Spring 2015 - University of Houstoncourses.egr.uh.edu/ECE/ECE6345/Class Notes/Topic 4... · Spring...
Transcript of Spring 2015 - University of Houstoncourses.egr.uh.edu/ECE/ECE6345/Class Notes/Topic 4... · Spring...
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Spring 2015
Notes 5
ECE 6345
Prof. David R. Jackson ECE Dept.
1
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Overview
This set of notes discusses improved models of the probe inductance of a coaxially-fed patch (accurate for thicker
substrates). A parallel-plate waveguide model is initially assumed (at the end of the notes we will also look at the actual finite patch).
z
h ,r rε µ2a
x2b
2
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Overview (cont.)
The following models are investigated:
Cosine-current model
Gap-source model
Frill model
H. Xu, D. R. Jackson, and J. T. Williams, “Comparison of Models for the Probe Inductance for a Parallel Plate Waveguide and a Microstrip Patch,” IEEE Trans. Antennas and Propagation, vol. 53, pp. 3229-3235, Oct. 2005.
Reference:
3
Derivations are given in the Appendix. Even more details may be found in the reference below.
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Cosine Current Model
cosI z k z h
Note: The derivative of the current is zero at the top conductor (PEC).
2
2
0c
inPZ
I
Pc = complex power radiated by probe current
z
h ,r rε µ2a
x
I (z)
We assume a tube of current (as in Notes 4) but with a z variation.
4
*12
p
c z szS
P E J dS= ∫
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5
( ) 222 (2)0 0 0 0 0 0
0
1 1( ) sec (1 ) ( ) ( )8
min r m m m mmr
Z k h k h I k H k a J k aρ ρ ρη ε δε
∞
=
= +
∑
Cosine Current Model (cont.)
Final result:
where
0
1, 00, 0m
mm
δ=
= ≠
2 2
2 ( ) sin( )1 ( ) ( )m
mo
khI khkh mδ π
= + −
1/2221m
mk khρπ = −
0/m mk k kρ ρ=
Note: The wavenumber kρm is chosen to be a positive real number or a
negative imaginary number.
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Gap Source Model
An ideal gap voltage source of height ∆ is assumed at the bottom of the probe.
z
h ,r rε µ2a
x1V + - ∆
1
inZI
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Gap Source Model (cont.)
7
(2)0
(2)00 0 0
( )1 24 sinc(1 ) ( )
min r
m m m m
H k aa mY jh k H k a h
ρ
ρ ρ
ππ εη δ
∞
=
′ ∆ = + ∑
Final result:
where 1/22
21m
mk khρπ = −
Note: The wavenumber kρm is chosen to be a positive real number or a
negative imaginary number.
0
1, 00, 0m
mm
δ=
= ≠
0/m mk k kρ ρ=
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A magnetic frill of radius b is assumed on the mouth of the coax.
s ρˆˆ ˆM z E z ρ E
z
h ,r rε µ2a
xb
1 1
lnρEρ b / a
sM Eφ ρ= −
(TEM mode of coax, assuming 1 V)
10inZ
I
Frill Model
8
Choose:
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Frill Model (cont.)
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( )
( ) ( )
( )
2 20 0
2200 0 0 0
( ) ( )1 1 1 4ln / ( ) (1 ) ( )
m min r
m m m m
H k b H k aY j
k h b a k H k aρ ρ
ρ ρ
πεη δ
∞
=
− = +
∑
Final result:
where
0
1, 00, 0m
mm
δ=
= ≠
Note: The wavenumber kρm is chosen to be a positive real number or a
negative imaginary number.
1/2221m
mk khρπ = −
0/m mk k kρ ρ=
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Comparison of Models
Next, we show results that compare the various models, especially as the substrate thickness increases.
z
h ,r rε µ2a
x2b
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Comparison of Models
0 0.005 0.01 0.015 0.02 0.025h (m)
-300
-200
-100
0
100
200
300
Xin
(Ω)
Uniform current modelFrill modelCosine current modelHFFS simulation
Models are compared for changing substrate thickness.
εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)
11
z
h ,r rε µ2a
x2b
(λ0 = 15 cm)
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Comparison of Models (cont.)
0 0.005 0.01 0.015 0.02 0.025h (m)
0
50
100
150
200
250
300
350
400
450
500
Rin
(Ω)
Uniform current modelFrill modelCosine current modelHFFS simulation
εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)
/ 4 0.025 [m]dλ =Note:
Models are compared for varying substrate thickness.
12
z
h ,r rε µ2a
x2b
(λ0 = 15 cm)
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For the gap-source model, the results depend on ∆.
0 0.005 0.01 0.015 0.02 0.025h (m)
-300
-200
-100
0
100
200
300
Xin
(Ω)
Delta=0.01mmDelta=0.1mmDelta=1mm
Comparison of Models (cont.)
εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)
13
z
h ,r rε µ2a
x1V+-∆
(λ0 = 15 cm)
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The gap-source model is compared with the frill model,
for varying ∆, for a fixed h.
0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.001
∆ (m)
-300
-200
-100
0
100
200
300
400
500
Rin
or X
in (Ω
)
Gap source model reactance Frill model reactance Gap source model resistance Frill model resistance
Comparison of Models (cont.)
εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)
0.52mm3
b a−=
h = 20 mm
R
X
14
z
h ,r rε µ2a
x1V+-∆
(λ0 = 15 cm)
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These results suggest the “1/3” rule: The best ∆ is chosen as
3b a−
∆ =
This rule applies for a coax feed that has a 50 Ω impedance.
Comparison of Models (cont.)
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z
h ,r rε µ2a
x1V+-∆
z
h ,r rε µ2a
x2b
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Comparison of Models (cont.)
0 0.005 0.01 0.015 0.02 0.025h (m)
-300
-200
-100
0
100
200
Xin
(Ω)
Gap source modelFrill model
The gap-source model is compared with the frill model, using the
optimum gap height (1/3 rule).
εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)
16
z
h ,r rε µ2a
x2b
Reactance
(λ0 = 15 cm)
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Comparison of Models (cont.)
The gap-source model is compared with the frill model, using the
optimum gap height (1/3 rule).
0 0.005 0.01 0.015 0.02 0.025h (m)
0
100
200
300
400
500
Rin
(Ω)
Gap source modelFrill model
εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)
17
z
h ,r rε µ2a
x2b
Resistance
(λ0 = 15 cm)
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Probe in Patch A probe in a patch does not see an infinite parallel-plate waveguide.
Exact calculation of probe reactance:
L
W
x
y
(x0, y0)
cavityin p inZ jX Z
0
Imp in fX Z
f0 = frequency at which Rin is maximum
Zin may be calculated by HFSS or any other software, or it may be measured.
0cavityinX
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Probe in Patch (cont.)
Using the cavity model, we can derive an expression for the probe reactance (derivation given later)
This formula assumes that there is no z variation of the probe current or cavity fields (thin-substrate approximation), but it does accurately account for the actual patch dimensions.
Le
We
x
y
PMC
(x0, y0)
Cavity Model
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2 2 20 0
0 2 21 0 20 0
cos cos sinc2
1 14
p
e ep r
m,n , e em n
e e
nπwmπx nπyL W
X ω μ μ hW L mπ nπδ δ k
L W
3 2/pw e a
a = probe radius
(x0, y0) = probe location
Probe in Patch (cont.)
20
Final result:
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Using image theory, we have an infinite set of “image probes.”
Probe in Patch (cont.)
Image theory can be used to improve the simple parallel-plate waveguide model when the probe gets close to the patch edge.
Image Theory
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A simple approximate formula is obtained by using two terms: the original probe current in a parallel-plate waveguide and one image. This should be an improvement when the probe is close to an edge.
