LEARNING DECIBELS AND THEIR APPLICATIONS

APPENDIX

C

LEARNING DECIBELS AND THEIR APPLICATIONS C.1

LEARNING DECIBEL BASICS

When working in the several disciplines of telecommunications, a clear understanding of the decibel (dB) is mandatory. The objective of this appendix is to facilitate this understanding and to encourage the reader to take advantage of this useful tool. The decibel relates to a ratio of two electrical quantities such as watts, volts, and amperes. If we pass a signal through some device, it will suffer a loss or achieve a gain. Such a device may be an attenuator, amplifier, mixer, transmission line, antenna, subscriber loop, trunk, or a telephone switch, among others. To simplify matters, let’s call this generic device a network, which has an input port and an output port, as shown:

The input and output can be characterized by a signal level, which can be measured in either watts (W), amperes (A), or volts (V). The decibel is a useful tool to compare inputto-output levels or vice versa. Certainly we can say that if the output level is greater than the input level, the device displays a gain. The signal has been amplified. If the output has a lower level than the input, the network displays a loss. In our discussion we will indicate a gain with a positive sign (+) such as +3 dB, +11 dB, +37 dB; and a loss with a negative sign (−): −3 dB, −11 dB, −43 dB. 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, for example, it is watts or any of its metric derivatives. Remember: 1 W = 1000 milliwatts (mW), 1 W = 1,000,000 (1 × 106 ) microwatts (µW), 1 W = 0.001 kilowatts (kW), 1000 mW = 1 W, 1 kW = 1000 W. Fundamentals of Telecommunications, Second Edition, by Roger L. Freeman ISBN 0-471-71045-8 Copyright © 2005 by Roger L. Freeman

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LEARNING DECIBELS AND THEIR APPLICATIONS

We will start off in the power domain (watts are in the power domain, so are milliwatts; volts and amperes are not). We will deal with volts and amperes later. Again, the decibel expresses a ratio. In the power domain (e.g., level is measured in watts or milliwatts), the decibel value of such a ratio is 10 × logarithm of the ratio. Consider this 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 ).

(C1.1)

Some readers may feel apprehensive about logarithms. The logarithm (log) used here is to the number base 10. A logarithm is an exponent. In our case it is the exponent of the number 10 such as 100 101 102 103 104

=1 = 10 = 100 = 1000 = 10,000

the the the the the

log log log log log

is is is is is

0 1 2 3 4, etc.

For numbers less than 1, we use decimal values, so 100 = 1 10−1 = 0.1 10−2 = 0.01 10−3 = 0.001 10−4 = 0.0001

the the the the the

log log log log log

is is is is is

0 −1 −2 −3 −4, etc.

Let us now express the decibel values of the same numbers: 100 = 1 101 = 10 102 = 100 103 = 1000 104 = 10,000 10−1 = 0.1 10−2 = 0.01 10−3 = 0.001 10−4 = 0.0001

log = 0 log = 1 log = 2 log = 3 log = 4 log = −1 log = −2 log = −3 log = −4

dB dB dB dB dB dB dB dB dB

value = 10 value = 10 value = 10 value = 10 value = 10 value = 10 value = 10 value = 10 value = 10

log 1 = 10 × 0 = 0 dB log 10 = 10 × 1 = 10 dB log 100 = 10 × 2 = 20 dB log 1000 = 10 × 3 = 30 dB log 10,000 = 10 × 4 = 40 dB, etc. log .1 = 10 × −1 = −10 dB log .01 = 10 × −2 = −20 dB log .001 = 10 × −3 = −30 dB log .0001 = 10 × −4 = −40 dB, etc.

We now have learned how to handle power ratios of 10, 100, 1000, and so on, and 0.1, 0.01, 0.001, and so on. These, of course, lead to dB values of +10 dB, +20 dB, and +30 dB; −10 dB, −20 dB, −30 dB, and so on. The next step we will take is to learn to derive dB values for power ratios that lie in between 1 and 10, 10 and 100, 0.1 and 0.01, and so on.

