This implies that 2 2 pr sinc 4 g0o 0o 0 w f

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Unformatted text preview: s still along our new z axis, therefore, we have 0 G(✓2 , 0 ) = G(✓2 45o , 2 ). This implies that 2 2 Pr = Sinc 4⇡ G(0o , 0o ) = 0 W. f) Now the short dipole is polarized along the y direction and our electrical field is in x-z plane. ˆ Due to the polarization mismatch, the dipole direction is perpendicular to the incident field polarization (cross-polarization), so Pr = 0 W. 2. a) The wavenumber is k = 2⇡ m. q p p (1/ 2)2 + (1/ 2)2 = 1 rad . Then, the wavelength is simply m = 2⇡ k = b) Considering that the wave is a right-hand circularly polarized plane TEM wave, we are given p the real part of the electric field phasor as Ey = y 2 and asked to find the imaginary part. ˆ Since this is a TEM wave, we can draw the imaginary components and the propagation vector as shown in the below figure where the possible unit vectors of the imaginary part of the electric field phasor are displayed in green and red. as 1 z k ˜ Exz x y ˜ Exz Imagine that we rotate the coordinate system in the above figure in a way that the propagation...
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