# Butterworth Filters MCQ [Free PDF] – Objective Question Answer for Butterworth Filters Quiz

1. Which of the following is true in the case of Butterworth filters?

A. Smooth passband
B. Wide transition band
C. Not so smooth stopband
D. All of the mentioned

Butterworth filters have a very smooth passband, which we pay for with a relatively wide transmission region.

2. What is the magnitude frequency response of a Butterworth filter of order N and cutoff frequency ΩC?

A. $$\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}$$

B. $$1+(\frac{Ω}{Ω_C})^{2N}$$

C. $$\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}$$

D. None of the mentioned

A Butterworth is characterized by the magnitude frequency response

|H(jΩ)|=$$\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}$$

where N is the order of the filter and ΩC is defined as the cutoff frequency.

3. What is the factor to be multiplied to the dc gain of the filter to obtain filter magnitude at cutoff frequency?
A. 1
B. √2
C. 1/√2
D. 1/2

The dc gain of the filter is the filter magnitude at Ω=0.
We know that the filter magnitude is given by the equation

|H(jΩ)|=$$\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}$$

At Ω=ΩC, |H(jΩC.|=1/√2=1/√2(|H(jΩ)|)

Thus the filter magnitude at the cutoff frequency is 1/√2 times the dc gain.

4. What is the value of magnitude frequency response of a Butterworth low pass filter at Ω=0?

A. 0
B. 1
C. 1/√2
D. None of the mentioned

The magnitude frequency response of a Butterworth low pass filter is given as

|H(jΩ)|=$$\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}$$

At Ω=0 => |H(jΩ)|=1 for all N.

5. As the value of the frequency Ω tends to ∞, then |H(jΩ)| tends to ____________

A. 0
B. 1
C. ∞
D. None of the mentioned

We know that the magnitude frequency response of a Butterworth filter of order N is given by the expression

|H(jΩ)|=$$\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}$$

In the above equation, if Ω→∞ then |H(jΩ)|→0.

6. |H(jΩ)| is a monotonically increasing function of frequency.

A. True
B. False

|H(jΩ)| is a monotonically decreasing function of frequency, i.e., |H(jΩ2)| < |H(jΩ1)| for any values of Ω1 and Ω2 such that 0 ≤ Ω1 < Ω2.

7. What is the magnitude squared response of the normalized low pass Butterworth filter?

A. $$\frac{1}{1+Ω^{-2N}}$$
B. 1+Ω-2N
C. 1+Ω2N
D. $$\frac{1}{1+Ω^{2N}}$$

We know that the magnitude response of a low pass Butterworth filter of order N is given as

|H(jΩ)|=$$\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}$$

For a normalized filter, ΩC =1

=> |H(jΩ)|=$$\frac{1}{\sqrt{1+(Ω)^{2N}}}$$ => |H(jΩ)|2=$$\frac{1}{1+Ω^{2N}}$$

Thus the magnitude squared response of the normalized low pass Butterworth filter of order N is given by the equation,

|H(jΩ)|2=$$\frac{1}{1+Ω^{2N}}$$.

8. What is the transfer function of magnitude squared frequency response of the normalized low pass Butterworth filter?

A. $$\frac{1}{1+(s/j)^{2N}}$$

B. $$1+(\frac{s}{j})^{-2N}$$

C. $$1+(\frac{s}{j})^{2N}$$

D. $$\frac{1}{1+(s/j)^{-2N}}$$

We know that the magnitude squared frequency response of a normalized low pass Butterworth filter is given as

H(jΩ)|2=$$\frac{1}{1+Ω^{2N}}$$ => HN(jΩ).HN(-jΩ)=$$\frac{1}{1+Ω^{2N}}$$

Replacing jΩ by ‘s’ and hence Ω by s/j in the above equation, we get

HN(s).HN(-s)=$$\frac{1}{1+(\frac{s}{j})^{2N}}$$ which is called the transfer function.

9. Where does the poles of the transfer function of normalized low pass Butterworth filter exists?

A. Inside unit circle
B. Outside unit circle
C. On unit circle
D. None of the mentioned

The transfer function of normalized low pass Butterworth filter is given as

HN(s).HN(-s)=$$\frac{1}{1+(\frac{s}{j})^{2N}}$$

The poles of the above equation is obtained by equating the denominator to zero.

=> $$1+(\frac{s}{j})^{2N}$$=0

=> s=(-1)1/2N.j

=> sk=$$e^{jπ(\frac{2k+1}{2N})} e^{jπ/2}$$, k=0,1,2…2N-1

The poles are therefore on a circle with radius unity.