C.1

LEARNING DECIBEL BASICS

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One excellent recourse is the scientific calculator. Here we apply a formula (C1.1). For example, let us deal with the following situation:

Because the output of this network is greater than the input, the network has a gain. Keep in mind we are in the power domain; we are dealing with mW. Thus: dB value = 10 log 4/2 = 10 log 2 = 10 × 0.3010 = +3.01 dB. We usually roundoff this dB value to +3 dB. If we were to do this on our scientific calculator, we enter 2 and press the log button. The value 0.3010—appears on the display. We then multiply (×) this value by 10, arriving at the +3.010 dB value. This relationship should be memorized. The amplifying network has a 3-dB gain because the output power was double the input power (i.e., the output is twice as great as the input). For the immediately following discussion, we are going to show that under many situations a scientific calculator is not needed and one can carry out these calculations in his or her head. We learned the 3-dB rule. We learned the +10, +20, +30 dB; −10, −20, −30 (etc.) rules. One should be aware that with the 3-dB rule, there is a small error that occurs two places to the right of the decimal point. It is so small that it is hard to measure. With the 3-dB rule, multiples of 3 are easy. If we have power ratios of 2, 4, and 8, we know that the equivalent (approximate) dB values are +3 dB, +6 dB, and +9 dB, respectively. 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?

One thing that is convenient about dBs is that when we have networks in series, each with a loss or gain given in dB, we can simply sum the values algebraically. Likewise, we can do the converse: We can break down a network into hypothetical networks in series, so long as the algebraic sum in dB of the gain/loss of each network making up the whole is the same as that of the original network. We have a good example with the preceding network displaying a gain of +9 dB. Obviously 3 × 3 = 9. 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:

We should be able to do this now by inspection. Remember that +3 dB is double the power; the power at the output of a network with +3-dB gain has 2× the power level at the input. Obviously, the output of the first network is 12 mW (point A above). The

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input to the second network is now 12 mW and this network again doubles the power. The power level at point B, the output of the second network, is 24 mW. The third network—double the power still again. The power level at point C is 48 mW. Thus we see that a network with an input of 6 mW and a 9-dB gain, will have an output of 48 mW. It multiplied the input by 8 times (8 × 6 = 48). That is what a 9-dB gain does. Let us remember: +3 dB is a two-times multiplier; +6 dB is a four-times multiplier, and +9 dB is an eight-times multiplier. Let us carry this thinking one step still further. We now know how to handle 3 dB, whether + or −, and 10 dB (+ or −), and all the multiples of 10 such as 100,000 and 0.000001. Here is a simple network. Let us see what we can do with it.

We can break this down into two networks using dB values that are familiar to us:

If we algebraically sum the +10 dB and the −3 dB of the two networks in series shown above, the result is +7 dB, which is the gain of the network in question. We have just restated it another way. Let us see what we have here. The first network multiplies its input by 10 times (+10 dB). The result is 15 × 10 or 150 mW. This is the value of the level at A. The second network has a 3-dB loss, which drops its input level in half. The input is 150 mW and the output of the second network is 150 × 0.5, or 75 mW. This thinking can be applied to nearly all dB values except those ending with a 2, 5, or 8. Even these values can be computed without a calculator, but with some increase in error. We encourage the use of a scientific calculator, which can provide much more accurate results, from 5 to 8 decimal places. Consider the following problem:

This can be broken down as follows:

Remember that +50 dB is a multiplier of 105 and −6 dB is a loss that drops the power to one quarter of the input to that second network. Now the input to the first network is 0.3 mW and so the output of the first network (A) is 0.3 mW × 100,000 or 30,000 mW (30 W). The output of the second network (B) is one-quarter of that value (i.e., −6 dB), or 7500 mW. Now we will do a practice problem for a number of networks in series, each with its own gain or loss given in dB. The idea is to show how we can combine these several

C.2

dBm AND dBW

619

networks into an equivalent single network regarding gain or loss. We are often faced with such a problem in the real world. Remember, we add the dB values in each network algebraically.

Look what happens when we combine these four networks into one equivalent network. We just sum: +12 − 28 + 7 − 11 = −20, and −20 dB is a number we can readily handle. Thus the equivalent network looks like the following:

To see really how well you can handle dBs, the instructor might pose a difficult problem with several networks in series. The output power of the last network will be given and the instructor will ask the input power to the first network. Let us try one like that so the instructor will not stump us.

