What is the meaning of transmission units? What is the units of transmission of signals?
The study of transmission units has a unique importance for communication engineer who has to maintain and install telecommunication equipments achieving the standards set up by international consultation committees.
In order to control the quality of wanted signal in the presence of many undesired signals, we should be able to specify the amount of wanted and unwanted signals at a point in the telecommunications network.
The components used in the telecommunication circuit either give loss or gain to the signals they handle. There are certain specific operating conditions to be satisfied for various components without which the optimum performance cannot be obtained from these components. For this, it is essential to define conditions that control those operating conditions. This can be done only if the conditions are specified in terms of certain units of the quantity the components are to handle.
With analog transmission systems using copper cable there are three major categories of impairments. They are attenuation, noise, and distortion.
- Attenuation: There are two commonly used processes to compensate (overcome) for attenuation or loss:
(a) Repeaters are the most commonly used devices to compensate for “Loss.” However, repeaters amplify the noise along with the signal resulting in a poor signal to noise ratio.
(b) Signal to Noise Ratio: The ratio of the average signal power (strength) to the average noise power (strength) at any point in a transmission path.
- Noise: Any random disturbance or unwanted signal on a transmission facility that obscures the original signal. Noise is generally caused by the environment in which the system is operating.
- Distortion: Inaccurate reproduction of a signal caused by changes in the signal’s waveform, either amplitude or frequency, to compensate for distortion equalizers may be used. One type of equalizer used in the analog environment is the load coil. Load coils are used to flatten the frequency response.
Note: Generally the higher the frequency the greater the distortion. That is, the higher voice frequencies attenuate at a higher rate than the lower voice frequencies.
Noise and distortion on a carrier facility can be separated into two types:
(a) Predictable impairments that are almost always present on our facilities.
(b) Unpredictable impairments those are transient in nature and difficult to overcome.
Historically speaking ‘attenuation’ was first of all defined in terms of the attenuation produced by a standard reference cable known as “mile of standard cable”. It consists of 88 ohms series impedance and 0.54 µF as shunt impedance.
The fundamental objection to this unit was the fact that the attenuation of the standard cable varied with frequency. With the introduction of systems operating over different frequency ranges, it became necessary to define a unit which was independent of frequency .The unit which represents the useful and convenient concepts in connection with the transmission of signals over telephone lines has been named and defined as “Bel”(which comes from the name Alexander Graham Bell -the inventor of Telephone). In practice ,however , a smaller and more convenient unit called decibel (abbreviated as dB) is used.
One tenth of the common logarithm of the ratio of relative powers, equal to 0.1 B (bel). The decibel is the conventional relative power ratio, rather than the bel, for expressing relative powers because the decibel is smaller and therefore more convenient than the bel. The ratio in dB is given by
X = log P2/P1 B i.e. = 10 log P2/P1 dB
where P 1 and P 2 are the actual powers. Power ratios may be expressed in terms of voltage and impedance, E and Z, or current and impedance, I and Z. Thus dB is also given by;
X = 20 log V2/ V1 dB. (when Z 1 = Z 2 )
Note: The dB is used rather than arithmetic ratios or percentages because when circuits are connected in tandem, expressions of power level, in dB, may be arithmetically added and subtracted. For example, in an optical link if a known amount of optical power, in dBm, is launched into a fiber, and the losses, in dB, of each component (e.g., connectors, splices, and lengths of fiber) are known, the overall link loss may be quickly calculated with simple addition and subtraction.
Let us look at the following network:
The input is 1W and its output 2W, therefore,
Gain = 10 log (output)/(input) dB.
= 10 log 2/1 dB= 10 (0.3010) dB=3.101 dB
= 3dB approximately
Let us look at another network:
Loss = 10 log Input/Output =10 log 1000/1 dB =10 log 103 dB
=30 log 10 dB
= 30 dB
Thus a network with an input of 5 W and output of 10 W is said to have
Gain = 10 log 10/5 dB
= 10 log 2 dB
= 3 dB.
Let us remember that doubling the power means a 3 dB gain; likewise halving the power means a 3dB loss.
Consider a network with a 13 dB gain:
0.1W——Network 13 db gain—–?
Gain = 10 log P2/P1 dB = 10 log P2/0.1 dB =13db
i.e., log P2/0.1 = 1.3 or P2/0.1 = antilog 1.3 or
P2 = 0.1 antilog 1.3
P2 = 2W
Consider the following network
1W——Network 27 db Loss—–?W
What is the power output of this network? To do this without pencil and paper, we would proceed as follows:
Suppose the network attenuated the signal by 30 dB. Then the output would be 1/1000 of the input or 1mW.
Now 27 dB loss is 3dB less than 30 dB.
Thus the output would be twice 1m W i.e., 2mW.
(Because the loss is less by 3 dB, the corresponding output will be more i.e. double but not half)
It is quite simple. Thus, if we have multiples of 10 or 3 up or 3 down from these multiples, we can work it out in our mind without pencil and paper.
Let us take the next example.