Page 442 - Engineering Electromagnetics, 8th Edition
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424                ENGINEERING ELECTROMAGNETICS











                                                              in          in










                                                Figure 12.5 A three-interface problem in which input
                                                impedance η in,a is transformed back to the front interface
                                                to form input impedance η in,b .

                                        The procedure in this section for evaluating wave reflection has involved calcu-
                                     lating an effective impedance at the first interface, η in , which is expressed in terms of
                                     the impedances that lie beyond the front surface. This process of impedance trans-
                                     formation is more apparent when we consider problems involving more than two
                                     interfaces.
                                        Forexample, consider the three-interface situation shown in Figure 12.5, where
                                     awaveis incident from the left in region 1. We wish to determine the fraction of the
                                     incident power that is reflected and back-propagates in region 1 and the fraction of the
                                     incident power that is transmitted into region 4. To do this, we need to find the input
                                     impedance at the front surface (the interface between regions 1 and 2). We start by
                                     transforming the impedance of region 4 to form the input impedance at the boundary
                                     between regions 2 and 3. This is shown as η in,b in Figure 12.5. Using (36), we have


                                                                η 4 cos β 3 l b + jη 3 sin β 3 l b
                                                        η in,b = η 3                                 (47)
                                                                η 3 cos β 3 l b + jη 4 sin β 3 l b
                                     We have now effectively reduced the situation to a two-interface problem in which
                                     η in,b is the impedance of all that lies beyond the second interface. The input impedance
                                     at the front interface, η in,a ,isnow found by transforming η in,b as follows:

                                                               η in,b cos β 2 l a + jη 2 sin β 2 l a
                                                       η in,a = η 2                                  (48)
                                                               η 2 cos β 2 l a + jη in,b sin β 2 l a
                                                                    2
                                     The reflected power fraction is now | | , where
                                                                     η in,a − η 1
                                                                  =
                                                                     η in,a + η 1
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