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Chapter 5                                                               243

            Here  we used the  well-known expression for the geometric progression sum and Euler’s

            formula associating the trigonometric and complex exponential function:     = cos + sin.
            We preferred to build the following discussion on the phasor interpretation,  not on (5.87)
            analysis seeing it as more vivid and understandable. Besides, the phasor diagrams are a very
            powerful tool for  screening  computer  model  fairness and controlling numerical simulation
            results. To interpret the expression (5.86) graphically, we will use a phasor diagram displaying
            each term in (5.86) as the vectors of length | | rotated with respect to one another at the angle .
                                               
            If so, the phasor diagram using traditional rules of vector addition can be built. We specify for
            certainty that  =  or   = 0.5 and N = 5 (i.e. 6 radiators in total). The procedure can be
                                 ⁄
            outlined as:




                                12    = 2











                 SLL = -12.43dB









                            Figure 5.4.7 Uniform array pattern and phasor diagrams

             = ° According to (5.86)  = 0 and all the phasors      = | | are collinear and add up
                                                         
                                                                  
            in phase as Figure 5.4.7 (black polygon) displays. The sum in (5.86) reaches its maximum and
            (90°) marked in black is clearly the peak of the main beam. Evidently, this peak can be
            increased by adding elements.
             =   and slightly less than 90°. That leads to the diagram shown in Figure 5.4.7 (green
                 
            polygon) where     5  = −   or 5 = ∓. We chose this case  to demonstrate how the
                           5        0
            radiation of some elements to be canceled in the far field zone and so why (  ) < (90°).
                                                                           1
            Since 5 = −5cos = −5cos = ∓ we can calculate cos = ∓0.2 or  = 78.465°
                               1          1                       1          1
            aka 101.54°.

             =   and  = −cos = ∓/3 meaning cos = ∓1/3 and   = 70.53° or 109.5°. The
                                 2
                 
                                                     2
                                                                   2
            corresponding phasor diagram in Figure 5.4.7 (red vectors) demonstrates that the vectors 
                                                                                      
                                                                                   3
            form a hexagon. Then (  ) = 0 since in (5.86) as it follows from the diagram    =
                                                                                 3
                                   2
                             
            − ,    4  = −   and   5  = −   2 . Therefore,   is the angular position of the first
                                                            2
                                             2
                                   5
                           1
              0
                 4
            pattern null. Clearly, the next null will be when  = ∓2/3 and so on. By generalizing this
            result, we can calculate the exact position of any null as
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