What Is The Transfer Function Of A Filter?

Transfer Office Analysis

This chapter discusses filter transfer functions and associated analysis. The transfer office provides an algebraic representation of a linear, time-invariant (LTI) filter in the frequency domain:

$\textstyle \parbox{0.8\textwidth}{The \emph{transfer function}\index{transfer f... ..., and $X(z)$\ denotes the {\it z} transform of the filter input signal $x(n)$.}$

The transfer function is as well called the system function [sixty].

Let announce the impulse response of the filter. It turns out (as we volition evidence) that the transfer function is equal to the z transform of the impulse response :

$\displaystyle \zbox {H(z) = \frac{Y(z)}{X(z)}}$

Since multiplying the input transform by the transfer office gives the output transform , we see that embodies the transfer characteristics of the filter--hence the name.

It remains to define ``z transform'', and to prove that the z transform of the impulse response always gives the transfer function, which we will practise by proving the convolution theorem for z transforms.

The Z Transform

The bilateral z transform of the discrete-time signal is defined to be

$\displaystyle X(z) \isdefs \sum_{n=-\infty}^\infty x(n) z^{-n} \qquad\hbox{(bilateral {\it z} transform)} \protect$

(seven.1)

where is a complex variable. Since signals are typically defined to brainstorm (get nonzero) at time , and since filters are often causeless to be causal,^7.ane the lower summation limit given above may exist written as 0 rather than $-\infty$ to yield the unilateral z transform:

$\displaystyle X(z) \isdefs \sum_{n=0}^\infty x(n) z^{-n} \qquad\hbox{(unilateral {\it z} transform)}$

(7.2)

The unilateral z transform is most commonly used. For inverting z transforms, see §6.8.

Recall (§4.1) that the mathematical representation of a discrete-time signal maps each integer $n\in{\bf Z}$ to a complex number ( $x(n)\in{\bf C}$ ) or real number ( $x(n)\in{\bf R}$ ). The z transform of , on the other manus, , maps every complex number $z\in{\bf C}$ to a new complex number $X(z)\in{\bf C}$ . On a higher level, the z transform, viewed as a linear operator, maps an entire signal to its z transform . Nosotros call back of this as a ``office to function'' mapping. Nosotros may say is the z transform of past writing

$\displaystyle \zbox {X \leftrightarrow x}$

or, using operator notation,

$\displaystyle X(z) = {\cal Z}_z\{x(\cdot)\}$

which can exist abbreviated equally

$\displaystyle X = {\cal Z}\{x\}.$

1 also sees the convenient but maybe misleading annotation

$X(z) \leftrightarrow x(n)$ , in which and must be understood as standing for the entire domains

$n\in{\bf Z}$ and

$z\in{\bf C}$ , as opposed to denoting particular fixed values.

The z transform of a betoken tin exist regarded as a polynomial in $z^{-1}$ , with coefficients given past the signal samples. For example, the signal

$\displaystyle x(n) = \left\{\begin{array}{ll} n+1, & 0\leq n \leq 2 \\ [5pt] 0, & \mbox{otherwise} \\ \end{array}\right.$

has the z transform

$X(z) = 1 + 2z^{-1}+ 3z^{-2} = 1 + 2z^{-1}+ 3(z^{-1})^2$ .

Beingness of the Z Transform

The z transform of a finite-amplitude signal volition always be provided (one) the betoken starts at a finite fourth dimension and (2) it is asymptotically exponentially divisional, i.eastward., there exists a finite integer , and finite real numbers $A\geq 0$ and $\sigma$ , such that $\left\vert ten(due north)\right\vert<A\exp(\sigma n)$ for all $n\geq n_f$ . The bounding exponential may fifty-fifty exist growing with ( $\sigma>0$ ). These are non the most general conditions for existence of the z transform, but they suffice for about applied purposes.

For a signal growing every bit $\exp(\sigma n)$ , for $\sigma>0$ , one would naturally expect the z transform to be defined only in the region $\left\vert z\right\vert>\exp(\sigma)$ of the complex plane. This is expected because the infinite series

$\displaystyle \sum_{n=0}^\infty e^{\sigma n} z^{-n} = \sum_{n=0}^\infty \left(\frac{e^{\sigma}}{z}\right)^n$

requires

$\left\vert z\right\vert>\exp(\sigma)$ to ensure convergence. Since

$\sigma<0\,\Leftrightarrow\,\exp(\sigma)<1$ for a decaying exponential, nosotros see that the region of convergence of the transform of a decomposable exponential ever includes the unit circle of the plane.