( ) ( ) ( ) ( )two 1 10 0 0 04 4 2inX = khY ka J ka khY k J ks aη η− −
0
0 /r
r
k k ε
η η ε
=
=
Probe in Patch (cont.)
Original
Image
two probe imagein in inX = X X+
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As shown on the next plot, the best overall approximation in obtained by using the following formula:
( )probe twomax ,in in inX X X=
Probe in Patch (cont.)
“modified CAD formula”
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Probe in Patch (cont.) Results show that the simple formula (“modified CAD formula”) works fairly well.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Xr
0
5
10
15
20
25
30
35
X f (Ω
)
Uniform current modelCavity modelUniform current model (with one image)Modified CAD formulaHFFS simulation
L
W
x
y
(x0, y0)
0 / 2/ 2r
x LxL−
=
24
( )probe twomax ,in in inX X X=
center edge
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Next, we investigate each of the improved probe models in more detail:
Cosine-current model Gap-source model Frill model
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Appendix
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Cosine Current Model
Assume that ( ) cos ( )I z k z h= −
Note: (0) cos ( )I kh=
z
h ,r rε µ2a
x
I (z)
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21 (0)2c inP Z I=
Cosine Current Model (cont.)
Circuit Model: (0)I
coax feed inZ
2
2
0c
inPZ
I
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Represent the probe current as: 0
( ) cosmm
m zI z Ihπ∞
=
=
∑
2
2
0c
inPZ
I
This will allow us to find the fields and hence the power radiated by the probe current.
z
h ,r rε µ2a
x
I (z)
28
Cosine Current Model (cont.)
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Hence
00 0
( ) cos cos cosh h
mm
m z m z m zI z dz I dzh h hπ π π∞
=
′ ′ =
∑∫ ∫
( )
( )
00
00
( )cos 12
2 ( )cos1
h
m m
h
mm
m z hI z dz Ih
m zI I z dzh h
π δ
πδ
= + ⇒ = +
∫
∫
Using Fourier-series theory:
Cosine Current Model (cont.)
The integral is zero unless m = m′.
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Result:
( ) 0
2 cos ( ) cos1
h
mmo
m zI k z h dzh h
πδ
= − + ∫or
2 2
2 ( ) sin( )1 ( ) ( )m
mo
khI khkh mδ π
= + −
(derivation omitted)
Cosine Current Model (cont.)
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(Time-Harmonic Fields)
Note: We have both Ez and Eρ To see this:
so
0E∇⋅ =
1 1( ) 0zE EEz
φρρ
ρ ρ ρ φ∂ ∂∂
+ + =∂ ∂ ∂
0Eρ ≠
Cosine Current Model (cont.)
31
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For Ez, we represent the field as follows:
where
( )00
cosz m mm
m zE A J kh ρπ ρ
∞− −
=
=
∑aρ <
aρ > ( )(2)0
0cosz m m
m
m zE A H kh ρπ ρ
∞+ +
=
=
∑
( )
1/ 222
1/ 22 2
m
zm
mk kh
k k
ρπ = −
= −
Cosine Current Model (cont.)
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At
so
aρ =
z zE E+ −= (BC 1)
( ) ( )(2)0 0m m m mA H k a A J k aρ ρ
+ −=
Cosine Current Model (cont.)
33
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Also we have
To solve for Eρ , use
(BC 2) 2 1 szH H Jφ φ− =
1 zE EHj z
ρφ ωµ ρ
∂ ∂−= − ∂ ∂
EjH ωε=×∇
Cosine Current Model (cont.)
where
34
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so 1
1
z HHj Ez
HE
j z
φρ
φρ
ωερ φ
ωε
∂∂= −
∂ ∂∂
= −∂
Hence we have 2
2
1 1 zH EHj j z
φφ ωµ ωε ρ
∂ ∂= − − −
∂ ∂
For the mth Fourier term:
( )( )
( ) 2 ( )1 1 mm m z
zmEH k H
j jφ φωµ ωε ρ ∂
= − − − − ∂
Cosine Current Model (cont.)
