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Calculating antenna directivity

  1. Feb 26, 2009 #1
    1. The problem statement, all variables and given/known data

    An antenna has a normalised E-filed pattern, En where [tex]\theta[/tex] = vertical angle as measured from z-axis and [tex]\phi[/tex] = azimuth angle measured from x-axis.
    Calculate the exact directivity

    En has a non-zero value whenever [tex]0 <= \theta <= \pi[/tex] and [tex]0 <= \phi <= \pi[/tex]. Elsewhere, En is zero


    The correct answer is 6 , but I cannot get this number

    2. Relevant equations

    [tex]En = sin(\theta)sin(\phi)[/tex]

    Direcitivity is calculating using

    [tex]D = 4\pi/ ( \int\intPn(\theta,\phi )d\phid\theta )[/tex]


    3. The attempt at a solution


    Ok, first of all [tex]Pn = En^{2}[/tex]

    Therefore [tex]Pn = sin(\theta)^{2}sin(\phi)^{2}[/tex]

    Now [tex]sin^{2}(\theta) = 0.5(1 - cos(2\theta))[/tex] with respect to [tex]\theta[/tex]

    If we perform the integration of the [tex]sin^{2}(\theta)[/tex] terms we should get

    [tex]\theta/2 - sin(2\theta)/4[/tex]

    Applying the [tex]0 <= \theta <= \pi[/tex] limits, I got [tex]\pi/2[/tex] for the first integral. If we integrate the [tex]sin(\phi)[/tex] term, we should get [tex]\pi/2[/tex] as well, making [tex]\pi^{2}/4[/tex] the actual answer for

    [tex]\int\intPn(\theta,\phi )d\phid\theta )[/tex]

    Substituting [tex]\pi^{2}/4[/tex] into the denomiator part of D, I get [tex]16/\pi[/tex] which obviously is not right. Im not sure where I could be going wrong
     
    Last edited: Feb 27, 2009
  2. jcsd
  3. Feb 26, 2009 #2
    If we perform the integration of the [tex]sin^{2}(\theta)[/tex] terms we should get

    [tex]\theta/2 - sin(2\theta)/4[/tex]

    Applying the 0 <= [tex]\theta[/tex] <= [tex]\pi[/tex] limits, I got [tex]\pi[/tex]/2 for the first integral. If we integrate the [tex]sin(\phi)[/tex] term, we should get [tex]\pi[/tex]/2 as well, making [tex]\pi^{2}/4[/tex] the actual answer for

    [tex]\int[/tex][tex]\int[/tex]Pn([tex]\theta[/tex],[tex]\phi[/tex] )[tex]d\phi[/tex][tex]d\theta[/tex]
     
  4. Feb 26, 2009 #3
    Guys, please refer to my 2nd most outlining the integration process

    cheers and thanks in advance
     
  5. Feb 27, 2009 #4
    1. The problem statement, all variables and given/known data

    An antenna has a normalised E-filed pattern, En where [tex]\theta[/tex] = vertical angle as measured from z-axis and [tex]\phi[/tex] = azimuth angle measured from x-axis.
    Calculate the exact directivity

    En has a non-zero value whenever [tex]0 <= \theta <= \pi[/tex] and [tex]0 <= \phi <= \pi[/tex]. Elsewhere, En is zero


    The correct answer is 6 , but I cannot get this number

    2. Relevant equations

    [tex]En = sin(\theta)sin(\phi)[/tex]

    Direcitivity is calculating using

    [tex]D = 4\pi/[/tex] [tex]\int[/tex][tex]\int[/tex][tex]Pn[/tex]([tex]\theta[/tex],[tex]\phi[/tex] )[tex]d\phi[/tex][tex]d\theta[/tex]


    3. The attempt at a solution


    Ok, first of all [tex]Pn = En^{2}[/tex]

    Therefore [tex]Pn = sin(\theta)^{2}sin(\phi)^{2}[/tex]

    Now [tex]sin^{2}(\theta) = 0.5(1 - cos(2\theta))[/tex] with respect to [tex]\theta[/tex]

    If we perform the integration of the [tex]sin^{2}(\theta)[/tex] terms we should get

    [tex]\theta/2 - sin(2\theta)/4[/tex]

    Applying the [tex]0 <= \theta <= \pi[/tex] limits, I got [tex]\pi/2[/tex] for the first integral. If we integrate the [tex]sin(\phi)[/tex] term, we should get [tex]\pi/2[/tex] as well, making [tex]\pi^{2}/4[/tex] the actual answer for

    [tex]\int[/tex][tex]\int[/tex][tex]Pn[/tex]([tex]\theta[/tex],[tex]\phi[/tex] )[tex]d\phi[/tex][tex]d\theta[/tex]

    Substituting [tex]\pi^{2}/4[/tex] into the denomiator part of D, I get [tex]16/\pi[/tex] which obviously is not right. Im not sure where I could be going wrong
     
    Last edited: Feb 27, 2009
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