Logarithmic Representation of signal Levels “Decibel Notation dB”

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Transcript Logarithmic Representation of signal Levels “Decibel Notation dB”

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