M x 2 R generate M y = f (x+ M x) ; f (x). Then ; My 0 f (x) = st Mx whenever this exists and is the same for each innitesimal M x 6= 0. The increment lemma states that if y = f (x) is dierentiable at x and M x 0, then M y = f 0 (x) M x + " M x for some innitesimal ". Proofs of the usual theorems of calculus are now easier. The following theorem is considered one of the most fundamental results of mathematical logic. We give a detailed proof. Theorem 12. (Robinson Consistency Theorem) Let L1 and L2 be two languages with L = L1 \L2 . Suppose T1 and T2 are satisable theories in L1 and L2 respectively. Then T1 T2 is satisable i there is no sentence
of L such that T1 j= and T2 j= : . Proof. The direction ) is easy and motivates the whole theorem. We begin the proof in the ( direction. Our goal is to show that T1 T2 is satisable. The following claim is a rst step. Claim. T1 f sentences of L : T2 j= g is satisable. Proof of Claim. Using the compactness theorem and considering conjunctions, it su!ces to show that if T1 j= 1 and T2 j= 2 with 2 a sentence of L, then f 1 2 g is satisable. But this is true, since otherwise we would have 1 j= : 2 and hence T1 j= : 2 and so : 2 would be a sentence of L contradicting our hypothesis.
The basic idea of the proof from now on is as follows. In order to construct a model of T1 T2 we construct models A j= T1 and B j= T2 and an isomorphism f : AjL ! BjL between the reducts of A and B to the language L, witnessing that AjL = BjL. We then use f to carry over interpretations of symbols in L1 n L from A to B , giving an expansion B of B to the language L1 L2 . Then, since B jL1 = A and B jL2 = B we get B j= T1 T2 . The remainder of the proof will be devoted to constructing such an A, B and f . A and B will be constructed as unions of elementary chains of An's and Bn 's while f will be the union of fn : An ,! Bn .
3. DIAGRAMS AND EMBEDDINGS
28
We begin with n = 0, the rst link in the elementary chain. Claim. There are models A0 j= T1 and B0 j= T2 with an elementary embedding f0 : A0 jL ,! B0 jL. Proof of Claim. Using the previous claim, let A0 j= T1 f sentences of L : T2 j= g We rst wish to show that Th(A0 jL)A0 T2 is satisable. Using the compactness theorem, it su!ces to prove that if 2 Th(A0 jL)A0 then T2 f g is satisable. For such a let ca0 : : : can be all the constant symbols from LA0 n L which appear in . Let ' be the formula of L obtained by replacing each constant symbol cai by a new variable ui . We have A0 jL j= ' a0 : : : an ] and so A0 jL j= 9u0 : : : 9un ' By the denition of A0 , it cannot happen that T2 j= :9u0 : : : 9un ' and so there is some model D for L2 such that D j= T2 and D j= 9u0 : : : 9un '. So there are elements d0 : : : dn of D such that D j= ' do : : : dn ]. Expand D to a model D for L2 LA0 , making sure to interpret each cai as di . Then D j= , and so D j= T2 f g. Let B0 j= Th(A0 jL)A0 T2 . Let B0 be the reduct of B0 to L2 clearly B0 j= T2 . Since B0 jL can be expanded to a model of Th(A0 jL)A0 , the Elementary Diagram Lemma gives an elementary embedding f0 : A0 jL ,! B0 jL and nishes the proof of the claim. The other links in the elementary chain are provided by the following result. Claim. For each n 0 there are models An+1 j= T1 and Bn+1 j= T2 with an elementary embedding fn+1 : An+1 jL ,! Bn+1 jL such that An An+1 Bn Bn+1 fn+1 extends fn and Bn range of fn+1 :
A0
B0
#f0
A1
B1
#f1
An
An+1
Bn
Bn+1
#fn
#fn+1
The proof of this claim will be discussed shortly. Assuming the claim, let S S S A = n2N An , B = n2N Bn and f = n2N fn . The Elementary Chain Theorem gives that A j= T1 and B j= T2 . The proof of the theorem is concluded by simply verifying that f : AjL ! BjL is an isomorphism. The proof of the claim is long and quite technical it would not be inappropriate to omit it on a rst reading. The proof, of course, must proceed by induction on
3. DIAGRAMS AND EMBEDDINGS
29
n. The case of a general n is no dierent from the case n = 0 which we state and prove in some detail. Claim. There are models A1 j= T1 and B1 j= T2 with an elementary embedding f1 : A1 jL ,! B1 jL such that A0 A1 , B0 B1 , f1 extends f0 and B0 range of f1. A0 A1 #f0
#f1
B0 B1 Proof of Claim. Let A+0 be the expansion of A0 to the language L+1 = L1 fca : a 2 A0 g formed by interpreting each ca as a;2 A0 A+ 0 is just another notation for (A0 )A0 . The elementary diagram of A+0 is Th A+0 A+0 . Let B0 be the expansion of B0 jL to the language L = L fca : a 2 A0 g fcb : b 2 B0 g formed by interpreting each ca as f0 (a) 2 B0 and each cb as b 2 B0 . ;
We wish to prove that Th A+0 A+0 ThB0 is satisable. By the compactness ; theorem it su!ces to prove that Th A+0 A+0 f g is satisable for each in ThB0 . For such a sentence , let ca0 : : : cam cb0 : : : cbn be all those constant symbols occuring in but not in L. Let '(u0 : : : um w0 : : : wm ) be the formula of L obtained from by replacing each constant symbol cai by a new variable ui and each constant symbol cbi by a new variable wi . We have B0 j= so B0 jL j= ' f0 (a0 ) : : : f0 (am ) b0 : : : bn ] So B0 jL j= 9w0 : : : 9wn ' f0 (a0 ) : : : f0 (am )] Since f0 is an elementary embedding we have : A0 jL j= 9w0 : : : 9wn ' a0 : : : am ] Let '^(w0 : : : wn ) be the formula of L+1 obtained by replacing occurances of ui in '(u0 : : : um w0 : : : wn ) by cai then A+0 j= 9w0 : : : 9wn '^. So, of course, ; + A0 A+0 j= 9w0 : : : 9wn '^ and this means that there are d0 : : : dn in A+0 = A0 such that (A+0 )A+0 j= '^ d0 : : : dn ]: ; + We can now expand A0 A+0 to a model D by interpreting each cbi as di to obtain ; D j= and so Th A+0 A+0 f g is satisable. ; Let E j= Th A+0 A+0 ThB0 . By the elementary diagram lemma A+0 is elementarily embedded into EjL+1 . So there is a model A+1 for L+1 with A+0 A+1 and an isomorphism g : A+1 ! EjL+1 . Using g we expand A+1 to a model A01 isomorphic to E. Let A1 denote A01 jL we have A1 j= ThB0 . ;
We now wish to prove that Th (A1 )A1 Th B+0 B+0 is satisable, where B+0 is the common expansion of B0 and B0 to the language L2+ = L2 fca : a 2 A0 g fcb : b 2 B0 g:
3. DIAGRAMS AND EMBEDDINGS
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By the compactness theorem, it su!ces to show that ; Th B+0 B+0 f g
is satisable for each in Th (A1 )A1 . Let cx0 : : : cxn be all those constant symbols which occur in but are not in L . Let (u0 : : : un ) be the formula of L obtained from by replacing each cxi with a new variable ui . Since (A1 )A1 j= we have A1 j= x0 : : : xn ], and so A1 j= 9u0 : : : 9un : Also A1 j= ThB0 and ThB0 is a complete theory in the language L hence 9u0 : : : 9un is in ThB0 . Thus B0 j= 9u0 : : : 9un and so ; + B0 B+0 j= 9u0 : : : 9un and therefore there are b0 : : : bn in B+0 = B0 such that ; + B0 B+0 j= b0 : : : bn]: ;
We can now expand B+0 B+0 to a model F by interpreting each cxi as bi then ; F j= and Th B+0 B+0 f g is satisable. ; Let G j= Th (A1 )A1 Th B+0 B+0 . By the elementary diagram lemma B+0 is elementarily embedded into GjL+2 . So there is a model B+1 for L+2 with B+0 B+1 and an isomorphism h : B+1 ! GjL+2 . Using h we expand B+1 to a model B01 isomorphic to G. Let B1 denote B01 jL . Again by the elementary diagram lemma A1 is elementarily embedded into B1 . Let this be denoted by f1 : A1 ,! B1 : Let a 2 A0 we will show that f0 (a) = f1 (a). By denition we have B0 j= (v0 = ca ) f0 (a)] and so B+0 j= (v0 = ca ) f0 (a)]: Since B+0 B+1 , B+1 j= (v0 = ca ) f0 (a)] and so B1 j= (v0 = ca ) f0 (a)]: Now A+0 j= (ca = v1 ) a] and A+0 A+1 so A+1 j= (ca = v1 ) a] so A1 j= (ca = v1 ) a]: Since f1 is elementary, B1 j= (ca = v1 ) f1 (a)] so B1 j= (v0 = v1 ) f0 (a) f1 (a)] and so f0 (a) = f1(a). Thus f1 extends f0. Let b 2 B0 we will prove that b = f1 (a) for some a 2 A1 . By denition we have: B0 j= (v0 = cb ) b] so B+0 j= (v0 = cb ) b]. Since B+0 B+1 B+1 j= (v0 = cb ) b] so B1 j= (v0 = cb ) b]. On the other hand, since (9v1 )(v1 = cb ) is always satised, we have: A1 j= (9v1 )(v1 = cb ) so there is a 2 A1 such that A1 j= (v1 = cb ) a]: Since f1 is elementary, B1 j= (v1 = cb ) f1 (a)] so B1 j= (v0 = v1 ) b f1 (a)] so b = f1(a). Thus B0 range of f1 . We now let A1 be A+1jL1 and let B1 be B+1 jL2 . We get A0 A1 and B0 B1 and f1 : A1 jL ! B1 jL remains an elementary embedding. This completes the proof of the claim. Exercise 16. The Robinson Consistency Theorem was originally stated as:
3. DIAGRAMS AND EMBEDDINGS
31
Let T1 and T2 be satisable theories in languages L1 and L2 respectively and let T T1 \T2 be a complete theory in the language L1 \ L2 . Then T1 T2 is satisable in the language L1 L2 . Show that this is essentially equivalent to our version in Theorem 12 by rst proving that this statement follows from Theorem 12 and then also proving that this statement implies Theorem 12. Of course, for this latter argument you are looking for a proof much shorter than our proof of Theorem 12 however it will help to use the rst claim of our proof in your own proof. Theorem 13. (Craig Interpolation Theorem) Let ' and be sentences such that ' j= . Then there exists a sentence , called the interpolant, such that ' j= and j= and every relation, function or constant symbol occuring in also occurs in both ' and . Exercise 17. Show that the Craig Interpolation Theorem follows quickly from the Robinson Consistency Theorem. Also, use the Compactness Theorem to show that Theorem 12 follows quickly from Theorem 13.
CHAPTER 4
Model Completeness The quantier 8 is sometimes said to be the universal quantier and the quantier 9 to be the existential quantier. A formula ' is said to be quantier free whenever no quantiers occur in '. A formula ' is said to be universal whenever it is of the form 8x0 : : : 8xk where is quantier free. A formula ' is said to be existential whenever it is of the form 9x0 : : : 9xk where is quantier free. A formula ' is said to be universal-existensial whenever it is of the form 8x0 : : : 8xk 9y0 : : : 9yk where is quantier free. We extend these notions to theories T whenever each axiom of T has the property. Remark. Note that each quantier free formula ' is trivially equialent to the existential formula 9vi ' where vi does not occur in '. Exercise 18. Let A and B be models for L with A B. Verify the following four statements: (i) A B i BA j= Th(AA ) i AA j= Th(BA ): (ii) A B i BA j= 4A i BA j= for each quantier free of Th(AA ): (iii) A B i BA j= for each existential of Th(AA ): (iv) A B i AA j= for each universal of Th(BA ): Definition 28. A model A of a theory T is said to be existentially closed if whenever A B and B j= T , we have AA j= for each existential of Th(BA ): Remark. If A is existentially closed and A0 = A then A0 is also existentially closed. Definition 29. A theory T is said to be model complete whenever T 4A is complete in the language LA for each model A of T . Theorem 14. ( A. Robinson ) Let T be a theory in the language L. The following are equivalent: (1) T is model complete, (2) T is existentially complete, i.e. each model of T is existentially closed. (3) for each formula '(v0 : : : vp ) of L there is a universal formula (v0 : : : vp ) such that T j= (8v0 : : : 8vp )(' $ ) (4) for all models A and B of T , A B implies A B. Proof. (1) ) (2): Let A j= T and B j= T with A B. Clearly AA j= 4A it is also easy to see that BA j= 4A. Now by (1), T 4A is complete and both AA and BA are models of this theory so they are elementarily equivalent. 32
4. MODEL COMPLETENESS
33
So let be any sentence of LA (existential or otherwise). If BA j= then AA j= and (2) follows. (2) ) (3): Lemma 4 shows that it su!ces to prove it for formulas ' in prenex normal form. We do this by induction on the prenex rank of ' which is the number of alternations of quantiers in '. The rst step is prenex rank 0. Where only universal quantiers are present the result is trivial. The existential formula case is non-trivial it is the following claim: Claim. For each existential formula '(v0 : : : vp ) of L there is a universal formula (v0 : : : vp ) such that T j= (8v0 ) : : : (8vp )(' $ ) Proof of Claim. Add new constant symbols c0 : : : cp to L to form L = L fc0 : : : cp g and to form a sentence ' of L obtained by replacing each instance of vi in ' with the corresponding ci ' is an existential sentence. It su!ces to prove that there is a universal sentence of L such that T j= ' $ . Let ; = funiversal sentences of L such that T j= ' ! g We hope to prove that there is some 2 ; such that T j= ! ' . Note, however, that any nite conjunction 1 ^ 2 ^ ^ n of sentences from ; is equivalent to a sentence in ; which is simply obtained from 1 ^ 2 ^ ^ n by moving all the quantiers to the front. Thus it su!ces to prove that there are nitely many sentences 1 2 : : : n from ; such that T j= 1 ^ 2 ^ ^ n ! ' : If no such nite set of sentences existed, then each T f1 2 : : : n g f:' g
would be satisable. By the compactness theorem, T ; f:' g would be satisable. Therefore it just su!ces to prove that T ; j= ' . In order to prove that T ; j= ' , let A be any model of T ;. Form LA as usual but ensure that fc0 : : : cp g fca : a 2 Ag so that L LA . Let " = T f' g 4A: We wish to show that " is satisable. By the compactness theorem it su!ces to consider T f' g where is a conjunction of nitely many sentences of 4A. Let be the formula obtained from by exchanging each constant symbol from LA n L occuring in for a new variable ua . So A j= 9ua0 : : : 9uam (ua0 : : : uam ): But then A is not a model of the universal sentence 8ua0 : : : 8uam :(ua0 : : : uam ). Recalling that A j= ;, we are forced to conclude that this universal sentence is not in ; and so not a consequence of T f' g. So T f' g f9ua0 : : : 9uam (ua0 : : : uam )g must be satisable, and any model of this can be expanded to a model of T f' g. Thus " is satisable for the language LA .
4. MODEL COMPLETENESS
34
Let C j= ". By the diagram lemma, there is a model B for L such that AjL B and B = CjL. Thus both AjL and B are models of T . Furthermore BA = C so that BA j= ' . We now use (2) to get that (AjL)A j= ' . But (AjL)A is just AA and so A j= ' . This means T ; j= ' and nishes the proof of the claim. We will now do the general cases for the proof of the induction on prenex rank. There are two cases, corresponding to the two methods available for increasing the number of alternations of quantiers: (a) the addition of universal quantiers (b) the addition of existential quantiers. For the case (a), suppose '(v0 : : : vp ) is 8w0 : : : 8wm (v0 : : : vp w0 : : : wm ) and has prenex rank lower than ' so that we have by the inductive hypothesis that there is a quantier free formula (v0 : : : vp w0 : : : wm x0 : : : xn ) with new variables x0 : : : xn such that T j= (8v0 : : : 8vp 8w0 : : : 8wm )( $ 8x0 : : : 8xn ) Therefore, case (a) is concluded by noticing that this gives us T j= (8v0 : : : 8vp )(8w0 : : : 8wm $ 8w0 : : : 8wm 8x0 : : : 8xn ): Exercise 19. Check this step using the denition of satisfaction. For case (b), suppose '(v0 : : : vp ) is 9w0 : : : 9wn (v0 : : : vp w0 : : : wm ) and has prenex rank less than '. Here we will use the inductive hypothesis on : which of course also has prenex rank less than '. We obtain a quantier free formula (v0 : : : vp w0 : : : wm x0 : : : xn ) with new variables x0 : : : xn such that T j= (8v0 : : : 8vp 8w0 : : : 8wm )(: $ 8x0 : : : 8xn ) So T j= (8v0 : : : 8vp )(8w0 : : : 8wm : $ 8w0 : : : 8wm 8x0 : : : 8xn ) And T j= (8v0 : : : 8vp)(9w0 : : : 9wm $ 9w0 : : : 9wm 9x0 : : : 9xn :) Now 9w0 : : : 9wm 9x0 : : : 9xn : is an existential formula, so by the claim there is a universal formula such that T j= (8v0 : : : 8vp )(9w0 : : : 9wm 9x0 : : : 9xn : $ ): Hence T j= (8v0 : : : 8vp )(9w0 : : : 9wn $ ) which is the nal result which we needed. (3) ) (4) Let A j= T and B j= T with A B. Let ' be a formula of L and let a0 : : : ap be in A such that B j= ' a0 : : : ap ] Obtain a universal formula such that T j= (8v0 : : : 8vp )(' $ ) so B j= a0 : : : ap ] Since A B A j= a0 : : : ap ]
4. MODEL COMPLETENESS
35
and so A j= ' a0 : : : ap ] and A B. (4) ) (1): Let A j= T . We will show that T 4A is complete by showing that for any model B for LA , B j= T 4A implies B j= Th(AA ). Letting B be such a model, we note that BjL, the restriction of B to L, can be expanded to B, a model of 4A. So by the diagram lemma A is isomorphically embedded in BjL. Furthermore, by checking the proof of the diagram lemma we can ensure that there is an f : A ,! BjL such that for each a 2 A, f (a) is the interpretation of ca in B. (Recall that LA = L fca : a 2 Ag). Moreover, as in Exercise 13, there is a model D for L such that A D and an isomorphism g : D ! BjL with the property that for each a 2 A, g(a) is the interpretation of ca in B. Now let 2 Th(AA ), so that AA j= . Let '(u0 : : : uk ) be the formula of L obtained by replacing each occurance of cai in by the new variable ui . We have A j= ' a0 : : : ak ] Since A D we can use (4) to get A D and so we have D j= ' a0 : : : ak ]. With the isomorphism g we get that BjL j= ' g(a0 ) : : : g(ak )] and since g(ai ) is the interpretation of cai in B we have B j= . Thus B j= Th(AA ) and this proves (1). Example 9. We will see later that the theory ACF is model complete. But ACF is not complete because the characteristic of the algebraically closed feild can vary among models of ACF and the assertion that \I have characteristic p" can easily be expressed as a sentence of the language of ACF. Exercise 20. Suppose that T is a model complete theory in L and that either 1. any two models of T are isomorphically embedded into a third or 2. there is a model of T which is isomorphically embedded in any other. Then prove that T is complete. Example 10. Let N be the usual natural numbers and < the usual ordering. Let B = hN
4. MODEL COMPLETENESS
36
Proof. W.L.O.G. we assume that T is satisable. We use conditions (1) and (2) to prove the following: Claim. T has existentially closed models of each innite size . Proof of Claim. By the Lowenheim-Skolem Theorems we get A0 j= T with jA0 j = . We recursively construct a chain of models of T of size A0 A1 : : : An An+1 with the property that if B j= T and An+1 B and is an existential sentence of Th(BAn ), then (An+1 )An j= . Suppose An is already constructed we will construct An+1. Let "n be a maximally large set of existential sentences of LAn such that for each nite "0 "n there is a model C for LAn such that C j= "0 T 4An By compactness T "n 4An has a model D and without loss of generosity An D. By the Downward Lowenheim-Skolem Theorem we get E such that An E, jEj = and E D. Let An+1 = EjL we will show that An+1 has the required properties. Since E D, E j= T 4An and so A An+1 (See Exercise 18). Let B j= T with An+1 B and be an existential sentence of Th(BAn ) we will show that (An+1 )An j= . Since "n consists of existential sentences and D E (An+1 )An BAn we have (see Exercise 18) that BAn j= "n . The maximal property of "n then forces to be in "n because if 2= "n then there must be some nite "0 "n for which there is no C such that C j= "0 f g T 4An but BAn is such a C! Now since 2 "n and E D j= "n we must have E = (An+1 )An j= : Now let A be the union of the chain. By hypothesis A j= T . It is easy to check that jAj = . To check that A is existentially closed, let B j= T with A B and let be an existential sentence of ThBA . Since can involve only nitely many constant symbols, is a sentence of LAn for some n 2 N . Thus An+1 A B gives that (An+1 )An j= . Since is existential (see Exercise 18 again) we get that A j= . This completes the proof of the claim. We now claim that T is model complete using Theorem 14 by showing that every model A of T is existentially closed. There are three cases to consider: 1. jAj = 2. jAj > 3. jAj < where T is -categorical. Case (1). Let A be an existentially closed model of T of size . Then there is an isomorphism f : A ! A . Hence A is existentially closed. Case (2). Let be an existential sentence of LA and B j= T such that A B and BA j= . Let X = fa 2 A : ca occurs in g. By the Downward LowenheimSkolem Theorem we can nd A0 such that A0 A, X A0 and jA0 j = . Now by Case (1) A0 is existentially closed and we have A0 B and in LA0 so A0A0 j= . But since 2 Th(A0A0 ) and A0 A we have AA j= .
