Digital Signal Processing Questions and Answers – Design of Low Pass Butterworth Filters – 2

This set of Digital Signal Processing test focuses on “Design of Low Pass Butterworth Filters-2”.

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

Answer: d
Explanation: We know that the equation for the cutoff frequency of a Butterworth filter is given as
ΩC = \(\frac{\Omega_P}{(10^{-K_P/10}-1)^{1/2N}}\)
We know that KP=-1 dB, ΩP=4 rad/sec and N=5
Upon substituting the values in the above equation, we get
ΩC=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?
a) \(\frac{1}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\)
b) \(\frac{1}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+1}\)
c) \(\frac{2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\)
d) None of the mentioned
View Answer

Answer: c
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 ΩC=4.5787 rad/sec
We know that for a 5th order normalized low pass Butterworth filter, system equation is given as
H5(s)=\(\frac{1}{(s+1)(s^2+0.618s+1)(s^2+1.618s+1)}\)
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
Ha(s)=\({2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\)

3. If H(s)=\(\frac{1}{s^2+s+1}\) 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?
a) \(\frac{100}{s^2+10s+100}\)
b) \(\frac{s^2}{s^2+s+1}\)
c) \(\frac{s^2}{s^2+10s+100}\)
d) None of the mentioned
View Answer

Answer: a
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
=\(\frac{100}{s^2+10s+100}\).
advertisement
advertisement

4. If H(s)=\(\frac{1}{s^2+s+1}\) 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?
a) \(\frac{100}{s^2+10s+100}\)
b) \(\frac{s^2}{s^2+s+1}\)
c) \(\frac{s^2}{s^2+10s+100}\)
d) None of the mentioned
View Answer

Answer: b
Explanation: The low pass-to-high pass transformation is
s→Ωu/s
Hence the required high pass filter is
Ha(s)= H(s)|s→1/s
=\(\frac{s^2}{s^2+s+1}\)

5. If H(s)=\(\frac{1}{s^2+s+1}\) 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?
a) \(\frac{100}{s^2+10s+100}\)
b) \(\frac{s^2}{s^2+s+1}\)
c) \(\frac{s^2}{s^2+10s+100}\)
d) None of the mentioned
View Answer

Answer: c
Explanation: The low pass-to-high pass transformation is
s→Ωu/s
Hence the required low pass filter is
Ha(s)=H(s)|s→10/s
=\(\frac{s^2}{s^2+10s+100}\)
Note: Join free Sanfoundry classes at Telegram or Youtube

6. If H(s)=\(\frac{1}{s^2+s+1}\) 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?
a) \(\frac{s^2}{s^4+10s^3+20100s^2+10^5 s+1}\)
b) \(\frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+1}\)
c) \(\frac{s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}\)
d) \(\frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}\)
View Answer

Answer: d
Explanation: The low pass-to-band pass transformation is
\(s\rightarrow\frac{s^2+\Omega_u \Omega_l}{s(\Omega_u-\Omega_l)}\)
Thus the required band pass filter has a transform function as
Ha(s)=\(\frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}\).

7. If H(s)=\(\frac{1}{s^2+s+1}\) 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?
a) \(\frac{(s^2+100)^2}{s^4+2s^3+204s^2+200s+10^4}\)
b) \(\frac{(s^2+10)^2}{s^4+2s^3+204s^2+200s+10^4}\)
c) \(\frac{(s^2+10)^2}{s^4+2s^3+400s^2+200s+10^4}\)
d) None of the mentioned
View Answer

Answer: a
Explanation: The low pass-to- band stop transformation is
\(s\rightarrow\frac{s(Ω_u-Ω_l)}{s^2+Ω_u Ω_l}\)
Hence the required band stop filter is
Ha(s)=\(\frac{(s^2+100)^2}{s^4+2s^3+204s^2+200s+10^4}\)
advertisement

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

Answer: c
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=\(\frac{-Ω_1^2+Ω_u Ω_l}{Ω_1 (Ω_u-Ω_l)}\) and B=\(\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}\)
=> 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

Answer: b
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=\(\frac{-Ω_1^2+Ω_u Ω_l}{Ω_1 (Ω_u-Ω_l)}\) and B=\(\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}\)
=> 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
N=\(\frac{log⁡[(10^{-K_P/10}-1)(10^{-K_s/10}-1)]}{2 log⁡(\frac{1}{Ω_S})}\)=2.83
Rounding off to the next large integer, we get, N=3.
advertisement

10. Which of the following condition is true?
a) N ≤ \(\frac{log⁡(\frac{1}{k})}{log⁡(\frac{1}{d})}\)
b) N ≤ \(\frac{log⁡(k)}{log⁡(d)}\)
c) N ≤ \(\frac{log⁡(d)}{log⁡(k)}\)
d) N ≤ \(\frac{log⁡(\frac{1}{d})}{log⁡(\frac{1}{k})}\)
View Answer

Answer: d
Explanation: If ‘d’ is the discrimination factor and ‘K’ is the selectivity factor, then
N ≤ \(\frac{log⁡(\frac{1}{d})}{log⁡(\frac{1}{k})}\)

Sanfoundry Global Education & Learning Series – Digital Signal Processing.

To practice all areas of Digital Signal Processing for various tests, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.