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Test: Butterworth Filters Design - 2


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Test: Butterworth Filters Design - 2 - Question 1

 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? 

Detailed Solution for Test: Butterworth Filters Design - 2 - Question 1

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.

Test: Butterworth Filters Design - 2 - Question 2

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?

Detailed Solution for Test: Butterworth Filters Design - 2 - Question 2

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.

Test: Butterworth Filters Design - 2 - Question 3

 If H(s)= 1/(s2+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?

Detailed Solution for Test: Butterworth Filters Design - 2 - Question 3

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).

Test: Butterworth Filters Design - 2 - Question 4

 If H(s)= 1/(s2+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?

Detailed Solution for Test: Butterworth Filters Design - 2 - Question 4

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

Test: Butterworth Filters Design - 2 - Question 5

 If H(s)= 1/(s2+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?

Detailed Solution for Test: Butterworth Filters Design - 2 - Question 5

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

Test: Butterworth Filters Design - 2 - Question 6

 If H(s)= 1/(s2+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?

Detailed Solution for Test: Butterworth Filters Design - 2 - Question 6

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

Thus the required band pass filter has a transform function as

Test: Butterworth Filters Design - 2 - Question 7

 If H(s)= 1/(s2+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?

Detailed Solution for Test: Butterworth Filters Design - 2 - Question 7

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

Hence the required band stop filter is

Test: Butterworth Filters Design - 2 - Question 8

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?

Detailed Solution for Test: Butterworth Filters Design - 2 - Question 8

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.

Test: Butterworth Filters Design - 2 - Question 9

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?

Detailed Solution for Test: Butterworth Filters Design - 2 - Question 9

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.

Test: Butterworth Filters Design - 2 - Question 10

 Which of the following condition is true?

Detailed Solution for Test: Butterworth Filters Design - 2 - Question 10

Explanation: If ‘d’ is the discrimination factor and ‘K’ is the selectivity factor, then

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