4. MODEL COMPLETENESS
37
Case (3). Let and B be as in case (2). By the Upward Lowenheim-Skolem Theorem we can nd A0 such that A A0 and jA0j = . By case (1) A0 is existentially closed. Claim. There is a model B0 such that A0 B0 and BA B0A . Assuming this claim, we have B0 j= T and B0A j= and by the fact that A0 is existentially closed we have A0A0 j= . Since A A0 we have AA j= . The following lemma implies the claim and completes the proof of the theorem. Lemma 8. Let A, B and A0 be models for L such that A B and A A0 . Then there is a model B0 for L such that A0 B0 and BA B0A . Proof. Let A, B, A0 and L be as above. Note that since A B we have
BA j= 4A and so AA BA . Let be a sentence from 4A0 . Let fdj : 0 j mg be the constant symbols from LA0 n LA appearing in . Obtain a quantier free formula '(u0 : : : um) of LA by exchanging each dj in with a new variable ui . Since A0A0 j= we have A0A j= 9u0 : : : 9um': Since A A0 we have AA A0A and so AA j= 9u0 : : : 9um ': Since AA BA , BA j= 9u0 : : : 9um ': Hence for some b0 : : : bm in B, BA j= ' b0 : : : bm ]. Expand BA to be a model BA for the w language LA fdj : 0 j mg by interpreting each dj as bj . Then BA j= and so Th(BA ) f g is satisable. This shows that ThBA " is satisable for each nite subset " 4A0 . By the Compactness Theorem there is a model C j= 4A0 ThBA . Using the Diagram Lemma for the language LA we obtain a model B0 for L such that A0A B0A and B0A = CjLA . Hence B0A j= ThBA and so B0A BA . Exercise 21. Suppose A A0 are models for L. Prove that for each sentence
of LA , if 4A0 j= then 4A j= .
Exercise 22. Prove that if T has a universal-existential set of axioms, then the union of a chain of models of T is also a model of T . Remark. The converse of this last exercise is also true it is called the ChangL os- Suszko Theorem. Theorem 16. The following theories are model complete: 1. dense linear orders without endpoints. (DLO) 2. algebraically closed elds. (ACF) Proof. (DLO): This theory has a universal existential set of axioms so that it is closed under unions of chains. It is @0 -categorical (by Exercise 11) so Lindstrom's test applies. (ACF): We rst prove that for any xed characteristic p, the theory of algebraically closed elds of characteristic p is model complete. The proof is similar to that for DLO, with @1 -categoricity (Lemma 7 ). Let A B be algebraically closed elds. They must have the same characteristic p. Therefore A B. Corollary 4. Any true statement about the rationals involving only the usual ordering is also true about the reals.
4. MODEL COMPLETENESS
38
<11 i and B = hR<<22 i where <1 and <2 are the usual Proof. Let A = hQ < orderings. The precise version of this corollary is: A B. This follows from Theorem 14 and Theorem 16 and the easy facts that A j= DLO, B j= DLO and A B. The reader will appreciate the power of these theorems by trying to prove A B directly, without using them. Corollary 5. (Hilbert's Nullstellensatz)
Let " be a nite system of polynomial equations and inequations in several variables with coecients in the eld A. If " has a solution in some eld extending A then " has a solution in the algebraic closure of A. Proof. Let be the existential sentence of the language LA which asserts the fact that there is a solution of ". Suppose " has a solution in a eld B with A B. Then BA j= . So B0A j= where B0 is the algebraic closure of B. Let A0 be the algebraic closure of A. Since A B, we have A0 B0 . By Theorem 16, ACF is model complete, so A0 B0 . Hence A0A B0A and 0 AA j= .
CHAPTER 5
The Seventeenth Problem We will give a complete proof later that RCF, the theory of real closed ordered elds, is model complete. However, by assuming this result now, we can give a solution to the seventeenth problem of the list of twenty-three problems of David Hilbert's famous address to the 1900 International Congress of Mathematicians in Paris. Corollary 6. (E. Artin) Let q(x1 : : : xn ) be a rational function with real coecients, which is positive definite. i.e. q(a1 : : : an ) 0 for all a1 : : : an 2 R Then there are nitely many rational functions with real coecients f1(x1 : : : xn ), : : : , fm (x1 : : : xn ) such that
q(x1 : : : xn ) =
m X j =1
(fj (x1 : : : xn ))2
We give a proof of this theorem after a sequence of lemmas. The rst lemma just uses calculus to prove the special case of the theorem in which q is a polynomial in only one variable. This result probably motivated the original question. Lemma 9. A positive denite real polynomial is the sum of squares of real polynomials. Proof. We prove this by induction on the degree of the polynomial. Let p(x) 2 R x] with degree deg(p) 2 and p(x) 0 for all real x. Let p(a) = minfp(x) : x 2 Rg, so p(x) = (x ; a)q(x) + p(a) and p0 (a) = 0 for some polynomial q. But p0 (a) = (x ; a)q0 (x) + q(x)]x=a = q(a) so q(a) = 0 and q(x) = r(x)(x ; a) for some polynomial r(x). So p(x) = p(a) + (x ; a)2 r(x): For all real x we have (x ; a)2 r(x) = p(x) ; p(a) 0: Since r is continuous, r(x) 0 for all real x, and deg(r) = deg(p) ; 2. So, by Pn induction r(x) = i=1 (ri (x))2 where each ri (x) 2 R x]. So p(x) = p(a) +
n X i=1 39
(x ; a)2 (ri (x))2 :
5. THE SEVENTEENTH PROBLEM
i.e. p(x) =
hp
i2
p(a) +
n X i=1
40
(x ; a)ri (x)]2 :
The following lemma shows why we deal with sums of rational functions rather than sums of polynomials. Lemma 10. x4 y2 + x2 y4 ; x2 y2 + 1 is positive denite, but not the sum of squares of polynomials. Proof. Let the polynomial be p(x y). A little calculus shows that the mini26 and conrms that p is positive denite. mum value of p is 27 Suppose
p(x y) =
l X i=1
(qi (x y))2
where qi (x y) are polynomials, each of which is the sum of terms of the form axm yn . First consider powers of x and the largest exponent m which can occur in any of the qi . Since no term of p contains x6 or higher powers of x, we see that we must have m 2. Considering powers of y similarly gives that each n 2. So each qi (x y) is of the form: ai x2 y2 + bi x2 y + ci xy2 + di x2 + ei y2 + fi xy + gi x + hi y + ki for some coe!cients ai bi ci di ei fi gi hi and ki . Comparing coe!cients of x4 y4 in p and the sum of the qi2 gives 0=
l X i=1
a2i
so each ai = 0. Comparing the coe!cients of x4 and y4 gives that each di = 0 = ei . Now comparing the coe!cients of x2 and y2 gives that each gi = 0 = hi . Now comparing the coe!cients of x2 y2 gives ;1 =
which is impossible.
l X i=1
fi2
The next lemma is easy but useful. Lemma 11. The reciprocal of a sum of squares is a sum of squares. Proof. For example A 2+ B 2 1 = A2 + B 2 = A2 + B 2 (A2 + B 2 )2 A2 + B 2 A2 + B 2 The following lemma is an algebraic result of E. Artin and O. Schreier, who invented the theory of real closed elds.
5. THE SEVENTEENTH PROBLEM
41
Let P0 = :
i=1
c2i ;
m X j =1
9 =
d2j b : l m 2 N ci 2 B dj 2 B not all zero
We claim that (1), (2), (3) P and (4) holdPfor P0 . (1) and (3) are obvious. In l 2 2 order to verify (2), note that if m j =1 dj b = i=1 ci , then by the previous lemma about reciprocals of sums of squares, b would be a sum of squares. Now (4) holds by denition of P0 , noting that c2i (;d2j b) = ;(ci dj )2 b and (;d2j b)(;d2k b) = (dj dk b)2 . We now construct larger and larger versions of P0 to take care of requirement (5). We do this in the following way. Suppose P0 P1 , P1 satises (1), (2), (3) and (4), and c 2= P1 . We dene P2 to be: fp(;c) : p is a polynomial with coe!cients in P1 g: It is easy to see that ;c 2 P2 , P1 P2 and that (1), (3) and (4) hold for P2 . To show that (2) holds for P2 we suppose that p(;c) = 0 and bring forth a contradiction. Considering even and odd exponents we obtain: p(x) = q(x2 ) + xr(x2 ) for some polynomials q and r with coe!cients in P1 . If r(c2 ) = 0 then q(c2 ) = 0. But q(c2 ) is in P1 , which is a contradiction. On the other hand, if r(c2 ) 6= 0 then 0 = p(;c) = q(c2 ) ; cr(c2 ) which means that 2 1 2 2 c = q(c ) r(c ) r(c2 ) and since each of the factors on the right hand side is in P1 we get a contradiction. Now we need:
5. THE SEVENTEENTH PROBLEM
42
Lemma 13. Every ordered eld can be embedded as a submodel of a real closed ordered eld. Proof. It su!ces to prove that for every ordered eld A there is an ordered eld B such that A B and for each natural number n 1, B j= n where n is the sentence in the language of eld theory which formally states: If p is a polynomial of degree at most n and w < y such that p(w) < 0 < p(y) then there is an x such that w < x < y and p(x) = 0. Consider the statement called IH(n): For any ordered eld E there is an ordered eld F such that E F and E j= n . IH(1) is true since any ordered eld E j= 1 . We will prove below that for each n, IH(n) implies IH(n + 1). Given our model A j= ORF, we will then be able to construct a chain of models:
A B1 B2 : : : Bn Bn+1 such that each Bn j= ORF f n g. Let B be the union of the chain. Since the theory ORF is preserved under unions of chains (see Exercise 22), B j= ORF. Furthermore, the nature of the sentences n allows us to conclude that for each n, B j= n and so B j= RCF. All that remains is to prove that for each n, IH(n) implies IH(n + 1). We rst make a claim: Claim. If E j= ORF f n g and p is a polynomial of degree at most n + 1 with coecients from E and a < d are in E such that p(a) < 0 < p(d) then there is a model F such that E F, F j= ORF and there is b 2 such that a < b < d and p(b) = 0. Let us rst see how this claim helps us to prove that IH(n) implies IH(n+1). Let E j= ORF we will use the claim to build a model F such that E F and F j= n+1 . We rst construct a chain of models of ORF E = E0 E1 : : : Em Em+1 such that for each m and each polynomial p of degree at most n +1 with coe!cients from Em and each pair of a, d of elements of Em such that p(a) < 0 < p(d) there is a b 2 Em+1 such that a < b < d and p(b) = 0. Suppose Em has been constructed we construct Em+1 as follows: let "m be the set of all existential sentences of LEm of the form (9x)(ca < x ^ x < cd ^ p(x) = 0) where p is a polynomial of degree at most n + 1 and such that ca , cd and the coe!cients of the polynomial p are constant symbols from LEm and (Em )Em j= p(ca ) < 0 ^ 0 < p(cd) We claim that ORF 4Em "m is satisable.
5. THE SEVENTEENTH PROBLEM
43
Using the Compactness Theorem, it su!ces to nd, for each nite subset of "m , a model C such that Em C and C j= ORF f1 : : : k g: By IH(n), obtain a model F1 such that Em F1 and F1 j= ORF f n g. By the claim, obtain a model F2 such that F1 F2 and F2 j= ORF f1 g. Again by IH(n), obtain F3 such that F2 F3 and F3 j= ORF f n g. Again by the claim, obtain F4 such that F3 F4 and F4 j= ORF f2 g. Continue in this manner, getting models of ORF Em F1 : : : F2k with each F2j j= j . Since each j is existential, we get that F2k is a model of each j (see Exercise 18). Let D j= ORF 4Em "m and then use the Diagram Lemma to get Em+1 such that Em Em+1 , Em+1 j= ORF and Em+1 j= "m , thus satisfying the required property concerning polynomials from Em . Let F be the union of the chain. Since ORF is a universal-existential theory, F j= ORF (see Exercise 22) and F j= n+1 by construction. So IH(n + 1) is proved. We now nish the entire proof by proving the claim. Proof of Claim. If p(x) = 0 for some x in E such that a < x < d then we can let F = E. Otherwise, introduce a new element b to E where the place of b in the ordering is given by: b = supremumft 2 E : t < d and p(t) < 0g: Note that continuity-style considerations show that b 6= d. We now show that since E j= n the polynomial p cannot be factored in E. Suppose p(x) = q(x) s(x). The denition of b allows us to nd a1 and d1 such that a1 < b < d1 and p(a1 ) < 0 < p(d1 ) and such that, other than possibly b, q and s have no roots in the interval of E between a1 and d1 . Now since p(a1 ) p(d1 ) < 0 either q(a1 ) q(d1 ) < 0 or s(a1 ) s(d1 ) < 0 and q and s each have degree n, forcing b to be an element of E. The fact that p is irreducible over E means that we can extend hE + 0 1i by quotients of polynomials in b of degree n to form a eld hF + 0 1i in the usual way. We leave the details to the reader. Note that the construction cannot force q(b) = 0 for any polynomial q(x) with coe!cients from E of degree n. This is because we could take such a q(x) of lowest degree and divide p(x) by it to get p(x) = q(x) s(x) + r(x) where degree of r < degree of q. This means that r(x) = 0 constantly and so p could have been factored over E. Now we must expand hF + 0 1 i to an ordered eld F while preserving the order of E. We are aided in this by the fact that if q is a polynomial of degree at most n with coe!cients from E then there are a1 and a2 in E such that a1 < b < a2 and q doesn't change sign between a1 and a2 this comes from the fact that E j= n . f1 : : : k g
5. THE SEVENTEENTH PROBLEM
44
Proof of the Corollary. Using Lemma 11 we see that it su!ces to prove the corollary for a polynomial p(x1 : : : xn ) such that p(a1 : : : an ) 0 for all a1 : : : an 2 R. Let B = hR(x1 : : : xn ) + 0 1i be the eld of \rational functions". Note that B contains the reduct of R to f+ 0 1g as a subeld, where R is dened as in Example 3 as the usual real numbers. By Lemma 12, if p is not the sum of squares in B, then we can nd an ordering
Hilbert also asked: If the coe!cients of a positive denite rational function are rational numbers (i.e. it is an element of Q (x1 : : : xn )) is it in fact the sum of squares of elements of Q (x1 : : : xn )? < 0 1i be The answer is \yes" and the proof is very similar. Let Q = hQ + : < the ordered eld of rationals as in Example 3. Lemma 12 holds for A = Q and B = hQ (x1 : : : xn ) + 0 1 i by Lemma 11 every positive rational number is the sum of squares since every positive integer is the sum of squares n = 1 + 1 + + 1. Exercise 23. Finish the answer to Hilbert's question by making any appropriate changes to the proof of the corollary. Hint: create an ordering on R(x1 : : : xn ) so that B0 and R are each submodels.
CHAPTER 6
Submodel Completeness Definition 30. A theory T is said to admit elimination of quantiers in L whenever for each formula '(v0 : : : vp ) of L there is a quantier free formula (v0 : : : vp ) such that: T j= (8v0 : : : 8vp )('(v0 : : : vp ) $ (v0 : : : vp )) Remark. There is a ne point with regard to the above denition. If ' is actually a sentence of L there are no free variables v0 : : : vp . So T j= ' $ for some quantier free formula with no free variables. But if L has no constant symbols, there are no quantier free formulas with no free variables. For this reason we assume that L has at least one constant symbol, or we restrict to those formulas ' with at least one free variable. This will become relevant in the proof of Theorem 17 for (2) ) (3). Exercise 24. If T admits elimination of quantiers in L and L has no constant symbols, show that for each sentence of L there is a quantier free formula (v0 ) such that T j= $ 8v0 $ 9v0 Definition 31. A theory T is said to be submodel complete whenever T 4A is complete in LA for each submodel A of a model of T . Exercise 25. Use Theorem 14 and the following theorem to nd four proofs that every submodel complete theory is model complete. Theorem 17. Let T be a theory of a language L. The following are equivalent: (1) T is submodel complete (2) If B and C are models of T and A is a submodel of both B and C, then every existential sentence which holds in BA also holds in CA . (3) T admits elimination of quantiers (4) whenever A B, A C, B j= T and C j= T there is a model D such that both BA and CA are elementarily embedded in DA . Proof. (1) ) (2) Let B j= T and C j= T with A B and A C. Then BA j= T 4A and CA j= T 4A. So (1) and Lemma 6 give BA CA . Thus (2) is in fact proved for all sentences, not just existential ones. (2) ) (3) Lemma 4 shows that it su!ces to prove (3) for formulas in prenex normal form. We do this by induction on the prenex rank of '. This claim is the rst step. Claim. For each existential formula '(v0 : : : vp ) of L there is a quantier free formula '(v0 : : : vp ) such that T j= (8v0 : : : 8vp )(' $ ) 45
6. SUBMODEL COMPLETENESS
46
Proof of Claim. Add new constant symbols c0 : : : cp to L to form L = L fc0 : : : cp g
and to form a sentence ' of L obtained by replacing each instance of vi in ' with the corresponding ci ' is an existential sentence. It su!ces to prove that there is a quantier free sentence of L such that T j= ' $ : Let S = f quantier free sentences of L : T j= ' ! g: It su!ces to nd some in S such that T j= ! ' . Since a nite conjunction of sentences of S is also in S , it su!ces to nd 1 : : : n in S such that T j= 1 ^ ^ n ! ' : If no such nite subset f 1 : : : n g of S exists, then each T f 1 : : : n g f:' g
would be satisable. So it su!ces to prove that T S j= ' . Let C j= T S with the intent of proving that C j= ' . Let A be the least submodel of C in the sense of the language L . That is, every element of A is the interpretation of a constant symbol from L fc0 : : : cp g or built from these using the functions of C. We can ensure that L LA . Let P = f of 4A : is a sentence of L g: We wish to show that T f' g P is satisable. By compactness, it su!ces to consider T f' g where is a sentence in P . If this set is not satisable then T j= ' ! : so that by denition of S we have : 2 S and hence C j= : . But this is impossible since A C means that CA j= 4A. Let B0 j= T f' g P . The interpretations of fc0 : : : cp g generate a submodel of B0 isomorphic to A. So there is a model B for L such that B = B0 and A B. In order to invoke (2) we use the restrictions AjL, BjL and CjL of A, B and C to the language L. We have BjL j= T , CjL j= T , AjL BjL and AjL CjL. ' is an existential sentence of L LA and since B0 j= ' we have (BjL)A j= ' . So by (2), (CjL)A j= ' and nally C j= ' which completes the proof of the claim. We now do the general cases for the proof of the induction on prenex rank. There are two cases, corresponding to the two methods available for increasing the number of alternations of quantiers: (a) the addition of universal quantiers (b) the addition of existential quantiers. For case (a), suppose '(v0 : : : vp ) is 8w0 : : : 8wm(v0 : : : vp w0 : : : wm ) and has prenex rank lower than '. Then : also has prenex rank lower than ' and we can use the inductive hypothesis on : to obtain a quantier free formula 1 (v0 : : : vp w0 : : : wm ) such that T j= (8v0 : : : 8vp )(8w0 : : : 8wm )(: $ 1 ) So T j= (8v0 : : : 8vp )(9w0 : : : 9wm : $ 9w0 : : : 9wm 1 )
6. SUBMODEL COMPLETENESS
47
By the claim there is a quantier free formula 2 (v0 : : : vp ) such that T j= (8v0 : : : 8vp )(9w0 : : : 9wm 1 $ 2 ) So T j= (8v0 : : : 8vp )(9w0 : : : 9wm : $ 2 ) So T j= (8v0 : : : 8vp )(8w0 : : : 8wm $ :2 ) and so :2 is the quantier free formula equivalent to '. For case (b), suppose '(v0 : : : vp ) is 9w0 : : : 9wm (v0 : : : vp w0 : : : wm ) and has prenex rank lower than '. We use the inductive hypothesis on to obtain a quantier free formula 1 (v0 : : : vp w0 : : : wm ) such that T j= (8v0 : : : 8vp )(8w0 : : : 8wm )( $ 1 ) So T j= (8v0 : : : 8vp )(9w0 : : : 9wm $ 9w0 : : : 9wm 1 ) By the claim there is a quantier free formula 2 (v0 : : : vp ) such that T j= (8v0 : : : 8vp )(9w0 : : : 9wm 1 $ 2 ) So T j= (8v0 : : : 8vp )(9w0 : : : 9wm $ 2 ) and so 2 is the quantier free formula equivalent to '. This completes the proof. (3) ) (4) Let A B, A C, B j= T and C j= T . Using the Elementary Diagram Lemma it will su!ce to show that Th(BB ) Th(CC ) is satisable. Without loss of generosity, we can ensure that LB \ LC = LA . By the Robinson Consistency Theorem, it su!ces to show that there is no sentence of LA such that both: Th(BB ) j= and Th(CC ) j= :
Suppose is such a sentence, Th(BB ) j= . Let fca0 : : : cap g be the set of constant symbols from LA n L appearing in . Let '(v0 : : : vp ) be obtained from by exchanging each cai for a new variable ui . Let (v0 : : : vp ) be the quantier free formula from (3): T j= (8v0 : : : 8vp )(' $ ) Let be the result of substituting cai for each ui in . is also quantier free. Since BB j= , B j= ' a0 : : : ap ]. Since B j= T , B j= a0 : : : ap ] and so BA j= . Since is quantier free and AA BA we have AA j= since AA CA we then get that CA j= . Hence C j= a0 : : : ap ] and then since C j= T we then get that C j= ' a0 : : : ap ]. But then this means that CA j= and so CC j= so is in Th(CC ) and we are done. (4) ) (1) Let B j= T and A B we show that T 4A is complete. Noting that BA j= T 4A , we see that it su!ces by Lemma 6 to show that BA C0 for each C0 j= T 4A . For each such C0 , by the Diagram Lemma, there is a model C for L such that A C and CA = C0 . Then C j= T so by (4) there is a D into which both BA and CA are elementarily embedded. In particular BA DA CA so we are done. Example 11. (Chang and Keisler) Let T be the theory in the language L = fU V W R S g where U , V and W are
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unary relation symbols and R and S are binary relation symbols having axioms which state that there are innitely many things, that U V W is everything, that U , V and W are pairwise disjoint, that R is a one-to-one function from U onto V and that S is a one-to-one function from U V onto W . Exercise 26. Show that T above is complete and model complete but not submodel complete. Hints: For completeness, use the L os-Vaught test and for model completeness use Lindstrom's test. For submodel completeness use (2) of the theorem with B j= T and A B where a 2 A = fb 2 B : B j= W (v0 ) b]g along with the sentence (9v0 )(U (v0 ) ^ S (v0 ca )): Remark. We will prove that each of the following theories admits elimination of quantiers: 1. dense linear orders with no end points (DLO) 2. algebraically closed elds (ACF) 3. real closed ordered elds (RCF) C. H. Langford proved elimination of quantiers for DLO in 1924. The cases of ACF and RCF were more di!cult and were done by A. Tarski. Thus, by Exercise 25, we will have model completeness of RCF which was promised at the beginning of Chapter 5. Exercise 27. Use part (4) of the previous theorem and the fact that RCF admits elimination of quantiers to prove that RCF is complete another result originally due to A. Tarski. Hint: Show that Q of Example 3 can be isomorphically embedded into any real closed eld and then use (4) from Theorem 17. Exercise 28. Let T be the theory DLO in the language L = f< c1 c2 g where c1 and c2 are constant symbols. Use the fact that DLO admits elimination of quantiers in its own language f
be a sentence of LC . Then let A be the nite subset of C consisting of these elements of C (other than 0 or 1 ) which are mentioned in . Let ' be the formula of L formed by exchanging each ca for a new variable. Then ACF j= 8v0 : : : 8vp(' $ ) for some quantier free . Hence C j= i C j= ' a0 : : : ap ] i C j= a0 : : : ap ] but checking this last statement amounts to evaluating nitely many polynomials in a0 : : : ap . Remark. In fact Tarski's original proof actually gave an explicit method for nding the quantier free formulas and this led, via the corollary above, to an effective decision proceedure for determining the truth of elementary algebraic statements about the reals or the complex numbers.
