1. Logarithmic Representation of signal 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.: V o /V i = Gain)  dB is dimensionless  Either way, a drop of 3dB represents half the power and vice versa.

2. The basic equations to calculate decibels I in IoIo VoVo PoPo V in P in

3. Adding it all up

4. Converting between dB and Gain notation  For dB = 20 log (V o /V in ) if it is needed to convert from dB to output-input ratio i.e. V o /V in  V o = V in 10 dB/20 or V o = V in EXP(dB/20) Ex: calculate the output voltage V o if the input voltage V in =1mV and an amplifier of +20 dB is used:  V o =(0.001V) 10 (20/20) =(0.001) (10) = 0.01V A v =20dB 1 mV Vo?Vo? V in

5. special decibel scales: dBm  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

6. Converting dBm to voltage or voltage to dBm  Converting voltage to dBm : Use the expression P=V 2 /R=V 2 /50 to find milliwatts, and then use the equation of dBm EX: Express a signal level of 800 μV rms in dBm  P=V 2 /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)(10 dBm/10 ) P =(1 mW)(10 -6 dBm/10 ) =(1 mW)(10 -0.6 ) =(1 mW)(0.25)=0.25 mW If P=V 2 /50, then V = (50P) 1/2 = 7.07(P 1/2 ),  V = (7.07)(P 1/2 ) = (7.07)(0.25 1/2 ) = 3.54 mV