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LECTURE 21: MICROSTRIP ANTENNAS – PART II (Transmission-line model. Design procedure for a rectangular patch. Cavity model for a rectangular patch.) 1. Transmission Line Model – The Rectangular Patch

The TL model is the simplest of all, representing the rectangular patch as a parallel-plate transmission line connecting two radiating slots (apertures), each of width W and height h. In the figure below, z is the direction of propagation of the transmission line.

W

h

Slot #2

Slot #1

L

yzx

The TL model is not accurate and lacks versatility. However, it gives a

relatively good physical insight into the nature of the patch antenna and the field distribution for all TM00n modes.

The slots represent very high-impedance terminations on both sides of the transmission line (almost an open circuit). Thus, the patch has highly resonant characteristics depending crucially on its length L along z. The resonant length of the patch, however, is not exactly equal to the physical length due to the fringing effect. The fringing effect makes the effective electrical length of the patch longer than its physical length, effL L> . Thus, the resonance condition

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( ) / 2neffL nβ π= ⋅ , 1,2,n = , depends on effL , not L. A sketch of the E-field

distribution for the first (dominant) resonant mode, 1n = , is shown in the figure below.

x

z

patch (side view)

h

ground

effL L>

L

001E

A. Computing the effective patch length

( )

( )

0.3 0.2640.412

0.258 0.8

eff

eff

r

r

WL h

Whh

ε

ε

+ + ∆ = − +

. (21.1)

For the computation of effrε , see previous Lecture. The effective length is

2effL L L= + ∆ . (21.2)

L∆LL∆

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B. Resonant frequency of the dominant TM001 mode

0 (001)(001) (001)2 2 2 2eff eff

eff rr eff r

v c cL ff f L

λε ε

= = = ⇒ = (21.3)

The resonant frequency of a patch depends strongly on L, therefore, the exact calculation of Leff is necessary to predict the antenna resonance:

( )

(001)

2 2eff

rr

cfL Lε

=+ ∆

. (21.4)

The field of the TM001 mode does not depend on the x and y coordinates but it strongly depends on the z coordinate, along which a standing wave is formed. The figure below shows the vertical E-field distribution along z when the patch is in resonance.

z

/ 2effz L=

cosxeff

E zLπ

-1-0.8-0.6-0.4-0.2

00.20.40.60.8

1

0 effz L=

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C. The patch width W

0 0

1 2 21 2 12 r r rr

cWff ε εµ ε

= =+ +

(21.5)

Expression (21.5) makes the width W equal to about half a wavelength. It leads to good radiation efficiencies and acceptable dimensions. D. Equivalent circuit of the patch

The dominant TM001 mode has a uniform field distribution along the y-axis at the slots formed at the front and end edges of the patch. The equivalent conductance G is obtained from the theory of uniform apertures while B is related to the fringe capacitance:

2

0 0 0

1 2 11 , for 120 24 10

W h hG πλ λ λ

= − <

, (21.6)

2

0 0 0

2 11 0.636ln , for 120 10

W h hB πλ λ λ

= − <

. (21.7)

The limitation 0( / ) 0.1h λ < is necessary since a uniform field distribution along the x-axis is assumed. The patch has two radiating slots (see the figure below).

x

zy

L

W(001)E

slot #

1

slot #

2

The equivalent circuit of a slot is constructed as a parallel R-C circuit, using the values computed by (21.6) and (21.7):

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B G

1 /G R= represents the radiation loss, while B j Cω= is the equivalent

susceptance, which represents the capacitance of the slot. More accurate values for the conductance G can be obtained through the

cavity model: 2/ (120 )G I π= , (21.8)

where

( ) 2

0 3

0

sin 0.5 cos sinsin 2 cos ( )cos ik W XI d X X S X

X

π θθ θ

θ

= = − + + ⋅ + ∫ , (21.9)

and X = k0W, 0 0 0k ω µ ε= . iS denotes the sine integral, ( )iS x =

0(sin ) /

xy ydy∫ .

The equivalent circuit representing the whole patch in the TM001 mode includes the two radiating slots as parallel R-C circuits and the patch connecting them as a transmission line, the characteristics of which are computed in the same way as those of a microstrip line.

B G BG

/ 2eff gL λ≈

0 0, effc g rZ β ω µ ε ε=

Here, cZ is the characteristic impedance of the line, and gβ is its phase constant. When the losses are not neglected, we must include also the attenuation constant α (see Lecture 20). For each slot, G represents the radiation loss and B Cω= represents the capacitance associated with the fringe effect.