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As an application of quantier elimination of RCF we have: Corollary 8. (The Tarski-Seidenberg Theorem) The projection of a semi-algebraic set in Rn to Rm for m < n is also semialgebraic. The semi-algebraic sets of Rn are dened to be all those subsets of Rn which can be obtained by repeatedly taking unions and intersections of sets of the form fhx1 : : : xn i 2 Rn : p(x1 : : : xn ) = 0g and fhx1 : : : xn i 2 Rn : q(x1 : : : xn ) < 0g where p and q are polynomials with real coecients. Proof. We rst need two simple results which we state as exercises. Let L be the language of RCF augumented with a constant symbol for each element of the reals R. Let T be RCF considered as a theory in this language. Exercise 29. T admits elimination of quantiers as a theory in the language L . < 0 1i be the usual model of the reals then RR is a model Let R = hR + < for the language L . Exercise 30. A set X Rn is semi-algebraic i there is a quantier free formula '(v1 : : : vn ) of L such that X = fhx1 : : : xn i : RR j= ' x1 : : : xn ]g: Now, in order to prove the corollary, let X Rn be semi-algebraic and let ' be its associated quantier free formula. The projection Y of X into Rm is Y = fhx1 : : : xm i : 9wm+1 : : : 9wn hx1 : : : xm xm+1 : : : xn i 2 X g = fhx1 : : : xm i : RR j= 9vm+1 : : : 9vn ' x1 : : : xm ]g Since T admits elimination of quantiers, there is a quantier free formula of L such that T j= (8v1 : : : 8vm )(9vm+1 : : : 9vn ' $ ) So RR j= 9vm+1 : : : 9vn ' x1 : : : xm ] i RR j= x1 : : : xm ]: So Y = fhx1 : : : xm i : RR j= x1 : : : xm ]g and by the exercise, Y is semi-algebraic.
CHAPTER 7
Model Completions Closely related to the notions of model completeness and submodel completeness is the idea of a model completion. Definition 32. Let T T be two theories in a language L. T is said to be a model completion of T whenever T 4A is satisable and complete in LA for each model A of T . Lemma 14. Let T be a theory in a language L. (1) If T is a model completion of T , then for each A j= T there is a B j= T such that A B. (2) If T is a model completion of T , then T is model complete. (3) If T is model complete, then it is a model completion of itself. (4) If T1 and T2 are both model completions of T , then T1 j= T2 and T2 j= T1 . Proof. (1) Easy. Just use the diagram lemma and the word \satisable" in the denition of model completion. (2) Easier. (3) Easiest. (4) This needs a proof. Let A j= T2 . It will su!ce to prove that A j= T1 . Let A0 = A. since A0 j= T and T1 is a model completion of T we obtain, from (1), a model A1 j= T1 such that A0 A1. Similarly, since A1 j= T and T2 is a model completion of T we obtain A2 j= T2 such that A1 A2 . Continuing in this manner we obtain a chain: A0 A1 A2 : : : An An+1 Let B be the union of the chain, fAn : n 2 N g. Each even A2n j= T2 . By (2) and (4) of Theorem 14 we get that for each n, A2n A2n+2 and by the Elementary Chain Theorem A0 B. Similarly A1 B. so A0 A1 and hence A j= T1 . Remark. Part (4) of the above lemma shows that model completions are essentially unique. That is, if model completions T1 and T2 of T are closed theories in the sense of Denition 12 then T1 = T2 . Since there is no loss in assuming that model completions are closed theories, we speak of the model completion of a theory T . Lemma 15. If T T are theories for a language L such that for each A j= T there is a B j= T such that A B, then the following are equivalent: (1) T is the model completion of T . (2) For each A j= T , B j= T and C j= T such that A B and A C we have a model D such that both BA and CA are elementarily embedded into DA . 50
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(3) For each A j= T , B j= T and C j= T such that A B and A C we have a model D such that BA DA and CA is elementarily embedded into DA . (4) For each A j= T , B j= T and C j= T such that A B and A C we have a model D such that BA is isomorphically embedded into DA and C D. Proof. (1) ) (2) By the Elementary Diagram Lemma it su!ces to prove that the union of the elementary diagrams of BA and CA , is satisable. By the Robinson Consistency Theorem it su!ces to show that there is no sentence of LA such that ThBA j= and ThCA j= : . Indeed, if ThBA j= then BA j= . Now T 4A is a complete theory in LA and both BA j= T 4A and CA j= T 4A so BA CA . Therefore CA j=
and hence is in ThCA . (2) ) (3) and (3) ) (4) easily follow from the denitions. (4) ) (1) We rst show that T is model complete using Theorem 14 we show that T is existentially complete. Let A j= T we show that A is existentially closed. Let B j= T such that A B and let be an existential sentence of LA with BA j= our aim is to prove that AA j= . We invoke (4) with C = A to get a model D such that A D and an isomorphic embedding f : BA ! DA . Since is existential it is of the form 9v0 : : : 9vp ' for some quantier free formula '(v0 : : : vp ) of LA . BA j= 9v0 : : : 9vp ' So for some b0 : : : bp in B we have BA j= ' b0 : : : bp ]: By Exercises 8 and 18 we have DA j= ' f (b0 ) : : : f (bp )] and so DA j= 9v0 : : : 9vp ': Now A D implies that AA DA so AA j= 9v0 : : : 9vp ': Hence AA j= and T is model complete. We now show that T is the model completion of T . Let A j= T by the hypothesis on T and T we have that T 4A is satisable. We show that T 4A is complete in LA by showing that for each B j= T and C j= T with A B and A C we have BA CA . Letting B and C be as above, we invoke (4) to obtain a model D such that BA is isomorphically embedded into DA and C D. C D gives that D j= T . The isomorphic embedding gives us a model E such that B E and DA = EA . So E j= T . Using model completeness of T and Theorem 14 we can conclude that B E. We have: BA EA DA CA and we are done. Let's compare the denitions of model completion and submodel complete. Let T be the model completion of T . Then T will be submodel complete provided
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that every submodel of a model of T is a model of T . Unfortunately this is not always the case. However this seems like a promising approach to show submodel completeness (and hence elimination of quantiers) of some theories T | we just need to show that T is the model completion of some theory T and T has the property that every submodel of a model of T is also a model of T . Since T T , it would be enough to show that every submodel of a model of T is again a model of T . And this is indeed the case whenever T is a universal theory, that is, whenever T has a set of axioms consisting of universal sentences. Our ultimate aim is to show that DLO, ACF and RCF are submodel complete. We will in fact show that these theories are the model completions of LOR, FEI and ORF respectively. See Example 5 to recall the axioms for these theories. Now LOR is a universal theory but FEI and ORF are not. The culprits are the existence axioms for inverses: 8x9y (x + y = 0) and 8x9y (y x = 1) In fact, a submodel A of a eld B is only a commutative ring, not necessarily a subeld. Nevertheless, A generates a subeld of B in a unique way. This motivates the following denition. Definition 33. A theory T is said to be almost universal whenever A B, B j= T and A C, C j= T imply there are models D and E such that D j= T , A D B and E j= T , A E C and DA = EA . Example 12. LOR is almost universal since any universal theory T is almost universal | just let D = E = A and note A j= T . Example 13. FEI is almost universal | just let D and E be the subelds of B and C, respectively, generated by A. The isomorphism DA = EA is the natural one obtained from the identity map on A. Example 14. ORF is almost universal | again just let D and E be the ordered subelds of B and C, respectively, generated by A. The extension of the identity map on A to the isomorphism DA = EA is aided by the fact that the order placement of the inverse of an element a is completely determined by the order placement of a. Theorem 18. Let T and T be theories of the language L such that T is almost universal and T is the model completion of T . Then T is submodel complete. Proof. We show that condition (2) of Theorem 17 is satised. Let B and C be models of T with A a submodel of both B and C. We will show that, in fact, BA CA . Now T T so B j= T and C j= T . Since T is almost universal there are models D and E of T such that A D B, A E C and DA = EA . So BD j= T 4D and CE j= T 4E. Now BD is a model for the language LD whereas CE is a model for LE . We wish to obtain a model C0 for LD which \looks exactly like" CE . We just let C0 be C and in fact let C0 jLA = CE jLA . The interpretation of a constant symbol cd 2 LD n LA is the interpretation of ce 2 LE n LA in CE where the isomorphism DA = EA takes d to e.
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Now D j= T and since T is the model completion of T , T 4D is complete. The isomorphism DA = EA ensures that C0 j= T 4D. So BD C0 . Hence 0 BD jLA C jLA that is, BA CA . The way to show that DLO, ACF and RCF admit elimination of quantiers is now clear. We will use Theorem 17 and 18. This reduces to showing that DLO, ACF and RCF are the model completions of LOR, FEI and ORF respectively. To do this we will use Lemma 15, so we rst need to show that each pair of these theories satisfy the general hypothesis of Lemma 15: if A j= T then there is a B j= T such that A B. For the case T = LOR and T = DLO is easy every linear order can be enlarged to a dense linear order without endpoints by judiciously placing copies of the rationals into the linear order. The case T = FEI and T = ACF is just the well known fact that every eld has an algebraic closure. The case T = ORF and T = RCF is just Lemma 13. So all that remains of the quest to prove elimination of quantiers for DLO, ACF and RCF is to verify condition (4) of Lemma 15 in each of these cases. We rephrase this condition slightly as: For each A j= T , B j= T and C j= T with A B and A C there is a D such that C D and an isomorphic embedding f : B ,! D such that f A is the identity on A. At this point the reader may already be able to verify this condition for one or more of the pairs T = LOR and T = DLO, T = FEI and T = ACF, or T = ORF and T = RCF. However the remainder of this chapter is devoted to a uniform method. Definition 34. Let L be a language and "(v0 ) a set of formulas of L in the free variable v0 . A model A for L is said to realize "(v0 ) whenever there is some a 2 A such that A j= ' a] for each '(v0 ) in "(v0 ). Definition 35. The set of formulas "(v0 ) in the free variable v0 , is said to be a type of the model A whenever (1) every nite subset of "(v0 ) is realized by A (2) "(v0 ) is maximal with respect to (1). Remark. Every set of formulas "(v0 ) having property (1) of the denition of type can be enlarged to also have property (2). Lemma 16. Suppose A is an innite model for a language L. Let X A and let "(v0 ) be a type of AX in the language LX . Then there is a B such that A B and BX realizes "(v0 ). Proof. Let T = ThAA "(c) where c is a new constant symbol and "(c) = f'(c) : ' 2 "(v0 )g and of course '(c) is '(v0 ) with c replacing v0 . By denition of type, for each nite T 0 T , there is an expansion A0 of A such that A0 j= T 0 . The Elementary Diagram Lemma and the Compactness Theorem will complete the proof.
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Lemma 17. Suppose A is an innite model for a language L. There is a model B for L such that A B and BA realizes each type of AA in the language LA . Proof. Let f" (v0 ) : 2 I g enumerate all types of AA in the language LA . For each 2 I introduce a new constant symbol c and let " (c ) = f'(c ) : ' 2 " (v0 )g: Let " = f" (c ) : 2 I g. Let "0 " be any nite subset. Claim. "0 ThAA is satisable for the language LA fc : 2 I g. Proof of Claim. Let "1 (v0 ) : : : "n (v0 ) be nitely many types such that "0 "1 (c0 ) "2 (c1 ) "n (cn ): By Lemma 16 there is a model A1 such that A A1 and (A1 )A realizes "1 (v0 ). Using Lemma 16 repeatedly, we can obtain A A1 A2 An such that each (Aj )A realizes "j (v0 ). Now A An so (An )A j= ThAA . It is easy to check that since each Aj An, An realizes each "j (v0 ) and furthermore so does (An )A . So we can expand (An )A to the language LA fc1 : : : cn g to satisfy "0 ThAA .
By the claim and the Compactness Theorem, there is a model C j= " ThAA . By the Elementary Diagram Lemma, A is elementarily embedded into CjL, the restriction of C to the language L. Therefore there is a model B for L such that A B and BA = CjLA . It is now straightforward to check that BA realizes each type " (v0 ). Definition 36. A model A for L is said to be -saturated whenever we have that for each X A with jX j < , AX realizes each type of AX . Recall that + is dened to be the cardinal number just larger than . So a model A will be + -saturated whenever we have that for each X A with jX j , AX realizes each type of AX . In particular, if B is any innite set, A will be jB j+ saturated whenever we have that for each X A with jX j jB j, AX realizes each type of AX . Remark. A model A is said to be saturated whenever it is jAj-saturated, < i is saturated to where jAj is the size of the universe of A. For example, hQ < < iX . By prove this let X be a nite subset of Q and let "(v0 ) be a type of hQ < Lemma 16 and the Downward Lowenheim-Skolem Theorem get a countable B such < iX BX and B realizes "(v0 ). Use the hint for Exercise 11 to show that that hQ < < iX < iX . hQ < = BX and then note that this means that "(v0 ) is realized in hQ < Lemma 18. (R. Vaught) Suppose C is an innite model for L and B is an innite set. There is a jBj+ saturated model D such that C D.
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Proof. We build an elementary chain
C = C0 C1 C2 Cn n 2 N such that for each n 2 N (Cn+1 )Cn realizes each type of (Cn )Cn . This comes immediately by repeatedly applying Lemma 17. Let D be the union of the chain the Elementary Chain Theorem assures us that C D and indeed each Cn D. Let X D with jX j jBj and let "(v0 ) be a type of DX . If X Cn for some n, then "(v0 ) is a type of (Cn )X since (Cn )X DX . Now "(v0 ) can be enlarged to a type of (Cn )Cn which is realized in (Cn+1 )Cn and so "(v0 ) is realized in (Cn+1 )Cn . Since (Cn+1 )Cn DCn , we can easily check that "(v0 ) is realized in DCn . Since "(v0 ) involves only constant symbols associated with X , we have that DX realizes "(v0 ). We have almost proved that D is jBj+ -saturated, but not quite, because there is no guarantee that if X D = fCn : n 2 N g and jX j jBj, then X Cn for some n. However, there would be no problem if X was nite. The problem with innite X is that the elementary chain is not long enough to catch X . The solution is to upgrade the notion of an elementary chain to include chains which are indexed by any well ordered set, not just the natural numbers. We sketch < i to the the appropriate generalization of the above argument from the case of hN < case of an arbitrary well ordered set hI jBj. Definition 37. We say that B is a simple extension of A whenever
1. A B and 2. there is some b 2 B such that no properly smaller submodel of B contains A fbg.
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Theorem 19. (Blum's Test) Suppose T T are theories of a language L. Suppose further that: (1) T is an almost universal theory (2) every model of T can be extended to a model of T and (3) for each A j= T , B a submodel of a model of T , C j= T such that C is jBj+ saturated, A C and B is a simple extension of A, there is an isomorphic embedding f : B ! C such that f A is the identity on A. Then: (4) for each A j= T , B j= T and C j= T with A B and A C there is a model D such that C D and an isomorphic embedding f : B ! D such that f A is the identity on A, (5) T is the model completion of T and (6) T admits elimination of quantiers. Proof. Because of Lemma 15, Theorem 17 and Theorem 18, (5) and (6) follow from (1), (2) and (4). We will therefore only need to prove (4). Let A, B and C be as in (4). Using Lemma 18 we obtain a jBj+ -saturated model D such that C D. We wish to nd an isomorphic embedding f : B ,! D which is the identity on A. Let f be a maximal element of fg : for some B0 B g is an isomorphic embedding from B0 into D such that g A is the identity on Ag in the sense that no other such g properly extends f . We will prove that this f satises condition (4). The proof is by contradiction suppose f : B0 ,! D and B0 6= B. Claim. There is a model G such that B0 G B, G j= T and a model H such that G H and an isomorphism j : H ! D such that j B0 = f . Proof of Claim. From the isomorphic embedding f : B0 ,! D we get a model E such that B0 E and an isomorphism e : E ! D such that e B0 = f . We have B0 E and B0 B with both E j= T and B j= T . By condition (1) there are models F and G of T such that B0 F E, B0 G B and FB0 = GB0 . This gives an isomorphic embedding g : G ,! E such that g B0 is the identity map. From g we get a model H such that G H and an isomorphism h : H ! E such that h G = g. Now let j = e h. Thus j is an isomorphism from H to D such that j B0 = e h B0 = e g B0 = e B0 = f which nishes the proof of the claim.
Once we have the claim, there are two cases: G 6= B and G = B. For the case G 6= B, there must be some b 2 B n G and we use b to form the simple extension B00 of G by b. Now use condition (3) on G, B00 and H to obtain an embedding k : B00 ,! H which is the identity map on G. Now j k extends f , which is a contradiction. For the case G = B, the function j G extends f and gives the contradiction.
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The following lemma completes the proofs that each of the theories DLO, ACF and RCF admit elimination of quantiers. Lemma 19. Each of the following three pairs of theories T and T satisfy condition (3) of Blum's Test. (1) T = LOR theory of linear orderings. T = DLO, theory of dense linear orderings without endpoints. (2) T = FEI, theory of elds. T = ACF, theory of algebraically closed elds. (3) T = ORF, theory of ordered elds. T = RCF, theory of real closed elds. Proof of (1). Let A and B be linear orders, with B = A fbg and A B. Let C be a jBj+ -saturated dense linear order without endpoints with A C. We wish to nd an isomorphic embedding f : B ! C which is the identity on A. Consider a type of CA containing the following formulas: ca < v0 for each a 2 A such that a < b v0 < ca for each a 2 A such that b < a Since C is a dense linear order without endpoints each nite subset of the type can be realized in CA . Saturation now gives some t 2 C realizing this type. We set f (b) = t and we are nished. Proof of (2). Let A be a eld and B a simple extension of A witnessed by b such that B is a submodel of a eld (a commutative ring). Let C be a jBj+ -saturated algebraically closed eld such that A C. We wish to nd an isomorphic embedding f : B ! C which is the identity on A. There are two cases: (I) b is algebraic over A, (II) b is transcedental over A. Case(I). Let p be a polynomial with coe!cients from A such that p(b) = 0 but b is not the root of any such polynomial of lower degree. Since C is algebraically closed there is a t 2 C such that p(t) = 0. We extend the identity map f on A to make f (b) = t. We extend f to the rest of B by letting f (r(b)) = r(t) for any polynomial r with coe!cients from A. It is straightforward to show that f is still a well-dened isomorphic embedding. las:
Case (II). Let us consider a type of CA containing the following set of formuf:(p(v0 ) = 0)g
where p is a polynomial with coe!cients in fca : a 2 Ag. Since C is algebraically closed, it is innite and hence each nite subset is realized in CA . Saturation will now give some t 2 C such that t realizes the type. We set f (b) = t. Since t is transcedental over A, the extension of f to all of B comes easily from the fact that every element of B n A is the value at b of some polynomial function with coe!cients from A.
