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

The peak sidelobe in the case of a rectangular window has a value of -13dB.

The peak sidelobe in the case of the Hanning window has a value of -32dB.

The frequency of oscillations in the passband of a low pass filter increases with an increase in the value of M, but they do not diminish in amplitude.

The multiplication of hd(n) with a rectangular window is identical to truncating the Fourier series representation of the desired filter characteristic Hd(ω).

The truncation of the Fourier series is known to introduce ripples in the frequency response characteristic H(ω) due to the non-uniform convergence of the Fourier series at a discontinuity. The oscillatory behavior near the band edge of the low pass filter is known as the Gibbs phenomenon.

The frequency response shown in the figure is the frequency response of a low pass filter designed using a Blackman window of length M=61.

The frequency response shown in the figure is the frequency response of a low pass filter designed using a Kaiser window of length M=61 and α=4.

The transition width of the main lobe in the case of the Blackman window is equal to 12π/M where M is the length of the window.

The Hanning window has a time domain sequence as

h(n)=\(\frac{1}{2}(1-cos⁡\frac{2πn}{M-1})\)

Since the width of the main lobe is inversely proportional to the value of M if the value of M increases then the main lobe becomes narrower.

The larger side lobes of W(ω) result in undesirable ringing effects in the FIR filter frequency response H(ω), and also in relatively large side lobes in H(ω).

In the frequency sampling method, we specify the frequency response Hd(ω) at a set of equally spaced frequencies, namely ωk=\(\frac{2π}{M}(k+\alphA.\)

To reduce the side lobes, it is desirable to optimize the frequency specification in the transition band of the filter. This optimization can be accomplished numerically on a digital computer by means of linear programming techniques.

The desired output of an FIR filter with an input h(n) and using a window of length M is given as

H(ω)=\(\sum_{n=0}^{M-1} h(n)e^{-jωn}\)

We know that

ωk=\(\frac{2π}{M}\)(k+α) and H(ω)=\(\sum_{n=0}^{M-1} h(n)e^{-jωn}\)

Thus from substituting the first in the second equation, we get

H(k+α)=\(\sum_{n=0}^{M-1} h(n)e^{-j2π(k+α)n/M}\); k=0,1,2…M-1

We know that

H(k+α)=\(\sum_{n=0}^{M-1} h(n)e^{-j2π(k+α)n/M}\); k=0,1,2…M-1

If we multiply the above equation on both sides by the exponential exp(j2πkm/M), m=0,1,2….M-1 and sum over k=0,1,….M-1, we get the equation

h(n)=\(\frac{1}{M} \sum_{k=0}^{M-1}H(k+α)e^{j2π(k+α)n/M}\); n=0,1,2…M-1

Since {h(n)} is real, we can easily show that the frequency samples {H(k+α)} satisfy the symmetry condition
H(k+α)= H*(M-k-α).

The symmetry condition, along with the symmetry conditions for {h(n)}, can be used to reduce the frequency specifications from M points to (M+1)/2 points for M odd and M/2 for M even. Thus the linear equations for determining {h(n)} from {H(k+α)} are considerably simplified.

Although the frequency sampling method provides us with another means for designing linear phase FIR filters, its major advantage lies in the efficient frequency sampling structure, which is obtained when most of the frequency samples are zero.

There is a potential problem with the frequency sampling realization of the FIR linear phase filter. The frequency sampling realization of the FIR filter introduces poles and zeros at equally spaced points on the unit circle.

In the ideal situation, the zeros cancel the poles and, consequently, the actual zeros of the H(z) are determined by the selection of the frequency samples H(k+α). In a practical implementation of the frequency sampling realization, however, quantization effects preclude a perfect cancellation of the poles and zeros.