Biblio5

616 LEARNING DECIBELS AND THEIR APPLICATIONS

Exercise 1c.

Exercise 1d.

(Answers: 1a: +13 dBm c 20 mW; 1b: +29 dBW, 1c: +32 dBm, and 1d: +7 dBm c 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:

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 c reading in dBm. Thus the answer is −11 VU − 1.4 dB c −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, of course, with gains and losses for a device or several devices (called networks) that are inserted in a circuit. Follow the thinking behind this series of equations:

Gain/ LossdB c 10 log P1/ P2 c 10 log I21R1/ I22R2 c 20 log I1/ I2 c 10 log R1/ R2.

If we let R1 c R2, then the term 10 log R1/ R2 c 0. (Hint: The log of 1 c 0.)

Remember from Ohm’s law that E c IR, and from the power law Pw c EI. Thus Pw c I2R c E2/ R.

To calculate gain or loss in dB when in the voltage/ current domain we derive the following two formulas from the reasoning just shown:

618 LEARNING DECIBELS AND THEIR APPLICATIONS

A similar network has a signal output of 40 V and a loss of 6 dB. What is the input signal voltage? (Equal impedances assumed.)

−6 dB c 20 log 40/ X.

We shortcut this procedure by remembering our 10 log values. With a voltage or amperage relationship, the dB value is double (20 is twice the magnitude of 10). The value of X is 80 V. Whereas in our 10 log regime 3 dB doubled (or halved), here 6 dB doubles or halves.

A more straightforward way of carrying out this procedure will be suggested in the next section. X can be directly calculated.

C.5 CALCULATING A NUMERIC VALUE GIVEN A dB VALUE

The essence of the problem of calculating a numeric value given a dB value can be stated as such: If we are given the logarithm of a number, what is the number? To express this, two types of notation are given in the literature as follows:

In the case of example 1, the logarithm is 0.3010, which corresponds to the number 2. If we were to take the log (base 10) of 2, the result is 0.3010. In example 2, the log of 100 is 2 or, if you will, 2 is the logarithm of 100.

For our direct application we may be given a decibel value and be required to convert to its equivalent numeric value. If we turn to our introductory comments, when dealing in the power domain, we know that if we are given a decibel value of 20 dB, we are working with a power gain or loss of 100; 23 dB, 200; 30 dB, 1000; 37 dB 5000, and so on.

A scientific calculator is particularly valuable when we are not working directly with multiples of 10. For instance, enter the logarithm of a number onto the calculator keypad and the calculator can output the equivalent numeric value. Many hand-held scientific calculators use the same button for the log as for the antilog. Usually one can access the antilog function by first pressing the “2nd” button, something analogous to upper case on a keyboard. Often printed directly above the log button is “10x.”

On most calculators we first enter the logarithm on the numerical keypad, being sure to use the proper signs (+ or −). Press the “2nd” button; then press the “log” button. After a short processing interval, the equivalent number is shown on the display.

Let us get to the crux of the matter. We are interested in dBs. Let us suppose we are given 13 dB and we are asked to find its numeric equivalent (power domain). This calculation is expressed by the following formula:

log−1(13/ 10) c 20.

Let us use a calculator and compute the following equivalent numeric values when given dB values:

1. −21.5 dB. Divide by 10 and we have −2.15. Enter this on the keyboard with the negative sign. Press 2nd F (function) to access the upper case, which is the same as the log button but marked right above “10x .” Press c button and the value 0.00708 appears on the display.

2. +26.8 dB. Enter this number on the keypad and divide by 10; press c. Press 2nd F; press log button (10x ) and press c. The equivalent numeric value appears on the display. It is 478.63.

When working in the voltage or current domain, we divide the dB value by 20 rather than 10. Remember we are carrying out the reverse process that we used calculating a dB value when given a number (numeric) (i.e., the result of dividing the two numbers making up the ratio). This is expressed by the following formula:

log−1(dB value/ 20) c equivalent numeric.

Consider this example. Convert 26 dB (voltage domain) to its equivalent numeric value. Enter 26 on the keypad and divide by 20. The result is 1.3. Press 2nd F button and press the log button (10x ) and press c. The value 19.952 appears on the display. The reader probably did this in his or her head and arrived at a value of 20.

Try the following six example problems, first in the power domain, and then in the voltage/ amperage domain. The correct answers appear just below.

Answers: 1: 0.251; 0.501. 2: 3,981,071.7; 1995.26. 3: 0.006309; 0.07943. 4: 50.118; 7.07945. 5: 0.001995; 0.044668. 6: 7.413; 2.7227.

C.5.1 Calculating Watt and Milliwatt Values When Given dBW and dBm Values

We will find that the process of calculating numeric values in watts and milliwatts is very similar to calculating the numeric value of a ratio when given the equivalent value in decibels. Likewise, the greater portion of these conversions can be carried out without a calculator to a first-order estimation. In the case where the dB value is 10 or a multiple thereof, the value will be exact.

