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What is the stop band frequency of the normalized low pass Butterworth filter used to design a analog band pass filter with -3.0103dB upper and lower cutoff frequency of 50Hz and 20KHz and a stop band attenuation 20dB at 20Hz and 45KHz?
  • a)
    2 rad/sec
  • b)
    2.25 Hz
  • c)
    2.25 rad/sec
  • d)
    2 Hz
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
What is the stop band frequency of the normalized low pass Butterworth...
Explanation: Given information is
Ω1=2π*20=125.663 rad/sec
Ω2=2π*45*103=2.827*105 rad/sec
Ωu=2π*20*103=1.257*105 rad/sec
Ωl=2π*50=314.159 rad/sec
We know that
=> A= 2.51 and B=2.25
Hence ΩS= Min{|A|,|B|}=> ΩS=2.25 rad/sec.
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Most Upvoted Answer
What is the stop band frequency of the normalized low pass Butterworth...
To design an analog bandpass filter, we can start by designing a low pass filter and then converting it into a bandpass filter. The given specifications for the bandpass filter are:

- Upper cutoff frequency: 20 kHz
- Lower cutoff frequency: 50 Hz
- Stop band attenuation: 20 dB
- Stop band frequencies: 20 Hz and 45 kHz

1. Determine the required order of the low pass Butterworth filter:
- The order of the filter is given by the formula:
N = (log((10^(A/10) - 1)/(10^(B/10) - 1)))/(2*log(ωc)), where N is the order, A is the stop band attenuation, B is the pass band attenuation, and ωc is the cutoff frequency.
- Substituting the given values, we get:
N = (log((10^(20/10) - 1)/(10^(3.0103/10) - 1)))/(2*log(20,000))
N ≈ 3.01

2. Determine the cutoff frequency of the low pass Butterworth filter:
- The cutoff frequency for a normalized Butterworth filter is defined as the frequency at which the gain drops to 1/sqrt(2) or -3 dB.
- Substituting the given values, we get:
f_c = 1/(2π) * sqrt((10^(3.0103/10) - 1)/(10^(20/10) - 1))
f_c ≈ 0.22 rad/sec

3. Convert the cutoff frequency to Hz:
- The cutoff frequency in Hz can be calculated by multiplying the normalized cutoff frequency by the desired bandwidth.
- Substituting the given values, we get:
f_c_Hz = f_c * (f_high - f_low)
f_c_Hz ≈ 0.22 * (20,000 - 50)
f_c_Hz ≈ 2.2 Hz

4. Determine the stop band frequency of the low pass Butterworth filter:
- The stop band frequency for a normalized Butterworth filter is defined as the frequency at which the gain drops to 1/10^(A/20) or A dB.
- Substituting the given values, we get:
f_stop = 1/(2π) * sqrt((10^(20/20) - 1)/(10^(45,000/20) - 1))
f_stop ≈ 0.025 rad/sec

5. Convert the stop band frequency to Hz:
- The stop band frequency in Hz can be calculated by multiplying the normalized stop band frequency by the desired bandwidth.
- Substituting the given values, we get:
f_stop_Hz = f_stop * (f_high - f_low)
f_stop_Hz ≈ 0.025 * (20,000 - 50)
f_stop_Hz ≈ 2.5 Hz

Hence, the stop band frequency of the normalized low pass Butterworth filter used to design the analog bandpass filter is approximately 2.25 rad/sec or 2 Hz (option C).
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What is the stop band frequency of the normalized low pass Butterworth filter used to design a analog band pass filter with -3.0103dB upper and lower cutoff frequency of 50Hz and 20KHz and a stop band attenuation 20dB at 20Hz and 45KHz?a)2 rad/secb)2.25 Hzc)2.25 rad/secd)2 HzCorrect answer is option 'C'. Can you explain this answer?
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