Lecture 16: radiation from antennas

Since 
 <S> = \frac{1}{2} \Re \{E* \times H \} = \frac{1}{2} E_+ H_+ \cos{\phi} <S>=12ℜ{E∗×H}=12E+H+cos⁡ϕ <S> = \frac{1}{2} \Re \{E* \times H \} = \frac{1}{2} E_+ H_+ \cos{\phi} <S>=21​ℜ{E∗×H}=21​E+​H+​cosϕ
, a wave with E and H fields 90 degrees out of phase will have no power.
In a transmission line, voltage causes an E field between the forward and reverse paths while current causes an H field. On forward propagation, as long as 
 Z_0 Z0 Z_0 Z0​
 is real, the E field is in phase with voltage and H field is in phase with current, and each are in phase with each other.
If there is a reflection/standing wave, however, voltage and current -- and correspondingly, E and H -- can move out of phase.
open t line
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In an open t line, which is half a wavelength long, at steady state, a half wavelength current pattern is observed due to the standing wave (what does this mean?). The equal and opposite current distributions on the top and bottom cancel out fields outside of the transmission line, so there is no radiation.
short dipole antenna
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Radiation can be created by bending a length 
 D/2 << \lambda/2 D/2<<λ/2 D/2 << \lambda/2 D/2<<λ/2
 upwards and downwards. Then we have an E field normal to the H field, resulting a poynting vector in positive x. The phases aren't exactly in quadrature because the E field generated by charge at  z = D/2, z = -D/2 z=D/2,z=−D/2 z = D/2, z = -D/2 z=D/2,z=−D/2
 is  kD/2 kD/2 kD/2 kD/2
 out of phase with the H field generated at  z = 0 z=0 z = 0 z=0
, so  \cos {\phi} cos⁡ϕ \cos {\phi} cosϕ
 can be  > 0 >0 > 0 >0
.
The antenna only really radiates at higher frequencies because there will be more current on the antenna.
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An antenna can be modeled as a resistor in series with some some reactance. The series resistance has a radiative component and a loss component.
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The current distribution on a short dipole is a triangle by small angle approximation. The voltage distribution is a step (why?).
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The reactance of a short dipole is negative, so it acts like a capacitor. A matching coil can therefore be created to match it, so the circuit looks like an RLC series circuit with bandwidth defined by 
 Q = \frac{\sqrt{L/C}}{R} Q=L/CR Q = \frac{\sqrt{L/C}}{R} Q=RL/C​​
.
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Half-wavelength dipoleInstead of 
 D << \lambda D<<λ D << \lambda D<<λ
, we can create a dipole antenna with  D = \lambda/2 D=λ/2 D = \lambda/2 D=λ/2
 to get  Z_\text{rad} \approx 73.1 \Omega + j42.5 Zrad≈73.1Ω+j42.5 Z_\text{rad} \approx 73.1 \Omega + j42.5 Zrad​≈73.1Ω+j42.5
. This is slightly inductive, which can be fixed by making it shorter or adding a "capacity hat." The bandwidth  \Delta f / f \approx 8\% Δf/f≈8% \Delta f / f \approx 8\% Δf/f≈8%
. Higher order modes like  3\lambda/2, 5\lambda/2, ... 3λ/2,5λ/2,... 3\lambda/2, 5\lambda/2, ... 3λ/2,5λ/2,...
also fit on the antenna.
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In the middle of the antenna for even modes, the driving point impedance looks infinite because 
 I = 0, |V| > 0 I=0,∣V∣>0 I = 0, |V| > 0 I=0,∣V∣>0
. To fix this, the feed of the antenna can be moved away from the center of the dipole.
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Patch antennasA patch antenna is like many dipoles in parallel, so it has low loss but high L and C. I don't understand how it works...
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The center frequency is given by 
 D = \lambda/2 D=λ/2 D = \lambda/2 D=λ/2
 and it has bandwidth  \Delta f / f \approx 3\% Δf/f≈3% \Delta f / f \approx 3\% Δf/f≈3%
. The width  z z z z
 would be  \approx 1.5 D ≈1.5D \approx 1.5 D ≈1.5D
 to suppress "modes in z direction."
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The high impedance problem can be fixed with an inset feed or poking in from the back -- though don't poke into the center, where impedance is very low.
Lecture 16: radiation from antennas

open t line

short dipole antenna

Half-wavelength dipole

Patch antennas

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e157: radio frequency circuit design
