Question 6.4: Decibels and Logarithmic Frequency Scales , The transfer fun...
Decibels and Logarithmic Frequency Scales
The transfer function magnitude of a certain filter is given by
|H(f)|=\frac{10}{\sqrt{1+(f/5000)^6}}
a. What is the value of the transfer function magnitude in decibels for very low frequencies?
b. At what frequency f_{\text{3dB}} is the transfer function magnitude 3 dB less than the value at very low frequencies?
c. At what frequency f_{\text{60dB}} is the transfer function magnitude 60 dB less than the value at very low frequencies?
d. How many decades are between f_{\text{3dB}} \text{ and }f_{\text{60dB}} ? How many octaves?
a. Very low frequencies are those approaching zero. For f = 0, we have | H(0) | = 10. Then, we have |H(0)|_{\text{dB}}=20 \log(10)=20 \text{ dB}.
b. Because –3 dB corresponds to 1/ \sqrt{2}, we have
|H(f_\text{3dB})|=\frac{10}{\sqrt{2}}=\frac{10}{\sqrt{1+(f_{\text{3dB}}/5000)^6}}
from which we find that f_{\text{3dB}} = 5000 \text{ Hz}.
c. Also, because –60 dB corresponds to 1/1000, we have
|H(f_{\text{60dB}})|=\frac{10}{1000}=\frac{10}{\sqrt{1+(f_{\text{60dB}}/5000)^6}}
from which we find that f_{\text{60dB}} = 50 \text{ kHz}.
d. Clearly, f_{\text{60dB}} = 50 \text{ kHz} is one decade higher than f_{\text{3dB}} = 5 \text{ Hz}. Using Equation 6.18, we find that the number of octaves between the two frequencies is
\text{ number of octaves } = \log_2 \left( \frac{f_2}{f_1} \right) = \frac{\log (f_2/f_1)}{\log (2)} (6.18)
\frac{\log(50/5)}{\log(2)} =\frac{1}{\log(2)}=3.32