More generally, it turns out that, in all cases of practical interest, the domain of tin be extended to include the entire complex plane, except at isolated ``singular'' points^7.ii at which $\vert X(z)\vert$ approaches infinity (such as at $z=\exp(\sigma)$ when $x(n)=\exp(\sigma n)$ ). The mathematical technique for doing this is called analytic continuation, and it is described in §D.i equally applied to the Laplace transform (the continuous-fourth dimension analogue of the z transform). A point to notation, however, is that in the extension region (all points such that $\left\vert z\right\vert<\exp(\sigma)$ in the above example), the bespeak component corresponding to each singularity inside the extension region is ``flipped'' in the time domain. That is, ``causal'' exponentials become ``anticausal'' exponentials, as discussed in §8.7.

The z transform is discussed more fully elsewhere [52,lx], and we will derive below only what we will demand.

Shift and Convolution Theorems

In this section, we bear witness the highly useful shift theorem and convolution theorem for unilateral z transforms. We consider the space of infinitely long, causal, complex sequences $x(n)\in{\bf C}$ , $n\in{\bf Z}$ , with for .

Shift Theorem

The shift theorem says that a filibuster of $\Delta$ samples in the time domain corresponds to a multiplication by $z^{-\Delta}$ in the frequency domain:

$\displaystyle {\cal Z}_z\{$ SouthHIFT $\displaystyle _\Delta\{x\}\} \;=\; z^{-\Delta} X(z), \; \Delta\ge 0,$

or, using more common note,

$\displaystyle \zbox {x(n-\Delta) \;\leftrightarrow\; z^{-\Delta} X(z), \; \Delta\ge 0.}$

Thus,

$x(\cdot - \Delta)$ , which is the waveform $x(\cdot)$ delayed by $\Delta$ samples, has the z transform

$z^{-\Delta}X(z)$ .

Proof:

$\begin{eqnarray*} {\cal Z}_z\{\mbox{{\sc Shift}}_\Delta\{x\}\} &\isdef & \sum_{n... ...ty}x(m) z^{-m} \\ &\isdef & z^{-\Delta} X(z), % \quad\pfendmath \end{eqnarray*}$

where we used the causality assumption for $ m<0$ .

Convolution Theorem

The convolution theorem for z transforms states that for any (existent or) complex causal signals and , convolution in the time domain is multiplication in the domain, i.e.,

$\displaystyle \zbox {x\ast y \;\leftrightarrow\; X\cdot Y}$

or, using operator note,

$\displaystyle {\cal Z}_z\{x \ast y\} \;=\; X(z)Y(z),$

where

$X(z)\isdef {\cal Z}_z(x)$ , and

$Y(z)\isdef {\cal Z}_z(y)$ . (See [84] for a evolution of the convolution theorem for discrete Fourier transforms.)

Proof:

$\begin{eqnarray*} {\cal Z}_z(x\ast y) &\isdef & \sum_{n=0}^{\infty}(x\ast y)_n z... ...(by the Shift Theorem)}\\ &\isdef & X(z)Y(z) % \quad\pfendmath \end{eqnarray*}$

The convolution theorem provides a major cornerstone of linear systems theory. It implies, for example, that any stable causal LTI filter (recursive or nonrecursive) can be implemented by convolving the input signal with the impulse response of the filter, as shown in the next section.

Z Transform of Convolution

From Eq. $\,$ (5.five), we take that the output from a linear time-invariant filter with input and impulse response is given past the convolution of and , i.e.,

$\displaystyle y(n) \eqsp (h \ast x)(n) \protect$

(7.3)

where `` $\ast$ '' ways convolution equally before. Taking the z transform of both sides of Eq. $\,$ (6.3) and applying the convolution theorem from the preceding department gives

$\displaystyle Y(z) \eqsp H(z)X(z) \protect$

(7.4)

where H(z) is the z transform of the filter impulse response. We may divide Eq. $\,$ (half dozen.4) by to obtain

$\displaystyle H(z) \eqsp \frac{Y(z)}{X(z)} \;\isdef \; \hbox{transfer function}.$

This shows that, as a directly result of the convolution theorem, the z transform of an impulse response is equal to the transfer role

of the filter, provided the filter is linear and time invariant.

Z Transform of Deviation Equations

Since z transforming the convolution representation for digital filters was so fruitful, let'south apply it now to the full general difference equation, Eq. $\,$ (5.1). To do this requires two properties of the z transform, linearity (easy to bear witness) and the shift theorem (derived in §6.three above). Using these two properties, we can write downwards the z transform of any difference equation by inspection, as we now evidence. In §half dozen.8.ii, we'll evidence how to invert by inspection as well.

