Electromagnetic Coupling to Thin Wire Structures

Electromagnetic Coupling to Thin Wire Structures inside Resonators S.V. Tkachenko1), J.B. Nitsch1) , R. Rambousky2) and R. Vick 1) 1))Otto-von-Guericke-University

Magdeburg, Magdeburg Germany 2)WIS, Munster, Germany

EUROEM 2016 London, 14 Julyy

Outline •Introduction Introduction •Approximate analytical solution of EFIE for electrically small antenna inside resonator. •Approximate analytical solution for thin-wire antenna (transmission line) inside resonator. •Exact analytical solution of EFIE for thin-wire structures keeping the symmetry of resonator. •Conclusion C l i

Objectives • Coupling with electronic equipment including electrically small (printed circuits, chips, etc.) as well as electrically large objects (long interconnectors) inside boxes and racks . • Interpretation of the correspondence between testing (scattering) data for the device under test in shielded rooms (reverberation chambers) and in free space. •Using well – developed transmission line (TL) models for the investigation of the coupling between TL and cavity modes modes. g obtained analytical y solutions it is p possible to p provide a •Using quick qualitative and quantitative analysis of the system for different parameter values.

   EFIE for the current J (l ) induced by the electric field E0 ( r ) in a wire

l inside the cavity

    E el  ( E0  GRES J)

 E GRES

l 

0

- is the tensor integral Green's function operator for the electric field in the resonator:

  E GRES  J 

G

 r2 l

  E GRES (r1 , r2 ,  )  E RES

    A r1 , r2 , , GRES (r1 , r2 ,  )

E RES

    (r1 , r2 (l ),  )  J (r2 (l ),  )  dl 

1 i 0 0

k

2 0

  A (r1 , r2 ,  )  grad (div ) GRES

where G are tensor (dyad) Green's functions in a resonator for the electric field and the vector potential, respectively

How to solve EFIE for thin wires in a cavity? I.

Numerical methods: MoM, TLM, etc.

II.

Analytical methods: 1. Approximate methods based on small physical parameters. a). ) Method of small antenna for electrically y small electric and magnetic scatterers (antennas). Small parameter: L/λ lilinear system t ffor modes d iin interval k 0    k v  k 0   (S. Tkachenko, J. Nitsch, R. Vick). 0,003

TEM, inside cavity A l ti l iinside Analytical, id cavity it Analytical, free space

Im(J(0)), ( ( )), A

0,002 0,001 0,000 -0,001

RES

-0,002

f, MHz -0,003 40

50

60

70

80

Comparison between analytical and TLM calculations for long antenna (L=4.9 m, h=0.5 m, a=0.5 mm) exited by a central voltage source V0=1 V in the large MSC (A 7 9m B (A=7.9m, B=6.4 64m m, H H=3.45 3 45 m)

Exact analytical solution of EFIE for thin-wire structures keeping the symmetry y y of the resonator. I. Solution of EFIE for symmetrical thin-wires in free space: 1. Straight wire parallel to PEC or several parallel wires. Translation symmetry with arbitrary translation vector. Method of solution - Fourier integral. integral Modal functions - e  jk1l representation of symmetry group.





2. A vertical semi-circular loop above PEC or severall co-axial i l wires. i A Axial i l rotation t ti symmetry. Method of solution – Fourier series. Modal functions - e  jm representation of symmetry group. group





II. Solution of EFIE for symmetrical thin-wires in resonator. Symmetry of rectangular

    t  m a e  m b e  m h e resonator – translation symmetry with fixed translations vector x x y y z z Method of solution – Fourier series. Modal functions - cos( . nz z / h) representation of symmetry group (taking into account the boundary conditions)

Exact analytical solution of EFIE for the short-circuit line Resonator Green’s Green s function for the electric field:

4 0 E   Gzz (r , r1 )  jkV v  m1 , m2 , m3

 m (k 2  kv2 ) sin( k xv x1 ) sin(k xv x) sin( k yv y1 ) sin( k yv y ) cos(k zv z1 ) cos(k zv z )  kv2  k 2  j m 1 

