Sites like this one, inform that in order to create a lowpass filter you can choose a small capacitance for the capacitor and then use the formula:
\[ R = \frac{1}{2πF_{c}C} \]
in order to figure out the value of R.
Choosing a low C for the capacitor makes sense since \( Z_{C} = \frac{1}{jωC} \) if C << 1 Z>infinity and acts as an open circuit, thus the output voltage is equal to the input. While ω>infinity Zc > 0 so the output becomes 0.
The equation \( R = \frac{1}{2πF_{c}C} \) comes from the fact that you assume that the magnitude of the transfer function of the low pass filter circuit is equal to 3db. or
\( H(s) = \frac{1}{\sqrt{2}} \) which leads to the above equation.
Setting the H(s) to that value which is approximately H(s) = 0.707, means that at that point \( V_{o} = 0.707 \cdot V_{in} \).
This doesn't make sense. Why is every article (even in books) use this cutoff point (cutoff frequency) for designing filters?
The filter should make the output 0 not 70.71% (0.707) of the original signal.
In the following example, I chose C = 4nF, Fc = 15kHz and then calculated \[ R = \frac{1}{2πF_{C}C}=\frac{1}{2π \cdot 15000 \cdot 4 \cdot 10^{9}} = 2.65kΩ \].
I thought this filter will allow all frequencies < Fc = 15kHz and reject >= Fc.
First I saw what happens for a low frequency of 10Hz:
The output is identical to the input, as it should be.
For 15kHz, the output magnitude is reduced by 70.71% which is correct, but this is not a filter, in my opinion, it should be zero from this point and forward:
For 20kHz, the output is still quite large:
And finally for 100kHz now the output approaches zero:
So why is everyone on the Internet using this way to calculate R? How is this a good filter?
I tried mathematically to set the transfer function H(s) = 0 instead of 1/sqrt(2) but you can't solve for f or R because the result is infinity.
What if I set H(s) = something very small = 0.0000001? Will this work?
And even if this works, what exactly do we mean by very small? How can we define this small number?
Thanks!
\[ R = \frac{1}{2πF_{c}C} \]
in order to figure out the value of R.
Choosing a low C for the capacitor makes sense since \( Z_{C} = \frac{1}{jωC} \) if C << 1 Z>infinity and acts as an open circuit, thus the output voltage is equal to the input. While ω>infinity Zc > 0 so the output becomes 0.
The equation \( R = \frac{1}{2πF_{c}C} \) comes from the fact that you assume that the magnitude of the transfer function of the low pass filter circuit is equal to 3db. or
\( H(s) = \frac{1}{\sqrt{2}} \) which leads to the above equation.
Setting the H(s) to that value which is approximately H(s) = 0.707, means that at that point \( V_{o} = 0.707 \cdot V_{in} \).
This doesn't make sense. Why is every article (even in books) use this cutoff point (cutoff frequency) for designing filters?
The filter should make the output 0 not 70.71% (0.707) of the original signal.
In the following example, I chose C = 4nF, Fc = 15kHz and then calculated \[ R = \frac{1}{2πF_{C}C}=\frac{1}{2π \cdot 15000 \cdot 4 \cdot 10^{9}} = 2.65kΩ \].
I thought this filter will allow all frequencies < Fc = 15kHz and reject >= Fc.
First I saw what happens for a low frequency of 10Hz:
The output is identical to the input, as it should be.
For 15kHz, the output magnitude is reduced by 70.71% which is correct, but this is not a filter, in my opinion, it should be zero from this point and forward:
For 20kHz, the output is still quite large:
And finally for 100kHz now the output approaches zero:
So why is everyone on the Internet using this way to calculate R? How is this a good filter?
I tried mathematically to set the transfer function H(s) = 0 instead of 1/sqrt(2) but you can't solve for f or R because the result is infinity.
What if I set H(s) = something very small = 0.0000001? Will this work?
And even if this works, what exactly do we mean by very small? How can we define this small number?
Thanks!
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