This set of Digital Signal Processing test focuses on “Design of Low Pass Butterworth Filters”.

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?

a) 3.5787 rad/sec

b) 1.069 rad/sec

c) 6 rad/sec

d) 4.5787 rad/sec

View Answer

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.

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?

d) None of the mentioned

View Answer

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.

3. 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?

d) None of the mentioned

View Answer

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/(s

^{2}+10s+100).

4. 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?

d) None of the mentioned

View Answer

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

s→Ωu/s

Hence the required high pass filter is

5. 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?

d) None of the mentioned

View Answer

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

s→Ωu/s

Hence the required low pass filter is

6. 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?

View Answer

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

Thus the required band pass filter has a transform function as

7. 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?

d) None of the mentioned

View Answer

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

Hence the required band stop filter is

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?

a) 2 rad/sec

b) 2.25 Hz

c) 2.25 rad/sec

d) 2 Hz

View Answer

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.

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?

a) 2

b) 3

c) 4

d) 5

View Answer

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.

10. Which of the following condition is true?

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