Digital Signal Processing MCQ [Free PDF] – Objective Question Answer for Digital Signal Processing Quiz

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.

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.

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.

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.

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.

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

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

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

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

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

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.

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