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E. Resonant input resistance When the patch is resonant, the susceptances of both slots cancel out at the

feed point regardless of the position of the feed along the patch. Thus, the input admittance is always purely real. This real value, however, strongly depends on the feed position along z. This is easily shown through the Smith chart for the admittance transformation through a transmission line. At the feed point, the impedance of each slot is transformed by the respective transmission line representing a portion of the patch:

B G BG

/ 2eff gL λ≈

inY

1L 2L

1Y 2Y

1 2inY Y Y= + (21.10)

The admittance transformation is given by

1tan( ),

tan( ) g

L c gin c L c cL

c L g

Y jY LY Y Y Y ZY jY L β π

ββ

−=

+= = = +

(21.11)

if the line is loss-free. Below, the Smith charts illustrate the slot-impedance transformations and their addition, which produces a real normalized admittance, in three cases: (1) the patch is fed at one edge ( 1 0L = , 2L L= ), (2) the patch is fed at the center ( 1 2 / 2L L L= = ), and (3) the patch is fed at a distance (feed inset) 0 0.165z λ= .

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feed-point at the edge of cavity

0.46L λ≈

slot #11Y

2Yslot #2

2Y ′

2 0.46L L λ= ≈

1 2 2 0.09 0.18inY Y Y ′= + = × =

line to slot #2

1(0.18/50) 278inZ −= ≈ Ω

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feed-point at the middle of cavity

0.46L λ≈

slot #1

slot #2

1Y

2Y

2Y ′ 1Y ′

1 2 / 2 0.23L L L λ= = ≈

1 2 2 15 30inY Y Y′ ′= + = × =1(30/50) 1.67inZ −= ≈ Ω

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slot #1

1 0 0.165L z λ= ≈

line to slot #2

0.46L λ≈

line to slot #1

2 0 (0.46 0.165) 0.315L L z λ λ= − ≈ − =

slot #2

1 2 2 0.5 1inY Y Y′ ′= + = × =

1Y

2Y

2Y ′1Y ′

0feed-point at inset z

50inZ ≈ Ω

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The edge feed and the inset feed are illustrated below.

0z

L

slot #1 slot #2

inY inY

The two slots are separated by an electrical distance of 180° . However, because of the fringe effect the physical length L is slightly less than λ/2. The reduction of the length is not much. Typically, it is 0.48λ ≤ L ≤ 0.49λ.

Ideally, the resonant input impedance of the patch for the dominant TM001 mode is entirely resistive and equal to half the transformed resistance of each slot:

1

1 12in in

inZ R

Y G= = =

′. (21.12)

In reality, there is some mutual influence between the two slots, described by a mutual conductance and it should be included for more accurate calculations:

1 12

12( )inR

G G=

′ ±, (21.13)

where the “+” sign relates to the odd modes, while the “–” sign relates to the even modes. Normally, 12 1G G′

. For most patch antennas fed at the edge, Rin is greater than the characteristic

impedance Zc of the microstrip feed line (typically Zc = 50 to 75 Ω). That is why, the inset-feed technique is widely used to achieve impedance match.

The figure below illustrates the normalized input impedance of a 1-D (along the y axis) loss-free open-ended transmission-line, the behavior of which is very close to that of the dominant mode of the patch.

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Fig 14.14, pp 735, C. Balanis

Using modal expansion, the input resistance for the inset-feed at 0z z= is given approximately by

( )

2 211 12 2

0 0 021 12

1 2cos sin sin2in

c c

G B BR z z zG G L Y L Y L

π π π+ = + − ± . (21.14)

Here, 1G and 1B are calculated using (21.6) and (21.7). For most feeding microstrips, 1 / 1cG Y and 1 / 1cB Y . Then,

( ) ( )

2 20 00

1 12

1 cos cos2in in zR y R y

G G L Lπ π

= = = ±

. (21.15)

Notice that the inset feeding technique for impedance match of the microstrip antennas is conceptually analogous to the off-center or asymmetrical feeding techniques for dipoles. In both cases, a position is sought along a resonant structure, where the current magnitude has the desired value.

