Antennas Balanis Solutions Chapter 5

Antennas Balanis Solutions Chapter 5 - 5.7 A resonant...

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Unformatted text preview: 5.7. A resonant 6-turn loop of closely spaced turns is operating at 50 MHz. The radius of the loop is M30, and the loop is connected to a SO-ohm transmission line. The radius of the wire is M300, its conductivity is 0‘ = 5.7 X 107 Sim, and the spacing between the turns is M 100. Determine the (a) directivity of the antenna (in dB) (b) radiation efficiency taking into account the proximity effects of the turns (c) reflection efficiency ' (d) gain of the antenna (in dB) '7 _ a 5 ' " K30 , b: A/300, 2C: fl/mo =,‘>C =Moo, N=gaf= SHOE a. Since G: 7L/3c) << 7\ D.= I-5 =L'761 d5 b. 1%.: .20 Trifif: zonz C-fi‘fl‘r =20 Tr‘Cl 324 x155)=o.3r!?8 ohms C = 2m=2fi (5-7;) =+§A Ry (Single turn): 0.37?8 :3th Rv- C 6 turns) = [3.676 clams PL: flg'Rs'CgE-H) C = 74.200 _ §_ __ A’ 1/30!) _ z“ 1'5 9%.:06'5 (Sin le) I? = Ni? 2F): (NT/Ho"! H 4“; -7 3 L A/Sooy/ QCSIWX‘O?) —101/5~_7f_xm = ZUCIOJ no“: t86.o?:?xto“t 5‘2. (6 turns) (-1: [36_O?;fi>.(g)a55)xw‘4: l,842.3|Xtoh+.5L a Er {3.973 ecd— 2 - ° Kloo - _6 PM“ 3513+ o4a4231 (f8 7 A C = l (RF'I‘RL —50 _ [3.857“50 __ -36 [4.277 _ ‘;1_ _|._-___ :o.566 (Rr-I-RL) +50 [3857-1-50 63.857 er = Cl—tf‘l‘)xloo =(I—- {0.566l") xloo = (t—o.sz)xloo=68z at G0 = ecdD‘, =(o.?867)D. = Co.?867)(/.5) = [, H) 3 .45 105; ' Gro = l-48005 , 6- Total Maximum 34m does not indude the refledlufi 5.13. Design a circular loop of constant current such that its field intensity vanishes only at 0 = 0° (0 = 180°) and 90°. Find its (a) radius (b) radiation resistance (c) directivity 5'13. Since Ep ~ J. ckaszne) a. E¢le=o=$Ckasine)le=°=3l(o)=o E53 leaf/2 =3 (kasinefleqo‘: mamzo => 4m: 8.84 Thus a=§fi= 3713;}? =06“:st b. Since azaems A) 035A, use 9.0va loop approximation. ACCovdinj m cs—esa> RY: some/A)=wear)=60v*<2v<o-ws>) = 2.27334 C The directivity is given by (13—631)). or D0 = 0,682 (—2) =o.682(2%)=0-682C2TU(016W5) :261? 5.17. A very small (:1 < A) circular loop of constant current is placed ’a distance 12 above an infinite electric ground plane. Assuming z is perpendicular to the ground plane, find the total far~zone field radiated by the loop when its plane is parallel to the (a) -x-z plane (b) y~z plane A veg smll loop of Constant current is equlunleni: to a magnetic dipole, Since the loop Is placeol for both ports (a and b) perpendicular to the xy—plcme C the plane of the loop is perpend— Iculor to the xyoplqne), the axis: of the linear magnetic. dipole will also be parallel to the xy-plane. Thereforce acrorollna "to Fljure. 4 . 12o, “the image, of the horizontal wajnefic diPole— will be as Shown infill: fijure. In turn the array factor for both port‘s (a and b) of this problem will be thescxme as tharof the Vertical. eleckrlc dipole of Fljurezlla or AF = 2-Co3 Clehcose) '8’ Actual Source Since the actual Source and the image l. are oriented In the Same direction. Therefore uCcordlnj to C5—2!{a)~(5~.2r{c) Cl. Plane of the loop is parallel to the x — I z pone 1" Ex = ’Z————‘““"I° 5’ only (AF), siml=ll~C°SW = l1 453‘“ 4r ' = \{l ~sln‘es‘m‘p! ; :4: =7 (lea) roe) '81”, [laugh-cow] ...
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This note was uploaded on 11/01/2009 for the course EEE 443 taught by Professor Balanis during the Spring '08 term at ASU.

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Antennas Balanis Solutions Chapter 5 - 5.7 A resonant...

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