A halfwave dipole with length of 05 m operates in FMTV broad

A half-wave dipole with length of 0.5 m operates in FM/TV broadcast band at 300 mhz. The dipole is made of copper wire with a radius of 1 mm. determine the radiation efficiency of the antenna. (the loss resistance of a circular wire of radius a and length l is given by 1/2pia squareroot of pi integral meau 0/sigma,the conductivity of copper is sigma=5.8 x10^3 s/m, meau o= 4 pi x 10^7H/M. determine the radiation efficiency of the antenna. (The loss resistance of the of a circular wire of radius a and length l is given by l is given by l/2 pia a squareroot of pi f meau o/sigma,the conductivity of copper is sigma =5.8x10^7S/m,meau = 4 pi x10^7H/m. calculate the antenna gain in decibels.

Solution

= c/ f =( 3 * 10⁸ m/s) / (300*10⁶ Hz) =1 m.

As l/ = (0.5 m) / (1 m)=0.5m , this antenna is a short (Hertzian) dipole. Thus,

R_rad= 80² (l/ )² = 80²(0.5)²= 197.192 ()

R_loss=(l/2a) (fµ_c/ c)^1/2 = (0.5/2*10^-3) ((300*10⁶)(4*10^-7)/(5.8*10⁷))^1/2=0.359

Radiation efficiency of the antenna () = R_rad/ (R_rad+ R_loss)  =197.192/(197.192+0.359)=99.81%

A Hertzian dipole has a directivity of 1.5. The gain is G = D =0.9981 * 1.5= 1.497