Re: 3 dB Bandwidth
Kevin,
In what concerns your questions about the relation between the –20dB/dec and –6 dB/oct, here goes:
Lets imagine a first order low pass filter, which has a pole at the frequency fp. For f<fp the gain of the filter is aprox constant and for f>fp it starts falling at a rate of 20 dB/dec (it decreases 20 dB for each increase of 10 times in frequency) or equivalently a rate of 6dB/octave (the gain decreases 6 dB for each increase of 2 times in frequency). Your question is the justification for this fact.
The gain of a low pass filter is given by
A(f)=A0/(1+jf/fp)
Where A0 is the low frequency gain, f is the frequency, fp is the pole frequency of the filter and j is the imaginary unit (j = sqrt(-1)).
From the equation shown above you can relate the magnitude and phase of the input and output sinewaves. To answer your question, we are interested on the magnitude response; the magnitude can be written has (remember your calculus classes ?):
|A(f)|=A0/sqrt(1+(f/fp)^2)
For f<< fp |A(f)| can be approximated by A0, which is why this is called the “low frequency gain”. For f>>fp |A(f)| can be approximated by
|A(f)|=aprox=A0/(f/fp)
Now you can use this simpler equation to see what happens to the magnitude response, when f>fp. When you compare the magnitude of the filter’s voltage gain for f1 and for f2=10 f1, you have:
(|A(f2)|/ |A(f1)|)dB =20 log10[(A0/(f2/fp))/(A0/(f1/fp))]=
=20 log10[(A0/(10f1/fp))/(A0/(f1/fp))]=
=20 log10[1/10]= -20 dB
Thus, when the frequency increases 10 times the gain of a first order low pass filter decreases 20 dB.
If you make the same calculations for f1 and f2=2 f1 (f2 is an octave above f1):
(|A(f2)|/ |A(f1)|)dB =20 log10[(A0/(f2/fp))/(A0/(f1/fp))]=
=20 log10[(A0/(2f1/fp))/(A0/(f1/fp))]=
=20 log10[1/2]=-6 dB
Therefore, when the frequency increases 2 times (an octave), the gain of the filter decreases 6 dB.
Finally, Kevin, I tell you again, sin45 has nothing to do with dBs …