5.5. NUMERICAL RESULTS 83
0 10 20 30
-60
-50
-40
-30
-20
-10
0
d [mm]
|E
my
/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
0 10 20 30
-70
-60
-50
-40
-30
-20
-10
0
d [mm]
|E
my
/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)
Γ
Γ
Γ
Γ
Γ
Γ
Γ
Γ
Figure 5.20: The first eight harmonics of dipole FSS layer at 15 GHz with oblique incidence (k
x
=
20 m
1
,k
y
= 10 m
1
), (a) Reflected components, (b) Transmitted components.
5 6 7 8 9 10 11 12 13 14 15
0
0.2
0.4
0.6
0.8
1
Frequency [GHz]
Coefficients Magnitude
Γ
Co
En ti re
Γ
Co
Casc.
Γ
x
Ent i re
Γ
x
Casc.
T
Co
Ent i re
T
x
Casc.
5 6 7 8 9 10 11 12 13 14 15
0
0.2
0.4
0.6
0.8
1
Frequency [GHz]
Coefficients Magnitude
Γ
Co
Ent i re
Γ
Co
Casc.
Γ
x
Ent i re
Γ
x
Casc.
T
Co
En ti re
T
Co
Casc.
(a) (b)
Figure 5.21: Reflection and transmission coefficients of two identical dipole FSS with oblique incidence
TE
z
case (k
x
= 20 m
1
,k
y
= 10 m
1
), (a) d = 10 mm, (b) d = 18 mm.
5.5.5 TEST CASE 5 (1:1 CASE, OBLIQUE INCIDENCE AND SMALL GAP)
The algorithm is used to analyze the same structure shown in Fig. 5.15. The structure is illuminated
by an obliquely incident plane wave k
x
=40m
1
and k
y
=0m
1
(for minimum frequency of almost
1.9 GHz); the frequency range of interest is 5-15 GHz, and d = 10 mm. The structure is to be
simulated using the cascading technique; the same procedure used in test case 4 was used, and it
was found that for a gap of 10 mm at a frequency of 15 GHz, only one harmonic needs to be added

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