10. What is the general formula that represents the phase of the poles of the transfer function of normalized low pass Butterworth filter of order N?

A. $$\frac{π}{N} k+\frac{π}{2N}$$ k=0,1,2…N-1

B. $$\frac{π}{N} k+\frac{π}{2N}+\frac{π}{2}$$ k=0,1,2…2N-1

C. $$\frac{π}{N} k+\frac{π}{2N}+\frac{π}{2}$$ k=0,1,2…N-1

D. $$\frac{π}{N} k+\frac{π}{2N}$$ k=0,1,2…2N-1

The transfer function of normalized low pass Butterworth filter is given as

HN(s).HN(-s)=$$\frac{1}{1+(\frac{s}{j})^{2N}}$$

The poles of the above equation is obtained by equating the denominator to zero.

=> $$1+(\frac{s}{j})^{2N}$$=0

=> s=(-1)1/2N.j

=> sk=$$e^{jπ(\frac{2k+1}{2N})} e^{jπ/2}$$, k=0,1,2…2N-1

The poles are therefore on a circle with radius unity and are placed at angles,

θk=$$\frac{π}{N} k+\frac{π}{2N}$$ k=0,1,2…2N-1

11. What is the Butterworth polynomial of order 3?

A. (s2+s+1)(s-1)
B. (s2-s+1)(s-1)
C. (s2-s+1)(s+1)
D. (s2+s+1)(s+1)

Given that the order of the Butterworth low pass filter is 3.

Therefore, for N=3 Butterworth polynomial is given as B3(s)=(s-s0) (s-s1) (s-s2)

We know that, sk=$$e^{jπ(\frac{2k+1}{2N})} e^{jπ/2}$$

=> s0=(-1/2)+j(√3/2), s1= -1, s2=(-1/2)-j(√3/2)

=> B3(s)= (s2+s+1)(s+1).

12. What is the Butterworth polynomial of order 1?

A. s-1
B. s+1
C. s
D. none of the mentioned

Given that the order of the Butterworth low pass filter is 1.
Therefore, for N=1 Butterworth polynomial is given as B3(s)=(s-s0).

We know that, sk=$$e^{jπ(\frac{2k+1}{2N})} e^{jπ/2}$$

=> s0=-1

=> B1(s)=s-(-1)=s+1.

13. What is the transfer function of Butterworth low pass filter of order 2?

A. $$\frac{1}{s^2+\sqrt{2} s+1}$$

B. $$\frac{1}{s^2-\sqrt{2} s+1}$$

C. $$s^2-\sqrt{2} s+1$$

D. $$s^2+\sqrt{2} s+1$$

We know that the Butterworth polynomial of a 2nd order low pass filter is

B2(s)=s2+√2 s+1

Thus the transfer function is given as $$\frac{1}{s^2+\sqrt{2} s+1}$$.

14. What is the passband edge frequency of an analog low pass normalized filter?

Let H(s) denote the transfer function of a low pass analog filter with a passband edge frequency ΩP equal to 1 rad/sec. This filter is known as the analog low pass normalized prototype.

15. If H(s) is the transfer function of an analog low pass normalized filter and Ωu is the desired passband edge frequency of the new low pass filter, then which of the following transformation has to be performed?

A. s → s/Ωu
B. s → s.Ωu
C. s → Ωu/s
D. none of the mentioned

If Ωu is the desired passband edge frequency of the new low pass filter, then the transfer function of this new low pass filter is obtained by using the transformation s → s/Ωu.

16. Which of the following is a low pass-to-high pass transformation?

A. s → s / Ωu
B. s → Ωu / s
C. s → Ωu.s
D. none of the mentioned

The low pass-to-high pass transformation is simply achieved by replacing s by 1/s. If the desired high pass filter has the passband edge frequency Ωu, then the transformation is
s → Ωu / s.

17. Which of the following is the backward design equation for a low pass-to-low pass transformation?

A. $$\Omega_S=\frac{\Omega_S}{\Omega_u}$$

B. $$\Omega_S=\frac{\Omega_u}{\Omega’_S}$$

C. $$\Omega’_S=\frac{\Omega_S}{\Omega_u}$$

D. $$\Omega_S=\frac{\Omega’_S}{\Omega_u}$$

If Ωu is the desired passband edge frequency of the new low pass filter, then the transfer function of this new low pass filter is obtained by using the transformation s → s/Ωu. If ΩS and Ω’S are the stopband frequencies of prototype and transformed filters respectively, then the backward design equation is given by
$$\Omega_S=\frac{\Omega’_S}{\Omega_u}$$.

