I need help with this assignment on gain structure. I have attached it for better viewing. Thx
0 dB References: Example 1
How about +4.5 dBV?
In this case, we simply use the voltage-decibel formula, and plug in our known variables:
dB = 20 log V/Vref => the dB is +4.5, the reference voltage (Vref) is 1V, and we are solving for V, that is, the voltage that is 4.5 dB greater than 1V.
+4.5 = 20 log V/1
4.5/20 = log V
100.225 = V
V = 1.68
Hence, 1.68 volts is 4.5 dB greater than 1 V. Does this make sense based on what we know about voltage levels and changes? It should.
*Here are important 0 dB reference levels to learn and remember:
0 dBu = 0.775 V
0 dBµ (pronounced "dB mi-ew") = 0.000001 V, or one microvolt
0 dBW = 1W (one watt)
0 dBm = 0.001W, or one milliwatt
0 dB SPL = 0.0002 dynes/cm2 or 0.00002N/m2 (newtons per meter squared) or Pa (pascals)
0 VU (0 on a VU meter) is calibrated to +4 dBu; what would that be as a voltage?
+4 = 20 log V/0.775
4/20 = log V/0.775
100.2 = V/0.775
0.775 x 100.2 = V
V= 1.228
0 dB References: Example 2
How about 60 dB SPL?
Acoustic sound pressure, like voltage, is a pressure and therefore we can use the same formula as voltage to calculate changes in sound pressure:
dB SPL = 20 log (P/Pref)
60 = 20 log P/0.0002
3 = log P/0.0002
103 = P/0.0002
P = 0.0002 x 103 = 0.2 dynes/cm2