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

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

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.

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.

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.

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.

|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.

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=\(\frac{1}{1+Ω^{2N}}\).

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.

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.

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

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).

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.

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 \(\frac{1}{s^2+\sqrt{2} s+1}\).

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.

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.

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.

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
\(\Omega_S=\frac{\Omega’_S}{\Omega_u}\).

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→\(\frac{s^2+Ω_u Ω_l}{s(Ω_u-Ω_l)}\)

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

\(\Omega_S=\frac{\Omega_u}{\Omega’_S}\).

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→\(\frac{s(Ω_u-Ω_l)}{s^2-Ω_u Ω_l}\)

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=\(\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}\).

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=\(\frac{Ω_2 (Ω_u-Ω_l)}{Ω_2^2-Ω_u Ω_l}\).

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.

An increase in the sampling rate by an integer factor of I can be accomplished by interpolating I-1 new samples between successive values of the signal.