Test: Butterworth Filters Design - 2 - Electronics and Communication Engineering (ECE) MCQ

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.

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 pass-to-low 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.

Explanation: The low pass-to-low pass transformation is
s→s/Ωu
Hence the required low pass filter is
Ha(s)= H(s)|s→s/10
= 100/(s²+10s+100).

Explanation: The low pass-to-high pass transformation is
s→Ωu/s
Hence the required high pass filter is

Explanation: The low pass-to-high pass transformation is
s→Ωu/s
Hence the required low pass filter is

Explanation: The low pass-to-band pass transformation is

Thus the required band pass filter has a transform function as

Explanation: The low pass-to- band stop transformation is

Hence the required band stop filter is

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.

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