Since the electric field intensity at is proportional to we can write r . (1-4) If we recall that the signal is really a sinusoid at a frequency of , the carrier frequency, we can write the electric field (E-field) at as (1-5) where is the time the signal takes to propagate from the source to the area . For the next step we invoke the relations , , and where is the speed of light and denotes c wavelength. With this, we can write the electric field at as . (1-6) Let us now turn our attention to determining the E-field at some when we have the two radiators of We will use the geometry of as an aid in our Figure 1-2. Figure 1-3 derivation. We denote the upper radiator (point) of as radiator 1 and the lower Figure 1-3 radiator as radiator 2. The distances from the individual radiators to are and . We assume that the electric field intensity of each radiator is equal to . Where is the total P power delivered to the radiators. The factor of 2 is included to denote the fact that the power is split evenly between the radiators (uniform weighting). In the above, we have made the tacit assumption that the radiation resistance is 1 ohm. The other terms that we will need are shown on Figure 1-3. Figure 1-3 Geometry for two element radiator problem – We can write the E-fields from the two radiators as (1-7) and . (1-8) From we can write and as Figure 1-3, (1-9) and . (1-10) As indicated earlier, we will assume that . With this we can write (1-11) and . (1-12) Where we have made use of the relation
Since the electric field intensity at r is proportional to we can write . (1-4) If we recall that the signal is really a sinusoid at a frequency of , the carrier frequency, we can write the electric field (E-field) at as (1-5) where is the time the signal takes to propagate from the source to the area . For the next step we invoke the relations , , and where c is the speed of light and denotes wavelength. With this, we can write the electric field at as . (1-6) Let us now turn our attention to determining the E-field at some when we have the two radiators of Figure 1-2. We will use the geometry of Figure 1-3 as an aid in our derivation. We denote the upper radiator (point) of Figure 1-3 as radiator 1 and the lower radiator as radiator 2. The distances from the individual radiators to are and . We assume that the electric field intensity of each radiator is equal to . Where P is the total power delivered to the radiators. The factor of 2 is included to denote the fact that the power is split evenly between the radiators (uniform weighting). In the above, we have made the tacit assumption that the radiation resistance is 1 ohm. The other terms that we will need are shown on Figure 1-3. Figure 1-3 – Geometry for two element radiator problem We can write the E-fields from the two radiators as (1-7) and . (1-8) From Figure 1-3, we can write and as (1-9) and . (1-10) As indicated earlier, we will assume that . With this we can write (1-11) and . (1-12) Where we have made use of the relation
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