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Logoperiódica: Diseño Práctico

May 4, 2009 No Comments

Tabla de contenido de Logoperiódicas: Parámetros de diseño para bastimento

  1. Logoperiódicas: Introducción
  2. Logoperiódicas: Parámetros de las Antenas Logoperiodicas de dipolos rectos
  3. Logoperiódica: Diseño Práctico

Se procede a diseñar una antena logoperiodica que cubra los canales desde el 7 hasta el 49 y con una directividad substancial de 9dB. Los cálculos y forma de diseño son los siguientes:

f_{sup} = 683 MHz

f_{inf} = 177 MHz AB = 506 MHz Directividad = 9 dBi \tau = 0.918 \ldots \sigma = 0.169 \alpha = 6.91^{o} \ldots k_{2} = 0.52 k = 1.1 + 7.7 \left( 1 - \tau\right) ^{2} \cot \alpha 1.1 + 7.7 \left( 1 -0.918\right) ^{2} \cot 6.91 k = 1.5272 k_{1} = \dfrac{k_{2}}{k} = \dfrac{0.52}{1.5272} = 0.34 L_{max} = k_{2} \lambda_{inf} = 0.52 \dfrac{3 \ast 10^{8}}{177 \ast ^{10^{6}}} L_{max} = 0.88 m = L_{1} L_{min} = k_{1} \lambda_{sup} = 0.34 \dfrac{3 \ast 10^{8}}{683 \ast 10^{6}} = 0.15 m \dfrac{L_{max}}{L_{min} \ast k} = \dfrac{0.88}{0.15 \ast 1.5272} = 3.841 N = - \dfrac{\log\left(k \ast B\right) }{\log\left( \tau \right) } + 1 N = - \dfrac{\log\left(1.5272 \ast 3.841\right) }{\log\left( 0.918 \right) } + 1 L_{n+1} = \tau \ast L_{n} L_{2} = \tau \ast L_{1} = 0.918 \ast 0.88 = 0.80 L_{3} = \tau \ast L_{2} = 0.918 \ast 0.80 = 0.73 L_{4} = \tau \ast L_{3} = 0.918 \ast 0.73 = 0.67 L_{5} = \tau \ast L_{4} = 0.918 \ast 0.67 = 0.61 L_{6} = \tau \ast L_{5} = 0.918 \ast 0.61 = 0.56 L_{7} = \tau \ast L_{6} = 0.918 \ast 0.56 = 0.51 L_{8} = \tau \ast L_{7} = 0.918 \ast 0.51 = 0.47 L_{9} = \tau \ast L_{8} = 0.918 \ast 0.47 = 0.43 L_{10} = \tau \ast L_{9} = 0.918 \ast 0.88 = 0.40 L_{11} = \tau \ast L_{10} = 0.918 \ast 0.40 = 0.36 L_{12} = \tau \ast L_{11} = 0.918 \ast 0.36 = 0.33 L_{13} = \tau \ast L_{12} = 0.918 \ast 0.33 = 0.31 L_{14} = \tau \ast L_{13} = 0.918 \ast 0.31 = 0.28 L_{15} = \tau \ast L_{14} = 0.918 \ast 0.28 = 0.26 L_{16} = \tau \ast L_{15} = 0.918 \ast 0.26 = 0.24 L_{17} = \tau \ast L_{16} = 0.918 \ast 0.24 = 0.22 L_{18} = \tau \ast L_{17} = 0.918 \ast 0.22 = 0.20 L_{19} = \tau \ast L_{18} = 0.918 \ast 0.20 = 0.18 L_{20} = \tau \ast L_{19} = 0.918 \ast 0.18 = 0.17 L_{21} = \tau \ast L_{20} = 0.918 \ast 0.17 = 0.15 L_{min} = L_{21} 0.15 = 0.15 \tan \alpha = \dfrac{L_{1}}{2 R_{1}} \dfrac{L_{1}}{2 \tan \alpha} = \dfrac{0.88}{2 \tan 6.91^{o}} = 3.63 R_{2} = \tau \ast R_{1} = 0.918 \ast 3.63 = 3.33 R_{3} = \tau \ast R_{2} = 0.918 \ast 3.33 = 3.06 R_{4} = \tau \ast R_{3} = 0.918 \ast 3.06 = 2.81 R_{5} = \tau \ast R_{4} = 0.918 \ast 2.81 = 2.58 R_{6} = \tau \ast R_{5} = 0.918 \ast 2.58 = 2.36 R_{7} = \tau \ast R_{6} = 0.918 \ast 2.36 = 2.17 R_{8} = \tau \ast R_{7} = 0.918 \ast 2.17 = 2.00 R_{9} = \tau \ast R_{8} = 0.918 \ast 2.00 = 1.83 R_{10} = \tau \ast R_{9} = 0.918 \ast 1.83 = 1.68 R_{11} = \tau \ast R_{10} = 0.918 \ast 1.68 = 1.54 R_{12} = \tau \ast R_{11} = 0.918 \ast 1.54 = 1.41 R_{13} = \tau \ast R_{12} = 0.918 \ast 1.41 = 1.30 R_{14} = \tau \ast R_{13} = 0.918 \ast 1.30 = 1.19 R_{15} = \tau \ast R_{14} = 0.918 \ast 1.19 = 1.09 R_{16} = \tau \ast R_{15} = 0.918 \ast 1.09 = 1.00 R_{17} = \tau \ast R_{16} = 0.918 \ast 1.00 = 0.92 R_{18} = \tau \ast R_{17} = 0.918 \ast 0.92 = 0.84 R_{19} = \tau \ast R_{18} = 0.918 \ast 0.84 = 0.78 R_{20} = \tau \ast R_{19} = 0.918 \ast 0.78 = 0.71 R_{21} = \tau \ast R_{20} = 0.918 \ast 0.71 = 0.65 R_{1} - R_{N} = \dfrac{L_{max} - L_{min}}{2 \tan \alpha} 2.99 \approx 3.01

Se debe tener presente que la alimentación, se la debe hacer desde la parte donde está el dipolo más pequeño y de ahí compartirla a los demás dipolos, proporcionando un desfase de 180^{o} entre dos dipolos consecutivos.

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