First, we will use the
∣T(jω)∣=1+(ωoω)2∣K∣ low pass Magnitude response equation.
AVdB=20∗log10(AV)
60=20∗log10(AV)
3=log10(AV)→AV=103=1000
Next, knowing that fof=ωoω, and that fo is given we can plug that in formula:
AV(f)=1+(1000f)21000= 995.95
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