Question 2 Answer the following parts about the fundamental...

Question 2 Answer the following parts about the fundamental parameters of antennas and radiation mechanisms: The formula of the electric field of finite length dipole is: $$E_\theta = \frac{jI_0 l e^{-jkr}}{2\pi r} \left| \frac{\cos \left(\frac{kl}{2}\cos\theta\right) - \cos \left(\frac{kl}{2}\right)}{\sin \theta} \right| $$ Write the electric filed formulas of $\lambda/2$ (i.e. $l = \lambda/2$) dipole and for a dipole (i.e. $l = \lambda$). If we are 0.8$\lambda$ away from both antenna, which antenna is in far field region? Justify your answer.

Transcript

00:01 So here in this question we have to show that for the case of the line of the sight propagation and no reflection the root mean square electric field at a radius r is powered as erms that is equals to 7 .02 which is multiplied by the p raised to the power 1 divided by 2r multiplied by the cos of 0 .5 pi multiplied by the cos of theta which from here is divided by r that is multiplied by the sin of theta.

00:29 So this is about the first part.

00:32 So from here this is the first part we can say that as we know that for electric field due to the half wave dipole which is given here we are having the half wave dipole.

00:44 So we are considering about half wave dipole that from here is represented as r of rod that is equals to 73 ohm.

00:52 So the value of pr from here is equal to summation that is from 0 to 2 of r for the rad that is divided by 2 which from here is equals to z naught to its whole square which is multiplied by the 73 that is divided by 2.

01:05 So the value of z naught from here is equals to under root 2 which is divided by 73 multiplied by the pr.

01:12 So the value of e naught from here is equals to 120 pi which is divided by the 2 pi r which is multiplied by sin of theta that is further multiplied by the 2 of pr which is divided by the 73 multiplied by the cos of 0 .5 pi of cos of theta.

01:29 Solving the term from here we get the value of e naught that become equals to 9 .93 that from here is equals to 9 .93 multiplied by the under root of pr which is divided by r sin of theta multiplied by the cos of 0 .5 multiplied by the cos of theta.

01:47 So the value of erms from here is equals to e naught which is divided by under root 2 that is equals to 7 .02 multiplied by the under root of pr which is divided by r sin of theta multiplied by the cos of 0 .5 of pi which is multiplied by the cos of theta.

02:05 So this is the value of erms hence the answer to the part a of the question...

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