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Proof of (3). Let A be an ordered eld and B be a simple extension of A witnessed by b such that B is a submodel of an ordered eld (an ordered commutative ring). Let C be a jBj+ -saturated real closed eld such that A C. We wish to nd an isomorphic embedding f : B ! C which is the identity on A. There are two cases: (I) b is algebraic over A. (II) b is transcedental over A. Case (I). Since b is algebraic over A we have a polynomial p with coe!cients in A such that p(b) = 0. All other elements of the universe of the simple extension B are of the form q(b) where q is a polynomial with coe!cients in A. Before beginning the main part of the proof we need some algebraic facts. Claim. Let D be a real closed ordered eld and q(x) be a polynomial over D of degree n. Then for any e 2 D we have: n q (m) (e) X (x ; e)m q(x) = m ! m=0
where q(m) stands for the polynomial which is the m-th derivative of q. Proof of Claim. This is Taylor's Theorem from Calculus unfortunately we cannot use Calculus to prove it because we are in D, not necessarily the reals R. However the reader can check that the Binomial Theorem gives the identity for the special cases of q(x) = xn and that these special cases readily give the full result.
Claim. Let D be a real closed ordered eld and q(x) a polynomial over D with e 2 D and q(e) = 0. If there is an a < e such that q(x) > 0 for all a < x < e then q0 (e) 0. If there is an a > e such that q(x) > 0 for all e < x < a then q0 (e) 0. Here q0 is the rst derivative of q. Proof of Claim. From the previous claim we get ! n q (m) (e) q(x) ; q(e) = q0 (e) + (x ; e) X m ; 2 (x ; e) x;e m=2 m!
for any x 6= e in D. By choosing x close enough to e we can ensure that the entire right hand side has the same sign as q0 (e). A proof by contradiction now follows readily. Claim. Let D be a real closed ordered eld and q(x) be a polynomial over D with e 2 D and q(e) = 0. If w and z are in D such that w < e < z and q(w) q(z ) > 0 then there is a d in D such that w < d < z and q0 (d) = 0.
7. MODEL COMPLETIONS
59
Proof of Claim. Without loss of generosity q(w) > 0 and q(z ) > 0. Since q has only nitely many roots, we can pick d1 to be the least x such that w < x e and q(x) = 0. Since q(x) 6= 0 for all w < x < d1 , the Intermediate Value Property of Real Closed Ordered Fields shows that q cannot change sign here and so q(x) > 0 for all w < x < d1 . By the previous claim, q0 (d1 ) 0. A similar argument with z shows that there is a d2 such that e d2 < z and q0 (d2 ) 0. If d1 = e = d2 take d = e. If d1 < d2 the Intermediate Value Property gives a d with the required properties. Claim. Let D be a real closed ordered eld with an ordered eld E D. Let f : E ! C be an isomorphic embedding into a real closed ordered eld. Let q be a polynomial with coecients in E such that fx 2 D : q0 (x) = 0g E. Let d 2 D n E be such that q(d) = 0 but d is not a root of a polynomial with coecients from E which has lower degree. Then f can be extended over the subeld of D generated by E fdg. Proof of Claim. Since the nitely many roots of q0 from D actually lie in E, we can get e1 and e2 in E such that e1 < d < e2 and q0(x) 6= 0 for all x in D such that e1 < x < e2 . Furthermore for all x in E we have q(x) 6= 0. We can now apply the previous claim to get that q(w) q(z ) < 0 for all w and z in E such that e1 < w < d < z < e2 . We now move to the real closed ordered eld C and the isomorphic embedding f . For each w and z in E such that e1 < w < d < z < e2 we have f (w) < f (z ) and q(f (w)) q(f (z )) < 0. By the Intermediate Value property of C we get, for each such w and z , a y 2 C such that f (w) < y < f (z ) and q(y) = 0. Since q has only nitely many roots there is some t 2 C such that q(t) = 0, f (w) < t for all e1 < w < d and t < f (z ) for all d < z < e2 . We now extend f by letting f (d) = t and f (r(d)) = r(t) for any polynomial r with coe!cients from E. It is straightforward to check that the extension is a well-dened isomorphic embedding of the simple extension of E by d into C. We
use the fact that ORF is almost universal to extend the isomorphic embedding to all of the subeld of D generated by E fdg, since we can rephrase the denition of almost universal as follows: Whenever C j= T , D j= T , E0 D and f : E0 ,! C is an isomorphic embedding, then there is a model E00 j= T such that E0 E00 D and f extends over E00 .
It is now time for the main part of the proof of this case. Using Lemma 13, let D be a real closed ordered eld with B D. We have a polynomial p with coe!cients from A such that p(b) = 0. By induction on the degree of p, we can show that there is a sequence of elements d0 : : : dm = b of elements of D, a sequence of subelds of D: A = E0 E1 : : : Em+1 with each dj 2 Ej+1 n Ej and corresponding isomorphic embeddings fj : Ej ! C
7. MODEL COMPLETIONS
60
coming from the previous claim and having the property that f0 is the identity and fj+1 extends fj . In this way we extend the identity map f0 : A0 ,! C until we reach fm+1 : Em+1 ,! C. We then note that since b 2 Em+1 we have B Em+1 and we are nished. Case (II). Let us consider a type of CA containing the following formulas: ca < v0 for all a 2 A with a < b
CA .
v0 < ca for all a 2 A with b < a :(p(v0 ) = 0) for all polynomials p with coe!cients in fca : a 2 Ag Since each interval of C is innite, each nite subset of this type is realized by
Saturation now gives t 2 C which realizes this type. We put f (b) = t. We can now extend f on the rest of B n A, since each such element is the value at b of a polynomial function with coe!cients from A.
Question. Suppose K is the reduct of p a real closed ordered eld to the language of eld theory. Can you show that K ;1] is algebraically closed using Model Theory and the Fundamental Theorem of Algebra?
Bibliography 1. J. Barwise (ed.), Handbook of Mathematical Logic, North Holland, 1997. 2. J. L. Bell and A. B. Slomson, Models and Ultraproducts, North Holland, 1969. 3. Felix E. Browder (ed.), Mathematical developments arising from Hilbert's problems, Proceedings of Symposia in Pure Mathematics, vol. XXVIII - Parts 1 and II, AMS, 1976. 4. S. Buechler, Essential Stability Theory, Perspectives in Mathematical Logic, Springer, 1996. 5. C. C. Chang and H. J. Keisler, Model Theory, second ed., North Holland, 1977. 6. G. Cherlin, Model Theoretic Algebra, Selected Topics, LNM521, Springer-Verlag, 1976. 7. N. Jacobson, Basic Algebra I, W. H. Freeman, 1985. 8. H. J. Keisler, Foundations of Innitesimal Calculus, Prindle, Weber and Schmidt, 1976. 9. M. D. Morley (ed.), Studies in Model Theory, MAA Studies in Mathematics, MAA, 1973. 10. A. Robinson, Introduction to Model Theory and the Metamathematics of Algebra, second ed., North-Holland, 1965. 11. , Non Standard Analysis, North Holland, 1966. 12. G. E. Sacks, Saturated Model Theory, W. A. Benjamin Inc., 1972. 13. D. H. Saracino and V. B. Weispfenning (eds.), Model theory and algebra: A Memorial Tribute to Abraham Robinson, LNM498, Springer Verlag, 1975. 14. S. Shelah, Classication Theory and the Numbers of Nonisomorphic models, second ed., North Holland, 1990. 15. J. R. Shoeneld, Mathematical Logic, Addison-Wesley, 1967. 16. A. Tarski, A Descision Method for Elementary Algebra and Geometry, second ed., The Rand Corporation, 1957.
61
Index A B, 18 A = fAn : n 2 Ng, 19 ThA, 14 ThAA , 27 j=, 14 , 18 A jL , 17 , 17 tx0 : : : xq ], 7 hC + 0 1i, 13 hQ << + 0 1i, 13 hR < + 0 1i, 13 hN + < 0 1i, 18 A. Robinson, 35 ACF, 14 submodel complete, 57 algebraically closed elds axioms,theory of, 14 almost universal, 57, 64 axioms, 14 0
elimination of quantiers, 49 existentially closed, 35 expansion language, 17 model, 17 FEI, 14 almost universal, 57 elds axioms,theory of, 14 formula, 5 free variable, 6 isomorphic models, 18 isomorphically embedded model, 27 language, 6 Leibniz Principle, 29 Lindstrom's Test, 38 linear orders axioms,theory of, 14 LOR, 14 almost universal, 57 L os-Vaught Test, 24 Lowenheim-Skolem Theorems Downward, 23 Upward, 23 model, 6 satises, 7 model complete theory, 35 model completion, 55 submodel complete, 57 Number Theory, 18 number theory non-standard models, 18 ordered eld, 45 ordered elds axioms,theory of, 14 ORF, 14 almost universal, 57 prenex normal form, 10 rational numbers, 13 RCF, 14 submodel complete, 57
0
Blum's Test, 60 bound variable, 6 categorical -categorical theory, 24 chain of models, 19 elementary, 19 Compactness Theorem, 17 complete theory, 24 Completeness Theorem, 17 complex, 13 Craig Interpolation Theorem, 33 dense linear orders without endpoints axioms,theory of, 14 diagram lemmas, 27 DLO, 14 submodel complete, 57 elementarily embedded model, 27 elementarily equivalent models, 18 Elementary Chain Theorem, 19 elementary diagram, 27 elementary extension, 18 elementary submodel, 18
62
INDEX
real closed ordered eld, 45 real closed ordered elds axioms, theory, 15 axioms,theory of, 14 Intermediate Value Property, 15 real numbers, 13 realize, 58 reduction language, 17 model, A jL , 17 Robinson Consistency Theorem, 30 satisfaction A j= , 14 saturated -saturated model, 59 sentence, 10 simple extension, 60 subformula, 5 submodel, 18 submodel complete, 49 submodel complete theory, 49 T. Skolem, 18 Tarski's Elementary Chain Theorem, 19 Tarski-Vaught Condition, 21 term, 5 theory, 14 almost universal, 57 model completion, 55 theory of A, 14 type, 58 variable, 5 0
0
63
Green’s functions for planarly layered media (continued) Massachusetts Institute of Technology 6.635 lecture notes
We shall here continue the treatment of multilayered media Green’s functions, starting from the TE/TM decomposition we have presented in the previous document. For the sake of illustration, let us consider a one layer medium with a reflection and transmission region, as shown in Fig. 1. z ¯ r0 ) J(¯
z0
#0: (²0 , µ0 )
PSfrag replacements
0
x #1: (²1 , µ1 )
−d
#2: (²0 , µ0 )
Figure 1: Geometry of the problem.
. We have shown before that the Green’s functions in various layers are expressed as: G`0
i = 2 8π
ZZ
¸ · 1 ¯ ¯ r0 ¯ r0 −iK·¯ −iK·¯ ¯ ˆ ¯ dk ⊥ Ke (k`z )ˆ e(−k0z ) e + Kh (k`z )h(−k0z ) e , k0z
(1)
where ` = 0 , z < z0 :
¯ ¯ r ¯ e (k0z ) = eˆ(−k0z ) eiK·¯ K + RT E eˆ(k0z ) eik·¯r ,
ˆ ¯ h (k0z ) = h(−k K 0z ) e `=1 :
`=2 :
¯ r iK·¯
+R
TMˆ
(2a) ,
(2b)
¯ e (k1z ) = A eˆ(k1z ) eik¯1 ·¯r + B eˆ(−k1z ) eiK¯ 1 ·¯r , K
(2c)
¯ 1 ·¯ iK r ˆ 1 ) eik¯1 ·¯r + D h(−k ˆ ¯ h (k1z ) = C h(k K , 1z ) e z
(2d)
¯ r ¯ e (k2z ) = T T E eˆ(−k0z ) eiK·¯ K ,
(2e)
ˆ ¯ h (k2z ) = T T M h(−k K 0z ) e
1
¯ r iK·¯
h(k0z ) e
¯r ik·¯
.
(2f)
2
Section 1. Transmission line analogy for multilayered media
By satisfying the boundary conditions at z = 0 and z = −d, we obtain the following system (for TE waves, TM waves can be solved similarly):
1 + RT E =A + B k1 k0z (−1 + RT E ) = z (A − B) , µ0 µ1 A e−ik1z d + Beik1z d = T T E eik0z d , k1z k0 (Ae−ik1z d − Beik1z d ) = − z T T E eik0z d . µ1 µ0
(3a) (3b) (3c) (3d)
Upon solving, we obtain:
RT E = TTE =
1 − e2ik1z d RT E , T E RT E e2ik1z d 01 1 + R01 10 (1 +
pT01E ) (1
4ei(k1z −k0z )d . T E RT E e2ik1z d ) + pT10E ) (1 + R01 10
(4b)
RT M =
1 − e2ik1z d RT M , T M RT M e2ik1z d 01 1 + R01 10
(5a)
TTM =
4ei(k1z −k0z )d . T M RT M e2ik1z d ) (1 + pT01M ) (1 + pT10M ) (1 + R01 10
(5b)
A=
T E RT E 1 − pT10E 1 + R01 10 e2ik1z d , T E RT E e2ik1z d 2 1 + R01 10
(6a)
B=
TE 1 + R01 , T E RT E e2ik1z d 1 + R01 10
(6b)
TM µ1 k 0 2R10 e2ik1z d , T M RT M e2ik1z d ) µ0 k1 (1 + pT01M ) (1 + R01 10 µ1 k 0 2 D= . T M RT M e2ik1z d ) µ0 k1 (1 + pT01M ) (1 + R01 10
C=
1
(4a)
(6c) (6d)
Transmission line analogy for multilayered media
Let us consider a plane wave incident from region 0, with its plane of incidence parallel to the (xz) plane. The medium it is incident upon is multilayered. ∂ = 0 in Maxwell’s equations. Thus, we All fields vectors are independent on y, so that ∂y can decompose the electromagnetic field into TE/TM components. We get, in region ` (for TE waves):
3
E`y =[A` eikz z + B` e−ikz z ] eikx x , kz H`x = − ` [A` eikz z − B` e−ikz z ] eikx x , ωµ` kx [A` eikz z − B` e−ikz z ] eikx x . H`z = ωµ`
(7a) (7b) (7c)
Following standard notation in transmission line theory, we shall use here the j notation!! From 6.630, we know that a transmission line is characterized by its length d, characteristic (p) PSfrag replacements impedance Zc and wavenumber kz , as defined in Fig. 2. (p)
(p)
d
I1
I2
(p)
[Zc , kz ]
(p)
V1
(p)
V2
Figure 2: Transmission line circuit.
In a source-free region, the transmission line equations are written as ∂ (p) V = − jkz Zc(p) I (p) , ∂z ∂ (p) 1 I = − jkz (p) V (p) , ∂z Zc
(8a) (8b)
where p refers to the polarization (TE or TM), and ωµ , kz kz . = ω²
ZcT E =
(9a)
ZcT M
(9b)
The solution to Eqs. (8) is:
V (p) =A(p) e−ikz z + B (p) eikz z , 1 I (p) = (p) [A(p) e−ikz z − B (p) eikz z ] , Zc
(10a) (10b)
As it can be seen, there is a direct analogy between the voltage/current in a transmission line and the components of the electric and magnetic fields. For example, referring back to Eq. (7), we write:
4
Section 1. Transmission line analogy for multilayered media
E`y =V`T E (z)eikx x ,
(11a)
H`x = − I`T E (z)eikx x .
(11b)
Therefore, the problem of computing the fields in multilayered media in the spectral domain comes down to determining the voltage/current in equivalent transmission line network. The analogy is illustrated in Fig. 3.
Figure 3: Transmission line analogy for horizontal electric source and vertical magnetic source. Other cases are obtained by duality.
The treatment of the source will not be demonstrated here and we shall just state the final results (details can be found in the literature). Thus, depending on the source type and orientation, different generators will have to be placed in the transmission line network. The various cases are (magnetic sources can be obtained by duality): Horizontal electric source Vertical electric source
current generator voltage generator
value: 1/2π value: 1/2π
The algorithm is therefore as follows: 1. Write the field components in terms of voltage and current. Locate the source and observation point. 2. Compute the equivalent transmission line network. Locate the source (type and position) and observation. 3. Starting from the source replace all the layers above and below the source by equivalent impedances (see Fig. 4). To do this, start with the extreme boundary conditions and propagate back to the source using:
5
Z up
PSfrag replacements Z down Figure 4: Equivalent upper and lower impedance.
Zin (p) = Zc(p)
(p)
(p)
(p)
(p)
ZL + jZc tan(kz d) Zc + jZL tan(kz d)
.
(12)
4. Using standard circuit theory, compute V (p) and I (p) at the upper and lower limits. 5. Propagate V (p) and I (p) until the observation point using à !à ! ! à (p) V2 V1 cos(kz d) −jZc sin(kz d) = . (p) −I2 I1 cos(kz d) −j/Zc sin(kz d)
(13)
6. Get the fields in the spectral domain.
2
Coming back to the space domain: Sommerfeld integral In the rest of these notes we come back to the i notation!!
To come back to the space domain, we need to evaluate the inverse Fourier transform. A typical integral we have to perform is: 1 f (¯ r) = (2π)2
ZZ
dkx dky f˜(kx , ky ) eikx x eiky y .
(14)
By symmetry of the problem (x and y axis are equivalent), we can make a change of variables and integrate one integral analytically. The proper change of variables is the following: kx =kρ cos kφ ,
x =ρ cos φ ,
(15)
ky =kρ sin kφ ,
y =ρ sin φ .
(16)
We can transform the exponential part as: eikx x eiky y = ei[kρ ρ cos kφ cos φ+kρ ρ sin kφ sin φ] = eikρ ρ cos(kφ −φ) , so that
1 f (¯ r) = (2π)2
Z
2π
dkφ 0
Z
(17)
∞
dkρ kρ f˜(kρ , kφ ) eikρ ρ cos(kφ −φ) . 0
(18)
6
Section 2. Coming back to the space domain: Sommerfeld integral
By rotational symmetry, f˜(kρ , kφ ) = f˜(kρ ). In addition, we can expand the exponential part using the following identity: e
ikρ ρ cos(kφ −φ)
=e
iβ cos θ
=
∞ X
(−i)m Jm (β) eimθ ,
(19)
m=−∞
(where β and θ have just been defined to simplify the notation in the identity and have no connection to physical parameters). As we can see, the exponential function is the only term depending on kφ . Performing the integration, we get: Z 2π−φ 0 if m 6= 0 , (20) eimθ = 2π if m = 0 . −φ Therefore, we end up with
Z ∞ 1 dkρ kρ f˜(kρ ) J0 (kρ ρ) , 2π 0 which is known as a Sommerfeld integral. f (¯ r) =
(21)
Note however that in Eq. (7b), the kernel is function of kx also, which adds a kρ cos kφ term in the integral, so that the integral in kφ cannot be performed exactly as shown above. However, the generalization to this case is straightforward. Upon performing the same expansion of the Bessel function, we see that this time the non-vanishing contribution will come from the m = 1 term (or ±1 depending on how the integral is written). Without further details, we generalize the definition of Sommerfeld integral to the n th order as: Z ∞ ˜ (22) dkρ Jn (kρ ρ)kρn+1 f˜(kρ ) , S n [f ] = 0
and the transformation from spectral to spatial can be summarized as follows (where A˜ is a function of kρ only): spectral domain
space domain
˜ = A˜ G
˜ G = S0 [A]
˜ = −ikx A˜ G
˜ G = − cos φ S1 [A]
˜ = −iky A˜ G
˜ G = − sin φ S1 [A] cos 2φ ˜ − cos2 φ S0 [kρ2 A] ˜ S1 [A] ρ
˜ = −kx2 A˜ G
G=
˜ = −ky2 A˜ G
G=−
˜ = −kx ky A˜ G
G=
cos 2φ ˜ − sin2 φ S0 [kρ2 A] ˜ S1 [A] ρ
sin 2φ ˜ − 1 sin 2φ S0 [kρ2 A] ˜ S1 [A] ρ 2
7
3
Numerical evaluation of Sommerfeld integrals
Sommerfeld integrals are difficult (but not impossible) to evaluate for two reasons: 1. The spectral kernel can present poles (and in general does). 2. They have an oscillatory tail. Fortunately these two problems appear in essentially two distinct regions.
