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What is the cutoff frequency of the Butterworth filter with a pass band gain KP= 1 dB at ΩP= 4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS= 8 rad/sec?
Explanation: We know that the equation for the cutoff frequency of a Butterworth filter is given as
We know that KP= 1 dB, ΩP= 4 rad/sec and N=5
Upon substituting the values in the above equation, we get
Ω_C = 4.5787 rad/sec.
What is the system function of the Butterworth filter with specifications as pass band gain KP= 1 dB at ΩP= 4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS= 8 rad/sec?
Explanation: From the given question,
KP= 1 dB, ΩP= 4 rad/sec, KS= 20 dB and ΩS= 8 rad/sec
We find out order as N=5 and Ω_C = 4.5787 rad/sec
We know that for a 5th order normalized low pass Butterworth filter, system equation is given as
The specified low pass filter is obtained by applying low passtolow pass transformation on the normalized low pass filter.
That is, Ha(s)=H5(s)s→s/Ωc
= H5(s)s→s/4.5787
upon calculating, we get option c.
If H(s)= 1/(s^{2}+s+1) represent the transfer function of a low pass filter(not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a low pass filter with a pass band 10 rad/sec?
Explanation: The low passtolow pass transformation is
s→s/Ωu
Hence the required low pass filter is
Ha(s)= H(s)s→s/10
= 100/(s^{2}+10s+100).
If H(s)= 1/(s^{2}+s+1) represent the transfer function of a low pass filter(not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a high pass filter with a cutoff frequency of 1rad/sec?
Explanation: The low passtohigh pass transformation is
s→Ωu/s
Hence the required high pass filter is
If H(s)= 1/(s^{2}+s+1) represent the transfer function of a low pass filter(not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a high pass filter with a cutoff frequency of 10 rad/sec?
Explanation: The low passtohigh pass transformation is
s→Ωu/s
Hence the required low pass filter is
If H(s)= 1/(s^{2}+s+1) represent the transfer function of a low pass filter(not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a band pass filter with a pass band of 10 rad/sec and a center frequency of 100 rad/sec?
Explanation: The low passtoband pass transformation is
Thus the required band pass filter has a transform function as
If H(s)= 1/(s^{2}+s+1) represent the transfer function of a low pass filter(not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a stop band filter with a stop band of 2 rad/sec and a center frequency of 10 rad/sec?
Explanation: The low passto band stop transformation is
Hence the required band stop filter is
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?
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.
What is the order 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?
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.
The order N of the normalized low pass Butterworth filter is computed as follows
= 2.83
Rounding off to the next large integer, we get, N=3.
Explanation: If ‘d’ is the discrimination factor and ‘K’ is the selectivity factor, then
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