When an adaptive null/ notch is placed in the antenna pattern - towards let's say in the direction of interference or any specific target - where does the original energy in the artificially created null go? Is it distributed in the side lobes? Or completely eliminated? In case of complete elimination, how do we justify the law of conservation of energy?
Many thanks
Anum
Dear Anum Ahmed Pirkani ,
welcome!
I think that I answered previous questions for you.
Assume that the radiated power is kept constant while distributing the radiation in the space then when nulls are created in some directions the rest of the power will be redistributed in the other directions.
From the antenna array point of view normally the radiation pattern contains a major lobe and many side lobes which are separated by dips in the space around the lobes. This is dictated by phase relationship between the different antennas.
Accordingly there will be space maxima and minima in the radiation.
Best wishes
It depends on the means you use to add the adaptive null.
Some techniques may result in an increase in the VSWR i.e. more reflection back into the power amplifier. This may be accompanied by a partial increase in the other lobes.
If the means absorbs the energy in the null direction, then mainly into the absorber.
If there is no absorption or impact on VSWR, then only into the other lobes.
Dear Anum Ahmed Pirkani - Positioning of nulls involves a redistribution of power into other lobes. Coupling between array antenna elements may imply some change in individual element VSWR as element excitation coefficients are changed, but this is a secondary effect (typical element coupling may be 15-20dB). The VSWR effect is generally small compared to effect on beams.
For conservation of energy, it is necessary that the 4 pi integral of the gain, expressed as
(1/4 pi) Integral [G(Omega) dOmega] = 1
where dOmega = sin(theta)dtheta dphi, theta is the elevation angle and phi is the azmuthal angle. This equation holds also when doing adaptive beamforming. So what this says is that when you squeeze the gain in one direction, it pops up elsewhere, somewhat like squeezing a baloon whose volume is constant. Exactly how the 4 pi gain changes depends on the algorithm. In general, you will find that the peak gain will be reduced by a little and that the average sidelobe level is a bit larger except in the direction of the null.
Many thanks for a very well explained answer...
Can I ask another thing. Why does the peak gain reduce?
I have actually developed a simulation model and gladly I am experiencing these and also observed a reduction on peak gain and wasn't exactly sure why its happening because the main lobe was well seperated from adaptive null....!!!
I'll assume a phased array. Then each element has a weight that is determined by uniform illumination or by Taylor weights, for example. Uniform weights provide the largest gain and narrowest beamwidth. Taylor weights provide a minimum beamwidth expansion for a given maximum sidelobe level.
So, with nulling weights, the main beam is no longer as sharp as before (i.e., the beamwidth expands). But the main beam is the primary contributor to the integral of Gain over 4 pi steradians. Thus it must be the case to keep that integral equal to unity, wider beamwidths are accompanied by reduced peak gain. A simple example is that a smaller array has lower peak gain and a broader beamwidth -- they go hand in hand in keeping the integrated gain constant.
Many thanks for a very clear and detailed explanation. Everything makes sense now... !!!!
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