Sorry, reviewing the extraction of the equation of Pload(dbm) in around 1:43 shouldn't the G(dbi) be a negative term when we solve to find field strength?
Hello. This is the reply from Frank Sanders, via the Publications Office of the Institute for Telecommunication Sciences (ITS). You can go either way with your sign (+/-) in your computation. Nature doesn’t care. BUT you DO need to keep your own signage straight as you go along, and you should state your signage convention if you are writing your results for others to see/read.
1 / (lambda^2) = f^2 / C^2. The frequency is 100 MHz, so 100*10^6 Hz. So the (10^6)^2 is the mega (because the frequency is in megahertz). But I think there should be a square over the (f, MHz) component as well, so (f, MHz)^2 because 100MHz^2 = (10^6)^2 * 100^2.
These are all great! I just found them. The instructor has obviously lived in both the theoretical and practical world.
No question...absolutely incredible
Sorry, reviewing the extraction of the equation of Pload(dbm) in around 1:43 shouldn't the G(dbi) be a negative term when we solve to find field strength?
Hello. This is the reply from Frank Sanders, via the Publications Office of the Institute for Telecommunication Sciences (ITS).
You can go either way with your sign (+/-) in your computation. Nature doesn’t care. BUT you DO need to keep your own signage straight as you go along, and you should state your signage convention if you are writing your results for others to see/read.
Sorry I missed it, but where does the (10^-6)^2 term in the numerator come from ?
1 / (lambda^2) = f^2 / C^2.
The frequency is 100 MHz, so 100*10^6 Hz.
So the (10^6)^2 is the mega (because the frequency is in megahertz).
But I think there should be a square over the (f, MHz) component as well, so (f, MHz)^2 because 100MHz^2 = (10^6)^2 * 100^2.