Find the half-power beam-width (HPBW), in degrees, for the following normalized radiation intensity: $U(\theta) = \cos^2(3\theta)$
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In this case, the maximum power is at $\theta = 0$, where $U(0) = \cos^2(0) = 1$. The half-power points occur when $U(\theta) = \frac{1}{2}$. Show more…
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