A hypothetical isotropic antenna

[Sze Wen]

Homework Statement

(1)A hypothetical isotropic antenna is radiating in free-space. At a distance of 100m from the antenna, the total electric field in the theta direction is measured to be 5V/m. Find the distance 150m for
(a) Power density
(b) Power radiation

Relevant Equations

Wrad = 0.5(Ex H*)
Prad = close integral of Wrad dS

At 100m:
(a) 0.03315 W/m
(b)4166 W

Since E is inversely proportional to 1/r^2, then E at 150m is 2.22 V/m.
(a) 2.22/377= 0.00654 W/m
(b) 4*pi*r^2*Wrad= 1665 W

Is this reasoning correct?

 

[Delta2]

I find it hard to comment on your work, however i believe there is one mistake in your work:

It is a well known fact that in the radiation zone (far-field zone) the fields are proportional to $\frac{1}{r}$ (NOT $\frac{1}{r^2}$) so you have to correct yourself a bit, I think the E-field at 150m is 3.33V/m.

 

[Sze Wen]

Since E is inversely proportional to 1/r, then E at 150m is 3.33 V/m.
H = E/377 = 3.33/377 = 0.00883 Am^(-1)

(a) Wrad = 0.5(ExH*) = 0.5*(3.33*2)/377= 0.0147 W/m
(b) Prad =4*pi*r^2*Wrad= 4156 W

Should I assume that the E will stilll be 50V/m at 150m?

 

[Delta2]

Since E is inversely proportional to 1/r, then E at 150m is 3.33 V/m.
H = E/377 = 3.33/377 = 0.00883 Am^(-1)

(a) Wrad = 0.5(ExH*) = 0.5*(3.33*2)/377= 0.0147 W/m
(b) Prad =4*pi*r^2*Wrad= 4156 W

I believe $P_{rad}$ should be the same in 100 and 150m and this is due to conservation of energy.

Should I assume that the E will stilll be 50V/m at 150m?

No.

 

[Sze Wen]

I believe $P_{rad}$ should be the same in 100 and 150m and this is due to conservation of energy.
Wrad = 0.5(ExH*) = 0.5*(3.33*3.33)/377= 0.0147 W/m
Thanks. I recalculated it. I got 4167 W as expected. Thanks for the correction.
No.

So I can claim that as distance increases, the Wrad must be decreasing. This is because the power density of the antenna will keep on decreasing with the spread of surface area? The surface area is spreading which causes the power that it have to spread out evenly.

The Prad should be constant as it is coming from the antenna source itself?

If the distance of the question is much smaller like 4m. Is it solvable with far field distance?

 

[Delta2]

So I can claim that as distance increases, the Wrad must be decreasing. This is because the power density of the antenna will keep on decreasing with the spread of surface area? The surface area is spreading which causes the power that it have to spread out evenly.

Yes the power density decreases as the distance increases because we have the same power the $P_{rad}$ power that spreads over a larger surface (a spherical surface of increasing radius). The product of power density x surface remains constant.

The Prad should be constant as it is coming from the antenna source itself?

Yes and due to conservation of energy it cannot be reduced or increased as the power radiates through space.

If the distance of the question is much smaller like 4m. Is it solvable with far field distance?

You can do the calculations at a distance of 4m and you will still get the result that $P_{rad}$ will be the same but $W_{rad}$ will be greater.

But in a distance of 4m and in real world applications we are in the near field of the antenna, and there are near field terms that are proportional to 1/r^2 and 1/r^3 for which the problem doesn't give us any data.

 

[rude man]

I

But in a distance of 4m and in real world applications we are in the near field of the antenna,

Nothing magic about 4m.
The near field is a fuction principally of wavelength and secondarily to antenna characteristics.

 

[Delta2]

I

Nothing magic about 4m.
The near field is a fuction principally of wavelength and secondarily to antenna characteristics.

Fine, it is just that if we include the near field terms, the fields are no longer mutually orthogonal, the calculation of the poynting vector and of radiating power is not so easy task, though it would still hold that the radiating power is independent of distance.

 

[rude man]

Fine, it is just that if we include the near field terms, the fields are no longer mutually orthogonal, the calculation of the poynting vector and of radiating power is not so easy task, though it would still hold that the radiating power is independent of distance.

Yes. But over the given distances the assumption is pretty obvious that far-field regions are assumed.