Abstract
Eq. 6.49 is \(\hat {S}=\varSigma ^N_{n=0} \mathrm {e}^{{\,\mathrm {i}\,} \, k \, n\, D \, \sin {(\phi )}}\) Set \(\mathrm {e}^{{\,\mathrm {i}\,} \, k \, n\, D \, \sin {(\phi )}}=q\). Then the difference \( \hat {S} - q \times \, \hat {S} \) is 1 − q N+1, since only the first an last terms of the series do not cancel. Solving for \( \hat {S} \), we have: \(\hat {S} = \frac { 1-q^{N+1} }{ 1-q }\). Next convert q to the quantity \(\mathrm {e}^{-{\,\mathrm {i}\,} \, k \, D \, \sin {(\phi )}}\) Then use the relation for \(\sin {x}=1/2 \times \, (\mathrm {e}^{{\,\mathrm {i}\,} x} - \mathrm {e}^{-{\,\mathrm {i}\,} x} ) \), to convert to trigonometric terms. Then factor a term \(\mathrm {e}^{-{\,\mathrm {i}\,} (N/2) \, k \, D/2 \, \sin {(\phi )}}\) out of the sum to convert to: The sum is
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Wilson, T.L., Hüttemeister, S. (2018). Solutions for Chapter 6: Fundamentals of Antenna Theory. In: Tools of Radio Astronomy - Problems and Solutions. Astronomy and Astrophysics Library. Springer, Cham. https://doi.org/10.1007/978-3-319-90820-5_22
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DOI: https://doi.org/10.1007/978-3-319-90820-5_22
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