Dfdzs Essay

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Tutorial Wireless Basics Revision of dB, dBW, dBm, dBV, etc. Q. Define the terms dB, dBW, dBm, dBi, and Signal to Noise Ratio (SNR). Q. Why do we use the dB scale? When do we use dB, dBW or dB? Q. Find the gain in dB if a 10 mW signal is inserted into a transmission line and the measured power some distance away is 5 mW? Repeat if the output is 20mW. Answer -3dB, +3dB NdB=10 Log(5/10)= 10(-0.3)= -3 dB Generally doubling power means 3 dB and halving power is -3 dB Q. A network has an input of 5mW and a gain of +9dB. What power level in mW would be expect to measure at the output port? Answer 40mW Given that each 3dB is doubling power then 9dB corresponds to 2 * 2 * 2 = 8 times So if the input is 5mW then output = 8 * 5 = 40 mW. Alternatively using the formulae 9 dB = 10 log10( Output/ 5 mW) Q. Fill the following table of approximate power ratios for various dB levels. Decibels ±1 ±2 ±3 ±4 ±5 Losses 0.5 0.8 0.63 0.4 0.32 Gains 2 1.25 1.6 2.5 3.2 ±6 0.25 4.0 ±7 0.2 5.0 ±8 0.16 6.3 ±9 0.125 8.0 ±10 0.1 10 You can use the calculator to work out the numbers in the table using dB = 10 log(Power Gain ) where Power Gain = Output Power / Input Power For + 2dB Power Gain = 10 ^( 2 /10) = 1.59 For -2 dB Power Gain = 10^(-2/10) = 0.63 Q. An antenna has an output of 10 W. What is its output in dBW and dBm? Repeat for 20W. Ans: 10 dBW and 40 dBm Answer Power (dBW) = 10 log (Power/1W) = 10 log10 = 10 dBW = 10 +30 dBm = 40 dBm Alternatively Power (dBm) = 10 log (Power in mW/1mW)= 10 log (10000mW/1mW)= 40 dBm Q. Prove that Power in dBm = Power in dBW + 30 Power (dBm) = 10 log (Power in mW/1mW) = 10 log (1000 * Power in W/1) using log (x * y) = log (x) + log (y) high school math = 10 log (1000) + 10 log (power in W/1) = 30 + Power in dBW Q. Fill following table with dBW and dBm values Decibels 1W 100mW 10mW dBW -20 0 -10 dBm 10 30 20 1mW -30dBW 0dBm 0.1mW -40 -10 0.01mW

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