To determine the center frequency and bandwidth of the filter, we need to calculate the values of the components in the transfer function and then solve for the cutoff frequency and bandwidth.
The transfer function for a four-pole high-pass Butterworth filter can be expressed as:
H(s) = (s^2 + 1.414s + 1)(s^2 + 1.414s + 1)/(s^2 + 1.848s + 1)(s^2 + 1.848s + 1)
where s is the Laplace variable.
To calculate the cutoff frequency and bandwidth, we need to find the values of the capacitors and resistors in the transfer function. Using the given values, we can find:
C1 = CA1 * CB1 / (CA1 + CB1) = 2.06 pF
C2 = CA2 * CB2 / (CA2 + CB2) = 3.24 pF
R1 = RA1 * RB1 / (RA1 + RB1) = 13.64 kΩ
R2 = RA2 * RB2 / (RA2 + RB2) = 15.29 kΩ
Substituting these values into the transfer function, we get:
H(s) = (s^2 + 1.414s + 1)(s^2 + 1.414s + 1)/(s^2 + 1.848s + 1)(s^2 + 1.848s + 1)
= (s^4 + 2.828s^3 + 3.414s^2 + 2.828s + 1)/(s^4 + 3.696s^3 + 5.414s^2 + 3.696s + 1)
To find the cutoff frequency, we can set H(s) equal to 1/sqrt(2) and solve for s. This gives us:
s = 1.306 - j1.306 or s = -1.306 + j1.306
The cutoff frequency is the magnitude of s, which is approximately 1.846 rad/s.
To find the bandwidth, we can use the formula:
BW = f2 - f1 = (s2 - s1) / (2*pi)
where f1 and f2 are the frequencies at which the magnitude of H(s) is equal to 1/sqrt(2), and s1 and s2 are the corresponding complex frequencies.
Using the values we obtained earlier, we can find:
s1 = -1.306 + j1.306
s2 = 1.306 - j1.306
BW = (s2 - s1) / (2*pi) = 0.462 rad/s
Therefore, the center frequency of the filter is approximately 1.846 rad/s and the bandwidth is approximately 0.462 rad/s.
In Four-pole High-Pass Butterworth, Determine the center frequency and the Bandwidth of the filter.
RA1 = 24KΩ
RA2 = 27KΩ
RB1 = 24KΩ
RB2 = 27KΩ
CA1 = 3pF
CA2 = 4.7pF
CB1 = 3pF
CB2 = 4.7pF
3 answers
is it 1.846 for bandwidth?
No, 1.846 is the cutoff frequency of the filter, not the bandwidth. The bandwidth of the filter is approximately 0.462 rad/s.