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.: 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

in

o

I

IdB log20

in

o

V

VdB log20

in

o

P

PdB log10

Iin Io

Vo

Po

Vin

Pin

Adding it all up

2001.0

2.011

inV

VA

5.02.0

1.0

1

2 V

VAtten

151.0

5.1

2

32 V

VA

45.1

6

33 V

VA o

600321 AAAttenAAV

6.55600log20 dBA

Converting between dB and Gain notationFor dB = 20 log (Vo/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

Av=20dB

1 mVVo?

Vin

special decibel scales: dBmdBmdBm: used in radio frequency

measurements (RF)0 dBm is defined as 1 mW of RF signal

dissipated in 50-Ω resistive loaddBm = 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 voltage to dBm

Converting voltage to dBm : Converting voltage to dBm : Use the expression P=V2/R=V2/50 to find milliwatts, and then use the equation of dBmEX: 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 : 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