History The decibel originates from methods used to quantify reductions in audio levels in telephone circuits. These losses were originally measured in units of Miles of Standard Cable (MSC), where 1 MSC corresponded to the loss of power over a 1 mile (approximately 1.6 km) length of standard telephone cable at a frequency of 5000 radians per second (795.8 Hz), and roughly matched the smallest attenuation detectable to the average listener. Standard telephone cable was defined as "a cable having uniformly distributed resistance of 88 ohms per loop mile and uniformly distributed shunt capacitance of .054 microfarad per mile" (approximately 19 gauge). The transmission unit (TU) was devised by engineers of the Bell Telephone Laboratories in the 1920s to replace the MSC. 1 TU was defined as ten times the base-10 logarithm of the ratio of measured power to a reference power level. The definitions were conveniently chosen such that 1 TU approximately equalled 1 MSC (specifically, 1.056 TU = 1 MSC). In 1928, the Bell system renamed the TU the decibel. Along with the decibel, the Bell System defined the bel, the base-10 logarithm of the power ratio, in honor of their founder and telecommunications pioneer Alexander Graham Bell. The bel is seldom used, as the decibel was the proposed working unit. Why logarithm? Decibels are expressed in a logarithmic manner which just happens to be the way the human ear and brain perceives sound levels. Also, it is easy to deal with large numbers. The decibel is used rather than arithmetic ratios or percentages because when certain types of circuits, such as amplifiers, attenuators, feed-lines and antennas are connected in series, expressions of power level in decibels may be arithmetically added and subtracted to easily obtain an overall figure for power gain or power loss. The decibel symbol is often qualified with a suffix, that indicates which reference quantity or frequency weighting function has been used. For example, dBm indicates a reference level of one milliwatt, while dBu is referenced to 0.775 volts RMS. 3-10 Rule y dBm = 10*log (x mw). Then 10*log(x/2 mw) = 10*(log(x mw) – log (2)) = y dBm – 10*0.3010 = y – 3.010 dBm (approximately y – 3 dBm) Then 10*log(x*2 mw) = 10*(log(x mw) + log (2)) = y dBm + 10*0.3010 = y + 3.010 dBm (approximately y + 3 dBm) y dBm = 10*log (x mw). Then 10*log(x/10 mw) = 10*(log(x mw) – log (10)) = y dBm – 10*1 = y – 10 dBm y dBm = 10*log (x mw). Then 10*log(x*10 mw) = 10*(log(x mw) + log (10)) = y dBm + 10*1= y +10 dBm Watts dBm dBw 0.001 0 -30 0.01 10 -20 0.1 20 -10 1 30 0 2 33 3 4 36 6 10 40 10 13 41 11 20 43 13 100 50 20 1000 60 30 Mathematical Problems based on RF Fundamentals 1. 2mW signal is fed through an amplifier that adds 16dB gain. Find the output power in terms of milliwatts. Mathematical Problems based on RF Fundamentals 2. 1. 2mW signal is fed through an amplifier that adds 5dB gain. Find the output power in terms of milliwatts. Mathematical Problems based on RF Fundamentals 3. An RF signal of power 2mW is sent through an amplifier that applies 5dB gain. find the output power in terms of dBm. Mathematical Problems based on RF Fundamentals 4. 1. 2mW signal is fed through an amplifier that adds 5dB gain. Find the output power in terms of milliwatts. Mathematical Problems based on RF Fundamentals 5. An access point sends a 30mW signal to a 12dBi antenna across a cable that cause 6dB loss; What is the power level of the radiated signal?