3.1
Poles of the Green’s function
It can be shown (out of the scope here) that the poles of the Green’s functions are associated with propagating waves (e.g. in single-mode regime, the Green’s function has only one pole). Therefore, we must have at least one value of ` (the index of the region) where k `z is real, i.e. k`z =
q
kl2 − kρ2 =
q k02 ²` µ` − kρ2
(23)
must be real. Since this needs to happen in at least one layer, it yields the condition: √ kρ < k0 max( ²` µ` ) ,
(24)
l
which puts an upper limit for the location of the poles. Although it can also be shown that poles have to correspond to kρ > k0 , we do not need this constraint here and we can limit ourselves √ to the interval [0, k0 maxl ( ²` µ` )]. In the lossless situation, the poles lie on the real axis, which renders the integral impossible to evaluate as is. We therefore need to deform the contour in the complex k ρ -plane: • At infinity, convergence is ensured by Sommerfeld’s radiation condition. √ • On [0, k0 maxl ( ²` µ` )], we perform the integration over an ellipse, as shown in Fig. 5.
=(kρ ) a b
<(kρ )
PSfrag replacements
Figure 5: Contour deformation in the complex kρ -plane.
8
3.2
3.2
Oscillatory tail
Oscillatory tail
The problem in this case is not the divergence of the integral (it does not diverge) bu the convergence, which is very slow because of the oscillatory behavior of the kernel. Yet, we can apply acceleration techniques to sum the series (these acceleration techniques belong to the family of extrapolation techniques). For the sake of illustration, we can mention Euler’s transformation, which is one of the best known acceleration technique: Sn + Sn+1 Sn0 = , (25) 2 P where Sn = ∞ i=0 ui is the partial sum (ui being the terms of the original series). By applying the formula repeatedly, we get: (k)
Snk+1
(k)
Sn + Sn+1 . = 2
(26)
A direct improvement of Euler technique is to weight the partial sums and write: Sn0 =
wn Sn + wn+1 Sn+1 . wn + wn+1
In our specific situation, we need to evaluate integrals of the form: Z ∞ S= f (x)dx ,
(27)
(28)
α
√ where α is related to k0 maxl ( ²` µ` ). In order to transform this integral in a series, we apply the approach of “integration then summation”. Thus, we define: Z xi f (x)dx , (29a) ui = xi−1
S= Sn =
∞ X
i=0 n X
ui ,
(29b)
ui .
(29c)
i=0
The break points xi have to be well chosen, and may for example be chosen based on the asymptotic behavior of f . If we refer back to Sommerfeld integrals, we can take the asymptotic expansion of the Bessel function: s 2 π π Jν (kρ ρ) ' cos(kρ ρ − ν − ) . (30) πkρ ρ 2 4 Hence, the easiest/simplest choice of break points will be π xn = k ρ n = x 0 + n , ρ
(31)
9
where x0 is the first break point greater than α. We will then approximate S by Sn as Z xn f (x)dx . Sn =
(32)
α
The problem of Sommerfeld’s integrals is that the remainder Z ∞ rn = S − S n = − f (x)dx
(33)
xn
decays slowly, so that we want to accelerate the series from which S n is evaluated (see Eq. (29c)). For Sommerfeld-type integrals, we write the generic form as: Z ∞ I= g(kρ ρ) f (kρ ) dkρ , (34) α
and the partial integral as I(n) =
Z
xn
g(kρ ρ) f (kρ ) dkρ .
(35)
α
The remainder is therefore I − I(n) =
Z
∞
g(kρ ρ) f (kρ ) dkρ .
(36)
xn
This integral can be expanded into an infinite series of inverse powers of ρ by integration by part. For example, with g(kρ ρ) = eikρ ρ (which is a generic form for Sommerfeld integrals), we can write: ¸ · i i i 00 I − I0 (n) = eian ρ fn + fn0 + ( )2 fn + . . . , (37) ρ ρ ρ where fn = f (xn ), fn0 = f 0 (xn ), etc. Note that if the right-hand side term converges (and it does in our case), the dominant term is O(ρ−1 ). Yet, if we now construct I1 (n) =
fn+1 I0 (n) + fn I0 (n + 1) , fn + fn+1
it appears to be a better estimate of I since the error is in O(ρ−2 ). If, in addition, f (kρ ) ∼ Ckρα e−βkρ ,
(38)
(39)
we can approximate fn and fn+1 and write I1 (n) =
I0 (n) + η0 I0 (n + 1) , 1 + η0
(40)
where η0 = [n/(n + 1)]α eβπ/ρ . At higher orders, a better approximation is given by: I2 (n) =
I1 (n) + η1 I1 (n + 1) , 1 + η1
(41)
where η1 = [(n − 1/2)/(n + 1/2)]α−2 eβπ/ρ . In practical applications, the parameters α and β may have to be adjusted for an optimum convergence. The technique presented briefly here is known as the “weighted average” method, and more details can be found in the literature under this keyword.
Integral Equations in Electromagnetics Massachusetts Institute of Technology 6.635 lecture notes
Most integral equations do not have a closed form solution. However, they can often be discretized and solved on a digital computer. Proof of the existence of the solution to an integral equation by discretization was first presented by Fredholm in 1903. In general, integral equations can be divided into two families: 1. When the unknown is in the integral only, the integral equation is called of the first kind. 2. When the unknown is both inside and outside the integral, the integral equation is called of the second kind. For electromagnetic applications, we can have both scalar and vector integral equations.
1
Scalar integral equations
Let us consider the situation of Fig. 1: two regions are defined in space, region 2 bounded by the closed surface S and region 1 being all the remaining space, bounded by S and S ∞ , in which sources are located. #1 J
n ˆ S #2
S∞
Figure 1: Definition of the geometry of the problem.
It is already known that we can write: (∇2 + k12 ) Φ1 (¯ r) = J(¯ r) ,
(1a)
(∇2 + k22 ) Φ2 (¯ r) = 0 ,
(1b)
1
2
Section 1. Scalar integral equations
and similarly for the Green’s functions (∇2 + k12 ) g1 (¯ r, r¯0 ) = −δ(¯ r − r¯0 )
in region 1,
(2a)
(∇ +
in region 2.
(2b)
2
k22 ) g2 (¯ r, r¯0 )
0
= −δ(¯ r − r¯ )
Upon performing (Eq. (1a)g1 (¯ r, r¯0 )−Eq. (2b)Φ1 (¯ r, r¯0 )), we get: [g1 (¯ r, r¯0 ) ∇2 Φ1 (¯ r) − Φ1 (¯ r) ∇2 g1 (¯ r, r¯0 )] = J(¯ r)g1 (¯ r, r¯0 ) + δ(¯ r − r¯0 )Φ1 (¯ r) .
(3)
Upon integrating Eq. (3) over the entire volume and using the identity ∇ · (g∇Φ − Φ∇g) = − Φ∇2 g, we get: Z Z 0 0 dvJ(¯ r) g(¯ r, r¯0 ) + Φ1 (¯ r0 ) . (4) dv ∇ · [g1 (¯ r, r¯ )Φ1 (¯ r) − Φ1 (¯ r)∇g1 (¯ r, r¯ )] =
g∇2 Φ
V
V
By Gauss theorem, we reduce the left-hand side integral to a surface integral. Also Z dvJ(¯ r)g1 (¯ r, r¯0 ) = −Φinc (¯ r0 ) .
(5)
V
We therefore obtain Z ds n ˆ · [g1 (¯ r, r¯0 )∇Φ1 (¯ r) − Φ1 (¯ r)∇g1 (¯ r, r¯0 )] = −Φinc (¯ r) + Φ1 (¯ r0 ) , −
r¯0 ∈ V .
S+S∞
(6)
By invoking the radiation condition, the integral over S∞ vanishes, leaving (and exchanging r¯ and r¯0 so that primed coordinates correspond to sources and unprimed ones to observation): Φ1 (¯ r) = Φinc (¯ r) −
Z
S
ds0 n ˆ · [g1 (¯ r, r¯0 )∇0 Φ1 (¯ r0 ) − Φ1 (¯ r0 )∇0 g1 (¯ r, r¯0 )]
r¯ ∈ V1 .
(7)
For r¯ ∈ V2 , the wave equation has no source and therefore the integration of the delta function yields a zero value. Performing the same steps for this second case, we get the generic relation: Φ (¯ ¯ ∈ V1 1 r) r ds0 n ˆ · [g1 (¯ r, r¯0 )∇0 Φ1 (¯ r0 ) − Φ1 (¯ r0 )∇0 g1 (¯ r, r¯0 )] = Φinc (¯ r) − 0 S r¯ ∈ V2 Z
(8)
This is directly evocative of Huygens’ principle:
• For r¯ ∈ V1 : the total field Φ1 (¯ r) is the sum of the incident field plus the field due to the surface currents on the surface S. • For r¯ ∈ V2 : the surface source on S produces a field that exactly opposes Φinc , yielding the extinction theorem. Applying the same reasoning to region 2, we write (where there is no incident field): Z
S
ds0 n ˆ · [g2 (¯ r, r¯0 )∇0 Φ2 (¯ r0 ) − Φ2 (¯ r0 )∇0 g2 (¯ r, r¯0 )] =
0
r¯ ∈ V1
Φ2 (¯ r) r¯ ∈ V2
(9)
3
Note that the sign reversal is due to the definition of the normal vector n ˆ which has to point outward from the surface. Here, since we use the same n ˆ as before, we have to take it as being negative. Eqs. (8) and (9) have four independent unknowns: Φ1 , Φ2 ; n ˆ · ∇Φ1 , n ˆ · ∇Φ2 ,
(10)
which can be related by the boundary conditions. Here, g(¯ r, r¯0 ) and n ˆ · ∇g(¯ r, r¯0 ) are the kernel of the integral equation.
2
Vector integral equation
For the sake of completeness, we shall write the vector wave equation as well, although we will not use it here directly. Considering the same situation as above, we know that the fields have to satisfy: ¯1 (¯ ¯1 (¯ ¯ r) , ∇×∇×E r ) − ω 2 ² 1 µ1 E r) =iωµ1 J(¯ ¯2 (¯ ¯2 (¯ ∇×∇×E r ) − ω 2 ² 2 µ2 E r) =0 ,
(11a) (11b)
and the Green’s functions: r, r¯0 ) − ω 2 ²1 µ1 G1 (¯ r, r¯0 ) = Iδ(¯ r − r¯0 ) , ∇ × ∇ × G1 (¯
∇ × ∇ × G2 (¯ r, r¯0 ) − ω 2 ²2 µ2 G2 (¯ r, r¯0 ) = Iδ(¯ r − r¯0 ) .
(12a) (12b)
R ¯1 (¯ r, r¯0 ) − Eq. (12a) · E r)]dv) By the same technique as before ( V dv [Eq. (11a) · G1 (¯ ¸ · Z Z ¯1 (¯ ¯ r) · G1 (¯ ¯1 (¯ ¯1 (¯ r, r¯0 ) − E r) · ∇ × ∇ × G1 (¯ r, r¯0 ) = iωµ1 dv J(¯ r, r¯0 ) − E r) . dv ∇ × ∇ × E r) · G1 (¯ V
V
(13) The left-hand side can be transformed into a surface integral (left as exercise) and the right-hand side written in terms of incident field, yielding: ½ ¾ Z 0 0 0 0 ¯ ¯ ¯ ¯ ds n ˆ · [∇× E1 (¯ r)]×G1 (¯ E1 (¯ r ) = Einc (¯ r )+ r, r¯ )+ E1 (¯ r)×∇×G1 (¯ r, r¯ ) , r¯0 ∈ V1 . (14) S+S∞
By reciprocity of the Green’s functions, we can transform ¯1 (¯ ¯1 (¯ n ˆ · [∇ × E r)] × G1 (¯ r, r¯0 ) = n ˆ × [∇ × E r)] · G1 (¯ r, r¯0 ) ¯ 1 (¯ = iωµ1 G1 (¯ r, r¯0 ) · n ˆ×H r) .
(15)
In addition, for an unbounded homogeneous medium (∇ × G(¯ r, r¯0 ) is reciprocal): ¯1 (¯ ¯1 (¯ n ˆ·E r) × ∇ × G1 (¯ r, r¯0 ) = n ˆ×E r) · ∇ × G1 (¯ r, r¯0 )
¯1 (¯ = −[∇ × G1 (¯ r, r¯0 )] · n ˆ×E r) .
(16)
4
Section 3. Problem with the internal resonance
For Green’s functions that satisfy the radiation condition, Eq. (14) becomes: ½ ¾ Z 0 0 0 0 ¯ ¯ ¯ ¯ ds n ˆ · iωµ1 G1 (¯ E1 (¯ r ) = Einc (¯ r)+ r, r¯ ) · n ˆ × H1 (¯ r) − [∇ × G1 (¯ r, r¯ )] · n ˆ × E1 (¯ r) .
(17)
S
We perform the same steps for the other region to eventually obtain (and again interchanging r¯ and r¯0 ): ¾ ¯ ½ E1 (¯ r) r¯ ∈ V1 ¯inc (¯ ¯ 1 (¯ ¯1 (¯ E r) + r, r¯0 ) · n ˆ×H r0 ) − [∇0 × G1 (¯ r, r¯0 )] · n ˆ×E r0 ) = ds0 iωµ1 G1 (¯ 0 S r¯ ∈ V2 Z
(18a)
−
Z
¾ ½ 0 0 0 0 0 ¯ 2 (¯ ¯2 (¯ r, r¯ ) · n ˆ×H r ) − [∇ × G2 (¯ r, r¯ )] · n ˆ×E r) = ds iωµ2 G2 (¯ 0
S
0
r¯ ∈ V1
E ¯2 (¯ r) r¯ ∈ V2
(18b)
Together with the boundary conditions ¯ 1 (¯ ¯ 2 (¯ n ˆ×H r) = n ˆ×H r) , ¯1 (¯ ¯2 (¯ n ˆ×E r) = n ˆ×E r) ,
(19a) (19b)
¯1 (¯ ¯ 1 (¯ this system can be solved for n ˆ×E r) and n ˆ×H r). Note also that Eqs. (18) can be written in terms of electric and magnetic currents, and magnetic Green’s functions: ¯ 1 (¯ J¯eq (¯ r0 ) = n ˆ×H r0 ) ;
¯ eq (¯ ¯1 (¯ −M r0 ) = n ˆ×E r0 ) ,
(20a)
so that for example: ½ ¾ ¯ E1 (¯ r) r¯ ∈ V1 ¯inc (¯ ¯ r0 ) + Gm (¯ ¯ (¯ E r) + ds0 iωµ1 Ge1 (¯ r, r¯0 ) · J(¯ ¯0 ) · M r0 ) = 1 r, r 0 S r¯ ∈ V2 Z
3
(21)
Problem with the internal resonance
A question arises: is the integral equation equivalent to Maxwell’s equations? Or asked differently, if we solve the integral equation and Maxwell’s equations, do we get the same solution? The answer is actually “no”, that is they are not always equivalent to each-other. The problem comes from spurious solutions at frequencies corresponding to the eigenfrequencies of the cavity enclosed by the surface S. This problem is generally referred to as the “internal resonance of the integral equation”. However, this lack of complete equivalence between the physical problem and its defining integral equation is rather minor and infrequent phenomenon, and is therefore often tolerated in practice.
5
4
Scattering by a rough surface
Let us consider the 2D problem (for which we shall use a scalar integral equation) depicted in Fig. 2. z
²0
f (x)
PSfrag replacements
x
²1
Figure 2: Rough surface S separating two media.
The integral equation in scalar form is given by: Φinc (¯ r) +
4.1
Z
S
ds0 n ˆ · [Φ(¯ r 0 )∇g(¯ r, r¯0 ) − g(¯ r, r¯0 )∇Φ(¯ r0 )] =
Dirichlet boundary conditions: EFIE
Φ(¯ r) 0
r¯ ∈ V0 r¯ ∈ V1
(22)
¯ = E yˆ) and PEC surface, the boundary condition is For a TE wave (E Φ(¯ r) = 0 ,
for r¯ ∈ S .
(23)
The integral equation becomes, for r¯ ∈ S , r¯0 ∈ S: Φinc (¯ r) −
Z
S
ds0 g(¯ r, r¯0 ) n ˆ · ∇Φ(¯ r0 ) =
Φ(¯ r)
(24)
0
This equation, in which Φ represent the electric field, is referred to as the electric field integral equation (EFIE). Note that as r¯ gets closer to the surface, Φ(¯ r) → 0 (from the boundary condition) so that we do not need to distinguish between approaching the surface from one side or the other. In fact, we can unify the equations and write: Z Φinc (¯ r) − ds0 g(¯ r, r¯0 ) n ˆ · ∇Φ(¯ r0 ) = 0 , r¯ ∈ S , r¯0 ∈ S . (25) S
g(¯ r, r¯0 )
In addition, has an integrable singularity as r¯ → r¯0 . Let us consider the surface depicted in Fig. 2, with z = f (x): s µ ¶2 p df 2 2 ds = dx + dz = dx 1 + , (26) dx such that the integral equation becomes: s µ ¶2 Z df dx0 1 + Φinc (¯ r) = g(x, f (x), x0 , f (x0 )) (ˆ n · ∇Φ(¯ r 0 )) , dx ∆x
at z 0 = f (x0 ) ,
(27)
6
4.2
Neumann boundary conditions: MFIE
where we can limit ∆x to [−L/2, L/2]. By letting s µ ¶2 df (ˆ n · ∇Φ(¯ r 0 ))|z 0 =f (x0 ) = U (x0 ) , 1+ dx
(28a)
Φinc (x, f (x)) = b(x) ,
(28b)
K(x, x0 ) = g(x, f (x), x0 , f (x0 )) ,
(28c)
we can rewrite the integral equation as Z L/2 dx0 K(x, x0 ) U (x0 ) = b(x) ,
(29)
−L/2
which has to be solved numerically. Before doing this, we shall write the integral equation with Neumann boundary conditions.
4.2
Neumann boundary conditions: MFIE
¯ = H yˆ with a PEC surface. In this case, the boundary This corresponds to a TM case with H condition is n ˆ · ∇Φ(¯ r) = 0 , (30) and the integral equation becomes Φ(¯ r) r¯ ∈ V0 Φinc (¯ r) + ds0 Φ(¯ r) n ˆ · ∇g(¯ r, r¯0 ) = 0 S r¯ ∈ V1 Z
(31)
In this case, it makes a difference if we approach the surface from the top or from the bottom. In fact Z Φinc (¯ r+ ) + ds0 Φ(¯ r0 ) n ˆ · ∇g(¯ r+ , r¯0 ) = Φ(¯ r+ ) , (32a) S Z Φinc (¯ r− ) + ds0 Φ(¯ r0 ) n ˆ · ∇g(¯ r− , r¯0 ) = 0 . (32b) S
These two equations seems inconsistent with one another. We shall show that in fact, they are actually consistent with each-other, due to the singularity of the Green’s functions. We shall examine what happens when we let r¯ approach the surface. Fig. 3 is an illustration of the situation at the immediate vicinity of the surface. The integral part of the equation can be written as: Z Z Z 0 0 0 0 0 0 ds0 Φ(¯ r0 ) n ˆ · ∇g(¯ r, r¯0 ) , (33) ds Φ(¯ r )n ˆ · ∇g(¯ r, r¯ ) + ds Φ(¯ r )n ˆ · ∇g(¯ r, r¯ ) = P V S
piece
S
where P V denotes the principal value and ‘piece’ refers to the integration over the local domain shown in Fig. 3. For this integral, we use the local coordinates: ds0 = dX 0 ,
n ˆ = Zˆ 0 , etc
(34)
7
r¯
PSfrag replacements
Z0 −a
r¯ = (0, Z) a X0 Figure 3: Zoom on the rough surface.
The integral becomes: Z Z 0 0 0 ds Φ(¯ r )n ˆ · ∇g(¯ r, r¯ ) = lim lim Φ(¯ r) a→0 |Z|→0
piece
a
dX 0 −a
∂ g(¯ r, r¯0 ) . ∂Z 0
(35)
Over the small piece, we have: |¯ r − r¯0 | =
p
X 02 + (Z − Z 0 )2 ,
(36a)
1 i (1) |¯ r− r − r¯0 |) → − g = H0 (|¯ ln( 2π 2 µ 4¶ Z ∂g . = ∂Z 0 Z 0 =0 2π(X 02 + Z 2 ) Thus, the integral becomes Z Z = lim lim Φ(¯ r) piece
r¯0 |
),
(36b) (36c)
a
1 Z 02 a→0 |Z|→0 2π X + Z 2 −a · ¸ 0 a Φ(¯ r) −1 X = lim lim tan a→0 |Z|→0 2π Z −a 1 Φ(¯ r) for Z > 0 = 2 − 1 Φ(¯ r) for Z < 0 dX 0
(37)
2
The two parts of the integral then become: Z Φinc (¯ r) + P V ds0 Φ(¯ r0 ) n ˆ · ∇g(¯ r, r¯0 ) + S Z ds0 Φ(¯ r0 ) n ˆ · ∇g(¯ r, r¯0 ) − Φinc (¯ r) + P V S
1 Φ(¯ r) = Φ(¯ r) , 2 1 Φ(¯ r) = 0 , 2
(38a) (38b) (38c)
which is cast into one equation: Φinc (¯ r) + P V
Z
1 r) . ds0 Φ(¯ r0 ) n ˆ · ∇g(¯ r, r¯0 ) = Φ(¯ 2 S
(39)
Eq. (39) is called the magnetic field integral equation (since Φ represents the magnetic field) and is an integral equation of the second kind.
