# Analog Filters Classification and Specification MCQ [Free PDF] – Objective Question Answer for Analog Filters Classification and Specification Quiz

1. What is the region between origin and the passband frequency in the magnitude frequency response of a low pass filter?

A. Stopband
B. Passband
C. Transition band
D. None of the mentioned

From the magnitude frequency response of a low pass filter, we can state that the region before passband frequency is known as ‘pass band’ where the signal is passed without huge losses.

2. What is the region between stopband and the passband frequencies in the magnitude frequency response of a low pass filter?

A. Stopband
B. Passband
C. Transition band
D. None of the mentioned

From the magnitude frequency response of a low pass filter, we can state that the region between passband and stopband frequencies is known as the ‘transition band’ where no specifications are provided.

3. What is the region after the stopband frequency in the magnitude frequency response of a low pass filter?

A. Stopband
B. Passband
C. Transition band
D. None of the mentioned

From the magnitude frequency response of a low pass filter, we can state that the region after stop band frequency is known as ‘stop band’ where the signal is stopped or restricted.

4. If δP is the forbidden magnitude value in the passband and δS is the forbidden magnitude value in the stopband, then which of the following is true in the passband region?

A. 1-δS≤|H(jΩ)|≤1
B. δP≤|H(jΩ)|≤1
C. 0≤|H(jΩ)|≤ δS
D. 1-δP≤|H(jΩ)|≤1

From the magnitude frequency response of the low pass filter, the hatched region in the passband indicates a forbidden magnitude value whose value is given as

1- δP≤|H(jΩ)|≤1.

5. If δP is the forbidden magnitude value in the passband and δS is the forbidden magnitude value in the stopband, then which of the following is true in the stopband region?

A. 1- δP≤|H(jΩ)|≤1
B. δP≤|H(jΩ)|≤1
C. 0≤|H(jΩ)|≤ δS
D. 1- δP≤|H(jΩ)|≤1

From the magnitude frequency response of the low pass filter, the hatched region in the stopband indicate a forbidden magnitude value whose value is given as
0≤|H(jΩ)|≤ δS.

6. What is the value of passband ripple in dB?

A. -20log(1- δP)
B. -20log(δP)
C. 20log(1- δP)
D. None of the mentioned

1-δP is known as the passband ripple or the passband attenuation, and its value in dB is given as -20log(1-δP).

7. What is the value of stopband ripple in dB?

A. -20log(1-δS)
B. -20log(δS)
C. 20log(1-δS)
D. None of the mentioned

δS is known as the stopband attenuation, and its value in dB is given as -20log(δS).

8. What is the passband gain of a low pass filter with 1- δP as the passband attenuation?

A. -20log(1- δP)
B. -20log(δP)
C. 20log(δP)
D. 20log(1- δP)

If 1-δP is the passband attenuation, then the passband gain is given by the formula 20log(1-δP).

9. What is the stopband gain of a low pass filter with δS as the passband attenuation?

A. -20log(1- δS)
B. -20log(δS)
C. 20log(δS)
D. 20log(1- δS)

If δS is the stopband attenuation, then the stopband gain is given by the formula 20log(δS).

10. What is the cutoff frequency of a normalized filter?

D. None of the mentioned

A filter is said to be normalized if the cutoff frequency of the filter, Ωc is 1 rad/sec.

11. The low pass, high pass, bandpass, and bandstop filters can be designed by applying a specific transformation to a normalized low pass filter.

A. True
B. False

It is known that the low pass, high pass, bandpass, and bandstop filters can be designed by applying a specific transformation to a normalized low pass filter. Therefore, a lot of importance is given to the design of a normalized low pass analog filter.

12. 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.

12. 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.

13. 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.

14. 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.

15. 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.

16. |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.

17. 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=$$\frac{1}{1+Ω^{2N}}$$.

18. 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.

19. Where do 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.

20. 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

21. 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).

22. 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.

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

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