Repeating the general difference equation for LTI filters, nosotros have (from Eq. $\,$ (5.ane)) $\begin{eqnarrayda} y(n) &=& b_0 x(n) &+& b_1 x(n - 1) + \cdots + b_M x(n - M)\\ & & &-& a_1 y(n - 1) - \cdots - a_N y(n - N).\\ \end{eqnarrayda}$

Allow's accept the z transform of both sides, denoting the transform by ${\cal Z}\{\}$ . Because ${\cal Z}\{\}$ is a linear operator, it may be distributed through the terms on the correct-hand side every bit follows:^7.3 $\begin{eqnarrayda} {\cal Z}_z\{y(\cdot)\} &=& {\cal Z}\{ b_0 x(n) &+& b_1 x(n ... ...-M} X(z)\\ & & &-& a_1 z^{-1}Y(z) - \cdots - a_N z^{-N} Y(z), \end{eqnarrayda}$ where we used the superposition and scaling properties of linearity given on page [*] , followed by use of the shift theorem, in that order. The terms in may be grouped together on the left-paw side to get

$\begin{eqnarray*} \lefteqn{Y(z) + a_1 z^{-1}Y(z) + \cdots + a_N z^{-N} Y(z) = }\... ... \\ & & b_0 X(z) + b_1 z^{-1}X(z) + \cdots + b_M z^{-M} X(z). \end{eqnarray*}$

Factoring out the mutual terms and gives

$\displaystyle Y(z)\left[1 + a_1 z^{-1}+ \cdots + a_N z^{-N}\right] = X(z)\left[b_0 + b_1 z^{-1}+ \cdots + b_M z^{-M}\right].$

Defining the polynomials

$\begin{eqnarray*} A(z) &\isdef & 1 + a_1\,z^{-1} + \,\cdots\, + a_N\,z^{-N}\\ B(z) &\isdef & b_0 + b_1\,z^{-1}+\,\cdots\,+b_M\,z^{-M}, \end{eqnarray*}$

the z transform of the departure equation yields

$\displaystyle A(z)\,Y(z) = B(z)\,X(z).$

Finally, solving for , which is by definition the transfer function , gives

$\displaystyle H(z) \isdefs \frac{Y(z)}{X(z)} \eqsp \frac{b_0 + b_1 z^{-1}+ \cdo... ...^{-M}}{1 + a_1 z^{-1}+ \cdots + a_N z^{-N}} \isdefs \frac{B(z)}{A(z)}. \protect$

(7.five)

Thus, taking the z transform of the general difference equation led to a new formula for the transfer office in terms of the difference equation coefficients. (Now the minus signs for the feedback coefficients in the departure equation Eq. $\,$ (v.1) are explained.)

Factored Grade

Past the cardinal theorem of algebra, every th order polynomial can be factored into a product of first-order polynomials. Therefore, Eq. $\,$ (6.5) higher up can be written in factored form as

$\displaystyle H(z) = b_0\frac{(1-q_1z^{-1})(1-q_2z^{-1})\cdots(1-q_Mz^{-1})}{(1-p_1z^{-1})(1-p_2z^{-1})\cdots(1-p_Nz^{-1})}. \protect$

(7.half-dozen)

The numerator roots

$\{q_1,\ldots,q_M\}$ are chosen the zeros of the transfer function, and the denominator roots

$\{p_1, \ldots, p_N\}$ are called the poles of the filter. Poles and zeros are discussed further in Chapter viii.

Series and Parallel Transfer Functions

The transfer role conveniently captures the algebraic structure of a filtering operation with respect to series or parallel combination. Specifically, nosotros have the following cases:

Transfer functions of filters in series multiply together.
Transfer functions of filters in parallel sum together.

Series Case

Figure 6.one illustrates the series connexion of two filters and . The output from filter 1 is used equally the input to filter two. Therefore, the overall transfer function is

$\displaystyle H(z) \isdefs \frac{Y(z)}{X(z)} \eqsp \frac{H_2(z)V(z)}{X(z)} \eqsp H_2(z)H_1(z).$

In summary, if the output of filter is given as input to filter (a series combination), as shown in Fig.six.1, the overall transfer function is

--transfer functions of filters connected in series multiply together.

Parallel Case

Figure vi.2 illustrates the parallel combination of two filters. The filters and are driven by the same input indicate , and their respective outputs and are summed. The transfer part of the parallel combination is therefore

$\displaystyle H(z) \isdefs \frac{Y(z)}{X(z)} \eqsp \frac{Y_1(z) + Y_2(z)}{X(z)} \eqsp \frac{Y_1(z)}{X(z)} + \frac{Y_2(z)}{X(z)} \isdefs H_1(z)+H_2(z).$

where we needed merely linearity of the z transform to have that

${\cal Z}\{y_1+y_2\} = {\cal Z}\{y_1\}+{\cal Z}\{y_2\}$ .