8
Section 5. Solving the EFIE
In the same way as before, we can define the kernel as 0
K(x, x ) =
s
1+
µ
df dx
¶2
n ˆ · ∇g(¯ r, r¯0 )|z=f (x),z 0 =f (x0 ) ,
(40)
and write the MFIE as Φinc (¯ r) + P V
5
Z
∞
dx0 Φ(x0 ) K(x, x0 ) = −∞
1 Φ(x) . 2
(41)
Solving the EFIE
Upon using the notation introduced in Section 4.1, the problem is reduced to solving the following integral equation: Z L/2 dx0 K(x, x0 ) U (x0 ) = b(x) . (42) −L/2
Let us subdivide the integration domain into N small elements, each of length ∆ = L/N , and centered at xm (m ∈ [1, N ]). Thus, constraining the observation at these discrete locations, the integral equation becomes Z
L/2
dx0 K(xm , x0 ) U (x0 ) = b(xm ) .
(43)
−L/2
Next, if we suppose that U (x) is constant in each interval, we replace the integral by a sum over all segments, excluding the singular term: Z X ∆x K(xm , xn ) U (xn ) + U (xm ) dx0 K(xm , x0 ) = b(xm ) . (44) m
n=1 n6=m
Note that we have to single out the singularity, i.e. the mth interval because K(xm , x0 ) is singular at x0 = xm . This part is known as the self-patch contribution. For a 2D problem, we have: K(xm , x0 ) =
i (1) p H (k (xm − x0 )2 + (f (xm ) − f (x0 ))2 ) . 4 0
(45)
Upon approximating f (x0 ) − f (xm ) ' f 0 (xm )(x0 − xm ), we write: K(xm , x0 ) = (1)
p i (1) H0 (k|x0 − xm | 1 + f 0 (xm )2 ) . 4
(46)
For small argument: H0 (α) ≈ 1 + i π2 ln(αγ/2) where γ is the Euler constant (γ ' 1.78). Therefore: µ · ¶¸ p γ i 1 0 0 02 k(x − xm ) 1 + f (xm ) . K(xm , x ) = 1 + i ln (47) 4 π 2
9
Therefore, the integral becomes: Z
0
0
dx K(xm , x ) = 2 m
Z
xm +∆x/2 0
0
dx K(xm , x ) = 2 xm
Z
∆x/2
dx0 K(xm , x0 + xm ) 0
· µ ¶¸ Z i ∆x/2 1 γ 0p 0 02 ' kx 1 + f (xm ) dx 1 + i ln 2 0 π 2 ½ µ ¶¾ p 2 γk i∆x 02 1 + i ln ∆x 1 + f (xm ) , = 4 π 4e
(48)
where ln(e) = 1. We can therefore cast the integral equation into a matrix equation of the form: N X
Amn Un = bm ,
(49)
n=1
where Un = U (xn ) is the unknown, bm = b(xm ) , ∆x K(xm , xn ) ½ µ ¶¾ Amn = i∆x p γk 2 02 4 1 + i π ln 4e ∆x 1 + f (xm )
The system can now easily be solved numerically.
(50a) (50b) for m 6= n for m = n
(50c)
The Method of Moments in Electromagnetics Massachusetts Institute of Technology 6.635 lecture notes
1
Introduction
In the previous lecture, we wrote the EFIE for an incident TE plane wave on a PEC surface. The solution was then obtained by some types of “intuitive” arguments, such as dividing the integration domain into small elements and supposing that the unknown does not vary too much over each elementary cell. We shall now see more rigorously what we actually did, and show that it was in fact a simple version of the Method of Moments. R. F. Harrington was the first to use the method of moments (MoM) in electromagnetics and his book remains a fundamental reference (and very easy to read!): R. F. Harrington, “Field Computation by Moment Method” (is now available from IEEE Press).
2
PEC surface with TE incident wave: EFIE
The situation we studied last time is depicted in Fig. 1. The integral equation (EFIE) we z
²0
f (x)
²1
x
Figure 1: Rough surface S separating two media.
eventually obtained was:
Φinc (¯ r) =
Z
L/2
dx −L/2
0
s
1+
µ
df dx
¶2
g(x, f (x), x0 , f (x0 )) [ˆ n · ∇Φ(¯ r 0 )]|z 0 =f (x0 ) ,
(1)
dx0 K(x, x0 ) U (x0 ) ,
(2)
which we rewrote as b(x) =
Z
L/2 −L/2
1
2
Section 2. PEC surface with TE incident wave: EFIE
where s
1+
µ
df dx
¶2
[ˆ n · ∇Φ(¯ r 0 )]|z 0 =f (x0 ) = U (x0 ) ,
Φinc (x, f (x)) = b(x) , 0
(3a) (3b)
0
0
K(x, x ) = g(x, f (x), x , f (x )) ,
(3c)
where U (x0 ) is the unknown we are solving for. The important step we did from this integral, although it probably appeared straightforward, was to say that x can only take discrete values on the surface, thus defining N intervals of length ∆x: x ∈ {xi } , i = 1, . . . , N, (4) The assumptions were therefore: 1. x ∈ {xi } ,
i = 1, . . . , N.
2. U (x0 ) is constant on each interval. which eventually yielded the following system of equations (supposing that the problem related to the singularity of the Green’s function has been accounted for): N X
Amn Un = bm .
(5)
n=1
This is a matrix equation with the two indices m and n corresponding to: m: observation point → unprimed coordinates. n: source point → primed coordinates. Physical interpretation: • element (m, n) represents the effect of cell n on cell m. • element (m, m) represents the self-term. Mathematically: The steps we had to perform to from Eq. (2) to Eq. (5) are 1. Write the unknown as U (x0 ) =
X
Un δ(xn )∆x ,
(6)
n
stating that now the unknowns become {Un } which are the amplitudes of the function. The integral equation becomes: Z L/2 X Un K(x, xn ) = b(x) . (7) dx0 K(x, x0 )U (x0 ) = ∆x −L/2
n
3
2. Dot-multiply both sides by δ(xm ): X ∆x Un K(xm , xn ) = b(xm ).
(8)
n
These two steps are at the basis of the method of moments: 1. First step can actually be decomposed into two steps: (a) Mesh the structure (i.e. choose the intervals over which Un will be defined). (b) Expand the unknown U (x) into basis functions. 2. The second steps concerns the observation: dot-multiply both sides of the equation by a test function (or weighting function). In the previous example: • basis functions: pulse basis function. • testing functions: we point-patch the integral equation at x = xm (the method is therefore called point matching). This is a very simple, yet very widely used version of the method of moments.
3
General considerations on MoM
Let us consider the inhomogeneous equation: L(f ) = g ,
(9)
where L is a linear operator, g is known, and f is to be determined. We shall now perform the two essential steps we have highlighted above. 1. Let f be expanded in a series of functions: X f= α n fn ,
(10)
n
where αn are constant. The set fn is called expansion function, or basis functions. Note that for an exact solution, the summation should be taken to ∞, but has to be truncated in practice. 2. It is assumed that a suitable inner product has been defined for the problem. Now, we define a set of weighting functions, or testing functions, w1 , w2 , . . . , wN in the range of L, and take the inner product of the previous equation with wm : X αn < wm , Lfn >=< wm , g > . (11) n
4
Section 4. A simple example for electrostatic
The system can now be written in matrix form as: [Amn ] [αn ] = [gm ] ,
(12)
where
< w1 , Lf1 > < w1 , Lf2 > . . . < w2 , Lf1 > < w2 , Lf2 > . . . , [Amn ] = .. .. .. . . .
α1 α [αn ] = 2 , .. .
< w1 , g > < w2 , g > . (13) [gm ] = .. .
If the matrix [Amn ] is not singular, the unknowns αn are simply given by: [αn ] = [Amn ]−1 [gm ] ,
(14)
and the original function f can be reconstructed using Eq. (10). We can now generalize the following definitions: • The basis functions used previously are defined as: 1 if x belongs to the interval n Pulse basis functions: fn = 0 otherwise
(15)
• The testing (or weighting functions): Point matching = taking Dirac δ functions as testing functions.
4
A simple example for electrostatic
The example is taken from the reference mentioned at the beginning of this document. Let us consider a square plate of side 2a lying on the z = 0 plane with its center at the origin (see Fig. 2. Let σ(x, y) represent the surface density on the plate, assumed to have zero thickness. The electrostatic potential Φ at any point in space is given by z y 2a
2a
PSfrag replacements
2b 2b
Figure 2: Discretized square plate.
x
5
Φ(x, y, z) =
Z
a
dx0 −a
Z
a
dy 0 σ(x0 , y 0 ) g(¯ r, r¯0 ) ,
(16a)
−a
1 g(¯ r, r¯0 ) = , 4π²R p R = (x − x0 )2 + (y − y 0 )2 + z 2 .
(16b) (16c)
The boundary condition is Φ = V = constant on the plate. The integral equation for the problem is therefore Z a Z a 1 σ(x0 , y 0 ) p dy 0 dx0 , (17) V = 4π² (x − x0 )2 + (y − y 0 )2 + z 2 −a −a
where the unknown to determine is σ(x0 , y 0 ). Let us perform the three steps mentioned before: 1. Mesh the structure: divide the plate into N squares of size 2b (see Fig. 2). 2. Basis functions: let us choose σ(x0 , y 0 ) '
N X
αn fn (x0 , y 0 )
with fn (x0 , y 0 ) =
n=1
1 0
on ∆Sn on ∆Sm , m 6= n.
(18)
3. Test functions: we choose to satisfy the integral equation at the mid-point (x m , ym ) of each ∆Sm : wm = δ(x − xm )δ(y − ym ). (19) With these three steps, we construct the matrix as (z = 0): Z Z 1 1 p . dy 0 dx0 [Amn ] = 0 2 4π² (xm − x ) + (ym − y 0 )2 ∆yn ∆xn
(20)
It is obvious to see that this integral is singular at (xm , ym ) ∈ ∆Sm . In this simple case fortunately, we can perform the integration analytically (this is not always the case), and write: Z b Z b √ 1 2b 1 0 p (21a) ln(1 + 2) , = Ann = dx dy 0 4π² x02 + y 02 π² −b −b Amn '
b2 ∆Sn = p , 4π²Rmn π² (xm − x0 )2 + (ym − y 0 )2
m 6= n .
(21b)
To rewrite this with the language of linear space: f (x, y) = σ(x, y) ,
(22a)
g(x, y) = V on the plate (the discretization gives: gm
L(f ) =
Z
a
dx −a
0
Z
a
dy 0 −a
f (x0 , y 0 ) 1 p . 4π² (x − x0 )2 + (y − y 0 )2
V = V ) .. .
(22b)
(22c)
Note: if we add another plate under the existing one, at z = −2d, with a potential −V , we build a new problem that can be analyzed in two ways:
6
Section 5. Vectorial MoM
1. By meshing both plates (i.e. meshing everything, which is of course always possible). This will yield matrix twice as big as the previous one, to solve for twice as many unknowns. 2. By using the image theory, and saying that the new problem is equivalent to the one of a unique plate on top of a ground plane at z = −d. In that case, we only have to change the Green’s function to take the ground plane into account, and we keep the same number of unknowns as in the initial problem. When possible, this solution is better because computationally less expensive (analytically more expensive). Basically, a general trend is to have a Green’s function that represents as much as possible the environment and to mesh only those parts that are external to the environment. This is in fact the reason why people are looking for Green’s functions in layered media, periodic media, etc.
5
Vectorial MoM
We can of course apply the MoM to the vectorial case, like for example the equation: Z ¯ r) = iωµ ¯ r0 ) . E(¯ ds0 Ge (¯ r, r¯0 ) · J(¯
(23)
S
The general expansion of the current will be: ¯ r0 ) = J(¯
X
αn f¯n (¯ r0 ) ,
(24)
Z
ds0 Ge (¯ r, r¯0 ) · f¯n (¯ r0 ) .
(25)
n
yielding ¯ r) = iωµ E(¯
X n
αn
Sn
¯ m and integrate over The third step is to dot-multiply the equation with a testing function h the cell surface (i.e. perform the inner product): Z Z Z X 0 ¯ m (¯ ¯ ¯ ds h r) · Ge (¯ ds r, r¯0 ) · f¯n (¯ r0 ) . (26) E(¯ r) · hm (¯ r) = iωµ αn Sm
n
Sn
Sm
The double double-integral on the right-hand side of this equation is known as the “impedance term” since we can cast this system of equation into a matrix representation as: [Em ] = [Zmn ] [αn ] .
6
(27)
Other basis and testing functions
The advantage of the MoM over purely numerical methods is that there is still a large part that remains analytic (like the Green’s functions for example). Yet, it remains a numerical method based on a matrix inversion technique and therefore, convergence issues need to be examined. The convergence of the MoM is closely related to the choice of basis functions and, although to a lesser extend, to the choice of testing functions. There are essentially two families of basis functions:
7
1. Entire domain basis functions: using these functions to expand the unknowns is analogous to a Fourier expansion or to a modal expansion. These types of functions yield a good convergence of the method but are not versatile since the geometry need be regular in order to have the modes defined. Note that in this case there is no use to mesh the geometry. 2. Sub-domain basis functions: they rely on a proper meshing of the geometry, which can be rectangular, triangular, etc. The choice of basis functions is here very wide, from Dirac δ (like for the weighting functions shown in this document), pulses (basis functions shown in this document), piecewise linear, etc. Finally, we can mention that point matching, which is easy to grasp and straightforward to implement, may not yield an optimal convergence. In most of the applications, the Galerkin technique is better, which consists in choosing the same testing functions as the basis functions. This applies to both sub-domain and entire domain functions.
Time Domain Method of Moments Massachusetts Institute of Technology 6.635 lecture notes
1
Introduction
The Method of Moments (MoM) introduced in the previous lecture is widely used for solving integral equations in the frequency domain. Yet, some attempts have been made recently at the use of the MoM in the time domain. We shall briefly expose this approach here.
2
Time domain equations
The first step is of course to write Maxwell’s equation and all other relations (constitutive relations and continuity) in time domain: ∂ ¯ ¯ (¯ B(¯ r, t) − M r, t) , ∂t ¯ r, t) = m(¯ ∇ · B(¯ r, t) , ¯ r, t) = ²E(¯ ¯ r, t) , D(¯
∂ ¯ ¯ r, t) , D(¯ r, t) + J(¯ ∂t ¯ r, t) = ρ(¯ ∇ · D(¯ r, t) , ¯ r, t) = µH(¯ ¯ r, t) , B(¯
(1b)
¯ r, t) + ∂ ρ(¯ ∇ · J(¯ r, t) = 0 , ∂t
∂ ¯ (¯ ∇·M r, t) + m(¯ r, t) = 0 . ∂t
(1d)
¯ r, t) = − ∇ × E(¯
¯ r, t) = ∇ × H(¯
(1a)
(1c)
For the time-domain MoM, it is easier to work with the potentials, and make use of the well-known “retarded potentials” theory. In view of doing this, we write the definition: ¯ r, t) = 1 ∇ × A(¯ ¯ r, t) H(¯ µ0 ∂ ¯ ¯ r, t) = −∇φ(¯ E(¯ r, t) − A(¯ r, t) . ∂t
(2a) (2b)
Both the vector potential A¯ and the scalar potential φ satisfy the wave equation which, in time-domain domain, writes: ∂2 ¯ ¯ r, t) , A(¯ r, t) = −µ0 J(¯ ∂t2 ∂2 ρ(¯ r, t) ∇2 φ(¯ r, t) − ²0 µ0 2 φ(¯ r, t) = − . ∂t ²0
¯ r, t) − ²0 µ0 ∇2 A(¯
(3a) (3b)
These potentials are linked by the time-domain Lorentz gauge: ¯ r, t) + ²0 µ0 ∇ · A(¯
1
∂ φ(¯ r, t) = 0 . ∂t
(4)
2
Section 2. Time domain equations
We can defined also a time-domain Green’s function which satisfies the time-domain scalar equation: 1 ∂2 (∇2 − 2 2 )g(¯ r, r¯0 , t, t0 ) = −δ(¯ r − r¯0 ) δ(t − t0 ) , (5) c ∂t which solution is (in free-space): |¯ r−¯ r0 | 1 0 0 c ) t>t, 4π|¯ r −¯ r 0 | δ(t − t − 0 0 (6) g(¯ r, r¯ , t, t ) = 0 t < t0 . From this, the solution to the wave equation for A¯ and φ can be written as: Z Z ∞ Z ¯ r0 , t − R/c) J(¯ 0 0¯ 0 0 0 0 ¯ dv dv 0 A(¯ r, t) =µ0 dt J(¯ r , t ) g(¯ r, r¯ , t, t ) = µ0 , 4πR V V −∞
φ(¯ r, t) =
Z
dv 0 V
ρ(¯ r, t − R/c) , 4π²0 R
(7a)
(7b)
where R = |¯ r − r¯0 |. These wave equations are known as the time retarded potentials, and essentially say that the potential (either A¯ or φ) can be calculated at a given point in space r¯ and given time t from all previous times. From these equations, we can calculate the space-time electromagnetic fields: Z ¯ r0 , τ ) J(¯ ¯ r, t) = 1 H(¯ dv 0 ∇ × , τ = t − R/c , (8a) 4π R Z Z ¯ r0 , τ ) ρ(¯ r , τ ) µ0 1 ∂ J(¯ 0 ¯ dv ∇ E(¯ r, t) = − − dv 0 . (8b) 4π²0 V R 4π ∂t R Let us continue with the electric field first: · ¸ Z Z 1 1 µ0 1 ∂ ¯ 0 ¯ r, t) = − 1 dv 0 ∇ρ(¯ r0 , τ ) + ρ(¯ r 0 , τ )∇ − E(¯ dv 0 J(¯ r , τ) . 4π²0 V R R 4π V R ∂t
(9)
At this point, we need to use the following relations: ¯ R , R ¯ 1 R ∇ =− 3, R R ¯ ∂ ∂ 1 ∂ 1R ∇ρ(¯ r0 , τ ) = ρ(¯ r0 , τ )∇τ = − ∇R ρ(¯ r0 , τ ) = − ρ(¯ r, τ ) . ∂τ c ∂τ c R ∂τ ∇R =
We can therefore continue with the electric field as: · ¯ ¸ Z Z ¯ 1 ∂ ¯ 0 µ0 1 R 0 1 R ∂ 0 0 ¯ dv 0 dv J(¯ r , τ) E(¯ r, t) = ρ(¯ r , τ ) + 3 ρ(¯ r , τ) − 2 4π²0 c R ∂τ R 4π R ∂τ ¸ ¯ · Z Z R 1 µ0 1 ∂ ¯ 0 1 1 ∂ J(¯ r , τ) . ρ(¯ r0 , τ ) + ρ(¯ r0 , τ ) 2 − dv 0 = dv 0 4π²0 c ∂τ R R 4π R ∂τ
(10a) (10b) (10c)
(11)
3
We can perform the same type of calculations for the magnetic field using the relation ¯ ¯ r0 , τ ) = − 1 R × ∂ J(¯ ¯ r0 , τ ) . ∇ × J(¯ c R ∂τ We get: ¯ r, t) = 1 H(¯ 4π
Z
· ¸ ¯ ¯ R 1 R ∂ ¯ 0 ¯ r0 , τ ) . dv 0 − × J(¯ r , τ ) − × J(¯ c R2 ∂τ R3
(12)
(13)
Upon gathering the expressions for the electric and magnetic field, we eventually get: ½· ¾ ¸ ¯ Z 1 1 ∂ 1 R µ0 ∂ ¯ 0 0 0 0 ¯ E(¯ r, t) = dv J(¯ r , τ) ρ(¯ r , τ ) + ρ(¯ r , τ) − 4π c ∂τ R ²0 R 2 R ∂τ · ¸ Z ¯ R 1 ¯ 0 1 ∂ ¯ 0 ¯ r, t) = 1 dv 0 J(¯ r , τ ) + J(¯ r , τ) × 2 . H(¯ 4π c ∂τ R R
(14a) (14b)
Upon using the boundary conditions for the electric and magnetic field, we construct the integral equations in a standard way: ¯i + E ¯ scat ) = 0 on PEC surface • EFIE: n ˆ × (E 1 ¯ i (¯ n ˆ× ⇒n ˆ×E r, t) + 4π
Z
ds0 [. . .]
(15)
¯i + H ¯ scat ) = J¯s . • MFIE: n ˆ × (H As we have seen before (in a previous class), this integral equation is expressed in terms of the principal value of the integral with a 1/2 additional factor. Thus: Z 1¯ 1 ¯ i (¯ J(¯ r, t) = n ˆ×H r, t) + n ˆ × PV ds0 [. . .] (16) 2 4π
For the sake of comparison, we can write the MFIE in the frequency domain and in the time domain: Z i ¯ ¯ ¯ r0 ) × ∇0 g(¯ J(¯ r) = 2ˆ n × H (¯ r) + 2ˆ n × PV ds0 J(¯ r, r¯0 ) r¯ ∈ S , (17a) · ¸ Z ¯ 1 ∂ ¯ 0 1 ¯ 0 R 1 ¯ r, t) = 2ˆ ¯ i (¯ n ˆ × PV ds0 (17b) J(¯ r , τ ) + J(¯ r , τ) × 2 . J(¯ n×H r) + 2π c ∂τ R R Note that in the principal value, we essentially exclude the part for which R = 0. Since τ = t − R/c and R 6= 0, we always have that τ < t. The time domain equations therefore ¯ i (¯ state that the current at location r¯ and time t is equal to a known term 2ˆ n×H r, t) plus a ¯ term (integral) known from the past history of J. This is the basis for solving the time domain integral equation by iterative methods, the most well-known one being the marching-on-in-time.
