Digital Filters Design MCQ [Free PDF] – Objective Question Answer for Digital Filters Design Quiz

1. Which of the following is a frequency domain specification?

A. 0 ≥ 20 log|H(jΩ)|
B. 20 log|H(jΩ)| ≥ KP
C. 20 log|H(jΩ)| ≤ KS
D. All of the mentioned

Answer: D

We are required to design a low-pass Butterworth filter to meet the following frequency domain specifications.

KP ≤ 20 log|H(jΩ)| ≤ 0
and 20 log|H(jΩ)| ≤ KS.

 

2. What is the value of gain at the pass band frequency, i.e., what is the value of KP?

A. -10 \(log⁡ [1-(\frac{\Omega_P}{\Omega_C})^{2N}]\)

B. -10 \(log⁡ [1+(\frac{\Omega_P}{\Omega_C})^{2N}]\)

C. 10 \(log⁡ [1-(\frac{\Omega_P}{\Omega_C})^{2N}]\)

D. 10 \(log⁡ [1+(\frac{\Omega_P}{\Omega_C})^{2N}]\)

Answer: B

We know that the formula for gain is K = 20 log|H(jΩ)|

We know that

\(|H(j\OmegA.|=\frac{1}{\sqrt{(1+(\frac{\Omega}{\Omega_C})^{2N}}}\)

By applying 20log on both sides of above equation, we get

K = 20 \(log|H(j \OmegA.|=-20 [log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]]^{1/2}\)

= -10 \(log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]\)

We know that K= KP at Ω=ΩP

=> KP=-10 \(log⁡[1+(\frac{\Omega_P}{\Omega_C})^{2N}]\).

 

3. What is the value of gain at the stop band frequency, i.e., what is the value of KS?

A. -10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\)

B. -10 \(log⁡[1-(\frac{\Omega_S}{\Omega_C})^{2N}]\)

C. 10 \(log⁡[1-(\frac{\Omega_S}{\Omega_C})^{2N}]\)

D. 10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\)

Answer: A

We know that the formula for gain is

K = 20 log|H(jΩ)|

We know that

\(|H(j \OmegA.|=\frac{1}{\sqrt{(1+(\frac{\Omega}{\Omega_C})^{2N}}}\)

By applying 20log on both sides of above equation, we get

K = 20 \(log|H(j\OmegA.|=-20 [log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]]^{1/2}\)

= -10 \(log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]\)

We know that K= KS at Ω=ΩS

=> KS=-10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\).

 

4. Which of the following equation is True?

A. \([\frac{\Omega_P}{\Omega_C}]^{2N} = 10^{-K_P/10}+1\)

B. \([\frac{\Omega_P}{\Omega_C}]^{2N} = 10^{K_P/10}+1\)

C. \([\frac{\Omega_P}{\Omega_C}]^{2N} = 10^{-K_P/10}-1\)

D. None of the mentioned

Answer: C

We know that,

KP=-10 \(log⁡[1+(\frac{\Omega_P}{\Omega_C})^{2N}]\)

=>\([\frac{Ω_P}{Ω_C}]^{2N} = 10^{\frac{-K_P}{10}}-1\)

 

5. Which of the following equation is True?

A. \([\frac{\Omega_S}{\Omega_C} ]^{2N} = 10^{-K_S/10}+1\)

B. \([\frac{\Omega_S}{\Omega_C} ]^{2N} = 10^{K_S/10}+1\)

C. \([\frac{\Omega_S}{\Omega_C} ]^{2N} = 10^{-K_S/10}-1\)

D. None of the mentioned

Answer: B

We know that,

KP=-10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\)

=>\([\frac{Ω_S}{Ω_C}]^{2N} = 10^{\frac{-K_S}{10}}-1\)

 

6. What is the order N of the low pass Butterworth filter in terms of KP and KS?

A. \(\frac{log⁡[(10^\frac{K_P}{10}-1)/(10^\frac{K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)

B. \(\frac{log⁡[(10^\frac{K_P}{10}+1)/(10^\frac{K_s}{10}+1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)

C. \(\frac{log⁡[(10^\frac{-K_P}{10}+1)/(10^\frac{-K_s}{10}+1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)

D. \(\frac{log⁡[(10^\frac{-K_P}{10}-1)/(10^\frac{-K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)

Answer: D

We know that, \([\frac{Ω_P}{Ω_C}]^{2N} = 10^{-K_P/10}-1\) and \([\frac{Ω_P}{Ω_C}]^{2N} = 10^{-K_S/10}-1\).

By dividing the above two equations, we get

=> \([Ω_P/Ω_S]^{2N} = (10^{-K_S/10}-1)(10^{-K_P/10}-1)\)

By taking log in both sides, we get

=> N=\(\frac{log⁡[(10^\frac{-K_P}{10}-1)/(10^\frac{-K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\).

