I am thinking of let us say AM or FM radio being transmitted by a vertical antenna. Would the they be polarized vertically or horizontally and wouldn't the angle of the receiving antenna then determine the strength of the signal.
Electrical – Are radio waves naturally polarized like light or is that a function of how they are produced
antennawave
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Yes. The 802.11 Medium Access Control, which is in charge of deciding on the data rate between two WiFi devices (which use IEEE 802.11), scales the data rate between the devices depending on a lot of things. One of them is signal strength. This correlation isn't necessarily linear, but it can be. Some packets are transmitted at 1Mbps while data is typically at the max speed. When the signal is not stable, it's common for the data rate to vary quickly. For more information see 802.11 Wireless Networks: The Definitive Guide, Second Edition
You can do it either way. First you need to get all your units together. In small RF circuits you might see the output noted in dBm, which is the absolute power referenced to one milliwatt. To convert from dBm to W or mW and in reverse, these equations from Wikipedia provide the answer,
$$ P = 1mW \times 10^{{dBm}\over{10}}\\ P = 1W \times 10^{{(dBm - 30)}\over 1}\\ dBm = 10log_{10} {P\over1mW}\\ dBm = 10log_{10} {P\over1W} + 30 $$
So, if your system is rated at an output 200mW, you'll calculate an output of 23dBm. This will be used in the equation at the bottom of this answer.
The gain of an antenna is not a literal boost in power output, it's a perceived gain in one direction or on one axis from what one would expect from either an isotropic radiator (a perfect sphere) or a dipole radiator (there are other less common reference points). That is, if you were standing a certain distance from a perfect isotropic radiator and measured a power of 0 dB it would read 0 dB at every point with the same distance from the antenna, but, a real antenna with a gain of 2dB will read 2dB at some point with the same distance from the antenna, but certainly not every point at that distance. If the antenna is listed with gain units of dB, it's likely with respect to an isotropic radiator, but it could be with respect to a dipole. Some manufacturers use the \$ dB_i\$ or \$dB_d\$ to denote this difference.
As an illustration between the two, in the image below all points on the line would read the same output power.
If only the non-isotropic antenna were on, and one measured the power output at the crossing point of the axis and the isotropic radiator the power would be higher than 0dB, say 3dB, then that antenna would be said to have a gain of 3dB. But, if the measurement was made on the opposite side the gain would be lower than 0dB, perhaps -5dB.
If you want to calculate how much power you will receive you can use the Friis transmission equation to get a rough estimate.
$$ P_r = P_t + G_t + G_r + 10log_{10}({\lambda \over {4\pi R}}) $$
The \$P_r\$ is the power received the \$P_t\$ is the power transmitted that you calculated from above, the \$G\$s are gains of the transmitting and receiving antennas respectively (in dB) and the final bit is the isotropic antenna equation, which will tell you how much power was 'lost' because it was not directed at your receiving antenna. So, as you can see, increasing either the gain of the antenna or the power transmitted will increase the power received.
So in summary, you can either get an antenna with a higher gain and point it correctly to increase signal strength or you can increase the power delivered to the antenna. Either one, done correctly, will increase the power output of your system (or at least the power received by the other end of the system).
Best Answer
Radio waves emitted by an antenna have a specific polarization, and receiving antennas are generally sensitive only to a specific polarization. So in principle if the transmit antenna were strictly vertical and your receive antenna were strictly horizontal, you would receive nothing. But there are a couple of complexities:
The key difference between radio waves and visible light is that most of the radio signals we are familiar with are produced by coherent emission processes, which (usually) produce fully-polarized radio waves. More, almost all detectors of radio waves coherently detect just one polarization; radio astronomers usually use pairs of crossed dipoles so we can record both polarizations and reconstruct the input signal's polarization state.
Most of the visible light sources we deal with are incoherent and produce unpolarized light (an even mixture of polarizations) and our detectors mostly aren't sensitive to polarization anyway. Lasers are coherent and indeed are polarized, but unless the laser is designed to have a stable polarization, you tend to get random jumping around on very short time scales, averaging out to unpolarized. The human eye is in fact very slightly sensitive to polarization, though we don't usually pay attention, and there are processes - like reflection - that readily add polarization to light, hence the utility of polarized sunglasses (to preferentially block light reflected off horizontal surfaces).