4
Section 3. The marching-on-in-time technique
3
The marching-on-in-time technique
3.1
General equations
The integral equation can often be cast in the following form: Z Z τ i 0 ¯ ¯ ¯ r 0 , t0 ) . J(¯ r, t) = J (¯ r, t) + dv dt0 K(¯ r, r¯0 , t − t0 ) · J(¯
(18) eq.10
0
S
eq.10 ¯ r0 , t0 ) has not yet been set to satisfy Note that we have a time integral also as in Eq. (18), J(¯ any causality condition. Hence, we must then impose J¯i (¯ r, t) = 0 for t < 0, r¯ ∈ S.
In order to apply the MoM, we discretize the current both in space and in time: ¯ r 0 , t0 ) = J(¯
N M X X
m0 =1
n0 =0
J¯p (m0 , n0 ) Ps (¯ r0 − r¯m0 ) Pt (t0 − tn0 ) ,
(19)
where P denotes the simple pulse function. In addition, we also apply point-matching, which means that we take the following testing functions: Wmn (¯ rm , tn ) = δ(¯ r − r¯m ) δ(t − n∆t) = δ(¯ r − r¯m ) δ(t − tn ) ,
(20)
0 |. The method is best illustrated on the rm − r¯m where we take ∆t = min{Rmm0 /c}, Rmm0 = |¯ following example.
3.2
Example
Let us consider a 1D example governed by the following integral equation: g(x, t) =
Z
x0
K(x, x0 ) f (x0 , τ )dx0 , −x0
x ∈ [−x0 , x0 ],
τ = τ (x, x0 , t) = t −
|x − x0 | . c
(21)
Let us chose the following expansion for f : f (x0 , τ ) '
N X J X
i0 =1
ai0 j 0 Pi0 j 0 (x0 , τ ) ,
(22)
j 0 =1
where the pulse basis functions are defined as 1 for x0 ∈ [x 0 − dx , x 0 + i i 2 P i0 j 0 = 0 elsewhere.
dx 2 ]
and t ∈ [tj 0 −
dt 0 2 , tj
+
dt 2]
(23)
Note that we use the definitions: xi0 = i0 dx , tj 0 = j 0 dt , and dx = cdt . In order to apply point matching, we take the following testing functions: wij (x, t) = δ(x − xi ) δ(t − tj ) .
(24)
5
Upon expanding and testing, we get: g(xi , tj ) = gij =
=
Z
x0 0
K(xi , x ) −x0
J N X X
i0 =1 j 0 =1
N Z (i0 + 1 )dx X 2
0 1 i0 =1 (i − 2 )dx
dx
0
J X
j 0 =1
ai0 j 0 Pi0 j 0 (x0 , τ )δ(t − tj )
ai0 j 0 Pi0 j 0 (x0 , τ )K(idx , x0 )δ(t − tj ) .
(25)
Coming back to the definition of τ , we write (with the test and the expansion): τ = tj −
dx |xi − xi0 | = jdt − |i − i0 | = (j − |i − i0 |)dt , c c
(26)
such that the coefficient ai0 j 0 becomes ai0 ,j−|i−i0 | . The integral equation becomes: gij =
N X
ai0 ,j−|i−i0 |
i0 =1
We can define the term Zii0 =
Z
Z
(i0 + 12 )dx (i0 − 21 )dx
(i0 + 21 )dx (i0 − 21 )dx
dx0 K(idx , x0 ) .
dx0 K(idx , x0 )
(27)
(28)
and rewrite the previous system as gij =
N X
Zii0 ai0 ,j−|i−i0 |
i0 =1
= Zii aij + Zi,i−1 ai−1,j−1 + Zi,i−2 ai−2,j−2 + . . . + Zi1 a1,j−i+1 + Zi,i+1 ai+1,j−1 + Zi,i+2 ai+2,j−2 + . . . + Z1,N aN,j−N +i .
(29)
In this equation, only the first term involves time step j, all the others terms being at j − 1, j − 2, . . . Therefore, we can solve for aij : ¸ · N X 1 aij = Zii0 ai0 ,j−|i−i0 | . gij − Zii 0
(30)
i =1 i0 6=i
The value of all aik are known for k < j, so that aij is completely specified in closed form by those and the present value of gij . This process is known as a 1D march-on-in-time approach. Time-domain MoM is nowadays in its early stage and, although it has been successfully applied to various simple situations, still suffers from numerical instabilities. More work is in progress...
Study of EM waves in Periodic Structures with addenda: “Study of EM waves in Periodic Structures (mathematical details)”
Massachusetts Institute of Technology 6.635 lecture notes
1
Introduction
We will study here the distribution of electromagnetic fields in dielectric periodic media. The main difference with the previous topic comes from the word “dielectric”. Obviously, even a 2D periodic dielectric medium cannot be studied with the Green’s functions presented in a previous lecture, since the Green’s function was for periodic metallic structures. In this topic, we will study the EM fields in media where: • The material is macroscopic and isotropic. • Since the material is constituted of real dielectric, we suppose we work in a small enough frequency band such that we can ignore the frequency dispersive behavior of ². • The dielectric are lossless, so that ² is purely real.
2
Wave equations
2.1
¯ Wave equations for H
Starting from Maxwell’s equations and using a permittivity ² = ²(¯ r), it is straightforward to show that we can write the following equations: ¸ µ ¶2 1 ¯ r) = ω H(¯ ¯ r) , ∇ × H(¯ ∇× ²(¯ r) c ¯ r) = 0 . ∇ · H(¯
·
In this approach, the strategy is therefore: ¯ r). 1. Find the modes of H(¯ ¯ r) by solving Maxwell’s equations. 2. Find those of E(¯
1
(1a) (1b)
2
2.2
¯ Wave equations for E
Note that we can rewrite Eq. (1a) as an eigenvalue problem:
where
µ ¶2 ¯ r) = ω H(¯ ¯ r) , Θ H(¯ c
(2)
¸ 1 ∇× . Θ=∇× ²(¯ r)
(3)
·
Upon solving, we get the eigenvectors which correspond to the field patterns of the harmonic modes, and the eigenvalues which are proportional to the squared frequencies of these modes. Note that Θ is a linear operator and that it is Hermitian. The demonstration of the last property is straightforward: R ¯ >= d¯ ¯ r) and show that (by Define a scalar product < F¯ , G r F¯ ? (¯ r) · G(¯ ¯ >=< ΘF¯ , G ¯ >, which is the integration by part for example): < F¯ , ΘG definition of a Hermitian operator. The consequence of this property is of course that Θ has real eigenvalues.
2.2
Wave equations for E¯
¯ instead Another approach to get the fields would be of course to write the wave equation for E ¯ of for H: µ ¶2 ¯ r) = ω ²(¯ ¯ r) . ∇ × ∇ × E(¯ r)E(¯ (4) c However, this system cannot be cast in a simple eigenvalue problem. Although it can still be solved for, it is far more complicated to get accurate results since the operator we would have to define would not be Hermitian. For this reason, this approach is in general avoided.
2.3
Bloch states
In a periodic medium, we know that the fields can be written as: ¯ ¯ k (¯ H r) = eik·¯r u ¯k (¯ r),
¯ =u u ¯k (¯ r + R) ¯k (¯ r) .
(5)
Inserting into Eq. (2) yields: (ik¯ + ∇) ×
·
¸ µ ¯ ¶2 1 ω(k) u ¯k (¯ r) , (ik¯ + ∇) × u ¯k (¯ r) = ²(¯ r) c
(6)
so that the operator becomes: ·
1 (ik¯ + ∇)× Θ = (ik¯ + ∇) × ²(¯ r)
¸
(7)
¯ =u Note that because u ¯k (¯ r + R) ¯k (¯ r), the eigenvalue problem can be restricted to a small zone in space, which would necessarily imply a discrete spectrum of eigenvalues. Therefore, we ¯ expect a set of discrete modes for each k.
3
3
Fundamentals of photonic crystals
We shall briefly explain some terminology here related essentially to solid state physics, but which is of prime importance for the study of the structures we are dealing with here.
3.1
Direct lattice (some details are given in the “mathematical details” addenda)
A photonic crystal is a periodic structure (that we will take to be dielectric here) in 1D, 2D or 3D. Any vector r¯0 in space can be written as ¯, r¯0 = r¯ + R
(8)
¯ is the translational vector in space defined by where R ¯ = α1 a R ¯ 1 + α2 a ¯ 2 + α3 a ¯3 ,
(9)
where α1,2,3 ∈ {. . . , −2, −1, 0, 1, 2, 3, . . .} and a ¯1 , a ¯2 and a ¯3 are the lattice vectors. From the lattice, we can construct the Wigner-Seitz cell as shown in Fig. 1.
Figure 1: Wigner-Seitz cell for an arbitrary position of points: the cell is constructed by joining the center element to its closest neighbors and drawing perpendicular lines from to the center of these segments. The polygon thus created is the smallest repeatable cell of the periodic lattice, and is defined as the Wigner-Seitz cell.
Note that there exist only one type of lattice for a 1D photonic crystal, five distinct types for 2D photonic crystals (rectangular, square, hexagonal or triangle, centered rectangular and oblique), and fourteen for 3D photonic crystals.
3.2
Reciprocal lattice (some details are given in the “mathematical details” addenda)
We will use here the same notation as [Joannopoulos, Meade, and Winn, “Photonic Crystals”] ¯ and write the reciprocal translational vector as G: ¯ = β1¯b1 + β2¯b2 + β3¯b3 , G
(10)
where β1,2,3 ∈ {. . . , −2, −1, 0, 1, 2, 3, . . .} and ¯b1 , ¯b2 and ¯b3 are the lattice vectors in the spectral domain. For the sake of illustration, Tab. 1 gives the definition of vectors a ¯ and ¯b for square and triangular lattices.
4
3.3
Square lattice
a ¯1 a ¯2 a ¯1 a ¯2
Triangular lattice
= aˆ x = aˆ y = aˆ x √ = (ˆ x + 3ˆ y)
¯b1 ¯b2 ¯b1 ¯b2
Bloch-Floquet theorem
= 2π/a x ˆ = 2π/a yˆ √ = 2π/a (ˆ x − 3/3 yˆ) √ = 2π/a 2 3/3 yˆ
Table 1: Definition of a ¯ and ¯b vectors for square and triangular (or hexagonal) lattice.
3.3
Bloch-Floquet theorem
From Bloch-Floquet theorem, we know that we can write the electric and magnetic fields as a ¯ (see the “mathematical details” part of the notes). This summation over reciprocal vectors G means that different k¯ do not necessarily correspond to different modes and that therefore there ¯ we can therefore reduce the study to what is called the first is a redundancy in the label k: Brillouin zone (which is the Wigner-Seitz cell in the reciprocal lattice). Some examples of Brillouin zones are given in Fig. 2 and Fig. 3.
¯b2
a ¯2
¯b1
a ¯1 PSfrag replacements
Figure 2: Direct square lattice and corresponding reciprocal lattice with highlighted Brillouin zone. ¯b2 a ¯2 a ¯1 PSfrag replacements
¯b1
Figure 3: Direct triangular (or hexagonal) lattice and corresponding reciprocal lattice with highlighted Brillouin zone.
4
Bragg-like diffraction
The standard Bragg diffraction is illustrated in Fig. 4. Here, we will derive another diffraction condition, equivalent to Bragg, and shall see that the diffraction is entirely governed by the ¯ reciprocal vector G.
5
θ
θ θ
a
PSfrag replacements
Figure 4: Schematic representation of Bragg diffraction. Maximal diffraction occurs at 2a sin θ = nλ where λ is the wavelength of the electromagnetic wave and n is an integer.
Referring to Fig. 5, we can write the scattering amplitude in terms of the reflection coefficient Γ at position r¯ times a phase factor.
r¯ O k¯0
k¯ PSfrag replacements
Figure 5: Diffraction from an elementary volume of a periodic medium: k¯ is the wavevector of the incident wave whereas k¯0 is the wavevector of the diffracted wave.
Upon integrating over the whole volume, we get: Z ¯ ¯0 0 ¯ ¯ F (k, k ) = Γ(¯ r)ei(k−k )·¯r dv .
(11)
Since the medium is periodic, we can write: ¯ = Γ(¯ Γ(¯ r + R) r) =
X
¯r ˜ G)e ¯ iG·¯ Γ( ,
(12)
¯ G
such that ¯ k¯0 ) = F (k,
XZ ¯ G
¯ r ¯ k)·¯ ˜ G)e ¯ i(G−∆ Γ( dv ,
(13)
6
Section 4. Bragg-like diffraction
¯ This amplitude is maximal when G ¯ − ∆k¯ = 2mπ or, when m = 0, where ∆k¯ = k¯0 − k. ¯ ¯ = ∆k. G
(14)
This is an important relation which, again, is a condition for maximal diffraction. Upon ex¯ = |k¯0 | = k): panding back in terms of k¯ and k¯0 and rising to the square, we write (noting that |k|
¯ or (taking −k¯ instead of k):
¯ + G2 , k 2 = k 2 + 2k¯ · G
(15)
¯ = G2 . 2k¯ · G
(16)
As an exercise, it is interesting to show that this condition is equivalent to the standard Bragg diffraction. Upon dividing both terms of Eq. (16) by 4, we eventually write ¯ ¯ G G k¯ · ( ) = ( )2 . 2 2
(17)
This last equation has a nice geometrical interpretation shown in Fig. 6 which shows that the vectors k¯ that satisfy the maximum diffraction condition are actually those which lie on the edge of the Brillouin zone. D
¯ D /2 G
O
¯ C /2 G
C
PSfrag replacements
Figure 6: Graphical representation of Eq. (17): each vector k¯ (black vector) with its tip on a dashed line (not all represented) will satisfy the equation. Graphically: all ¯ (red vector). those k¯ have the same projection on the generating vector G/2
Therefore: ¯ = 0) satisfy the maximum The edge of the Brillouin zone plus its center (G diffraction condition. This condition can also be rewritten in terms of group velocity: for those k¯ which tip lie on the edge of the Brillouin zone and k¯ = 0, the component of the group velocity normal to the
7
Bragg diffraction planes tends to zero since the electromagnetic wave tends to be completely ¯ reflected for these k: µ ¶norm norm ¯ ¯ vg (k ∈ BZ tip) = ∇k ω(k) (k¯ ∈ BZ tip) → 0 . (18) For the symmetry points, the diffracted wave is reflected in the direction of the incident wave so that for these points, the total group velocity is zero. This can be directly seen on the dispersion curves where, at the symmetry points of the crystal, the tangent to the curve is horizontal (except possibly for those points corresponding to a zero frequency).
5
Mathematical details
¯ and then Using all the principles shown before, we can construct the eigenvalue system for H ¯ The detailed mathematical manipulations are given in the annex document “Study solve for E. of EM waves in Periodic Structures (mathematical details)”. Note that to build the system, we need to evaluate the Fourier coefficients of the permittivity (or the inverse of the permittivity, κ). We shall show how to get these coefficients for the case of infinite dielectric rods ²a of circular cross-section organized in a square lattice, embedded in ¯ vector will be a background of ²b . We therefore place ourselves in a 2D situation where the G ¯ ¯ ¯ ρ ). written Gρ to denote that it does not depend on z (and similarly R will be noted R As a reminder, we write the permittivity as: ²(ρ) =
X
¯ ρ )eiG¯ ρ ·ρ , ²˜(G
¯ρ G
¯ ρ) = 1 ²˜(G Ω
Z
¯
²(ρ)e−iGρ ·ρ ,
(19)
Ω
where Ω denotes the surface of the elementary cell. The idea is to write the permittivity as ²(ρ) = ²b + (²a − ²b )
X ¯ρ R
¯ ρ |) , S(Rc − |ρ − R
(20)
¯ ρ denotes a dependency on x and y only, Rc is the radius of the where again the subscript ρ in R dielectric rods and S denotes the step function. Merging these two equations, we get: ¸ Z · X ¯ ρ |) e−iG¯ ρ ·ρ dρ , ¯ρ) = 1 S(Rc − |ρ − R ²˜(G ²b + (²a − ²b ) Ω Ω ¯ρ R Z Z X 1 1 ¯ ρ ·ρ −iG ¯ ρ |)e−iG¯ ρ ·ρ dρ . = ²b e dρ + (²a − ²b ) S(Rc − |ρ − R (21) Ω Ω Ω Ω ¯ Rρ
¯ ρ ) and the second integral I2 (G ¯ ρ ), and evaluate them separately. Let us call the first integral I1 (G Evaluation of I1 (ρ)
8
Section 5. Mathematical details
The first integral is easily evaluated as: ¯ ρ) = I1 (G
²
b
0
¯ρ = 0 if G elsewhere
(22)
Evaluation of I2 (ρ) ¯ ρ . Since ρ spans the For the second integral, we can make the change of variable ρ0 = ρ − R ¯ ρ is the translational vector, ρ0 spans the whole space. We can therefore whole domain Ω and R replace the sum of integrals over Ω by a single integral over the whole 2D space. We write then: ¯ ρ = 0: If G ¯ ρ ) = ²a − ² b I2 (G Ω where fr =
πRc2 Ω
ZZ
2 ¯ ρ |) = (²a − ²b ) πRc = fr (²a − ²b ), dρ0 S(Rc − |ρ − R Ω
(23)
is the fractional volume.
¯ρ = If G 6 0: ZZ ∞ 1 ¯ ρ |) e−iG¯ ρ ·ρ0 ¯ dρ0 S(Rc − |ρ − R I2 (Gρ ) = (²a − ²b ) Ω −∞ Z Rc Z 2π ²a − ² b 0 0 0 = dρ ρ dφe−iGρ ρ cos(φ−θ) Ω 0 0 Z Rc ²a − ² b dρ0 ρ0 J0 (ρ0 Gρ ), 2π = Ω 0
(24)
where we have used the change of variable (now standard) x0 = ρ0 sin φ, y 0 = ρ0 cos φ, Gx = Gρ sin θ, Gy = Gρ cos θ, and the well-known identity for the Bessel function. Upon R using xJ0 (αx)dx = x/α J1 (αx), we continue with ¸ Rc · 0 ρ 2π J1 (Gρ ρ0 ) (²a − ²b ) Ω Gρ 0 2J1 (Gρ Rc ) = fr (²a − ²b ) . Gρ Rc
¯ ρ) = I2 (G
(25)
This lead to the final result that: ² f + ² (1 − f ) a r r b ²˜(Gρ ) = (²a − ²b )fr 2J1 (Gρ Rc ) Gρ Rc
if Gρ = 0 elsewhere.
(26)
The reconstruction of the permittivity is then straightforward, and examples of two lattice are given in Fig. 7.
9
(a) Side view.
(b) Top view.
Figure 7: Reconstruction of the permittivity for cylindrical rods in a square lattice ²a = 10, ²b = 1, Rc /a = 0.2 (yielding a fractional volume fr = 12.5%).
10
6
Section 6. Dispersion curves
Dispersion curves
At this point, we have everything to build the eigensystem (34) given in the additional document “Study of EM waves in periodic structures (mathematical details)”. Solving it gives a set of eigenvalues that are directly related to the dispersion curve of the material. An example is given in Fig. 8. Square lattice: Rc/a = 0.2 (f=12.5664%), εa=10, εb=1. 0.8
0.7
Frequency ω a/2π c
0.6
0.5
0.4
0.3
0.2 TE TM
PSfrag replacements 0.1
0
Γ
X
M
Figure 8: Dispersion diagram for the photonic crystal of Fig. 7.
Γ
Study of EM waves in Periodic Structures (mathematical details)
Massachusetts Institute of Technology 6.635 partial lecture notes
1
Introduction: periodic media nomenclature 1. The space domain is defined by a basis,(a1 , a2 , a3 ), where any vector can be written as r0 = r + R = r + α1 a1 + α2 a2 + α3 a3 ,
(1)
where R is the translation vector, with α1 , α2 , α3 integers. 2. The spectral domain is defined by a basis, (b1 , b2 , b3 ), and similarly, the translational vector is written as G = β1 b1 + β2 b2 + β3 b3 ,
(2)
where β1 , β2 , β3 are integers. 3. The two basis are linked since the functions (fields, permittivity) are periodic. For example, if we write the permittivity: ZZZ X 1 iG·r Fourier expansion: ²(r) = ²˜(G) e where ²˜(G) = dr3 ²(r) e−iG·r . (3) Vcell G
Periodicity: ²(r + R) = =
X
G X
²˜(G) eiG·(r+R) ²˜(G) eiG·r eiG·R = ²(r)
(4)
G
so that eiG·R = 1 and G · R = 2mπ
where m ∈ {. . . , −1, 0, 1, 2, . . .} .
(5)
We can see that condition (5) is immediately verified if we impose: bj · ai = 2πδij .