18. Which of the following is a low pass-to-band pass transformation?

A. s→$$\frac{s^2+Ω_u Ω_l}{s(Ω_u+Ω_l)}$$
B. s→$$\frac{s^2-Ω_u Ω_l}{s(Ω_u-Ω_l)}$$
C. s→$$\frac{s^2+Ω_u Ω_l}{s(Ω_u-Ω_l)}$$
D. s→$$\frac{s^2-Ω_u Ω_l}{s(Ω_u+Ω_l)}$$

If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired bandpass filter, then the transformation to be performed on the normalized low pass filter is

s→$$\frac{s^2+Ω_u Ω_l}{s(Ω_u-Ω_l)}$$

19. Which of the following is the backward design equation for a low pass-to-high pass transformation?

A. $$\Omega_S=\frac{\Omega_S}{\Omega_u}$$

B. $$\Omega_S=\frac{\Omega_u}{\Omega’_S}$$

C. $$\Omega’_S=\frac{\Omega_S}{\Omega_u}$$

D. $$\Omega_S=\frac{\Omega’_S}{\Omega_u}$$

If Ωu is the desired passband edge frequency of the new high pass filter, then the transfer function of this new high pass filter is obtained by using the transformation s → Ωu /s. If ΩS and Ω’S are the stopband frequencies of prototype and transformed filters respectively, then the backward design equation is given by

$$\Omega_S=\frac{\Omega_u}{\Omega’_S}$$.

20. Which of the following is a low pass-to-band stop transformation?

A. s→$$\frac{s(Ω_u-Ω_l)}{s^2+Ω_u Ω_l}$$

B. s→$$\frac{s(Ω_u+Ω_l)}{s^2+Ω_u Ω_l}$$

C. s→$$\frac{s(Ω_u-Ω_l)}{s^2-Ω_u Ω_l}$$

D. none of the mentioned

If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band stop filter, then the transformation to be performed on the normalized low pass filter is

s→$$\frac{s(Ω_u-Ω_l)}{s^2-Ω_u Ω_l}$$

21. If A=$$\frac{-Ω_1^2+Ω_u Ω_l}{Ω_1 (Ω_u-Ω_l)}$$ and B=$$\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}$$, then which of the following is the backward design equation for a low pass-to-band pass transformation?

A. ΩS=|B|
B. ΩS=|A|
C. ΩS=Max{|A|,|B|}
D. ΩS=Min{|A|,|B|}

If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired bandpass filter and Ω1 and Ω2 are the lower and upper cutoff stopband frequencies of the desired bandpass filter, then the backward design equation is

ΩS=Min{|A|,|B|}
where,

A=$$\frac{-Ω_1^2+Ω_u Ω_l}{Ω_1 (Ω_u-Ω_l)}$$ and B=$$\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}$$.

22. If A=$$\frac{Ω_1 (Ω_u-Ω_l)}{-Ω_1^2+Ω_u Ω_l}$$ and B=$$\frac{Ω_2 (Ω_u-Ω_l)}{Ω_2^2-Ω_u Ω_l}$$, then which of the following is the backward design equation for a low pass-to-band stop transformation?

A. ΩS=Max{|A|,|B|}
B. ΩS=Min{|A|,|B|}
C. ΩS=|B|
D. ΩS=|A|

If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band stop filter and Ω1 and Ω2 are the lower and upper cutoff stopband frequencies of the desired band stop filter, then the backward design equation is
ΩS=Min{|A|,|B|}
where, A=$$\frac{Ω_1 (Ω_u-Ω_l)}{-Ω_1^2+Ω_u Ω_l}$$ and B=$$\frac{Ω_2 (Ω_u-Ω_l)}{Ω_2^2-Ω_u Ω_l}$$.

23. Which of the following is a low pass-to-high pass transformation?

A. s → s / Ωu
B. s → Ωu/s
C. s → Ωu.s
D. none of the mentioned

The low pass-to-high pass transformation is simply achieved by replacing s with 1/s. If the desired high pass filter has the passband edge frequency Ωu, then the transformation is
s → Ωu/s.

24. Which of the following operation has to be performed to increase the sampling rate by an integer factor I?

A. Interpolating I+1 new samples
B. Interpolating I-1 new samples
C. Extrapolating I+1 new samples
D. Extrapolating I-1 new samples