Nama : Nida Dusturia NIM
: 1101124314
Kelas : TT-36-02
Tugas Antena Balanis No. 37 – 38 (Berdasarkan Kocokan/Undian)
37. A circularly polarized wave, traveling in the positive z-direction, is incident upon a circularly polarized antenna. Find the polarization loss factor PLF (dimensionless and in dB) for righthand (CW) and left-hand (CCW) wave and antenna. Jawab :
Misal : = + (, , )
=
+ √2
− = √2
1 = | . | = |1 + 1| = 1 = 0 $% 4 38. The electric field radiated by a rectangular aperture, mounted on an infinite ground plane with z perpendicular to the aperture, is given by
= & cos * − + sin * cos . (, , *)
where (, , *) is a scalar function which describes the field variation of the antenna. Assuming that the receiving antenna is linearly polarized along the x-axis, find the polarization loss factor (PLF). Jawab :
= & cos * − + sin * cos . (, , *) atau
cos * − + sin * cos =/ 6 0123 * + 345 * 123 . (, , *) 0123 * + 345 * 123
=
cos * − + sin * cos
0123 * + 345 * 123
= | . | = 7
cos * − + sin * cos
0123 * + 345 * 123
. 8 7
Transform dari rectangular unit vector ke spherical menggunakan : 8 = 9 sin cos * + cos cos * − + sin * cos = 123 * + 345 * 123
Tugas Antena Balanis (diluar kocokan/undian) 9. In target-search ground-mapping radars it is desirable to have echo power received from a target, of constant cross section, to be independent of its range. For one such application, the desirable radiation intensity of the antenna is given by
Find the directivity (in dB) using the exact formula. Jawab : 9: = ;
>
=
=
; <(, ) sin $ $ >
= 2? @A>
>F
E>F
sin $ + A >F 0,342 123C1 D sin $G =
=
=
= 2? @H− cos I + 1J + 0,342 ( K − I )G = 1,87912
11. The normalized radiation intensity of a given antenna is given by: (a) < = sin sin *
(b) < = sin sin *
(e) < = sin sin *
(f) < = sin sinK *
(c) < = sin sinK *
(d) < = sin sin *
The intensity exists only in the 0 ≤ θ ≤ π, 0 ≤ φ ≤ π region, and it is zero elsewhere. Find the
(a) exact directivity (dimensionless and in dB). (b) azimuthal and elevation plane half-power beamwidths (in degrees). Jawab:
a. < = sin sin * ; 9: = 3,1318 <M8 = 1 N> =
4? <M8 = 4,0125 = 6,034 $% 3,1318
b. < = sin sin * ; 9: = 2,4590 <M8 = 1 N> =
4? <M8 = 5,1104 = 7,0845 $% 2,4590
c. < = sin sinK * ; 9: = 2,0870 <M8 = 1 N> =
4? <M8 = 6,0212 = 7,80 $% 2,0870
d. < = sin sin * ; 9: = 2,6579 <M8 = 1 N> =
4? <M8 = 4,728 = 6,746 $% 2,6579
e. < = sin sin * ; 9: = 2,0870 <M8 = 1 N> =
4? <M8 = 6,021 = 7,797 $% 2,0870
f. < = sin sinK * ; 9: = 1,7714 <M8 = 1 N> =
4? <M8 = 7,094 = 8,51 $% 1,7714
30. A uniform plane wave, of a form similar to (2-55), is traveling in the positive z-direction. Find the polarization (linear, circular, or elliptical), sense of rotation (CW or CCW), axial ratio(AR), and til tangle τ (in degrees) when (a) Ex = Ey ,3φ = φy − φx = 0 (b) Ex = Ey,3φ = φy − φx = 0 (c) Ex = Ey ,3φ = φy − φx = π/2 (d) Ex = Ey ,3φ = φy − φx =−π/2 (e) Ex = Ey ,3φ = φy − φx = π/4 (f) Ex = Ey ,3φ = φy − φx =−π/4 (g) Ex = 0.5Ey ,3φ = φy − φx = π/2 (h) Ex = 0.5Ey ,3φ = φy − φx =−π/2 In all cases, justify the answer. Jawab :
a. Linear, karena ∆φ = 0
b. Linear, karena ∆φ = 0 c. Circular, karena :
8 = U dan ∆φ =
V
CCW, karena U mendahului 8 . AR=1, τ = 900
d. Circular, karena :
8 = U dan ∆φ = −
V
CW, karena U tertinggal dari 8 . AR=1, τ = 900
e. Elliptical, karena :
∆φ bukan kelipatan
V
CCW, karena U mendahului 8
f. Elliptical, karena :
∆φ bukan kelipatan
V
CW, karena U tertinggal dari 8
g. Elliptical, karena :
8 ≠ U dan ∆φ bukan nol atau kelipatan π
CCW, karena U mendahului 8
h. Elliptical, karena :
8 ≠ U dan ∆φ bukan nol atau kelipatan π
CW, karena U tertinggal dari 8
39. A circularly polarized wave, traveling in the +z-direction, is received by an elliptically polarized antenna whose reception characteristics near the main lobe are given approximately by :
≃ &28 + U . (, , *)
Find the polarization loss factor PLF (dimensionless and in dB) when the incident wave is (a) right-hand (CW) (b) left-hand (CCW) Circularly polarized. Repeat the problem when
≃ &28 − U . (, , *)
In each case, what is the polarization of the antenna? How does it match with that of the wave?
