. 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 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. -6 dBm +19 dB Exercise 1b. ? dBW -2 dB -4 dB +35 dB Exercise 1c. ? dBm 40 mW -28 dB +12 dB Exercise 1d. -19 dBm -6 dB +32 dB (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 adBm=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, LEARNING DECIBELS AND THEIR APPLICATIONS 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/Lossab = 10 log P2=10 log 12R1/12R2 = 20 log I1/I2 = 10 log R1/R2. If we let R = R2, then the term 10 log R1/R2 = 0. (Hint: The log of 1 = 0.) Remember from Ohm's law that E = IR, and from the power law Pw = EI. Thus Pw = I R = 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: Gain/Lossda : 20 log E1/E2 