• In the book, the example given uses the speed of light for the example, which looks like this.

    wavelength = 300,000,000 m/s / 2,400,000,000 Hz
          wavelength = 0.125 m = 12.5 cm

    Yet directly preceding that, the book states that RF energy propogates at around 300,000 m/s in the earth's atmosphere. Which would yield the following result.

    wavelength = 300,000 m/s / 2,400,000,000 Hz
          wavelength = .000125 = .0125 cm

    If the second is the more realistic figure, why is the speed of light used in the example? Are antenna elements commonly .0125 cm in length?

  • to find the wavelength you need to divide the speed of light by the frequency of the wave. That is a basic physic. The correct speed of light is 300,000,000 m/s. Frequency of 11b/g is 2.4Ghz thus the wavelength is as stated. Antennas are most responsive if its element is multiple of wavelength

  • The book states you're using the speed of the wave. It then goes on to say that the speed of light is not a realistic figure and that the wave propogates at around 300,000 m/s in the earth's atmosphere. Hence, wouldn't you use 300,000 m/s?

  • please look at the errat for the book. Btw, what version of the book are you using? I know for fact that in the latest version, there has been a mistake in the number they use for the speed of light. Please refer to the errata

  • I'm using the latest version. Do you have a link, by chance?


    When you get to the CWSP and AP, it is a good idea to first take the printed errata and fix all of the errors. In some cases, I physically cut and paste the errata changes into my book (brings back good memories of second grade :)). It is really nice to know that you have the latest version when reading it with the included errata.

    GT Hill

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