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

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.

The interpolation process can be accomplished in a variety of ways. Among them, there is a process that preserves the spectral shape of the signal sequence x(n).

If v(m) denote a sequence with a rate Fy=I.Fx which is obtained from x(n) by adding I-1 zeros between successive values of x(n). Thus

v(m)= x(m/I), m=0,±I,±2I…. 0, otherwise.

By taking the z-transform of the signal v(m), we get

V(z)=(sum_{m=-∞}^∞ v(m)z^{-m})

=(sum_{m=-∞}^∞ x(m)z^{-mI})

= X(z-I)

The spectrum of v(m) is obtained by evaluating V(z) = X(zI) on the unit circle. Thus V(ωy)= X(ωyI), where ωy denotes the frequency variable relative to the new sampling rate.

We know that the relationship between sampling rates is Fy=IFx and hence the frequency variables ωx and ωy are related according to the formula

ωy= ωx/I.

From the diagram, the values are interpolated between two successive values of x(n), thus it is called as sampling rate conversion using interpolation by a factor I.