Decibels Debugged
July 1964 Radio-Electronics

July 1964 Radio-Electronics [Table of Contents] Wax nostalgic about and learn from the history of early electronics. See articles from Radio-Electronics, published 1930-1988. All copyrights hereby acknowledged.

Articles like "Decibels Debugged" from the July 1964 issue of Radio-Electronics magazine will always be useful, which is why similar articles appear regularly in electronics magazines over all time. When I was first introduced to logarithms in high school, my eyes rolled back in my head and I was completely lost. The same thing happened with factoring polynomials. Maybe it was because there was no apparent direct link to my everyday life. Electrical concepts were easily grasped, but the math behind it - other than various forms of Ohm's law - had me running for cover. I barely passed 9th grade due to poor grades in math and science, and would still be waiting to graduate high school (half a century later) if not for being able to spend three years in high school in an electrical vocational program. Rather than taking math and science courses at my regular high school, I spent half the day at the Lincoln Technical Vocational Center learning about electrical wiring, motors, control, lighting, etc. That counted toward credits for graduation.  At some point whilst serving in the USAF as a radar technician, I experienced what I call an epiphany, after which I developed a craving for mathematics and the foundational scientific principles it unlocked. I loved and excelled at geometry, calculus, differential equations, etc., because it made the other engineering courses easier to tackle. Fellow students who did not have a firm grasp of mathematics struggled with circuits, physics, electromagnetics, communications, and other math-intensive courses because they had to deal with not being able to work problems. By the time I was introduced to decibels, the concept made total sense since I understood the relationship between logarithmic and linear operation and addition / subtraction versus multiplication / division. The same thing occurred with using a Smith chart, once the concept was full grasped regarding what was going on with the graph.

Decibels Debugged

By H. R. Holtz

Converting a decibel figure to a power or voltage gain (or loss) is a simple mathematical proposition. But it assumes that you have a slide rule or table of logarithms handy. That may not be true, particularly in the field. If you need only normal accuracy, as is usually the case, there is a simple method that gives pretty good accuracy, in most cases, and may be performed mentally or, at worst, with a hasty scrawl on any convenient surface.

To master the system you need to know only that any 2:1 power ratio is equal to 3 db or a 2:1 voltage ratio to 6 db, and that 100 = 1,101 = 10,102 = 100, etc.

Suppose you have an amplifier that puts out a 100-watt output signal for a 1-watt input signal. You say that the output is "up 20 db." This means, simply, that the gain of the amplifier is 100:1. Let us first convert the power ratio, 100:1, to decibels:

db = 10 log₁₀ P₁/P₂

If P₁ represents the output and P₂ the input,

db = 10 log₁₀ 100/1

"Log₁₀ 100" simply means the logarithm, to the base 10, of the number 100 or, to put it another way, the power of 10 that equals 100. This is 2, since 10² is equal to 100. Therefore

db = 10 x 2 = 20 db

Observe that 100 contains two 0's, 1,000 three 0's, etc. The number of 0's is the logarithm of 10. Multiply the number of 0's by 10 and you have the db figure! 1,000:1 is equivalent to 30 db, 10,000:1 to 40 db, 1,000,000:1 to 60 db. (See Table 1.)

For voltage or current ratios, the formula is slightly different:

db = 20 log₁₀ (E₁ (or I₁) )/( (E₂ (or I₂) )

Therefore, for voltage or current, multiply the number of 0's by 20. Then, 1,000:1 equals 60 db, 1,000,000:1 120 db, etc. This holds only if the impedances across which E₁ and E₂ (or I₁ and I₂) are being measured are the same.

To go from decibels to ratio, simply reverse the process: Given a power gain or loss of 30 db, divide by 10 and get 3. This means a 1 followed by three 0's, or 1,000. A voltage gain of 60 db is converted by dividing 60 by 20 and getting 3, again resulting in 1,000.

Until now, our db figures have been nice round numbers, easily divisible by 10 or 20. In practice, the numbers we have to work with are not always so cooperative. Suppose we must transpose a power gain of 27 db to a ratio. Dividing by 10 yields 2.7, which is 102.7, or 10 followed by 2.7 zeros!