3

1

m2 1 m3  0

k xv   m1 / a k yv   m2 / b k zv   m3 / h

 E ( r ) Exciting EM field in the resonator : ex z

k v2  k xv   k yv   k zv  2

2

2



ex E  z ( x, y , m )  cos( m3 z / h)

m3  0

sc This field induces scattered current J , which is, in turn, the source of a scattered field E



h

  E (  , z )   GzzE (  , z ,  0 , z1 ) J sc ( z1 )dz1 sc z

0

    E zsc  0  r0 e y , z   E zex  0  r0 e y , z   0

The scattered current is defined by the EFIE: We are looking for the solution in the form :

J ( z)  sc



sc J  (m3 )  cos(m3 z / h)

m3  0

sc sc J sc ( z ) satisfies short   J ( z)  z zL  0 short-circuit circuit boundary conditions J ( z )  z z 0

sc

Exact analytical solution of EFIE for the short-circuit line ex E ( x , y , m ) jk J sc ( m3 )  z 02 0 2 3  0 (kv  k )  S

- an exact equation for the current Fourier amplitude Value S 2 v v v characterizes 4  sin(k y ( y0  r0 )) sin(k y y0 ) sin (k x x0 ) S  S (k , m3 , x0 , y0 )  two-dimensional ab m1 1 kv2  k 2  j scattering m2 1



What is connection of exact solution with the TL solution? If the th summation ti is i d done over th the n2 index i d after ft some calculations l l ti one obtains: bt i

 2 sinh(~v (b  y0 )) sinh(~v y0 ) 1  1  a | sin( x0 / a ) |  1  2 S   sin ( kvx x0 )  ln     ~ ~ a n1 1  v sinh( v a ) kvx  2   r0   If the frequency is far from the cavity resonances:

|   v | 

4 sin (k x ) sin (k y0 ) 2

where v ~

v x 0

2

abk ln2 x0 / r0  2 v

v y

v

is a shift of eigenfrequency from the TL in resonator

and the wire is close to one of the walls: r0  x0  y0 , a  S 

1 ln 2 x0 / r0  2

Connection with TL solution (cont)

jk  E zex ( x0 , y0 , m3 ) sc sc J ( z )  J TL ( z )   Z c m3 0 kv2  k 2

where

ZC 

 0  2 x0  (h  x )  ln TL 0 2  r0 

sc J TL ( z ) is a solution of the TL equations with short-circuit boundary conditions

obtained by Fourier transform sc sc dU TL dz  jLJ TL  E zex  sc sc  dJ TL dz  jC U TL  0

  2x  L  0 ln 0  2  r0 

C 

2 0 ln2 x0 r0 

sc sc U TL (0)  U TL ( L)  0 h

J

sc TL

  E zex ( z1 ) gTL ( z , z1 , k , h)dz1

- Solution of TL system with Green’s function

0

gTL ( z , z1 , k , h)  

cosk (h  z )  cos(k z1 ), z1  z j  Z C sin( kh) cosk (h  z1 )  cos(k z ), z1  z

Exact analytical solution of EFIE. Lumped excitations and loads. A lumped excitation from the left side:

E zleft ( x0 , y0 , z )  U 0 ( z   ),   0

jkU 0   m3 cos( m3 z / h) U0 sc : U 0Yleft ( z ) E ( x0 , y0 , m3 )   m3 cosm3 z / h   J left ( z )   2 2 h 0 m3 0 (kv  k )  S h ex z

right A lumped excitation from the right side: E z ( x0 , y0 , z )  U 0 ( z  L   ),

0

m jkU 0   m3 ( 1) 3 cos( m3 z / h) J ( z)  : U 0Yright ( z )  h 0 m3 0 ( kv2  k 2 )  S sc left

How to take into account lumped loads?

The lumped loads are considered as lumped sources with unknown amplitudes. ex.tot sc sc ex . For the total exciting field: E z ( z )   Z1 J (0) ( z   )  Z 2 J ( h) ( z  h   )  E z ( z ) sc sc Then the total current is a sum of three terms, where the amplitudes J tot (0) and J tot ( h) are found by the solution of a system of linear equations.