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2. Designing a Rectangular Patch Using the Transmission Line Model Input data: εr, h, fr

1) Calculate W using (21.5). 2) Calculate εreff using (21.5) and equation (6) from Lecture 20. 3) Calculate the extension ∆L due to the fringing effect using (21.1). 4) Calculate the actual (physical) length of the patch using

0 22

L Lλ= − ∆ or

0 0

1 22 r reff

L Lf ε µ ε

= − ∆ . (21.16)

5) Calculate radiating slot admittance using (21.6) and (21.7). 6) Calculate resonant input resistance at patch edge using (21.12) or

(21.13) with 1G G′ = from (21.6). 7) If Rin calculated in step 6 is too large, calculate the inset distance 0z

using (21.14) or (21.15). 3. Cavity Model for the Rectangular Patch

The TL model is very limited in its description of the real processes taking place when a patch is excited. It takes into account only the 00TMx

n modes where the energy propagates only in the longitudinal z direction. The field distribution along the x and y axes is assumed uniform. It is true that the dominant 001TMx is prevalent but the performance of the patch is also affected by higher-order modes.

The cavity model is a more general model of the patch which imposes open-end conditions at the side edges of the patch. It represents the patch as a dielectric-loaded cavity with:

- electrical walls (above and below), and - magnetic walls (around the perimeter of the patch.

The magnetic wall is a wall at which

ˆ 0 (the -field is purely normal)ˆ 0 (the -field is purely tangential)× =⋅ =

n H Hn E E

It is analogous to the open end termination in the theory of transmission lines. If we treat the microstrip antenna only as a cavity, we can not represent

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radiation because an ideal loss-free cavity does not radiate and its input impedance is purely reactive. To account for the radiation, a loss mechanism is introduced. This is done by introducing an effective loss tangent, δeff.

The thickness of the substrate is very small. The waves generated and propagating beneath the patch undergo considerable reflection at the edges of the patch. Only a very small fraction of them is being radiated. Thus, the antenna is quite inefficient. The cavity model assumes that the E field is purely tangential to the slots formed between the ground plane and the patch edges (magnetic walls). Moreover, it considers only TMx modes, i.e., modes with no Hx component. These assumptions are, basically, very much true.

W

hL

yzx

magnetic wall

magnetic wallelectric wall

The TMx modes are fully described by a single scalar function Ax – the x-component of the magnetic vector potential: ˆxA=A x . (21.17) In a homogeneous source-free medium, Ax satisfies the wave equation: 2 2 0x xA k A∇ + = . (21.18) For regular shapes (like the rectangular cavity), it is advantageous to use the separation of variables:

2 2 2

22 2 2

0x x xx

A A A k Ax y z

∂ ∂ ∂+ + + =

∂ ∂ ∂ (21.19)

( ) ( ) ( )xA X x Y y Z z= (21.20)

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2 2 2

22 2 2

X Y ZYZ XZ XY k XYZx x x

∂ ∂ ∂+ + = −

∂ ∂ ∂

2 2 2

22 2 2

1 1 1X Y Z kX x Y y Z z∂ ∂ ∂

+ + = −∂ ∂ ∂

(21.21)

2 2 2

2 2 22 2 2

0, 0, 0x y zd X d Y d Zk X k Y k Zdx dy dz

+ = + = + = (21.22)

The eigenvalue equation is 2 2 2 2

x y zk k k k+ + = . (21.23)

The solutions of (21.22) are harmonic functions: ( ) cos( ) sin( )c s

n xn n xnn

X x A k x A k x= +∑ ,

( ) cos( ) sin( )c sn yn n yn

nY y B k y B k y= +∑ , (21.24)

( ) cos( ) sin( )c sn zn n zn

nZ z C k z C k z= +∑ .

When the functions in (21.24) are substituted in (21.20), they give the general solution of (21.18). The particular solution of (21.18) depends on the boundary conditions.

In our case, there are electric walls at 0x = and x h= . There, the tangential E-field components must vanish, i.e.,

0,0y z x h

E E=

= = . Having in mind that

2 2 2

22

1 1 1, , x x zx x y z

A A AE k A E Ej x j x y j x zωµε ωµε ωµε

∂ ∂ ∂ = + = = ∂ ∂ ∂ ∂ ∂ ,(21.25)

we set xA at the top and bottom walls as

0,

0x

x h

Ax =

∂=

∂. (21.26)

At all side walls, we set a vanishing normal derivative for xA :

0, 0,

0, 0x x

z L y W

A Az y= =

∂ ∂= =

∂ ∂. (21.27)

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This ensures vanishing xH and yH at 0z = and z L= , as well as vanishing xH and zH at 0y = and y W= (magnetic walls), as follows from the relation

between the H-field and xA ,

1 10, , x xx y x

A AH H Hz yµ µ

∂ ∂ = = = ∂ ∂ . (21.28)