Jawab:
&28 ± U . ≃ &28 ± U . (, , *) = √5 (, , *) √5 a. = H =
[ \]^ √
J Wave Right-Hand (CW)
28 ± U √5
= | . | =
I
_>
= −0,4576 $% (Antena Left Hand di Receiving mode dan Right Hand di Transmitting mode)
_
= | . | = = −10 $% (Antena Right Hand di Receiving mode dan Left Hand _> di Transmitting mode) b. = H =
[ `]^ √
J Wave Left-Hand (CCW)
28 ± U √5
= | . | =
_
_>
= −10 $% (Antena Left Hand di Receiving mode dan Right Hand di Transmitting mode)
I
= | . | = _> = −0,4545 $% (Antena Right Hand di Receiving mode dan Left Hand di Transmitting mode)
45. A wave traveling normally outward from the page (toward the reader) is the resultant of two elliptically polarized waves, one with components of E, given by: U a = 3 cos bc
? 8 a = 7 cos Hdc + J 2
And the other with components given by:
U aa = 2 cos bc
? 8 a = 3 cos Hdc − J 2 a. What is the axial ratio of the resultant wave? b. Does the resultant vector E rotate clock wise or counter clock wise?
Jawab :
a. U = U a + U aa = 3 cos bc + 2 cos bc = 5 cos bc =
=
8 = 8 a + 8 aa = 7 cos Hbc + J + 3 cos Hbc − J
ef =
5 = 1,25 4
= −7 sin bc + 3 sin bc = −4 sin bc
b. ec bc = 0, gh = 5U
= ec bc = , gh = −48 Rotasinya CCW (Counter Clock wise)
PR Antena (Soal Bapak Heroe) Untuk susunan 8 elemen dipole
i
yang berjarak $ =
i j
antar elemennya. Distribusi Dolph
Chebychev, R = 26 dB. Hitung Distribusi Arusnya dan FNBW ! Jawab : n=8 R = 26 dB diubah menjadi numerik, R=20 m
m
m q\m q\m kl = oHp + 0pn − mJ + Hp − 0pn − mJ r n _
_
1 s\_ s\_ = oH20 + 020 − 1J + H20 − 020 − 1J r 2 = 1,15
t=
y\m
k kl
uqv = n w xy z{| H}ny + m~ J n yl
=
5 8 = =4 2 2
K
s = 2 w e cos H}2 + 1~ J 2 >
s = e> cos H J + e_ cos H3 J + e cos H5 J +eK cos H7 J
Jika t = z{| n
s(
) = e> b + e_ (4ωK − 3ω) +e (16ω − 20ωK + 5ω) +eK (64ω − 112ω + 56ωK − 7ω)
s(
) = (64eK )ω + (16e − 112eK )ω + (4e_ − 20e + 56eK )ωK + (e> − 3e_ + 5e − 7eK )ω
u(t) = q\m () s(
) = (D)
(64eK )ω + (16e − 112eK )ω + (4e_ − 20e + 56eK )ωK + (e> − 3e_ + 5e − 7eK )ω = 64x − 112x + 56x K − 7x
(64eK )ω = 64x
D (64eK ) = 64x D> eK = D>
eK = 1,15 = 2,66
(16e − 112eK )ω = −112x
D (16e − 112eK ) = −112x D> (16e − 112eK ) = −112D> e = 4,54
(4e_ − 20e + 56eK )ωK = 56x K
D K (4e_ − 20e + 56eK ) = 56x K D> (4e_ − 20e + 56eK ) = 56D> K e_ = 6,75
(e> − 3e_ + 5e − 7eK )ω = −7x D (e> − 3e_ + 5e − 7eK ) = −7x D> (e> − 3e_ + 5e − 7eK ) = −7D>
e> = 8,12
Distribusi Arusnya :
2,66
FNBW :
4,54
6,75
8,12
8,12
6,75
180> − 2(9,24> ) = 161,52>
4,54
2,66
