# 7.3. Structure of IIR filters – Digital Filters Design for Signal and Image Processing

## 7.3. Structure of IIR filters

### 7.3.1. Direct structures

In this section, we will refer to the direct canonic structure that is characterized by the following transfer function: First, we introduce an intermediary output x1(k), so that:  The transfer function in equation (7.1) can then be decomposed into a product of two transfer functions: All this occurs as if the filter Hz(z) has been obtained by cascading an FIR filter H1(z) and an IIR filter H2(z). Figure 7.2. Putting filters H1 and H2 into cascade

We end up with the most spontaneous realization, called the direct form structure 1. Figure 7.3. Direct form structure 1 of an IIR filter

We see that this structure requires N+M−2 delay cells. We can then ask the following question: can we share some cells and only use M-l delay cells, knowing that M>N? The answer is yes, leading us to the direct form structure 2. Figure 7.4. Direct form 2 of an IIR filter

An alternative approach consists of decomposing the transfer function shown in equation (7.1) into an IIR, then an FIR filter. The intermediate output x1(k) then satisfies:  Here we have  So by using equations (7.5) and (7.6), we can establish the structure shown in Figure 7.5. Figure 7.5. Direct form structure before sharing delay elements

We can nevertheless distribute some delay cells. We then have the structure shown in Figure 7.6. Figure 7.6. Canonic structure of an IIR filter

When M = N = 2, we end up with a second-order cell (see Figure 7.7). Figure 7.7. Canonic structure of a second-order cell

Taking into account the role played by the second-order cell, we need to understand what happens when we consider a linear system guided by a pole near the unity disk. To generate a resonance at the frequency/0, we consider a pole inside the unity disk with R < 1. is also a pole of the transfer function of the second-order cell. From here, the transfer function can be written in the two equivalent forms that follow: and The frequency response of the filter corresponds to the Fourier transform of the impulse response, which can be obtained by taking the transfer function Hz(z) as so that: According to the values of R, the resonance is more or less strong and tends towards infinity when R tends towards 1. So, Figure 7.8 shows the position of the poles and the corresponding frequency response in different situations. The normalized frequencies linked to the poles are equal to and the values of R are equal successively to 0.7, 0.9 and 0.99.

We then normalize H(f) so that |H(f0)| = 1, which conditions the gain value G as follows: By definition, the cut-off frequencies. fc of this filter at −3 dB are:  Figure 7.8. Frequential representation of a second-order filter, according to values of R (0.7, 0.9 and 0.99)

Equation (7.13) is reduced to: both by using equation (7.11) and from the fact that |H(f0)| = 1: This equation admits two solutions f1 and f2 which characterize the bandwidth Δf = f2f1. To avoid complex algebraic calculations, using a geometric approach helps us to easily find the result. Indeed, let ZA, ZB, and zQ be the complex numbers associated with points A, B, and Q in the complex z-plane and at frequencies f1, f2, and f0 (as in Figure 7.9). Moreover pi is associated with point P and pi* to P*.

We thus have:  By combining equations (7.16) and (7.17), we obtain: Now, when R tends towards 1, the poles approach the unity circle while the points P, Q, and A become very close and their distance in relation to O, the source in the complex plane, is approximately the same, i.e. . Therefore, equation (7.18) becomes:  Figure 7.9. Geometric resolution of the passband of a second-order filter

Since distances between points Q, A, and P* are approximately the same, this means that: Taking into account the equalities in equations (7.14) and (7.20), equation (7.18) becomes: The angle represented by segments AP and PQ is therefore approximately equal to , which means that the angle represented by segments PA and AQ is as well. The triangle PAQ is almost an isosceles.

This means that: When R tends towards 1, we assimilate the arc AB to the tangent vector to the unity circle in Q. We deduce that: from which we get: We can demonstrate in this way that the bandwidth is proportional to (1−R). The more the poles are close to the unity circle in the z-plane, the more the filter is selective around the frequencies associated with the poles.

Now we place a pair of zeros and close to the poles inside the disk and associated at the same frequencies. Equation (7.9) becomes: or The frequency response of the filter then satisfies: According to the values of r, we will observe an accentuation effect of the resonance at normalized frequencies when r < R and weakening if r > R. The accentuation or weakening level is controlled by the proximity of r to R. The spike width is always related to the proximity of the poles in relation to the unity circle (see Figures 7.10 to 7.13). Figure 7.10. Representation of the frequency response of a second-order filter without zeros. R = 0.7 Figure 7.11. Representation of the frequency response of a second-order filter with zeros close to the poles, R=0.7, r takes respectively the values of 0.65 and 0.75 Figure 7.12. Representation of the frequency response of a second order filter without zeros R=0.9 Figure 7.13. Representation of the frequency response of a second-order filter with zeros close to the poles R=0.9. r taking respectively the values 0.85 and 0.9

If we choose the value of r equal to 1, the expression of the transfer function given in equations (7.25) and (7.26) becomes:  and also  Figure 7.14. Frequential representation of a second-order filter. R equals 0.9 and r equals 0. then 1

Whether it has two poles or two poles and two zeros, we learn several facts from studying this kind of filter:

– we learn that the resonance acuity is regulated by the proximity degree of the pole in relation to the unity circle;

– we learn the association of zeros in the transfer function allows us to modify the response curve of the filter. There will be positions of zeros that will lessen the curve, and there will be others that instead favor resonance, as shown in the diagrams in Figure 7.15. Figure 7.15. Frequential representation of a second-order filter with zeros more or less close to the poles. R equals 0.9 and r varies from 0 to 1 by steps of 0.05

This technique can be generalized to synthesize a filter having a finite number of slots. It is enough to generate a numerator N(z) whose zeros are situated on the unity circle and correspond to the desired slots. We then construct the denominator which must equal N(R-1z) to deduce the transfer function.

We thus obtain a generalized expression of the following formula: APPLICATION.- with a signal sampled at 500 Hz, we want to eliminate a periodic signal whose fundamental frequency equals 50 Hz.

The normalized frequencies to be eliminated are represented as: The zeros of Hz(z) thus correspond to: and correspond to the tenth-degree root of the unity. Thus, we have: and we take, for example:  Figure 7.16. Frequential representation of a tenth-order filter. p equals 0.98

Slot filters allow us to remove harmonies, while comb filters can increase periodicities; that is, they increase the signals containing harmonic frequencies (multiples of the fundamental). These filters are used in audio applications for devices that create reverberations. The filters act as reflectors of sound waves and favor certain periodic signals.

The cascade structure decomposes the filter of the transfer function Hz(z) with a succession of first and second-order cells Hi(z). However, we can see that a first order cell is a specific example of a second-order cell by taking the values associated with z−2 that are equal to zero: or the following formula: or:   Figure 7.18. Cascade structure with distributed gain

### 7.3.3. Parallel structures

A parallel structure decomposes the filter of the transfer function Hz(Z) with a parallel interconnection of filters the transfer function of which is the transfer function Hi(z). We have, as a result, the following formula for the transfer function:  Figure 7.19. Parallel structure

In the following section, we will look more closely at direct and cascade structures.