It is now obvious that the solution must appear in terms of the functions

( ) cos( ), ,

( ) cos( ), ,

( ) cos( ), .

cn xn xn

n

cn yn yn

n

cn zn zn

n

X x A k x k nh

Y y B k y k nW

Z z C k z k nL

π

π

π

= =

= =

= =

∑

∑

∑

(21.29)

The spectrum of the eigenmodes in the cavity is discrete. The frequencies of those modes (the resonant frequencies) can be calculated from (21.23) as

( )2 2 2 2( )mnp

rm n ph W Lπ π π ω µε + + =

, (21.30)

2 2 2

( ) 12

mnpr

m n pfh W Lπ π π

π µε = + +

. (21.31)

The mode with the lowest resonant frequency is the dominant mode. Since usually L > W, the lowest-frequency mode is the 001TMx mode, for which

(001) 12 2

rr

cfL Lπ

π µε ε= = . (21.32)

The dominant 001TMx mode is exactly the mode considered by the transmission-line model (see previous sections). The field distribution of some low-order modes is given in the following figure.

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Fig. 14.13, pp. 741, Balanis

The general solution for the ( )mnp

xA [see (21.20) and (21.24)] is

( ) cos cos cosmnp c c cx x x xA A m x B n y C p z

h W Lπ π π =

, (21.33)

or

( ) cos cos cosmnpx mnpA A m x n y p z

h L Wπ π π = ⋅ ⋅ ⋅

. (21.34)

The respective field solution for the (m,n,p) mode is

2 2( ) cos( ) cos( ) cos( )x

x mnp x y zk kE j A k x k y k zωµε−

= − ⋅ ⋅ ⋅ , (21.35)

sin( ) sin( ) cos( )x yy mnp x y z

k kE j A k x k y k z

ωµε= − ⋅ ⋅ ⋅ , (21.36)

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sin( ) cos( ) sin( )x zz mnp x y z

k kE j A k x k y k zωµε

= − ⋅ ⋅ ⋅ , (21.37)

0xH = , (21.38)

cos( ) cos( ) sin( )zy mnp x y z

kH A k x k y k zµ

= − ⋅ ⋅ ⋅ , (21.39)

cos( ) sin( ) cos( )yz mnp x y z

kH A k x k y k z

µ= ⋅ ⋅ ⋅ . (21.40)

For the dominant 001TMx mode,

2 2 2001( / ) / ( ) cos( / ), 0,x y zE j k h A z L E Eπ ωµε π= − − = = (21.41)

001( / ) sin( / ), 0.y x zH L A z L H Hπ µ π= − = = (21.42)

4. Cavity Model for the Radiated Field of a Rectangular Patch

The microstrip patch is represented by the cavity model reasonably well assuming that the material of the substrate is truncated and does not extend beyond the edges of the patch. The four side walls (the magnetic walls) represent four narrow apertures (slots) through which radiation takes place.

The equivalence principle is used to calculate the radiation fields. The field inside the cavity is assumed equal to zero, and its influence on the field in the infinite region outside is represented by the equivalent surface currents on the surface of the cavity.

yzx

electric wall

,s sJ M

ˆˆ

s

s

= ×= − ×

J n HM n E

,s sJ M

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Because of the very small height h of the substrate, the field is concentrated beneath the patch. There is some actual electrical current at the top metallic plate, however, its contribution to radiation is negligible. That is because: (1) it is backed by a conductor, and (2) it is very weak compared to the equivalent currents at the slots. The actual electrical current density of the top patch is maximum at the edges of the patch.

In the cavity model, the side walls employ magnetic-wall boundary condition, which sets the tangential H components at the slots equal to zero. Therefore, ˆ 0s = × =J n H . (21.43)

Only the equivalent magnetic current density ˆs = − ×M n E (21.44)

has substantial contribution to the radiated field.

sMy

zx

electric wall

0ˆ2

s

s

== − ×

JM n E

sM

The influence of the infinite ground plane is accounted for by the image theory, according to which the currents sM in the presence of the infinite plane radiate as if magnetic currents of double strength radiate in free space: ˆ2s = − ×M n E . (21.45)

Note that an xE field at the slots corresponds to sM density vector, which is tangential to the ground plane. Thus, its image is of the same direction. The equivalent magnetic current densities for the dominant 001TMx mode are sketched below.

