Thanks for the response Audioguru... What does it mean that the gain is at -3dB? Can you articulate?
Decibels are another means of stating the ratio of the output to input. If the output is greater than the input then the term "gain" is used. If the output is smaller than the input then the term "attenuation" is used. For example, "The attenuation of the filter at 1 MHz is 100 dB".
The difference is the ratio is expressed in exponential terms and it is also weighted. The difference in the weighting between comparing voltages and power (i.e Watts) is such that regardless of whether you are referring to voltage or power the number of decibels will be the same. An example will help.
To convert between decibels and the ratio of voltages use
1) dB = 20 X log (Vout/Vin)
To convert between decibels and a ratio of power (i.e. Watts) use
2) dB = 10 X log (Pout/Pin)
If you have trouble with the term "log" read it as "the power of 10 that equals". For instance if Vout = Vin then Vout/Vin = 1. So the above would read 20 X the power of 10 that equals 1? Notice we use equation 1) since we are dealing with voltages not power levels. Well, since the power of 10 that equals 1 is 0, this becomes 20 X 0 = 0. So if the output voltage equals the input voltage this is 0 dB.
If Vout = 10 X Vin then Vout/Vin = 10. So the above would read 20 X the power of 10 that equals 10. Well, since the power of 10 that equals 10 is 1 this becomes 20 X 1 = 20 dB's.
If Vout = 100 X Vin then Vout/Vin = 100. So the above would read 20 X the power of 10 that equals 100. Well, since the power of 10 that equals 100 is 2 this becomes 20 X 2 = 40 dB's.
Note that if the output signal is smaller than the input signal the equation will result in a negative value. For example:
Vout = Vin/10 then Vout/Vin = 1/10th. So the above would read 20 X the power of 10 that equals 1/10th. Well since the power of 10 that equals 1/10th is -1 this becomes 20 X (-1) = -20 dB's.
In such cases you will often hear it stated as some number of dB's of attenuation. At other times it is stated as negative dB's. Still others will leave this out altogether if the speaker assumes the listener is aware it must be attenuation.
Every +20 dB increase is a ten-fold increase in the ratio of output voltage to input voltage (i.e. 20 dB means Vout = 10 X Vin, 40 dB means Vout = 100 X Vin, 60 dB = 1000 X Vin. Likewise every -20 dB's is a one-tenth decrease in the ratio of output voltage to input voltage (i.e. -20 dB means Vout = Vin/10, -40 dB means Vout = Vin/100, -60 dB means Vout = Vin/1000).
Another interesting and often used decibel increment is 6 dB. If Vout = 2 X Vin then the number of decibels is 20 X the power of 10 that equals 2. Since the power of 10 that equals 2 is 0.3 this becomes 20 X 0.3 = 6 dB. So every 6 dB increase corresponds to a further doubling of the ratio of output to input voltage.
From the above you should be able to recognize that +18 dB's is an 8-fold ratio between Vout and Vin since every 6 dB increase is a doubling of the voltage ratio (i.e 6 dB is X 2, 12 dB is X 4, and 18 dB is X 8 .
Likewise +26 dB means Vout is 20 times Vin. (Hint 26 is 20 + 6).
One final thing to note is 20 dB is a 10-fold increase in voltage. But it is a 100 fold increase in power since this will result in 10 times the current and power is Voltage X Current. Using equation 2) 10 dB's is a 10-fold increase in power. 20 dB's is a 100-fold increase in power. Likewise -10 dB's means Pout = Pin/10 and -20 dB's means Pout = Pin/100. Power level is actually the more common concern and hence what was used to define the units of decibels.
So when considering power the output to input power ratio increases 10-fold for every 10 dB increase and it doubles for every 3 dB increase. The +3 dB and -3 dB points are of particular interest as 3 dB's is a doubling in the power level while -3 dB's is a reduction in the output to input power by 1/2 (i.e. the half power level).
It is the -3 dB points on a filter which are used to define the bandwidth of the filter. If the -3 dB points on either side of a band-pass filter occur at 100 kHz and 105 kHz it is said to have a bandwidth of 5 kHz.
Why use decibels? Well as it turns out filters tend to have a linear drop-off in decibels when frequency is plotted logarithmically. In other words, if the x-axis is frequency with each tick defining a 10-fold increase in frequency then the gain or attenuation will often be a straight line. This is in fact what a Bode plot is for. On a Bode plot the ticks of the x-axis will be say 10Hz, 100Hz, 1KHz, 10KHz, etc resulting in a gain plot which is a line with a slope of say -20 dB's per decade of frequency. A slope of minus 20 dB's/decade of course means the voltage gain drops to 1/10 of its value every 10-fold increase (i.e. each x-axis tick) in frequency.