5.5. NUMERICAL RESULTS 77

8 10 12 14 16 18 20

0

0.2

0.4

0.6

0.8

1

Frequency [GHz]

Magnitude

Γ

FDTD Casc

T FDTD Cas c

Γ

Analytical Enti re

T Anal yt i c al Enti r e

8 10 12 14 16

-200

-150

-100

-50

0

50

100

150

200

Frequency [GHz]

Phase [deg]

Γ

FDTD Casc

T FDTD Cas c

Γ

Analytical Enti re

T Analyti cal Enti re

(a) (b)

Γ

Γ

Γ

Γ

Figure 5.13: Reﬂection and transmission coefﬁcients of inﬁnite dielectric slab with oblique incidence

k

x

= 104.8 m

−1

TE

z

, (a) Magnitude, (b) Phase.

Γ

Γ

Γ

Γ

8 10 12 14 16 18 20

0

0.2

0.4

0.6

0.8

1

Frequency [GHz]

Magnitude

Γ

FDTD Casc

T FDTD Casc

Γ

Analyti cal Enti re

T Anal yt i c al Ent i re

8 10 12 14 16 18 20

-200

-150

-100

-50

0

50

100

150

200

Frequency [GHz]

Phase [deg]

Γ

FDTD Casc

T F DTD Casc

Γ

Analyti cal Enti re

T Anal yti cal Enti re

(a) (b)

Figure 5.14: Reﬂection and transmission coefﬁcients of inﬁnite dielectric slab with oblique incidence

k

x

= 104.8 m

−1

TM

z

, (a) Magnitude, (b) Phase.

5.5.2 TEST CASE 2 (1:1 CASE, NORMAL INCIDENCE AND LARGE GAP)

In this test case, the multi-layer geometry consists of two identical FSS structures consisting of

dipole elements (1:1 case) separated by an air gap of width d. The dipole length is 12 mm and width

is 3 mm. The per iodicity is 15 mm in both x- and y-directions. The substrate has a thickness of

6 mm and relative permittivity ε

r

= 2.2, as shown in Fig. 5.15. The structure is illuminated by a TE

z

78 5. MULTILAYERED PERIODIC STRUCTURES

Figure 5.15: Two identical dipole FSS geometry (all dimensions are in mm).

normally incident plane wave (with polarization along y-axis). The frequency range of interest is

0-16 GHz. The FDTD grid cell size is x = y = z = 0.5 mm and 2,500 time steps and a 0.9

reduction factor of CFL time step are used. The CPML is used as the absorbing boundaries at the

top and the bottom of the computational domain. The ﬁrst step is to determine the distance d after

which the level of all the harmonics are less than −40 dB relative to the corresponding magnitude

of the incident ﬁeld components. Using the gap determination procedure:

1) The two layers are identical; analyzing the harmonics of one layer is enough. The reﬂection

and transmission harmonics must be calculated.

2) The frequency range of interest as speciﬁed by the problem is 0-16 GHz (as shown in Fig. 5.9,

at the highest frequency the effect of harmonics is maximum).

3) k

i

x

and k

i

y

are equal to zero (normal incidence). Determine the cut-off frequencies for the ﬁrst

eight harmonics as follows:

M

0,1

,M

0,−1

→ f

0,1

cut-off

= f

0,−1

cut-off

= 20GHz

M

1,0

,M

−1,0

→ f

1,0

cut-off

= f

−1,0

cut-off

= 20GHz

M

1,1

,M

1,−1

→ f

1,1

cut-off

= f

1,−1

cut-off

= 28.3GHz

M

−1,1

,M

−1,−1

→ f

−1,1

cut-off

= f

−1,−1

cut-off

= 28.3GHz

4) Use the harmonic analysis to calculate the magnitude coefﬁcient of the eight harmonics and

plot the behavior of these harmonics versus frequency, as shown in Figs. 5.16 and 5.17.

5.5. NUMERICAL RESULTS 79

M-1,-1 M-1,1 M-1,0 M0,-1 M0,0 M0,1 M1,0 M1,1 M1,-1

-60

-50

-40

-30

-20

-10

0

Harmonics in x- and y- directions

|E

ty

/E

i

| [dB]

0 5 10 15 20

-60

-50

-40

-30

-20

-10

0

d [mm]

|E

m

/E

i

|[dB]

M

0,1

& M

0,-1

M

1,0

& M

-1,0

M

-1-,1

& M

-1,1

& M

1,1

& M

1,-1

(a) (b)

Figure 5.16: The eight transmitted harmonics at 16 GHz: (a) Magnitude compared to incident electric

ﬁeld, (b) Decaying relative magnitude versus gap distance.

M-1,-1 M-1,1 M-1,0 M0,-1 M0,0 M0,1 M1,0 M1,1 M1,-1

-60

-50

-40

-30

-20

-10

0

Harmonics in x and y directions

|E

r

/E

i

| [dB]

0 5 10 15 20

-60

-50

-40

-30

-20

-10

0

d [mm]

|E

m

/E

i

|[dB]

M

0,1

&M

0,-1

M

1,0

&M

-1,0

M

-1,-1

&M

-1,1

&M

1,1

&M

1,-1

(a) (b)

Figure 5.17: The eight reﬂected harmonics at 16 GHz: (a) Magnitude compared to incident electric

ﬁeld, (b) Decaying relative magnitude versus gap distance.

As can be noticed from Figs. 5.16 and 5.17, almost 95% of the dominant mode will be

transmitted.In addition,a distance d = 15.5 mm between the two layers for this range of frequencies is

considered enough to neglect all the higher harmonics effects (the magnitude of all higher harmonics

are less than −40 dB compared to the incident ﬁeld magnitude).To validate the cascading technique

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