logarithmic representation of signal levels “decibel notation db” original unit was “bel” ...
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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.
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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
![Page 3: Logarithmic Representation of signal Levels “Decibel Notation dB” Original unit was “bel” The prefix “deci” means one tenth Hence, the “decible”](https://reader035.vdocument.in/reader035/viewer/2022071806/56649d935503460f94a7a3f3/html5/thumbnails/3.jpg)
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
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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
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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
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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