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Published byByron Peters Modified over 4 years ago

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For ‘Rule of 10s and 3s Example 4’ we have an access point transmitting at 30 mW. The cable and connector between the access point and the antenna create 3 dB of signal loss, and the antenna provides 20 dBi of gain.

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**Start off with the template.**

+ 3 2 * - ÷ 10 10 dBm mW 1 Start off with the template. Rule of 10s and 3s Example 4: 30 mW access point, 3 dB cable loss, 20 dBi antenna gain

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+ 3 2 * - ÷ 10 10 dBm mW Unknown 1 30 No matter how hard you try, you will not be able to multiply 1 by 2 and 10 to come up with 30. This does not mean that you can’t calculate the IR or EIRP. It only means that using the rule of 10s and 3s, you cannot set the dBm to equal 30 mW. In these cases, you can calculate the mW portion of the chart only. So enter the starting mW value, and set the dBm to unknown. You need to determine if by using * and ÷, and 2 and 10, you can change from 1 to 30 mW. Rule of 10s and 3s Example 4: 30 mW access point, 3 dB cable loss, 20 dBi antenna gain

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**+ * - ÷ 3 2 10 10 dBm mW -3 Unknown Unknown-3 30**

Even though you don’t know the dBm value, you can still calculate the dB change, and apply the correlative mathematical commands on the mW column. Since the cable and connector creates a 3 dB loss, enter it in the chart as usual and perform the calculations. Rule of 10s and 3s Example 4: 30 mW access point, 3 dB cable loss, 20 dBi antenna gain

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**+ * - ÷ 3 2 10 10 dBm mW -3 Unknown Unknown-3 30 =15 ÷2**

Even though you can’t calculate the dBm value, you can still calculate the effect of the dB change and perform the correlative mathematical commands on the mW column. Since the cable and connector creates a 3 dB loss, the mW column must be divided by 2. Rule of 10s and 3s Example 4: 30 mW access point, 3 dB cable loss, 20 dBi antenna gain

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**+ * - ÷ 3 2 10 10 dBm mW -3 Unknown Unknown-3 30 15 ÷2**

You now know that the IR is 15 mW. Since you did not know the original dBm value, you cannot give the IR value in dBms, only mWs. Rule of 10s and 3s Example 4: 30 mW access point, 3 dB cable loss, 20 dBi antenna gain

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**+ * - ÷ 3 2 10 10 dBm mW -3 +10 Unknown Unknown-3 Unknown+7 Unknown+17**

30 15 150 1500 ÷2 *10 And then perform the correlative mathematics on the mW column. Adding 10 twice in the dBm column means that you must multiply the mW column by 10 twice. Now you can add the 20 dBi gain of the antenna. You do this by adding 10 twice. Rule of 10s and 3s Example 4: 30 mW access point, 3 dB cable loss, 20 dBi antenna gain

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**+ * - ÷ 3 2 10 10 dBm mW -3 +10 Unknown Unknown-3 Unknown+7 Unknown+17**

30 15 150 1500 ÷2 *10 So the EIRP is 1500 mW or 1.5 watts. So even though you could not calculate the actual value for dBms using the rule of 10s and 3s, you were still able to calculate the mW value. Rule of 10s and 3s Example 4: 30 mW access point, 3 dB cable loss, 20 dBi antenna gain

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**+ * - ÷ 3 2 10 10 dBm mW 14.7712 30 dBm = 10 * log10(PmW)**

Don’t forget, instead of using the rule of 10s and 3s, you can always use the logarithmic formula to calculate the dBm value. dBm = 10 * log10(PmW) = 10 * log10(30) Rule of 10s and 3s Example 4: 30 mW access point, 3 dB cable loss, 20 dBi antenna gain

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+ 3 2 * - ÷ 10 10 dBm mW -3 +10 30 If you had used the logarithmic formula dBm = 10 * log10(PmW) to calculate the dBm value for 30 mW, the value would be If you then added and subtracted the dB gains and losses and calculated the corresponding mW values, this is what you would have come up with. Rule of 10s and 3s Example 4: 30 mW access point, 3 dB cable loss, 20 dBi antenna gain

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