RF Math

6 posts by 5 authors in: Forums > CWNA - Enterprise Wi-Fi Admin
Last Post: April 20, 2005:
  • The study guide specifically states that students are not required to know various equations, such as logarythms for the exam. So, I learned the principles of 10's and 3's.
    Below are 3 questions from Boson for the CWNA exam.
    I am unable to grasp these questions and answers. Are these the types of questions we will see on the exam, or are the Boson questions just really tough? Is there an easier way to figure these out?

    1.If a radar has a gain of 98 dB through its entire receive chain. A signal at the receive chain input (the antenna) has a power of 5.2·10-9 mW, what would the power be at the output of the receive chain? A.) 32.72 mW B.) 32.74 mW C.) 32.70 mW D.) 32.76 mW Gain = 98 dB implies the power ratio is = 10(98/10) = 10(9.8) = 6.3·109 Hence power at the output of the receive chain = 5.2·10-9 · 6.3·109 = 32.76 mW

    2.Radar A produces 5000 W peak power, while radar B produces 2000 W peak power. How many dB better is radar A? A.) 4 dB B.) 5 dB C.) 6 dB D.) 3 dB dB = 10·log10(P1 / P2) P2 is the reference, so use the 2000 W of radar B. dB = 10·log10(5000 / 2000) = 10·log10 (2.5) = 4 dB (Since the power of Radar A is greater than that of radar B, the dB value must come out positive.)

    3.If resistance remains constant, Decibel in terms of voltage ratio is equivalent to, A.) 20 log (V2/V1) B.) (10) 2 log (V2/V1) C.) 10 log (V2/V1) D.) (20) 2 log (V2/V1) Voltage and current ratios can also be expressed in terms of decibels, provided the resistance remains constant. First we substitute for P in terms of voltage, V, or current, I. Since P=VI and V=IR we have: P = I2R = V2/R Thus for a voltage ratio we have dB = 10 log[(V22/R)/(V12/R)] = 10 log [(V22)/(V12)] = 10 log [(V2)/(V1)] 2 = 20 log [(V2)/(V1)]

    Thanks in advance![/img]

  • boson are known to always be a little more difficult. However they are great for review.

  • Hi Mary of Toronto:

    These questions are harder than what I recall on both the CWNA and both the Cisco wireless certifications. On all three tests the rules of 10's and 3's got me through even if indirectly.

    Of these three questions the first seems too precise for 10's and 3's, the second is ok, and the third is over the top.

    I hope this helps. Thanks. /criss

  • Thanks for your input. I can breathe a little easier now.....

  • Mary, I concur with criss. the test should only cover the rules for 10 and 3.

    Question 3. you listed is using Ohms Laws to calculate power. This is more used for troubleshooting electronic circuitry.

  • We only test on the 10's and 3's on the CWNA exam. No formulas, and hence no calculators are required. Boson's exams are not endorsed or supported by The CWNP Program.


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