10 th OCT 201110 th OCT 201110-10-2011 1

Deltastar power project services Pvt.Ltd

Name of the Trg. Prg. : Fast Track Training programmefor Experienced Engineers

on Power system Protection

Topic : Carrier Communication basics, Communication Protocol and carrier protection

Date : 10 th OCT. 2011

By : Er .R.SURENDRAN(Former SE/P&C/TNEB)

10 th OCT 2011

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.

10 th OCT 2011

The basic equations to calculate decibels

in

o

II

dB log20

in

o

VV

dB log20

in

o

PP

dB log10

Iin Io

Vo

Po

Vin

Pin

10 th OCT 2011

Adding it all up

2001.02.01

1 inVV

A

5.02.01.0

1

2 VV

Atten

151.05.1

2

32 VV

A

45.1

6

33 VV

A o

600321 AAAttenAAV

6.55600log20 dBA

10 th OCT 2011

Converting between dB and Gain notation

For dB = 20 log (Vo/Vin)if it is needed to convert from dB to output-input ratio i.e. Vo/VinVo = 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 mV Vo?

Vin

10 th OCT 2011

special decibel scales: dBm

dBmdBm: 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

10 th OCT 2011

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 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 : 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

10 th OCT 2011