1
(6)
2
Section 1. Introduction: periodic media nomenclature
4. Bloch-Floquet theorem: Since EM fields are periodic, we can write them as a propagating function times a function with the same periodicity as the medium: ξ (r) = eik ·r ζ (r) k k
where
ζ (r + R) = ζ (r) , k k
(7)
and where ξ can represent either the electric or magnetic fields, E or H. Since ζ(r) is periodic, we can Fourier expand it: ζ (r) = k
ζ˜ eiG·r , G
(8)
e ei(k +G)·r , G
(9a)
h ei(k +G)·r . G
(9b)
X G
so that we shall write: E (r) = k H (r) = k
X
G X G
5. Wave equation in source-free region: From Maxwell’s equation, we can easily obtain the following wave equations in source-free regions (with ² = ²(r)):
µ ¶2 ω ∇ × ∇ × E(r) = µr ²r (r) E(r) , c · ¸ µ ¶2 1 ω ∇× ∇ × H(r) = µr H(r) , ²r (r) c
(10a) (10b)
To make these equations more symmetrical, we shall work with 1/²r (r) instead of ²r (r) directly, so that we define κr (r) =
X 1 = κ ˜ r (G) eiG·r . ²r (r) G
(11)
The wave equations are rewritten as:
µ ¶2 ω µr E(r) , c · ¸ µ ¶2 ω ∇ × κr (r)∇ × H(r) = µr H(r) . c
κr (r)∇ × ∇ × E(r) =
(12a) (12b)
3
2
Treatment of the E field
2.1
Method 1: direct expansion of the permittivity
We want to write Eq. (10a) with the decomposition of Eq. (9a). First, let us compute the first 0 curl (taking G as the variable for the expansion): ∇ × E k (r) =
· ¸ 0 0 X 0 ∇ × e 0 ei(k +G )·r = i (k + G ) × e 0 ei(k +G )·r . G G 0 0 G G
X
(13)
Taking the curl one more time gives · ¸ 0 X 0 0 ∇ × ∇ × E (r) = − (k + G ) × (k + G ) × e 0 ei(k +G )·r . k G 0 G
(14)
00
Upon using Eq. (3) but changing the index G into G , we write ²r (r)E(r) =
XX 0
G G
0
00
0
00
00 ²˜r (G ) e 0 ei(k +G +G )·r . G
(15)
²˜r (G − G ) e 0 ei(k +G)·r . G
(16)
00
By changing the variables G = G + G : XX
²r (r)E(r) =
G G
0
0
The wave equation (see Eq. (10a)) can therefore be rewritten as:
−
G
·
¸
0
(k + G ) × (k + G ) × e 0 ei(k +G )·r = G 0 0
X
0
µ ¶2 X X ω 0 µr ²˜r (G − G ) e 0 ei(k +G)·r . (17) G c 0 G G 00
We can simplify by exp (ik · r) and multiply by exp (−iG · r) to get:
−
X G
0
0
·
0
(k + G ) × (k + G ) × e
G
0
¸
0
e
00
i(G −G )·r
µ ¶2 X X 00 ω 0 = µr ²˜r (G − G ) e 0 ei(G−G )·r . (18) G c 0 G G
If we integrate this equation over the entire space, we can pull all the terms out of the 0 00 00 integral, except ei(G −G )·r on the left-hand side and. ei(G−G )·r on the right-hand side. Yet, we have ZZZ
00
dr3 ei(G−G )·r = V
1 00 δ(G − G ) , 3 (2π)
(19)
00
so that Eq. (18) becomes (upon substituting G by G since these are dummy variables): ·
−(k + G) × (k + G) × e
G
¸
µ ¶2 X ω 0 = µr ²˜r (G − G ) e 0 , G c 0 G
∀G .
(20)
4
2.2
2.2
Method 2: expansion of the inverse of the permittivity
Method 2: expansion of the inverse of the permittivity
Instead of working with Eq. (10a), we can also use Eq. (12a), which would need the expansion of Eq. (11). 00 Applying the same method (and transforming the index of Eq. (11) from G to G ), we get:
−
XX 00
G G
0
00
·
0
¸
0
0
κ ˜ r (G )(k + G ) × (k + G ) × e 0 e G
00
i(k +G +G )·r
0
µ ¶2 X 0 ω = µr e 0 ei(k +G )·r . (21) G c 0 G
00
which, upon substituting G = G + G , simplifying by exp (ik · r), multiplying by 00 00 exp (−iG · r), integrating over the whole space, using Eq. (19) and finally substituting G by G, becomes: −
X G
3
0
0
·
0
0
κ ˜ r (G − G )(k + G ) × (k + G ) × e 0 G
¸
µ ¶2 ω = µr e , G c
∀G .
(22)
Treatment of the H field
The H field is treated in an exactly similar way to eventually obtain very similar equations. However, these equations can still be pushed further by using the fact that ∇ · H (r) = 0. k 0 Upon using this equality, we see from Eq. (9b) that (using G for the expansion of the field): 0
(k + G ) · h 0 = 0 . G We can therefore define three vectors (ˆ e1 , eˆ2 , eˆ3 ) such that 0
(23)
0
k + G = |k + G | eˆ3 ,
(24a)
eˆ1 · eˆ3 = eˆ2 · eˆ3 = 0 ,
(24b)
and (ˆ e1 , eˆ2 , eˆ3 ) for an orthonormal tryad. In that case, we can decompose h
G
0
= h1
G
0
eˆ1 + h2
G
0
eˆ2 =
X
λ=1,2
hλ
G
0
eˆλ .
(25)
We need now to introduce this expression into Eq. (12b). First, we compute ∇ × H (r) = i k
XX G
0
hλ
λ
G
0
·
¸ 0 0 (k + G ) × eˆλ ei(k +G )·r ,
(26)
so that κr (r)∇ × H (r) = i k = i
XXX
G G λ XXX 00
G G
Taking the next curl, we write:
hλ
0
0
λ
hλ
¸
0
00
κ ˜ r (G ) (k + G ) × eˆλ ei(k +G +G )·r
0
· ¸ 0 0 κ ˜ r (G − G ) (k + G ) × eˆλ ei(k +G)·r .
G
G
·
0
00
0
(27)
5
·
¸
∇× κr (r)∇×H (r) = − k
XXX G G
0
hλ
λ
·
G
0
¸
κ ˜ r (G−G ) (k+G)×[(k+G )׈ eλ ] ei(k +G)·r , (28) 0
0
so that the wave equation (see Eq. (12b)) becomes:
µ ¶2 X X 0 ω − hλ 0 κ ˜ r (G−G ) (k+G)×[(k+G )׈ eλ ] e = µr hλ 0 ei(k +G )·r eˆλ . G G c 0 0 G G λ G λ (29) Always by the same token (multiplying by the proper functions and integrating over whole space), we write:
XXX
−
XX G
hλ
λ
0
·
0
0
G
0
¸
0
·
0
κ ˜ r (G − G ) (k + G) × [(k + G ) × eˆλ ]
i(k +G)·r
¸
µ ¶2 X ω = µr hλ00 eˆλ00 G c 00 λ
∀G . (30)
We can further simplify this expression by dot-multiplying the equation by eˆλ0 and noting that (using C · (A × B) = B · (C × A)) ·
¸
·
¸ · ¸ (k + G) × [(k + G ) × eˆλ ] · eˆλ0 = − (k + G ) × eˆλ · (k + G) × eˆλ0 0
0
(31)
Therefore, dot-multiplying Eq. (30) by eˆλ0 , we get the final result:
X X½· G
0
λ
¸ · ¸¾ 0 (k + G ) × eˆλ · (k + G) × eˆλ0 κ ˜ r (G − G ) hλ 0
0
00
G
=
0
0
µ ¶2 ω µr h λ 0 . G c 00
(32)
0
Upon exchanging G and G (transformations: G → G , G → G, G → G ), we obtain X X½· G
λ
¸ · ¸¾ 0 0 (k + G) × eˆλ · (k + G ) × eˆλ0 κ ˜ r (G − G) hλ
G
=
µ ¶2 ω µr h λ 0 0 . G c
(33)
which is the relation given in [Joannopoulos et al., 1995, p. 129]. Upon using the same notation, we rewrite Eq. (33) as: X λG
µ ¶2 ω h = µr h , (λG)0 (λG),(λG)0 (λG) c
Θk
(34a)
where Θk
(λG),(λG)0
· ¸ · ¸ 0 0 =κ ˜ r (G − G) (k + G) × eˆλ · (k + G ) × eˆλ0 .
(34b)
6
3.1
3.1
Matrix form
Matrix form
We can cast Eq. (33) in matrix form. First, we rewrite the kernel of the operator of Eq. (34b) as: ·
¯¯ ¯ ¸ · ¸ ¯ ¯ ¯¯ 0 0 ¯¯ ¯ ¯ ¯ (k + G) × eˆλ · (k + G ) × eˆλ0 = ¯(k + G)¯ ¯(k + G )¯ [ˆ e3 × eˆλ ] · [ˆ e3 × eˆλ0 ] .
(35)
Remembering that eˆ3 × eˆ1 = eˆ2 and eˆ3 × eˆ2 = −ˆ e1 , we can write: [ˆ e3 × eˆλ ] · [ˆ e3 × eˆλ0 ] =
Ã
eˆ2 · eˆ2 −ˆ e2 · eˆ1 −ˆ e1 · eˆ2 eˆ1 · eˆ1
!
,
(36)
so that we write the operator as: ·
¸ · ¸ 0 0 = κ ˜ ( G − G) ( k + G) × e ˆ · ( k + G ) × e ˆ r λ λ (λG),(λG)0 à ! ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ e ˆ · e ˆ −ˆ e · e ˆ 0 0 2 2 2 1 =κ ˜ r (G − G) ¯¯(k + G)¯¯ · ¯¯(k + G )¯¯ , −ˆ e1 · eˆ2 eˆ1 · eˆ1
Θk
used in Eq. (34a).
0
(37)
Study of EM waves in Periodic Structures: Photonic Crystals and Negative refraction Massachusetts Institute of Technology 6.635 lecture notes
1
Introduction
In the previous class, we have introduced various concepts necessary for the study of EM waves in photonic crystal structures. We shall now use these concepts to explain various results such as: • Reconstruction of the permittivity profile. • The band diagrams for rectangular and triangular lattices. • k-surfaces for various eigenvalues. In particular, we will show an example of how a periodic structure can exhibit k-surfaces typical of a negative refraction material (the concept of k-surface for left-handed materials was first introduced in the class of February 24, 2003). This class is based on the following two references:: [1] [2]
2
C. Luo, S. G. Johnson, and J. D. Joannopoulos, “All-angle negative refraction without negative effective index”, Physical Review B, vol. 65, 2002, #201104. J. D. Joannopoulos, R. D. Meade, and J. N. Winn, “Photonic Crystals, Molding the Flow of Light”, Princeton University Press, 1995.
Retrieving the permittivity
We have presented already the way to calculate the Fourier coefficients of the permittivity for cylindrical inclusions, with circular cross-section (see section 5 of March 19 2003 notes). An example was also given in Fig. 7. We shall just briefly comment on this. It is known already that a photonic crystal is defined by a lattice (rectangular, triangular, . . . ) and a basis (the shape of the inclusion). Both have to be incorporated in the retrieval of ² for example. In the treatment presented last time (section 5 of the reference mentioned above) we obtained the Fourier coefficients of the permittivity for a specific lattice (square). The result was:
1
2
Section 2. Retrieving the permittivity
² f + ² (1 − f ) a r r b ²˜(Gρ ) = 2J (G (²a − ²b )fr 1 ρ Rc ) Gρ Rc
if Gρ = 0 elsewhere.
(1)
Yet, the information on the basis is also included in Eq. (1), since both Gρ and fr will depend on it. It is straightforward to show that: ☞ Square lattice: fr = π
µ
Rc a
¶2
(where Rc is the radius of the inclusions and a the
lattice constant). µ ¶ 2π Rc 2 ☞ Triangular lattice: fr = √ . 3 a An illustration for square and triangular lattices is given in Fig. 1 (note that since the permittivity is obtained from its Fourier coefficients, a unavoidable Gibbs phenomenon will occur).
(a) Square lattice (fr = 38.48%).
(b) Triangular lattice (fr = 44.43%).
Figure 1: Reconstruction of the permittivity for cylindrical rods of circular crosssection for (a) a square lattice and (b) a rectangular lattice. Other parameters are: ²a = 1, ²b = 12, Rc /a = 0.35.
3
3
Band diagrams
The purpose of all the mathematical developments presented so far (getting the Fourier coefficients of the permittivity, building the eigensystem, etc) is to eventually obtain the eigenvalues and eigenvectors of the problem. Eigenvalues correspond to dispersion diagrams whereas eigenvectors correspond to the actual field distributions. We shall limit ourselves here to the consideration of eigenvalues only. Upon solving the eigensystem: ¶ ωk 2 ¯ H(¯ r) , c · ¸ 1 Θ=∇× ∇× , ²(¯ r)
¯ r) = Θ H(¯
µ
(2a) (2b)
¯ r) = 0 to reduce the size of the system), we get a set of eigenvalues (with the condition ∇ · H(¯ ¯ The band diagrams are then constructed by sweeping all possible k. ¯ for each incident k. Because of the periodicity of the medium the “all possible” k¯ can be reduced to the first Brillouin zone and, by symmetry, further reduced to the irreducible Brillouin zone. In addition, as we have seen, it is also enough to span the edge of the Brillouin zone since it corresponds to the maximum diffraction condition. Again by symmetry, we further reduce the domain to the edge of the irreducile Brillouin zone. This zone will of course depend on the lattice, as is shown in Figs 2 and 3 of the March 19 2003 notes. For both cases, the irreducible Brillouin zone is depicted in Fig. 2.
M
K
X
Γ
PSfrag replacements
M
Γ
PSfrag replacements
(a) Square lattice.
(b) Triangular lattice.
Figure 2: Irreducible Brillouin zones (red region) for (a) a square lattice and (b) a triangular lattice. Each region is defined between symmetry points of the crystal.
In order to span the edge of the irreducible Brillouin zone, we therefore need to know the coordinates of the symmetry points of the crystal. It is easy to show that:
4
Section 3. Band diagrams
• Square lattice: Γ → (kx = 0, ky = 0) , π X → (kx = , ky = 0) , a π π M → (kx = , ky = ) . a a
(3a) (3b) (3c)
• Triangular lattice: Γ → (kx = 0, ky = 0) , 2π 2π K → (kx = , ky = √ ) , 3a 3a 2π M → (kx = 0, ky = √ ) . 3a
(4a) (4b) (4c)
¯ solve the eigensystem and obtain the eigenvalues. Having these limit points, we can sweep k, An example is given in Fig. 3.
5
Square lattice: Rc/a = 0.48 (f=72.3823%), εa=1, εb=13. 1 TE TM 0.9
0.8
Frequency ω a/2π c
0.7
0.6
0.5
0.4
0.3
0.2
PSfrag replacements 0.1
0
Γ
X
Γ
M
(a) Square lattice.
Hexagonal lattice: Rc/a = 0.48 (f=83.5799%), εa=1, εb=13. 1 TE TM 0.9
0.8
Frequency ω a/2π c
0.7
0.6
0.5
0.4
0.3
0.2
PSfrag replacements 0.1
0
Γ
M
K
Γ
(b) Triangular lattice.
Figure 3: Band diagram for TE and TM modes as function of the normalized frequency for (a) a square lattice and (b) a triangular lattice. Notice the absolute band gap for TM modes. Parameters are: ²a = 1, ²b = 13, Rc /a = 0.48. Important note: TE modes are here defined as transverse to the axis of the crystal (z axis), therefore with an in-plane electric field!! Note also that these curves have not yet fully reached convergence.
6
Section 4. k-surfaces
4
k-surfaces
4.1
Well-chosen example
We can generalize the band diagram of the previous section (which, again, spans only the edge of the Brillouin zone), to the entire zone or even to the entire photonic crystal. The example we shall consider from now on is taken from [1] and the parameters are summarized in Tab. 1. In addition, we shall work with TE modes (remember that in this notation, TE corresponds to an in-plane electric field). Lattice: Inclusions: Background: Radius :
square ²a = 1 ²b = 12 Rc /a = 0.35
Table 1: Parameters for our example [1].
The band diagram obtained for this case is depicted in Fig. 4. Square lattice: Rc/a = 0.35 (f=38.4845%), ε =1, ε =12. a
b
0.5
Frequency ω a/2π c
0.4
0.3
0.2
TE 0.1
PSfrag replacements Γ Γ 0
0
X
M
1
Figure 4: Band diagram for TE modes for the structure defined in Tab. 1. The horizontal line is at the frequency corresponding to a change of curvature of the ksurface, and the green line is the light-line shifted to M.
.
7
We can also extend the plot from the edge of the irreducible Brillouin zone to the entire structure, for various eigenstates. This is represented in Figs. 5 to 7.
4.2
Negative refraction
From the k-surfaces shown in the figures of this section, we see that there is the potential of negative refraction (the iso-frequency surfaces converge to a point as frequency increases). Yet, we need to make sure that: 1. We can couple to one of these surfaces from free-space. 2. The power is bending on the same side of the normal. Answer to both points is shown in Fig. 4. 1. Light-line: the light-line gives the radius of the k-surface of an EM wave impinging on the crystal from free-space. In order to have a possible coupling, whole (or part) of the free-space k-surface has to be included in one of the k-surfaces of the crystal. The light-line is represented in Fig. 4 by the green line (it is actually the translation of the light-line to M ). Its intersection with say the curve of the first eigenvalue gives the maximal frequency for which the condition mentioned above (total inclusion of the free-space k-surface into a k-surface of the crystal) is satisfied. Note also that in order to have a negative refraction, the free-space k-surface has to be included in one of the k-surfaces converging to M and therefore, the actual crystal needs to be rotated by 45◦ . 2. Power bending: around M , the power is converging to a single point when frequency is increased. Yet, the direction of the power can be deduced from the gradient of the k-curves and therefore, directly from their radius of curvature. Hence, if the radius of curvature is such that the gradient is pointing toward M , the refraction will be negative. The frequency at which the radius of curvature diverge is given by the horizontal line in Fig. 4, which can be obtained by a direct inspection of Fig. 5(c).
As it can be seen, the power can bend on the opposite side of the normal. However, the phase is still propagating forward, which justifies the title of the paper: “All-angle negative refraction without negative index”. We can also examine the second eigenvalue and see if a similar phenomenon can appear. The k-surface has been depicted in Fig. 6(b) and is shown again in Fig. 9 with less curves represented. It is clear from this figure that again, there is a point at which the radius of curvature of the k-curves changes, and energy converges to a single point (Γ this time) as frequency increases. It is therefore again possible to have a negative index of refraction. Notice that in this case
8
4.2
Negative refraction
Squ. lattice. Band of eigenvalue nb 1. Rc/a = 0.35 (f=38.4845%), εa=1, εb=12.
0.25
0.2
0.15
0.1
0.05
0 4000 4000
2000 2000
0 0 −2000
−2000 −4000
−4000
(a) Brillouin cell.
(b) All structure.
Squ. lattice. Band of eigenvalue nb 1. Rc/a = 0.35 (f=38.4845%), εa=1, εb=12. 0.22 160 0.2 140 0.18 120
0.16 0.14
100
0.12
80
0.1 60 0.08 40 0.06 20
0.04 0.02 20
40
60
80
100
120
140
160
(c) 2D k-surface.
Figure 5: k-surfaces for the first eigenvalue of the system.
9
Squ. lattice. Band of eigenvalue nb 2. Rc/a = 0.35 (f=38.4845%), εa=1, εb=12.
0.36
0.34
0.32
0.3
0.28
0.26 4000 4000
2000 2000
0 0 −2000
−2000 −4000
−4000
(a) Brillouin cell.
Squ. lattice. Band of eigenvalue nb 2. Rc/a = 0.35 (f=38.4845%), εa=1, εb=12.
160
0.34
140 0.33 120 0.32 100 0.31 80
60
0.3
40
0.29
20 0.28 20
40
60
80
100
120
140
160
(b) 2D k-surface.
Figure 6: k-surfaces for the second eigenvalue of the system.
however, because of phase matching, power and phase are in directions that make an angle greater than π/2. Although it is not necessarily π like in a pure left-handed regime, it is still in the regime of left-handed behavior. Fig. 10 can be completed to see this phenomenon.
10
4.2
Negative refraction
Squ. lattice. Band of eigenvalue nb 3. Rc/a = 0.35 (f=38.4845%), ε =1, ε =12. a
b
0.45 0.44 0.43 0.42 0.41 0.4 0.39 0.38 4000 4000
2000 2000
0 0 −2000
−2000 −4000
−4000
(a) Brillouin cell.
Squ. lattice. Band of eigenvalue nb 3. Rc/a = 0.35 (f=38.4845%), εa=1, εb=12.
160
0.435
140
0.43
120
0.425
100
0.42
80
0.415
60
0.41
0.405
40
0.4
20
0.395 20
40
60
80
100
120
140
160
(b) 2D k-surface.
Figure 7: k-surfaces for the third eigenvalue of the system.
11
Squ. lattice. Band of eigenvalue nb 2. Rc/a = 0.35 (f=38.4845%), εa=1, εb=12. 0.31
55
50
0.3
45
40
0.29 35
30
0.28 25
20
0.27
15
0.26
10
5
5
10
15
20
25
30
35
40
45
50
55
Figure 9: k-surface for the second eigenvalue.
0.25
12
4.2
Negative refraction
Squ. lattice. Band of eigenvalue nb 2. Rc/a = 0.35 (f=38.4845%), εa=1, εb=12. 1.2 34 1 32 0.8
0.6
30
0.4 28 0.2 26
0
−0.2
24
−0.4 22 −0.6 22
24
26
28
30
32
34
Figure 10: k-surface for the second eigenvalue for the specific frequency where the radius of curvature is positive.