35
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Hence
( )2 ( ) 2 ( )
mm m z
zmEk H k H jφ φ ωερ
∂− = −
∂
2 2 2zm mk k kρ− =
so that
( )( )
2
mm z
m
EjHkφρ
ωερ
∂= −
∂
where
Cosine Current Model (cont.)
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where ( )( )
2
mm z
m
EjHkφρ
ωερ
∂= −
∂
2 1 2szIH H J
aφ φ π− = =
( ) ( ) ( )2 1 2m m m m
szIH H J
aφ φ π− = =
For the mth Fourier term:
Cosine Current Model (cont.)
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Hence
we have
( ) ( )(2)0 02 ( )
2m
m m m m mm
Ij k A H k a A J k ak aρ ρ ρρ
ωεπ
+ − − ′ ′− =
(2)0(2)
0 00
( )( ) ( )
( ) 2m mm
m m m pmm
H k a kIA H k a A J k aJ k a a j
ρ ρρ
ρ π ωε+ +
′ ′− = −
Cosine Current Model (cont.)
( ) ( )(2)0 0m m m mA H k a A J k aρ ρ
+ −=Using (BC 1)
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(2) (2)0 0 0 0 0( ) ( ) ( ) ( ) ( )
2mm
m m m m m m
kIA J k a H k a H k a J k a J k aa j
ρρ ρ ρ ρ ρπ ωε
+ ′ ′− = −
02 ( )
2mm
m mm
kIA j J k ak a a j
ρρ
ρπ π ωε+ − = −
( )2
014
mm m m
kA I J k aρ
ρωε+
= −
Hence
or
or
(using the Wronskian identity)
Cosine Current Model (cont.)
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*
2*
0 0
*
0
*
0
(2) *00
0 ' 0
12
1 ( )2
( )
( ) ( )21 ( )cos cos2
scs
h
z sz
h
z sz
h
z
h
m m mm m
P E J dS
E a J a dz d
a E a J dz
a E a I z dza
m z m zA H k a I dzh h
π
ρ
φ
π
ππ
π π∞ ∞+
′= =
−= ⋅
−=
= −
= −
′ = − ⋅
∫
∫ ∫
∫
∫
∑ ∑∫
We now find the complex power radiated by the probe:
Cosine Current Model (cont.)
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( )
( )
* (2)0 0
0
2(2) *
0 0 00
1 ( ) 12 2
11 ( ) ( )4 4
c m m m mm
mm m m m m
m
hP A I H k a
kh H k a I J k a I
ρ
ρρ ρ
δ
δωε
∞+
=
∞
=
= − +
= − + −
∑
∑
Integrating in z and using orthogonality, we have:
Hence, we have:
2 2 (2)0 0 0
0
1 (1 ) ( ) ( )16c m pm m m m
m
hP I k H k a J k aρ ρδωε
∞
=
= + +
∑
Am+ coefficient
Cosine Current Model (cont.)
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Therefore,
2
2cos ( )
cin
PZkh
=
( )22 2 (2)0 0 0
0
1 sec ( ) 1 ( ) ( )8in m m m m m
m
hZ kh I k H k a J k aρ ρ ρδωε
∞
=
= +
∑
Define: 0
2
0
mm
r r
kk
k
mk h
ρρ
πε µ
=
= −
42
Cosine Current Model (cont.)
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We then have
Also, use r
kεη
ωεεµω
ωε0000 ==
( ) 222 (2)0 0 0 0 0 0
0
1 1( ) sec (1 ) ( ) ( )8
min r m m m mmr
Z k h k h I k H k a J k aρ ρ ρη ε δε
∞
=
= +
∑
The probe reactance is: Im( )p inX Z=
43
Cosine Current Model (cont.)
