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Learning Decibel Basics:

When working in several disciplines of Telecommunication,a clear understanding of the decibel (dB) is mandatory.

The decibel relates to a ratio of two electrical quantitiessuch as watts ,volts, and amperes.

If we pass a signal through some device ,it will suffer a loss orachieve a gain.Such a device may be attenuator. amplifier, mixer, transmission line

, antenna , subscriber loop , trunk . or a telephone switch amongothers. To simplify matters , let us call this generic device a network.which has an input port and an output as shown:

The input and output can be characterized by a signal level, whichcan be measured in watts, amperes or Volts.The decibel is a useful tool to compare input-to-output levels or vice-versa.Certainly we can say that if the out put level is > the input level, thedevice displays a gain. The signal has been amplified.If the output has a lower level than the input , the network display aloss.

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Learning Decibel Basics: In our discussion we will indicate a gain with +ve sign such as +3 dB, +11 dB etc. and a loss

with a ve sign -3dB,-11db etc. At the outset it will be more convenient to use the same unit at the output of a network as at

the input , such as watts. If we use watts 1W= 1000mW: 1W= 1000000 microwatts 1kW=1000W We will start off in the power domain (watts, mW) Volts and amperes we will deal later. Again

decibel expresses a ratio. In the power domain ,the decibel value of such a ratio is 10Xlogarithm of the ratio.

Consider the network ,

we are concerned about the ratio of P1/P2 or vice versa.Algebraically , we express the decibel by this formula :

dB value = 10 log(P1/P2) or 10 log(P2/P1). 100 = the log is 2 , 1000 = the log is 3 etc.

0.1= the log is -1 , 0.01= the log is -2 Decibel values =1 dB value = 10 log 1= 10 X0= 0dB =10 dB value = 10 log 10= 10 X1=10 dB = 100 dB value = 10 log 100= 10 X2= 20 dB = 0.1 dB value = 10 log 0.1= 10 X(-1)=-10dB =0.01 dB value = 10 log0.01= 10 X(-2)=-20dB

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Learning Decibel Basics:

Let us deal with the following situation

Because the output of this network is greater than the input , networkhas a gain.

Thus dB value = 10 log 4/2= 10 log2=10X 0.3010=+3.01 dB. We usually round off this dB value to +3 dB This relationship should be memorized. The amplifying network has a 3 dB gain because the output power was

double the input power.With 3-dB rule multiple of 3 are easy. If we have power ratios of 2,4, and8, that the equivalent dB values are + 3dB,+6 dB and +9 dB respectively.

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Learning Decibel Basics:

Let us take the +9 dB as an example problem. A network has an input of 6 mW and a gain of+9 dB. What power level in mW would we expect to measure at the output port?

We break down the +9-dB network into three networks in series, each with a gain of +3 dB.This is shown in the following diagram:

Remember that +3 dB is double the power; the power at the output of a networkwith +3-dB gain has 2 the power level at the input. Obviously, the output of thefirst network is 12 mW (point A above).

The input to the second network is now 12 mW and this network again doubles thepower. The power level at point B, the output of the second network, is 24 mW. Thethird networkdouble the power still again. The power level at point C is 48 mW.

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dBm and dBW Once we have a grasp of dBm and dBW, we will find it easier to work

problems with networks in series. Example:

First we convert the input, 8 mW to dBm. Look how simple it is: 2 mW = +3dBm, 4 mW = +6 dBm, and 8 mW = +9 dBm.

To get the answer, the power level at the output is +9 dBm +23 dB = +32

dBm.

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dBm and dBW In this case the unknown will be the input to a

network.

In each case like this we ask ourselves, is theoutput greater than the input?

Because the network is lossy, the input is 17dB greater than the output. Convert the outputto dBm.

It is +10 dBm. The input is 17 dB greater, or

+27 dBm.