Exact analytical solution of EFIE. General solution. J totsc ( z )   Z1 J totsc (0)Yleft ( z )  Z 2 J totsc (h)Yright ( z )  J msc ( z ) jk   m3 cos(( m3 z / h) Yleft ( z )  ,  2 2 h 0 m3 0 (kv  k )  S

jk Yright ( z )  h 0

E zex (m3 ) cos( m3 z / h) J ( z)   (k 2  k 2 )  S  0 m3 0 v sc m

jk



J msc (0) 1  Z 2 Z 00   J msc (h) Z 2 Z 01 J ( 0)  1  Z1 Z 00 1  Z 2 Z 00   Z1Z 2 /( Z 01 ) 2 sc tot

J msc (h) 1  Z1 Z 00   J msc (0) Z1 Z 01 J ( h)  1  Z1 Z 00 1  Z 2 Z 00   Z1Z 2 /( Z 01 ) 2 sc tot

( Z 00 ) 1  Yleft (0)  Yright (h) ( Z 01 ) 1  Yleft (h)  Yright (0)





m3  0

 m (1) m cos( m3 z / h) 3

3

(kv2  k 2 )  S

This equations yield solution of EFIE for arbitrary excitation of loaded symmetrical line in resonator. After complex calculations (sometimes numerically) it is possible to show that for

S

1 ln(2 x0 / r0 ) 2

this equation is equal to the TL solution.

Example 1. Lumped voltage source excitation: input impedance and transfer admittance Analytical, resonator Analytical, TL Analytical TL-approximation approximation in resonator PROTHEUS, resonator

100000

Z1=0 Z2=0.1 M 10000

|Zin(j)|, 

1000

100

10

TL ( h)  J SC

U1 Z C j sin kh  Z 2 cos kh

1 0

100

150

200

250

300

350

400

450

f, MHz

Analytical result PROTHEUS code TL approximation

0,010

50

TL, free space Resonator

0,004 0,003

x0= 9 cm Z2=ZC=311 

x0=9cm

0,002

I(h,t), A

|I(h,j)|, A*s

0 001 0,001 0,005

0,000 -0,001

Matched load Modulated sinus source: U(t)=U0*exp(-t)sin(2f0t)h(t)

-0,002 -0,003 0,000 150

200

250

300

350

400

450

7 -1

U0=1 V, f0=370 MHz, =1*10

s

-0,004 0,0

-8

5,0x10

f, MHz

A=1.2 m, b=h=0.8m, r0=1mm, y0=0.4 m, x0=0.2 m, Zc=311 Ω

-7

1,0x10

t, s

-7

1,5x10

-7

2,0x10

Example 2. Distributed Field Excitation by an Interior Small Antenna: Current Transfer Ratio. Parameters of the resonator: a=30 cm, b=53 cm, h=79 cm, Q=104 Parameters of transmission line: r0=1 mm, LTL=h, hTL=x0=a/2, =a/2 hTL=x0=a/2, =a/2 y0=b/2 Parameters of the field generating small antenna: x1=0.118 m, y1=0.424 m, z1=0 m, LA=dl=4.1 cm, I0=1A. Loads: Z1=Z2=50 Ω Classical TL, Z1=Z2=342 

Classical TL Analytical solution PROTHEUS

z=h-1cm

10

Analytical solution, Z1=Z2=342 

z = 1cm

PROTHEUS simulation, Z1=Z2=342 

10

|I(z,f)/I0|

1

|I(z,f)/I0|

1

0,1

0,1 0,01

0,01

500

750

1000

1250

1500

Titel X-Achse

500

750

1000

1250

1500

f MHz f,

Magnitude of current ratio near the end of the conductor at z=h-1cm. Both ends are terminated with 50 Ohm.

Magnitude of current transfer ratio at z=1cm. Both ends are terminated with 342 Ohm. 18

Example 3. Distributed Field Excitation by an Exterior Plane Wave: Transfer Function “External Field->Current” Excitation of a cavity through a circular hole on the top by a perpendicularly incoming wave. Parameters of the resonator: a=30 cm, b=53 cm, h=79 cm, Q=104 Parameters of transmission line: r0=1 mm, LTL=h, hTL=x0=a/2, hTL=x0=a/2, y0=b/2 Loads: Z1=Z2=50 Ω Parameters of the hole: x1=a, y1=0.1m, z1=0.3m, d=5 cm. Classical TL TL + space Fourier transform Exact solution

Excitation through the aperture z=1cm

1 E -5 5

|I(z,f)/E0|, A/ V/m

|I(z,f)/E E0|, A/ V/m

1E-5

1E-6

1E-7

C la s s ic a l T L T L + s p a c e F o u rie r tra n s fo rm E x a c t s o lu tio n

E x c ita tio n th ro u g h th e a p e rtu re z = h -1 c m , Z 1 = Z 2 = 3 4 2  = ~ Z C

1 E -6 1 E -7 1 E -8 8 1 E -9

1E-8

1 E -1 0 1E-9

250 0

500

1000

f, MHz

Z1=Z2=50 Ohm.