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Keep only the m = 0 term
( ) 222 (2)0 0 0 0 0 0
0
1 1( ) sec (1 ) ( ) ( )8
min r m m m mmr
Z k h k h I k H k a J k aρ ρ ρη ε δε
∞
=
= +
∑
0 1:k h <<
(2)0 0 0 0
1 ( ) ( ) ( )4in rZ k h J ka H kaη µ≈
(same as previous result using uniform model)
Thin substrate approximation
2 2
2 ( ) sin( )1 ( ) ( )m
mo
khI khkh mδ π
= + −
The result is
44
Cosine Current Model (cont.)
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Gap Model z
h ,r rε µ2a
x1V + - ∆
( ) 1/ , 0,
0, otherwise.z
zE z a
− ∆ < < ∆=
( ) ( ) ( )20
0, cosz m m
m
m zE z B H khρπρ ρ
∞
=
=
∑
Note: It is not clear how best to choose ∆, but this will be re-visited later.
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Gap Model (cont.) z
h ,r rε µ2a
x1V + - ∆
( ) ( ) ( )20 0
2 sinc1m
m m
mBhh H k aρ
πδ
− ∆ = +
From Fourier series analysis (details omitted):
( ) ( ) ( )20
0
1 / , 0, cos
0, otherwise.z m mm
zm zE z a B H k ahρπ∞
=
− ∆ < < ∆ = =
∑
At ρ = a:
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Gap Model (cont.)
( )2in szY a Jπ= ∆ ( ) ( )szJ z H zφ=where
The magnetic field is found from Ez , with the help of the magnetic vector potential Az (the field is TMz):
1 zAHφ µ ρ∂
= −∂
22
2
1z zE k A
j zωµε ∂
= + ∂
( ) ( ) ( )20
0, cosz m m
m
m zA z A H khρπρ ρ
∞
=
=
∑
where
z
h ,r rε µ2a
x1V + - ∆
Use:
Setting ρ = a allows us to solve for the coefficients Am.
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Gap Model (cont.)
(2)0
(2)0 0 0
( )1 24 sinc(1 ) ( )
min
m m m m
H k aa mY j kh k H k a h
ρ
ρ ρ
ππη δ
∞
=
′ ∆ = + ∑
Final result:
z
h ,r rε µ2a
x1V + - ∆
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z
h ,r rε µ2a
xb
Frill Model
To find the current I (z) , use reciprocity.
10inZ
I
Introduce a ring of magnetic current K = 1 in the φ direction at z (the testing current “B”).
( )1 1
ln /sMb aφ ρ
= −
F
a b
V
b a
V
bs
S
I z H M dV A,B B,A
H M dV
H M dS
A
B SF
z
1V frill
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Frill Model (cont.)
( )
( )
( )
( ) ( )
( ) ( )
2
0
,0
2 ,0
1 12 ,0ln /
2 ,0ln /
F
bs
S
bb
sa
bb
sa
bb
a
bb
a
I z H M dS
H M d d
H M d
H db a
H db a
π
φ φ
φ φ
φ
φ
ρ ρ ρ φ
π ρ ρ ρ
π ρ ρ ρρ
π ρ ρ
= ⋅
=
=
= −
= −
∫
∫ ∫
∫
∫
∫
A B
SF z
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( ) ( ) ( )2 ,0ln /
bgap
a
I z H db a φπ ρ ρ= − ∫
z
b
The magnetic current ring B may be replaced by a 1V gap source of zero height (by the equivalence principle).
Let z → 0:
The field of the gap source is then calculated as was done in the gap-source model, using ∆ = 0.
Frill Model (cont.)
z
b
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Final result:
( )
( ) ( )
( )
2 20 0
220 0 0
( ) ( )1 1 4ln / ( ) (1 ) ( )
m min
m m m m
H k b H k aY j k
h b a k H k aρ ρ
ρ ρ
πη δ
∞
=
− = +
∑
Frill Model (cont.)
z
h ,r rε µ2a
xb
52