Half power beamwidth of antenna

Question

I was solving a problem regarding HPBW. From what I understood, HPBW of antenna is the angular separation on the beam where the radiation intensity is half of its maximum value. That means U(radiation intensity)=Umax/2 at HPBW Now in the solution of this question they have written U(at HPBW)=0.5. Isn’t it wrong? I mean this equation dictates that the value is half (0.5) but it should be half of its peak value, according to the actual definition.

U at HPBW should be Umax/2 not just 1/2. In the given solution they have written U=0.5, doesnt that mean that the actual value is 0.5, but according to the definition it should drop to half of its peak?? — user305532, Oct 22 '22 at 14:17

score 2 · Answer 1 · answered Oct 22 '22 at 15:35

2

Power is proportional to the square of intensity; so for half power, you need \$1/\sqrt2\$ intensity; and \$1/\sqrt2\approx 0.707\$.

And 0.5 is the correct absolute power, since the \$\max\limits_x \,\cos^2(x)\cos^2(3x) = 1\$.

answered Oct 22 '22 at 15:35

Marcus Müller

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score 1 · Accepted Answer · answered Oct 22 '22 at 15:59

They defined \$U(\theta)\$ as the "normalized radiation intensity" or "power pattern", not the radiation intensity. The formula they gave contains only unitless factors, so it is surely not the actual radiation intensity.

I take this to mean that \$U(\theta)\$ is normalized to have a value of 1 at its peak. This is confirmed by the fact that the formula given is a product of cosine terms, which can't have a maximum value greater than 1.

Then it is correct that at the angle where the intensity is half the peak value, we'd have \$U(\theta)=0.5\$.

Half power beamwidth of antenna

2 Answers2