1500

500

750

1000

1250

1500

f, M H z

Z1=Z2=342 Ohm=~ZC

Magnitude g of the transfer function “external field->current” 19

Example 4. Excitation of multiloaded transmission line by lumped voltage source.

.

Line in resonator, PROTHEUS Line in resonator, analytical method Line in free space, TL approxim ation

0,005

Currents in the first load

|I(z(1),f)|

0,004

|I(z(5),f)|

Resonator: a=30cm, b=53cm, h=79cm. Length and position of the transmission line: L=h=79cm, , x0=a/2, y0=b/2, and r0=1mm. Loads: N=5. Z1=Z2=Z3=Z4=5ZC, Z5=ZC=~333 Ohm Source: Uo=1 V Definition of the currents for each frequency can be reduced to the algebraic equation of Nth order. order Currents in the fifth, matched load (transfer coefficient)

Line in resonator, PROTHEUS Line in resonator, analytical method Line in free space, TL approximation

1E-3

0,003

0,002

1E-4

0,001

1E-5

0,000

500

1000

1500

2000

2500

f, MHz

Magnitude of the current spectrum in the load Z1

Allowed zones 0

500

1000

1500

2000

f, MHz

Magnitude of the current spectrum in the load Z5

Filtering effect in the periodically loaded transmission line is disappear in resonator! 20

Example 5. Excitation of two parallel transmission lines by lumped voltage source. Resonator: a=30cm, b=53cm, h=79cm. Length and position of the transmission lines: Conductor 1: x01=a/2, y01=b/2. Conductor 2:x02=a/4, =a/4 y02=b/2. =b/2 Conductor one is fed by voltage source with U0=1 V. Both conductors are short circuited. Definition of the currents for each frequency can be reduced to the algebraic equation of second order. Two wires in resonator One wire in resonator One wire, TL approximation

1

0 ,1

|I2(f,0)|, A

0,1

|I1(f,0)|, A

PROTHEUS A n a ly tic a l c a lc u la tio n

1

0 ,0 1

0,01

1 E -3 3

1E-3 1 E -4

1E-4 1 E -5 200

1E-5 0

200

400

600

800

1000

400

600

800

1000

f, M H z

f, MHz

Magnitude of the current at the left terminal of the first wire wire, with and without the second one

Magnitude of the current at the left terminal of the second wire. 21

Example 6. Excitation of two perpendicular transmission lines by lumped voltage source. Resonator: a=1.5 m, b=1.2 m, h=0.9cm. Length and position of the transmission lines: Conductor 1, z-direction: x01=9 cm, y02=37 cm. Conductor 2 2, x - direction:y02=0.6 =0 6 m m, z02=0.4 =0 4 m m. Conductor one is fed by voltage source with U0=1 V. Both conductors are short circuited. Definition D fi iti off th the currents t ffor each h frequency f can be b reduced to the linear system of Mmax order, where Mmax is a number of resonator modes taken into account (Mmax=500). Two wires in resonator One wire in resonator 1

1

0,1

|I2(k,z=0)|, A

0,1

|I1(k,z=0)|, A

1st wire 2nd wire

0,01

1E-3

0 01 0,01

1E-3

1E-4 1E-4 1E-5 1E-5 -2

0

2

4

6

8

k, m

10

12

14

16

-1

Current in the first wire at the point z=0 with and without another wire.

1E-6 0

2

4

6

8

k, m

10

12

14

-1

Comparison of the currents in the perpendicular wires at the points z=0 and x=0, correspondingly.

22

The investigation of the multiconductor structures in resonator allows making the next conclusions: 1. Our method and PROTHEUS calculations have a good agreement for the considered system of two wires. 2. Far from the cavity resonances (|ω-ωv|