 

7. What is the expression for cutoff frequency in terms of passband gain?

A. \(\frac{\Omega_P}{(10^{-K_P/10}-1)^{1/2N}}\)

B. \(\frac{\Omega_P}{(10^{-K_P/10}+1)^{1/2N}}\)

C. \(\frac{\Omega_P}{(10^{K_P/10}-1)^{1/2N}}\)

D. None of the mentioned

Answer: A

We know that,

\([\frac{Ω_P}{Ω_C}]^{2N} = 10^{-K_P/10}-1\)

=> \(Ω_C = \frac{Ω_P}{(10^{-K_P/10}-1)^{1/2N}}\).

 

8. What is the expression for cutoff frequency in terms of stopband gain?

A. \(\frac{\Omega_S}{(10^{-K_S/10}-1)^{1/2N}}\)

B. \(\frac{\Omega_S}{(10^{-K_S/10}+1)^{1/2N}}\)

C. \(\frac{\Omega_S}{(10^{K_S/10}-1)^{1/2N}}\)

D. None of the mentioned

Answer: C

We know that,

\([\frac{Ω_S}{Ω_C}]^{2N} = 10^{-K_S/10}-1\)

=> \(Ω_C = \frac{Ω_S}{(10^{-K_S/10}-1)^{1/2N}}\).

 

9. The cutoff frequency of the low pass Butterworth filter is the arithmetic mean of the two cutoff frequencies as found above.

A. True
B. False

Answer: A

The arithmetic mean of the two cutoff frequencies as found above is the final cutoff frequency of the low pass Butterworth filter.

 

10. What is the lowest order of the Butterworth filter with a passband gain KP=-1 dB at ΩP=4 rad/sec and stopband attenuation greater than or equal to 20dB at ΩS = 8 rad/sec?

A. 4
B. 5
C. 6
D. 3

Answer: B

We know that the equation for the order of the Butterworth filter is given as

N=\(\frac{log⁡[(10^{-K_P/10}-1)/(10^{-K_s/10}-1)]}{2 log⁡(\frac{Ω_P}{Ω_S})}\)

From the given question,

KP=-1 dB,
ΩP= 4 rad/sec,
KS=-20 dB
ΩS= 8 rad/sec

Upon substituting the values in the above equation, we get
N=4.289
Rounding off to the next largest integer, we get N=5.

 

11. What is the cutoff frequency of the Butterworth filter with a passband gain KP=-1 dB at ΩP=4 rad/sec and stopband 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

Answer: D

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.

 

12. What is the system function of the Butterworth filter with specifications as passband gain KP=-1 dB at ΩP=4 rad/sec and stopband 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

Answer: C

From the given question,

KP=-1 dB, ΩP=4 rad/sec
KS=-20 dB and ΩS=8 rad/sec

We find out an order as N=5 and ΩC=4.5787 rad/sec

We know that for a 5th order normalized low pass Butterworth filter, the 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)=\({2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\)

 

13. If H(s)=\(\frac{1}{s^2+s+1}\) represents the transfer function of a low pass filter (not Butterworth) with a passband of 1 rad/sec, then what is the system function of a low pass filter with a passband 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

Answer: A

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

=\(\frac{100}{s^2+10s+100}\).

 

14. If H(s)=\(\frac{1}{s^2+s+1}\) represents the transfer function of a low pass filter (not Butterworth) with a passband 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

Answer: B

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

=\(\frac{s^2}{s^2+s+1}\)

 

15. If H(s)=\(\frac{1}{s^2+s+1}\) represents the transfer function of a low pass filter (not Butterworth) with a passband 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

Answer: C

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

=\(\frac{s^2}{s^2+10s+100}\)

 

16. If H(s)=\(\frac{1}{s^2+s+1}\) represents the transfer function of a low pass filter (not Butterworth) with a passband of 1 rad/sec, then what is the system function of a bandpass filter with a passband 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. \(\frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}\)

Answer: D

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)=\(\frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}\).

 

17. If H(s)=\(\frac{1}{s^2+s+1}\) represent the transfer function of a low pass filter (not Butterworth) with a passband 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

Answer: A

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)=\(\frac{(s^2+100)^2}{s^4+2s^3+204s^2+200s+10^4}\)

 

18. What is the stopband frequency of the normalized low pass Butterworth filter used to design an analog bandpass filter with -3.0103dB upper and lower cutoff frequency of 50Hz and 20KHz and a stopband attenuation 20dB at 20Hz and 45KHz?

A. 2 rad/sec
B. 2.25 Hz
C. 2.25 rad/sec
D. 2 Hz

Answer: C

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.

 

19. What is the order of the normalized low pass Butterworth filter used to design an analog bandpass filter with -3.0103dB upper and lower cutoff frequency of 50Hz and 20KHz and a stopband attenuation of 20dB at 20Hz and 45KHz?

A. 2
B. 3
C. 4
D. 5

Answer: B

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.

 

20. 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 ≤ \(\frac{log⁡(\frac{1}{d})}{log⁡(\frac{1}{k})}\)

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

N ≤ \(\frac{log⁡(\frac{1}{d})}{log⁡(\frac{1}{k})}\)

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