First sum the values to have an equivalent single network: +23 + 15 − 12 = +26 dB. Thus,

We first must learn to ask ourselves: Is the input greater or smaller than the output? This network has gain, thus we know that the input must be smaller than the output. By how much? It is smaller by 26 dB. What is the numeric value of 26 dB? Remember, 20 dB is 100; 23 dB is 200, and 26 dB is 400. So the input is 1/400 of the output or 40/400 (mW) = 0.1 mW. C.2

dBm AND dBW

These are the first derived decibel units that we will learn. They are probably the most important. The dBm is also a ratio. It is a decibel value related to one milliwatt (1 mW). The dBW is a decibel value related to one watt (1 W). Remember the little m in dBm refers to milliwatt and the big W in dBW refers to watt. The values dBm and dBW are measures of real levels. But first we should write the familiar dB formulas for dBm and dBW: Value (dBm) = 10 log P1 /(1 mW), Value (dBW) = 10 log P1 /(1 W).

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Here are a few good relationships to fix in our memories: 1 mW = 0 dBm

(by definition),

1 W = 0 dBW

(by definition),

+30 dBm = 0 dBW = 1 W, −30 dBW = 0 dBm = 1 mW. Who will hazard a guess what +3 dBm is in mW? Of course, it is 3 dB greater than 0 dBm. Therefore it must be 2 mW. Of course, +6 dBm is 4 mW, and −3 dBm is half of 0 dBm or 0.5 mW. A table is often helpful for the powers of 10: 1 mW = 100 mW = 0 dBm, 10 mW = 101 mW = +10 dBm, 100 mW = 102 mW = +20 dBm, 1000 mW = 103 mW = +30 dBm = 0 dBW, 10 W = 104 mW = +40 dBm = +10 dBW(etc.). Likewise, 0.1 mW = 10−1 = −10 dBm, 0.01 mW = 10−2 = −20 dBm, 0.001 mW = 10−3 = −30 dBm, 0.0001 mW = 10−4 = −40 dBm. Once we have a grasp of dBm and dBW, we will find it easier to work problems with networks in series. We now will give some examples.

First we convert the input, 8 mW to dBm. Look how simple it is: 2 mW = +3 dBm, 4 mW = +6 dBm, and 8 mW = +9 dBm. Now watch this! To get the answer, the power level at the output is +9 dBm +23 dB = +32 dBm. Another problem will be helpful. In this case the unknown will be the input to a network.

In each case like this we ask ourselves, is the output greater than the input? Because the network is lossy, the input is 17 dB greater than the output. Convert the output to dBm. It is +10 dBm. The input is 17 dB greater, or +27 dBm. We should also be able to

C.4

USING DECIBELS WITH SIGNAL CURRENTS AND VOLTAGES

621

say: “that’s half a watt.” Remember, +30 dBm = 1 W = 0 dBW. Then +27 dBm (“3 dB down”) is half that value. Several exercises are in order. The answers appear after the four exercises. Exercise 1a.

Exercise 1b.

Exercise 1c.

Exercise 1d .

(Answers: 1a: +13 dBm = 20 mW; 1b: +29 dBW, 1c: +32 dBm, and 1d: +7 dBm = 0.005 W). C.3

VOLUME UNIT (VU)

The VU is the conventional unit for measurement of speech level. A VU can be related to a dBm only with a sinusoidal tone (a simple tone of one frequency) between 35 Hz and 10,000 Hz. The following relationship will be helpful: Power level in dBm = VU − 1.4dB

(for complex audio signals).

A complex audio signal is an audio signal composed of many sine waves (sinusoidal tones) or, if you will, many tones and their harmonics. One might ask: If the level reading on a broadcaster’s program channel is −11 VU, what would the equivalent be in dBm? Reading in VU − 1.4 dB = reading in dBm. Thus the answer is −11 VU − 1.4 dB = −12.4 dBm. C.4

USING DECIBELS WITH SIGNAL CURRENTS AND VOLTAGES

The dB is based upon a power ratio, as discussed. We can also relate decibels to signal voltages and to signal currents. The case for signal currents is treated first. We are dealing,