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x

zy

L

sM(001)E

slot #

1

slot #

2sM

At slots #1 and #2, the equivalent sM currents are co-directed and with equal amplitudes. They are constant along x and y. Radiation from a slot with constant current density

W

h

( , )Q x z′ ′

( , , )P x y z

P′

PQr

rz

y

x

φsM′r θ

The radiation from an (x-y) slot of constant sM currents is found using the

electric vector potential F. Since sM has only a y component, so does F: ˆyF=F y .

0

/2 /2

/2 /2

( , , )4

PQ

h Wy jk r

yPQh W

MF r e dx dy

rεθ φπ

−

− −

′ ′= ∫ ∫ . (21.46)

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Here, 02yM E= − , 0E being the phasor of the E-field at the radiating slot, and ˆ sin cos sin sinPQr r r x yθ φ θ φ′ ′ ′= − ⋅ = − −r r .

0

/2 /2

0 0 0/2 /2

2 exp( sin cos ) exp( sin sin )4

h Wjk ry

h W

eF E jk x dx jk y dyr

ε θ φ θ φπ

−

− −

′ ′ ′ ′=− ⋅∫ ∫ (21.47)

00 sin sin2

jk ry

E Wh X YF er X Y

επ

−⇒ = − ⋅ ⋅ ⋅ (21.48)

where

0 sin cos2

k hX θ φ= ,

0 sin sin2

k WY θ φ= .

According to the relation between the far-zone E-field and the vector potential, 0, , rE E j F E j Fφ θ θ φωη ωη≈ = = − , (21.49)

where 0 0/η µ ε= , cos sinyF Fθ θ φ= , and cosyF Fφ φ= .

000

sin sincos sin2

jk rWhE X YE j er X Yφ ωηε θ φ

π−⇒ = , (21.50)

000

sin sincos2

jk rWhE X YE j er X Yθ ωηε φ

π−⇒ = − . (21.51)

Since 0 0kωηε = ,

000

sin sincos sin2

jk rV X YE jk W er X Yφ θ φ

π− =

, (21.52)

000

sin sincos2

jk rV X YE jk W er X Yθ φ

π− = −

. (21.53)

Here 0 0V hE= is the voltage between the patch edge and the ground plane. Slots #1 and #2 form an array of two elements with excitation of equal

magnitude and phase, separated by the physical distance L. Their AF is

012 2cos cos

2effk L

AF θ =

. (21.54)

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Here 2effL L L= + ∆ is the effective patch length. Thus, the total radiation field is

000 0 sin sincos sin cos cos2

efft jk r k Lk WV X YE j er X Yφ θ φ θ

π− = ×

, (21.55)

000 0 sin sincos cos cos2

efft jk r k Lk WV X YE j er X Yθ φ θ

π− = − ×

. (21.56)

Introducing 0( / 2)coseffZ k L θ= , the pattern of the patch is obtained as

( ) 2 2 2 2 sin sin, 1 sin sin cosX Yf E E ZX Yφ θθ φ φ θ= + = − ⋅ ⋅ . (21.57)

E-plane pattern (xz plane, φ = 0°, 0° ≤ θ ≤ 180°)

( )0

0

0

sin sin2 cos cos

2sin2

effE

k hk L

f k h

θθ θ

θ

= ⋅

. (21.58)

0.2

0.4

0.6

0.8

1

30

6090

120

150

180 0

x

z

0

0

0.1; 2.20.5 /

r

eff r

k hL

ελ ε

= ==

E-plane

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0.2

0.4

0.6

0.8

1

30

6090

120

150

180 0

0

0

0.1; 4.00.5 /

r

eff r

k hL

ελ ε

= ==

E-plane

x

z

H-plane pattern (xy plane, θ = 90°, 0°≤ φ ≤ 90° and 270°≤ φ ≤ 360°)

( )0 0

0 0

sin cos sin sin2 2coscos sin

2 2

H

k h k W

f k h k W

φ φθ φ

φ φ

= ⋅ ⋅ (21.59)

0.2 0.4 0.6 0.8 1

30

60

90270

300

3300

x

yH-plane

0

0

0.1; 2.20.5 /

r

r

k hW

ελ ε

= ==

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0.2 0.4 0.6 0.8 1

30

60

90270

300

3300

H-plane0

0

0.1; 4.00.5 /

r

r

k hW

ελ ε

= ==

x

y

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Fig. 14.17, p. 746, Balanis

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Fig. 14.18, p. 747, Balanis

Non-radiating slots: It can be shown that the slots at / 2y W= − and / 2y W= do no radiate in the principle E- and H-planes. In general, these two

slots do radiate away from the principle planes, but their field intensity is everywhere small compared to that radiated by slots #1 and #2.