While modern test instruments can compute decibels for us, we should know what a decibel is and how to manually calculate gain or loss.
We all have a direct, personal interest in the decibel (dB) since our ears respond to all sound in a logarithmic fashion; thus our ears' response can be described using decibels, since they also are logarithmically based units (using a base of 10). But, this isn't the only reason to be knowledgeable about decibels.
The decibel is also used in electrical measurements throughout our industry in innumerable ways. For example, the ability of an isolation transformer or a power-line filter to reduce (attenuate) electrical noise over some range of frequencies is but one example of this, since the performance is generally described using decibels. Another example is how much of a signal is lost over a transport path such as a coaxial cable or similar metallic path. Also, the gain of an amplifier is generally expressed in decibels.
Even though we now enjoy having dB computed for us by sophisticated test instruments such as solid-state oscilloscopes, we do need to know what the dB is and how to work manually with the dB. This will prevent us from being snowed by what our instrument displays when we push the button.
The dB is typically expressed in relation to the electrical unit it's to be used with. For example, dBV and dBmV are used for decibels expressed in terms of voltage or millivolts; dBA and dBmA are used for decibels expressed in terms of amperes and milliamperes; and dBW is used for decibels expressed in terms off watts. Oddly enough, dB expressed in terms of mW is simply abbreviated as dBm, and the "W" is just left off. (It's very important to express dB in this manner if confusion is to be avoided.)
Prefixing dB with a minus sign (-) means a loss, and either no sign or a positive one (+) means a gain.
A lot of signal processes are nonlinear and actually are best described as being logarithmic in nature. Hence, if you try to use simple ratios to describe them, the results get either unrealistically spread-out at one end and are bunched-up at the other end of a chart or graph. This makes the information very hard to use.
An example of this problem can be cited using the ear again. When a linear resistance potentiometer is used as an audio volume control, all of the control is confined to the last few degrees of shaft rotation and the majority of the rotation doesn't appear to have much effect at all. However, when a logarithmically tapered control is used, the adjustment of volume is nearly uniform throughout the full rotational range of the potentiometer.
Another example is graphing a signal on linear marked graph paper. Because the signal is logarithmic in nature, its graph here either will be confusing due to the curvature of the plot or unusable due to the contraction and expansion of the data plot (similar to that encountered with the volume control and sound). Using logarithmic marked graph paper allows the "curved" plot to be generated using straight or almost straight lines. This makes for easier interpretation and general usage. An example of this is the charging or discharging of a capacitor over time.
A description of the general equation for determining the dB of gain (+dB) or loss (-dB) for any set of two voltages on a path of equal impedance is shown below.
dB =20 log ([E.sub.1]/[E.sub.2]) (equation 1)
Solving this form of equation is not difficult with a calculator that has a "log" key on it. All you have to do is first compute the ratio of the input to output voltage ([E.sub.1]/[E.sub.2]), press the "log" key, and then multiply the whole thing by 20. This gives the result in terms of -dB or +dB depending upon whether signal was lost or added to in the circuit. You can substitute current (I) for voltage using this equation if you desire.
Power calculations are a different matter. They're calculated in much the same manner as above, except the equation uses a factor of 10 as opposed to 20. Shown below is the basic power equation for computing dB in a circuit of the same impedance.
dB = 10 log ([P.sub.1]/[P.sub.2]) (equation 2)
The values of dB (calculated using equations 1 and 2) based on varying ratios of power and voltage/current are shown in the accompanying table. You might want to commit a few of the really important relationships in dB to memory since you may want to quickly estimate something. For example, the following common values should be remembered.
* Voltage or current:
Doubling or halving of voltage (or current) is a [+ or -]6 dB change.
A numerical ratio of 10 is 20 dB, 100 is 40 dB, 1000 is 60 dB, and so on.
Doubling or halving of power is a [+ or -]3 dB change.
A numerical ratio of 2 is 3 dB, 4 is 6 dB, 10 is 10 dB, 100 is 20 dB, and so on.
The nice thing about dB notation is that gains and losses in any given circuit are simply added and subtracted arithmetically to find the final value of gain or loss. Then, with a little algebra or with the anti-log key (10x) on your calculator, you determine the voltage, current, or power ratio from the resulting answer.
Let's see how the gains and losses of a given signal transport (losses) and amplifier (gain) system work together to produce a given output from a specified input.
Suppose you know the [+ or -]dB at any point in a system. You then can compute the voltage or current ratio. This is a simple algebra problem that's easy to do with a calculator. Let's say that the above "system" has 12 dB of gain. How do we compute the output voltage (or current) based upon knowing the input and the dB of gain? The answer lies in the following equation.
([E.sub.1]/[E.sub.2]) = [10.sup.(dB/20)] (equation 3)
This is the basic equation for computing the voltage ratio in a circuit of the same impedance when the [+ or -]dB is known. To use it, we first divide the known dB value by 20. The result is the power we wish to raise 10 to. Using our trusty calculator, we push the Xy key. The resulting value is the ratio between the input voltages.
Getting back to our example, we have 12 dB of gain after the signal is "processed and transported" from the source to the load. Using equation 3 and the procedure previously described, we find that this 12dB works out to a ratio of 3.98:1. This value times 10 = 39.8V output.
The ratio for power is computed in the same manner as in equation 3 except that 10 is used in place of 20 in the exponent's divisor. This is shown in the following basic equation for computing the power ratio in a circuit of the same impedance when the [+ or -]dB is known.
([P.sub.1]/[P.sub.2]) = [10.sup.(dB/10)] (equation 4)
Warren H. Lewis is President of Lewis Consulting Services, Inc., San Juan Capistrano, Calif. and Honorary Chairman of EC&M's Harmonics and Power Quality Steering Committee.