What is the total power radiated from an antenna element in Matlab phased array toolbox ?

Looking at the power pattern of an isotropic element to be one, it gives me a hint that the total power input to an element is 4*pi Watts and I get the same number when I integrate the power pattern of an isotropic element over a sphere covering 4*pi Steradians. However, when i take any element other than isotropic, the total power radiated is not equal to 4*pi !!! Can somebody may guide, what is the assumption of input power to an element in Matlab phased array toolbox. I am using the following code to find the total power of e.g. a cosine element:

sCos = phased.CosineAntennaElement('FrequencyRange',[3e8 1e9]);
az = -180:179;
el = -90:89;
Prad = 0;
 for m = 1:numel(el)

      MyAng=[-180:179;el(m)*ones(1,360)];

      resp = step(sCos,3e8,MyAng);   % voltage response of sensorElement

      resp_pow = (abs(resp)).^2;

      temp2 = resp_pow'*cosd(el(m));

      Prad = Prad+sum(temp2);

 end
 Prad2 = (Prad)*2*pi^2/(360*180);

Answer 1

[Honglei Chen]

Hi Raza,

Prad is the total power computed from the pattern but there is no guarantee that the sum needs to be 4*pi. In fact, if you look at the definition of the directivity, it is defined as D=4*pi*U/Prad, so the Prad varies with what element you chose.

Since you are mentioning input, my guess is maybe you mean how to see the power relation using phased.Radiator? That is a bit subtle since phased.Radiator by models the field relation toward a given direction, so the input is considered as the field needs to be propagated for that particular direction. Taking the following code as an example

ant = phased.CosineAntennaElement;
txant = phased.Radiator('Sensor',ant);
x = 1;
y = step(txant,x,[0;0])

You can see that y=1 because the field toward [0;0] is preserved.

Now, if you think the input, x, as the signal with a given input power to the antenna and want to see the field at the given direction to reflect the directivty gain, then you need to figure out how much of the field/power is radiated toward that direction. The relation is basically

y = sqrt(D)*x

where D is the directivity gain toward that particular direction. If you do some manipulation, you will end up with

y = F*x/sqrt(4*pi/Prad)

where F*x is what modeled by phased.Radiator and Prad is what you computed as the radiated power. Hence, if you do

% for cosine antenna, the Prad is about pi/2
Prad = pi/2;
y = step(txant,x*sqrt(4*pi/Prad),[0;0])

Compare to the directivity at that direction, you can see the two results match.

db2mag(directivity(ant,txant.OperatingFrequency,[0;0]))

Is this something you are looking for? If not, please reply and I can explain more.

[Raza Umar]

Thanks dear Honglei for your response. I have two concerns regarding what you explained, if you may kindly clarify:

(1) According to Gauss law, the power input to an element is same as the total power radiated over a sphere, and that should be independent of its radiation pattern, since the radiation pattern determines the power level in a particular direction but the total power radiated should not alter and it should be same as the total input power to the element. Do you find flaw in my argument?

Elaborating my point based on your example:

ant = phased.CosineAntennaElement;
txant = phased.Radiator('Sensor',ant);
x = 1;
y = step(txant,x,[0;0])

Total input power is available at the output of radiator in boresight direction, but in reality power would be radiated out of radiator not just in boresight direction, rather in all sphere enclosing the radiator, and this would come out to be more than the input power... This is what bothering me. Kindly guide me what I am missing!!!

Is it right that my input signal "x" should be scaled by 1/sqrt(Prad) to make sure that the total radiated power from radiator is same as the input power and also scaled by sqrt(4*pi) to make sure that the input power is the one that is supposed to be same as on the sphere spanning 4*pi steradians, not the power to be transmitted in one direction !!!

(2) As you rightly pointed out that the power output from radiator is

y=F*x

My concern is while simulating a radar: Refer to example: http://www.mathworks.com/help/phased/examples/scan-radar-using-a-uniform-rectangular-array.html

We are infact implementing FRIIS equation, so should not the radiator output be scaled by sqrt(4*pi/Prad), to get

y=sqrt(D)*x

Thanks in advance.

[Honglei Chen]

Hi Reza,

For (1), it depends on how you interpret the input, x. The radiator right now implements the field relation so the input in this case is considered as the field, not the power. It is true that if you want to see the power relation, you'll have to scale it.

I think your point (2) is related. I agree that if you want to simulate Friis equation, then you should use y = sqrt(D)*x. The example probably needs to be updated to reflect the correct power relation.

HTH

[santhosh santhosh]

Hi Honglei Chen

Why total radiated power (integrated radiation pattern) varies with element spacing in simulation? Since simulation or phased array equation does not consider any losses as such, why total radiated power varies? where that extra power is dissipatated? What am I missing here?

[奥 刘]

@santhosh santhosh Hi, may be I can explain a little bit here. Suppose you have two same ideal isotropic antennas, and both of them are radiating the same power but in opposite phase. Then if you vary the distance between this two ideal antennas, The total radiated power we can in space will also vary. To see that, we can consider two extreme cases. First, if we keep decreasing the distance, all the way to 0 (two antennas just coincide), due to the opposite phase here, we will get 0 power. Then, if we keep increasing the distance, to infinite, then the two antennas can be regarded as two independent antennas, with no mutual, and no beampattern effect, then we will get double radiated power from the integration of the whole space. There, we will find out that, between 0 and infinite distance of these two antennas, the radiated power will also vary between 0 and double of the power. As for wht it's that, I think it's because of all the mutual effects and coherent addition of signal from different antennas, which I haven't carefully researched, so I can't give you a sound answer here.