Finite Impulse Response (FIR) System Structure MCQ Quiz – Objective Question with Answer for FIR System Structure

The difference equation y(n)=\(\sum_{k=0}^{M-1}b_k x(n-k)\) describes the FIR system.

We know that the difference equation of an FIR system is given by

y(n)=\(\sum_{k=0}^{M-1}b_k x(n-k)\).

=>h(n)=bk=>\(\sum_{k=0}^{M-1}b_k z^{-k}\).

There are several structures for implementing an FIR system, beginning with the simplest structure, called the direct form. There are several other methods like cascade form realization, frequency sampling realization, and lattice realization which are used for implementing an FIR system.

The direct form realization follows immediately from the non-recursive difference equation given by

y(n)=\(\sum_{k=0}^{M-1}b_k x(n-k)\).

We observe that this structure requires M-1 memory locations for storing the M-1 previous inputs.

The structure of the direct form realization resembles a tapped delay line or a transversal system.

When the FIR system has a linear phase, the unit sample response of the system satisfies either the symmetry or asymmetry condition, h(n)=±h(M-1-n)

For such a system the number of multiplications is reduced from M to M/2 for M even and to (M-1)/2 for M odd. Thus for the structure given in the question, M is odd.

We have to do 3 multiplications for every second-order equation. So, we have to do 6 multiplications if we combine two second-order equations and we have to perform 3 multiplications by directly calculating the fourth-order equation. Thus the number of multiplications is reduced by a factor of 50%.

To derive the frequency sampling structure, we specify the desired frequency response at a set of equally spaced frequencies, namely

ωk=\(\frac{2\pi}{M}\)(k+α), k=0,1…(M-1)/2 for M odd

k=0,1….(M/2)-1 for M even

α=0 or 0.5.

In frequency sampling realization, the system function H(z) is characterized by the set of frequency samples {H(k+ α)} instead of {h(n)}. We view this FIR filter realization as a cascade of two filters. One is an all-zero or a comb filter and the other consists of a parallel bank of single-pole filters with resonant frequencies.

The system function H(z) which is characterized by the set of frequency samples is obtained as

H(z)=\(\frac{1}{M}(1-z^{-M} e^{j2πα})\sum_{k=0}^{M-1}\frac{H(k+α)}{1-e^{j2π(k+α)/M} z^{-1}}\)

We view this FIR realization as a cascade of two filters, H(z)=H1(z).H2(z)

Here H1(z) represents the all-zero filter or comb filter whose system function is given by the equation

H1(z)=\(\frac{1}{M}(1-z^{-M} e^{j2πα})\).

The system function of the comb filter is given by the equation

H1(z)=\(\frac{1}{M}(1-z^{-M} e^{j2πα})\)

Its zeros are located at equally spaced points on the unit circle at

zk=ej2π(k+α)/M k=0,1,2….M-1

The system function H(z) which is characterized by the set of frequency samples is obtained as

H(z)=\(\frac{1}{M}(1-z^{-M} e^{j2πα})\sum_{k=0}^{M-1}\frac{H(k+α)}{1-e^{\frac{j2π(k+α)}{M}}z^{-1}}\)

We view this FIR realization as a cascade of two filters, H(z)=H1(z).H2(z)
Here H1(z) represents the all-zero filter or comb filter, and the system function of the other filter is given by the equation

H2(z)=\(\sum_{k=0}^{M-1} \frac{H(k+α)}{1-e^{\frac{j2π(k+α)}{M}} z^{-1}}\)

H2(z)=\(\sum_{k=0}^{M-1} \frac{H(k+α)}{1-e^{\frac{j2π(k+α)}{M}} z^{-1}}\)

We obtain the poles of the above system function by equating the denominator of the above equation to zero.

=>\(1-e^{\frac{j2π(k+α)}{M}} z^{-1}\)=0

=>z=pk=\(e^{\frac{j2π(k+α)}{M}}\), k=0,1….M-1

When the desired frequency response characteristic of the FIR filter is narrowband, most of the gain parameters {H(k+α)} are zero. Consequently, the corresponding resonant filters can be eliminated and only the filters with nonzero gains need to be retained.

Lattice filters are used extensively in digital signal processing and in the implementation of adaptive filters.

Consider a sequence of FIR filer with system function Hm(z)=Am(z), m=0,1,2…M-1
where, by definition, Am(z) is the polynomial

Am(z)=\(1+\sum_{k=1}^m α_m (k)z^{-k}\), m≥1 and A0(z)=1.

We know that Hm(z)=Am(z) and Am(z) is a polynomial whose equation is given as

Am(z)=\(1+\sum_{k=1}^m α_m (k)z^{-k}\), m≤1 and A0(z)=1

A0(z)=1 => hm(0)=1 and

Am(z)=\(\sum_{k=1}^m α_m (k)z^{-k}\)(m≤1)=> hm(k)=αm(k) for k=1,2…m.

The FIR structure shown in the above figure is intimately related to the topic of linear prediction. Thus the top filter structure shown in the above figure is called a prediction error filter.

The single-stage lattice filter is shown below.

Thus the output of the single-stage lattice filter is given by y(n)= x(n)+Kx(n-1).

When two single stage lattice filters are cascaded, then the output from the first filter is given by the equation

f1(n)= x(n)+K1x(n-1)
g1(n)=K1x(n)+x(n-1)

The output from the second filter is obtained as

f2(n)=f1(n)+K2g1(n-1)
=x(n)+K2[K1x(n-1)+x(n-2)]+ K1x(n-1)
= x(n)+K1(1+K2)x(n-1)+K2x(n-2).

The equation for the output of an FIR filter represented in the direct form structure is given as
y(n)=x(n)+ α2(1)x(n-1)+ α2(2)x(n-2)

The output from the double stage lattice structure is given by the equation,
f2(n)= x(n)+K2(1+K2)x(n-1)+K2x(n-2)
By comparing the coefficients of both the equations, we get

α2(1)= K1(1+K2).

The equation of the output from the second stage lattice filter is given by
f2(n)= x(n)+K1(1+K2)x(n-1)+K2x(n-2)
In the above equation, the constants K1 and K2 are called reflection coefficients.

We get the output from the third stage lattice filter as
A3(z)=1+(13/24)z-1+(5/8)z-2+(1/3)z-3.

Thus the FIR filter coefficients for the direct form structure are (1,13/24,5/8,1/3).

Given the system function of the FIR filter is
H(z)= 1+(13/24)z-1+(5/8)z-2+(1/3)z-3

Thus the lattice coefficients corresponding to the given filter are (1/4,1/2,1/3).