Remember: 0 dBm c 1 mW; 0 dBW c 1 W by definition. Further, lest we forget: +3 dBm is twice as large as the equivalent 0 dBm value, thus where 0 dBm c 1 mW, +3 dBm c 2 mW.

Also, +10 dBm numeric value is 10 times the equivalent 0 dBm value (i.e., it is 10 dB larger). So +10 dBm c 10 mW; −10 dBm c 0.1 mW; −20 dBm c 0.01 mW.

In addition, −17 dBm is twice the numeric magnitude of −20 dBm. So −17 dBm c 0.02 mW, and so forth.

Try calculating the numeric equivalents of these dBm and dBW values without using a calculator.

Answers: 1: 20 mW. 2: 0.05 mW. 3: +14 dBW, 25 W. 4: 0.008 mW. 5: 500 W. 6: 40 mW. 7: 0.08 mW. 8: 50 kW.

C.6 ADDITION OF dBs AND DERIVED UNITS

Suppose we have a combiner, a device that combines signals from two or more sources. This combiner has two signal inputs: +3 dBm and +6 dBm. Our combiner is an ideal combiner in that it displays no insertion loss. In other words, there is no deleterious effect on the combining action, it is “lossless.” What we want to find out is the output of the combiner in dBm. It is not +9 dBm. The problem is shown diagrammatically as:

Some texts provide a nomogram to solve such a problem. We believe the following method is more accurate and, with the advent of affordable scientific calculators, easier. It is simple: Convert the input values to their respective numeric values in mW; add and convert the sum to its equivalent value in dBm.

The +3 and +6 dBm values are so familiar that we convert them by inspection, namely 2 and 4 mW. The sum is 6 mW. Now we take 10 log 6 to convert back to dBm again and the answer is +7.78 dBm. Remembering that there is an error when we work “3s” (3, 6, 9, 1, 4 and 7 values), we recalculated using a scientific calculator throughout. The answer was +7.76 dBm showing a 0.02-dB error.

On occasion, we will have to combine a large number of input/ outputs where each is of the same level. This is commonly done with frequency division multiplex equipment or with multitone telegraphy or data.

Suppose we have an FDM group (12 voice channel inputs), where each input was −16 dBm. What is the composite output? This is stated as:

Composite powerdBm c −16 dBm + 10 log 12

c−16 dBm + 10.79 dB

c−5.21 dBm.

The problem of adding two or more inputs in a combiner is pretty straightforward if we keep in the power domain. If we delve into the voltage or current domain with equivalent dB values, such as dBmV (which we cover in Section 15.3.2), we recommend returning to the power domain if at all possible. If we do not, we can open Pandora’s box, because of the phase relationship(s) of the inputs. In the next section we will carry out some interesting exercises in power addition.

622 LEARNING DECIBELS AND THEIR APPLICATIONS

Figure C.1 Line weightings for telephone (voice) channel noise.

The human ear is a filter, as is the telephone receiver (earpiece). The two are in tandem, as we would expect. For the telephone listener, noise is an annoyance. Interestingly we find that noise annoys a listener more near the reference frequencies of a voice channel than at other frequencies. When using the North American 500-type telephone set with average listeners, a simple 0-dBm tone at 1000 Hz causes a certain level of annoyance. To cause the same level of annoyance, a 300-Hz tone would have to be at a level of about +17 dBm; a 400-Hz tone at about +11.5 dBm; a 600-Hz tone at about +4 dBm, and a 3000-Hz tone also at about +4 dBm for equal annoyance levels for a population of average listeners.

The question arose of why should the transmission engineer be penalized in design of a system for noise of equal level across the voice channel? We therefore have “shaped” the voice channel as a function of frequency and “annoyance.” This shaping is called a weighting curve.

For the voice channel we will be dealing with two types of weighting: (1) C-message, used in North America, and (2) psophometric weighting as recommended by CCITT. Figure C.1 shows these weighting curves.

Weighting networks have been developed to simulate the corresponding response of C-message and psophometric weighting. Now we want to distinguish between flat response and weighted response. Of course, the curves in Figure C.1 show weighted response. Flat response, regarding a voice channel, has a low-pass response down 3 dB at 3 kHz and rolls off at 12 dB per octave. An octave means twice the frequency, so that it would be down 15 dB at 6 kHz and 27 dB at 12 kHz, and so on.

The term flat means equal response across a band of frequencies. Suppose a flat network has a loss of 3 dB. We insert a broad spectrum uniform signal at the input to the network. In the laboratory we generally use “white noise.” White noise is a signal that contains components of all frequencies inside a certain passband. We now measure the output of our network at discrete frequencies and at whatever frequency we measure