The number 102 represents 100; 103 represents 1,000. Therefore, 102.7 represents a number between 100 and 1,000. We are dealing with an exponential quality, like the charging curve of a capacitor. Therefore, we cannot take seven-tenths of the difference. Instead, we shall resort to a trick used in grade-school arithmetic: we shall borrow.

Let us raise our 27-db figure to 30 db by borrowing 3 db. Now we have an easy figure to work With, since 30 can be divided evenly by 10. As shown in an earlier example, 30 db represents a power ratio of 1,000: 1. Remember, we borrowed 3 db to make the conversion easy. Now we must pay it back. It was pointed out that 3 db is equal to a power ratio of 2:1. Raising 27 db to 30 db is, therefore, doubling the ratio. If 30 db is 1,000:1, then it follows that 27 db must be 500:1.

Suppose the db figure given is 23 db. To convert this to a figure that is amenable to the method, let us lend 3 db. The resultant 20 db is half the ratio represented by 23 db. Since 20 db is 102, or 100, and is one-half 23 db, 23 db must represent 200.

The object, then, is to use decibel figures evenly divisible by 10 for power ratios, or by 20 for voltage and current ratios. In lending or borrowing db to convert to such figures, use increments of 3 or 6 db (for power or voltage and current, respectively) because these represent simple multiplications or divisions by 2 in ratio.

To illustrate the method for voltage, let us convert a 52-db gain to voltage: 46 db is one-half 52 db, 40 db is one-half 46 db. Thus 40 db is one-fourth 52 db.

Forty db (voltage) is 10² (20 db is 10¹ or just 10; 40 db - double the number of db - is 10², or 100). Doubling twice or multiplying by 4 yields 400 (40 db = 100, 46 db = 200, 52 db = 400).

Suppose, now, that we run across a really recalcitrant number: 25 db power gain. Since this number does not lend itself readily to conversion, we have to interpolate. First, we determine that 23 db represents the ratio 200. You may have observed by now that adding decibels is equivalent to multiplying the quantities they represent (20 db = 100, 3 db = 2; 20 db plus 3 db = 100 x 2 = 200). Although we are dealing with an exponential, not a linear function, for practical purposes we can treat a small remainder (2 db in this case) as though it were linear. That is, we shall say that the final ratio is 2/3 the difference between 23 db (200) and 26 db (400). This gives us a final answer of about 332. The exact answer is slightly less than 317. Our accuracy is about 5%, more than adequate.

Reversing the process, consider the ratio 1,700. We split this into 1,000 and 700. The first, 1,000, we know to be 30 db, 2,000 would be 33 db. We know, then, that our figure is between 30 and 33 db. Interpolating the difference, again assuming linearity, we calculate seven-tenths of 3 db (3 db represents here the difference between 30 and 33 db, or 1,000 and 2,000): 0.7 x 3 = 2.1 db. Rounding off and adding to our earlier figure of 30 db, we have a final answer of 32 db. Since the exact answer is 32.3 db, we have come very close, especially since the commonly accepted ratio of 2: 1 for 3 db is approximate (it is exactly 1.995:1).

Expanding and combining Tables I, II and III and rounding off the numbers, we can easily evolve Table IV. Again, this table does not reflect the exact values of all the quantities shown, but it can be easily made up in a few minutes, if there are a number of conversions to be made under improvised conditions, and is accurate enough to be useful. Generating such a table is also good practice in understanding decibels. It was created entirely by combining the earlier figures and continuing the process of halving and dividing the ratios, subtracting and adding decibels in 3-db units. Every ratio shown is either half or twice the value of another entry.

To verify the accuracy of Table IV:

15 db = 9 db + 6 db = 8 x 4 (from the table) = 32

Verify this by finding the exact ratio for 15 db.

If further proof is desired, find the ratios for 3 db and 12 db and verify that 12 db + 3 db = 15 db = 16 x 2 = 32 Multiplying the ratios of any combination of decibels that adds up to 15 db will result in a ratio of 32, within the accuracy of the table. This is of course true for any other figures. If you wished to find the ratio of 45 db, not given in the table, 45 db = 35 db + 10 db = 3,125 x 10 = 31,250.

The procedure may be reversed. Assume you wish to find a db figure for the ratio 40,000, not given in the table. Factoring 40,000 into convenient numbers,

40,000 = 10,000 x 4 = (from the table) 40 db + 6 db = 46 db.