2
$\begingroup$

I am a bit disturbed lately since I don't know the answer this basic problem.

Say we have a standard isotropic antenna with some fixed parameters (load impedance, etc), and we feed this antenna with a sinusoidal current of the form: \begin{equation} I(t) = A\cos(2\pi f_{c}t) \end{equation} where $f_{c}$ is the carrier frequency (typically in the GHz range). Assuming no circuit mismatches or losses, the power delivered to the antenna (and radiated) is $P \propto (A^2)/2$. Since the antenna is isotropic this gives a electric far field strength of $|E| \propto I$ in every angle at some distance $d$ from the antenna.

Now let's say that we have 2 similar isotropic antennas "really" close together (compared to the wavelength), but we feed each one with sinusoidal current half of the amplitude, i.e. \begin{equation} I_d(t) = \frac{A}{2}\cos(2\pi f_{c}t) \end{equation} and the power delivered to each one antenna will be $P_d\propto (A^2)/8$. The electric fields will sum up constructively in all angles at distance $d$ from this 2 element antenna array the field strength is $|\frac{E}{2}+\frac{E}{2}|=|E|$, i.e. equal to the single antenna case. The problem is that the sum of the powers feed to the 2 antennas is $(A^2)/8 + (A^2)/8 = (A^2)/4$ which is smaller than in the first case.

Thus there might be something wrong here, because if one takes this approach one step forward, I would a field with infinite magnitude (hence infinite power) using an infinite amount of antennas for a given input power $P$. Where is the error in my approach?

I agree that in a practical system, there will be coupling between antennas thus their efficiencies will decrease, etc. But this cannot be the fundamental explanation for this because in standard textbook in antennas as the one of Balanis, the superposition principle is assumed and everything should be coherent from this point-of-view.

Thanks in advance for your help.

Joao

$\endgroup$
1
  • $\begingroup$ Its true that when the antenna becomes much bigger than the wavelength, then you can have huge gain factors. For example a satellite dish can have a gain of 60db which is a factor of a million ($10^6$). The practical limitation to increasing the size even more is the 1) phase coherency 2) aiming accuracy. If you have a 60db gain, already you must aim the sat dish to a fraction of a degree in accuracy. $\endgroup$ – Kostas Dec 5 '19 at 13:43
2
$\begingroup$

When antennas are placed very close together all bets are off. The near field coupling between antennas elements change the impedance's therefore currents and fields the far field patterns don't follow simple algorithms any more. in fact if they become close enough and fed in phase they asymptote towards a single element with two sources in parallel the far field pattern is as for a single element, and you will have a high standing wave ratio and the system will not be amenable to the previous simple analysis. Inter element coupling is a big issue in large array antennas and particularly troublesome for steered arrays. There is no stable fixed matching solution to steered beams. Independent multiple fixed orthogonal switched beams are a better solution (Butler Matrix, Luneberg lens etc.).

Regarding array gain violating laws of physics. The total radiated power over all space for a directed narrow beam antenna will be the same as a dipole for a given input power. the radiated power at specified beam areas (square degrees or steradians) will be different, no violation here. A large radio astronomy dish with a gain of 70 dBi (at beam peak), the beam-width will be a fraction of a degree. Such a dish can typically have a gain near the back of the dish of approx 0dBi the average gain of the dish over all space should be exactly 0dBi.

I think you problem is this statement:--"The electric fields will sum up constructively in all angles at distance d from this 2 element antenna array."

for two isotropes at half wave spacing: The electric fields will sum up constructively for two opposite angles at the far field distance, the field will sum destructively in the plane at right angles to a line between the two isotrope's, and partially for all other angles. if you integrate over all 3D space the directivity will be zero as Maxwell demands. I'm sure you have missed something in Balanis, I've a couple of his books and like them very much. Balanis may be referring to pattern multiplication where you multiply the array factor by the element factor?

$\endgroup$
1
$\begingroup$

Of course something is wrong: if you want to preserve the total intensity you feed each antenna with a current

$$I(t) = \frac {A}{\sqrt {2}} \cos(2\pi f_c t). \tag{i}$$

About the electric field beware, it is as you say, $\vec E = \vec E_1 + \vec E_2$, but if you follow my $\text {(i)}$, it becomes

$$\vec E = \frac {\vec E_1 + \vec E_2}{\sqrt {2}}. \tag{ii}$$

The superposition of the two fields will produce interference, minima and maxima, but the total intensity mediating over the minima and maxima, will be as with one single antenna.

$\endgroup$
1
  • $\begingroup$ I cannot accept that answer. Let me tell you why: The currents at the 2 antennas have the same phase, and the antennas are very closely spaced. Thus the two fields generated will always add up constructively in every direction. Hence to have the same field (and thus intensity) at some spatial position, the currents peaks to each antenna should be A/2 and not A/√2 as your said. $\endgroup$ – Peter Feb 20 '15 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.