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yungman
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The definition of Beam solid angle: For an antenna with single main lobe, the Beam solid angle defines as:
The solid angle [itex]\; \Omega_A\;[/itex] where all the radiated power would flow with radiation intensity equal to maximum and constant inside the Beam solid angle [itex]\; \Omega_A\;[/itex].
The book gave:
[tex]\Omega_A\;=\; \int F(\theta,\phi) d \Omega \;\hbox{ where }\; F(\theta,\phi)=\frac { P_{av}(\theta, \phi)}{P_{max}}[/tex]
But from the definition, the [itex]\; \Omega_A\;[/itex] depend on the total power radiated divid by the [itex]\;P_{max}\;[/itex] times the total steradians of the sphere.
Total power radiated = [itex] P_{Rad}=\int P_{av}(\theta,\phi) dS= \int R^2\;P_{av}(\theta,\phi)\;d\Omega [/itex]
[tex]\Omega_A = 4\pi \frac { P_{Rad}(\theta,\phi)}{P_{max}} = 4\pi R^2 \frac{\int P_{av}(\theta,\phi) d\Omega} {P_{max}} = 4\pi R^2 \int \frac { P_{av}(\theta,\phi)}{P_{max}} d \Omega = 4\pi R^2 \int F(\theta,\phi) d\Omega[/tex]
You see there is a [itex]\;4\pi R^2\;[/itex] difference between the definition and the given formula, please help.
Thanks
Alan
The solid angle [itex]\; \Omega_A\;[/itex] where all the radiated power would flow with radiation intensity equal to maximum and constant inside the Beam solid angle [itex]\; \Omega_A\;[/itex].
The book gave:
[tex]\Omega_A\;=\; \int F(\theta,\phi) d \Omega \;\hbox{ where }\; F(\theta,\phi)=\frac { P_{av}(\theta, \phi)}{P_{max}}[/tex]
But from the definition, the [itex]\; \Omega_A\;[/itex] depend on the total power radiated divid by the [itex]\;P_{max}\;[/itex] times the total steradians of the sphere.
Total power radiated = [itex] P_{Rad}=\int P_{av}(\theta,\phi) dS= \int R^2\;P_{av}(\theta,\phi)\;d\Omega [/itex]
[tex]\Omega_A = 4\pi \frac { P_{Rad}(\theta,\phi)}{P_{max}} = 4\pi R^2 \frac{\int P_{av}(\theta,\phi) d\Omega} {P_{max}} = 4\pi R^2 \int \frac { P_{av}(\theta,\phi)}{P_{max}} d \Omega = 4\pi R^2 \int F(\theta,\phi) d\Omega[/tex]
You see there is a [itex]\;4\pi R^2\;[/itex] difference between the definition and the given formula, please help.
Thanks
Alan
Last edited: