Respuesta :
Answer:
[tex] G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1[/tex]
If we square both sides we got:
[tex] G^2 (1+\frac{f}{f_c})^{2n}= 1[/tex]
We divide both sides by [tex] G^2[/tex] and we got:
[tex] (1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}[/tex]
Now we can apply log on both sides and we got:
[tex] 2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})[/tex]
And solving for n we got:
[tex]n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}[/tex]
And replacing we got:
[tex]n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}[/tex]
[tex] n = \frac{4.60517}{3.8918}=1.18[/tex]
And since n needs to be an integer the correct answer would be n=2 for the filter order.
Explanation:
For this case we can use the formula for the Butterworth filter gain given by:
[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]
Where:
G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value
[tex] f_c = 10 Hz[/tex] represent the corner frequency
[tex] f= 60 Hz[/tex] represent the original frequency
n represent the filter order and that's the variable that we need to find
[tex] G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1[/tex]
If we square both sides we got:
[tex] G^2 (1+\frac{f}{f_c})^{2n}= 1[/tex]
We divide both sides by [tex] G^2[/tex] and we got:
[tex] (1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}[/tex]
Now we can apply log on both sides and we got:
[tex] 2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})[/tex]
And solving for n we got:
[tex]n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}[/tex]
And replacing we got:
[tex]n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}[/tex]
[tex] n = \frac{4.60517}{3.8918}=1.18[/tex]
And since n needs to be an integer the correct answer would be n=2 for the filter order.
