The maximum voltage gain of a passive low pass filter is
a) 1 dB
b) 0 dB
C) depends on the value of the circuit components
D) -3 dB
D) -3 dB
Calculate the quality factor of a Band Pass Filter, if the centre frequency is 10.7 MHz and the band width is 200 kHz.
a) 1
b) 10
c) 53.5
d) 107
c) 53.5
The quality factor, Q, of a band pass filter can be calculated using the formula:
Q = f_c / BW
Where f_c is the center frequency and BW is the bandwidth.
Substituting the given values, we get:
Q = 10.7 MHz / 200 kHz
Q = 53.5
Therefore, the quality factor of the band pass filter is 53.5.
in a Butterworth HPF (6th Order), What is the roll-off rate at frequencies outside the passband?
a) cannot be determined
b) -60 dB/sec
c) -120 dB/sec
d) -40 dB/sec
c) -120 dB/sec
The roll-off rate of a Butterworth filter is based on the number of filter stages or poles. For a 6th order Butterworth high pass filter, the roll-off rate at frequencies outside the passband is 6 x 20 = 120 dB per decade or -120 dB/sec.
Therefore, the correct option is c) -120 dB/sec.
use the Butterworth response table and determine the closed-loop gain in dB to achieve a 3-pole Butterworth response.
a) 4 dB
b) 1.2 dB
c) 6 dB
D) 2 dB
The Butterworth response table gives the 3-dB frequency, or the frequency at which the filter attenuates the signal to half its original power. To determine the closed-loop gain in dB to achieve a 3-pole Butterworth response, we need to know the cutoff frequency of the filter and its passband gain.
Since the question does not provide any values for these parameters, we cannot determine the closed-loop gain in dB with the available information.
Therefore, the answer is a) cannot be determined.
Given the damping factor is 1.848 and
r2 = 6.8 kΩ. Determine the value of R1 to obtain a 4-pole Butterworth response.
For a 4-pole Butterworth response, the transfer function of the filter can be expressed as:
H(s) = K / [(s^2 + 0.765s + 1) (s^2 + 1.848s + 1)]
Where K is the gain constant and s is the Laplace variable.
Comparing this with the standard second-order low-pass filter transfer function, we can see that the damping factor for the second-order stage with roots at -0.3825 ± 0.9238j must be equal to 0.765.
We are given that the damping factor for the second-order stage with roots at -0.924 ± 0.383j is 1.848. This means that this stage does not achieve Butterworth response.
To correct this, we can add a resistor R1 to the circuit in parallel with one of the capacitors. The value of R1 can be calculated using the formula:
R1 = R2 / √(1 - ζ^2)
Where R2 is the resistor with which the capacitor is already in parallel and ζ is the damping factor of the second-order stage that needs to be corrected.
Substituting the given values, we get:
ζ = 0.383
R2 = 6.8 kΩ
R1 = R2 / √(1 - ζ^2) = 6.8 kΩ / √(1 - 0.383^2) = 10.85 kΩ
Therefore, the value of R1 to obtain a 4-pole Butterworth response is 10.85 kΩ.
Given the damping factor is 1.848 and
r2 = 6.8 kΩ. Determine the value of R1 to obtain a 4-pole Butterworth response.
a) 1 kΩ
b) 44.7 kΩ
c) 12.6 kΩ
d) 3.7 kΩ