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Download Logarithmic Representation of signal Levels “Decibel Notation dB”
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Levels “Decibel Notation dB” Original unit was “bel” The prefix “deci” means one tenth Hence, the “decible” is one tenth of a “bel” dB expresses logarithmically the ratio between two signal levels (ex.: Vo/Vi = Gain) dB is dimensionless Either way, a drop of 3dB represents half the power and vice versa. The basic equations to calculate decibels I Io dB 20 log I in Vo dB 20 log Vin Po dB 10 log Pin Iin o Vin Vo Pin Po Adding it all up V1 0.2 A1 20 Vin 0.01 V2 0.1 Atten 0.5 V1 0.2 V3 1.5 A2 15 V2 0.1 Vo 6 A3 4 V3 1.5 AV A1 Atten A2 A3 600 AdB 20 log 600 55.6 Converting between dB and Gain For dB = 20 log (Vnotation o/Vin) if it is needed to convert from dB to output-input ratio i.e. Vo/Vin Vo = Vin 10dB/20 or Vo = Vin EXP(dB/20) Ex: calculate the output voltage Vo if the input voltage Vin=1mV and an amplifier of +20 dB is used: Vo=(0.001V) 10(20/20) =(0.001) (10) = 0.01V Vin 1 mV Av=20dB Vo ? special decibel scales: dBm dBm: used in radio frequency measurements (RF) 0 dBm is defined as 1 mW of RF signal dissipated in 50-Ω resistive load dBm = 10 log (P/1 mW) EX: What is the signal level 9 mW as expressed in dBm? dBm = 10 log (P/1 mW) dBm = 10 log (9 mW/1 mW) = 9.54 dBm Converting dBm to voltage or Converting voltage to dBm : voltage dBm 2/50 to find Use the expression P=V2to /R=V milliwatts, and then use the equation of dBm EX: Express a signal level of 800 μV rms in dBm P=V2/50 P=0.00000064 V / 50 Ω→p=0.0000128 mW dBm = 10log(P/1mW)= -48.9 Converting dBm to voltage : Find the power level represented by the dBm level, and then calculate the voltage using 50 Ω as the load. EX: what voltage exists across a 50- Ω resistive load when -6 dBm is dissipated in the load? P=(1 mW)(10dBm/10) P =(1 mW)(10-6 dBm/10) =(1 mW)(10-0.6) =(1 mW)(0.25)=0.25 mW If P=V2/50, then V = (50P)1/2 = 7.07(P1/2), V = (7.07)(P1/2) = (7.07)(0.251/2) = 3.54 mV