Series Combination is Commutative

Since multiplication of circuitous numbers is commutative, nosotros have

$\displaystyle H_1(z)H_2(z)=H_2(z)H_1(z),$

which implies that any ordering of filters in serial results in the same overall transfer part. Note, nonetheless, that the numerical operation of the overall filter is usually affected by the ordering of filter stages in a serial combination [103]. Chapter 9 further considers numerical performance of filter implementation structures.

By the convolution theorem for z transforms, commutativity of a product of transfer functions implies that convolution is commutative:

$\displaystyle h_1 \ast h_2 \;\leftrightarrow\; H_1\cdot H_2 \;=\; H_2\cdot H_1 \;\leftrightarrow\; h_2 \ast h_1$

Fractional Fraction Expansion

An important tool for inverting the z transform and converting among digital filter implementation structures is the partial fraction expansion (PFE). The term ``partial fraction expansion'' refers to the expansion of a rational transfer function into a sum of kickoff and/or 2nd-order terms. The instance of first-club terms is the simplest and nigh fundamental:

$\displaystyle H(z) \isdefs \frac{B(z)}{A(z)} \eqsp \sum_{i=1}^{N} \frac{r_i}{1-p_iz^{-1}} \protect$

(7.7)

where

$\begin{eqnarray*} B(z) &=& b_0 + b_1 z^{-1}+ b_2z^{-2}+ \cdots + b_M z^{-M}\\ A(z) &=& 1 + a_1 z^{-1}+ a_2z^{-2}+ \cdots + a_N z^{-N} \end{eqnarray*}$

and . (The instance $M\geq N$ is addressed in the adjacent section.) The denominator coefficients are called the poles of the transfer function, and each numerator is called the residue of pole . Equation (6.7) is full general just if the poles are distinct. (Repeated poles are addressed in §half dozen.eight.v below.) Both the poles and their residues may be complex. The poles may be found by factoring the polynomial into first-lodge terms,^vii.4 e.g., using the roots function in matlab. The residue respective to pole may exist establish analytically as

$\displaystyle r_i \eqsp \left.(1-p_iz^{-1})H(z)\right\vert _{z=p_i} \protect$

(7.8)

when the poles are distinct. The matlab function

residuez

^7.5 will find poles and residues computationally, given the difference-equation (transfer-part) coefficients.

Note that in Eq. $\,$ (6.8), at that place is always a pole-zero cancellation at . That is, the term $(1-p_iz^{-1})$ is always cancelled past an identical term in the denominator of , which must exist because has a pole at . The rest is simply the coefficient of the ane-pole term $1/(1-p_i z^{-1})$ in the partial fraction expansion of at . The transfer function is $r_i/(1-p_i z^{-1})$ , in the limit, as $z\to p_i$ .

Example

Consider the two-pole filter

$\displaystyle H(z) \eqsp \frac{1}{(1-z^{-1})(1-0.5z^{-1})}.$

The poles are and . The corresponding residues are then

$\begin{eqnarray*} r_1 &=& \left.(1-z^{-1})H(z)\right\vert _{z=1} \eqsp \left.\f... ...\ &=& \left.\frac{1}{1-z^{-1}}\right\vert _{z=0.5} \eqsp -1\,. \end{eqnarray*}$

We thus conclude that

$\displaystyle H(z) \eqsp \frac{2}{1-z^{-1}} - \frac{1}{1-0.5z^{-1}}.$

As a bank check, we can add the two one-pole terms to a higher place to get

$\displaystyle \frac{2}{1-z^{-1}} - \frac{1}{1-0.5z^{-1}} \eqsp \frac{2-z^{-1}- 1 + z^{-1}}{(1-z^{-1})(1-0.5z^{-1})} \eqsp \frac{1}{(1-z^{-1})(1-0.5z^{-1})}$

as expected.

Complex Example

To illustrate an example involving complex poles, consider the filter

$\displaystyle H(z) \eqsp \frac{g}{1+z^{-2}},$

where can exist any real or complex value. (When is real, the filter as a whole is real also.) The poles are then and (or vice versa), and the factored class can be written as

$\displaystyle H(z) \eqsp \frac{g}{(1-jz^{-1})(1+jz^{-1})}.$

Using Eq. $\,$ (6.8), the residues are institute to be

$\begin{eqnarray*} r_1 &=& \left.(1-jz^{-1})H(z)\right\vert _{z=j} \eqsp \left.\... ...eft.\frac{g}{1-jz^{-1}}\right\vert _{z=-j} \eqsp \frac{g}{2}\,. \end{eqnarray*}$

Thus,

$\displaystyle H(z) \eqsp \frac{g/2}{1-jz^{-1}} + \frac{g/2}{1+jz^{-1}}.$

A more elaborate case of a partial fraction expansion into complex i-pole sections is given in §3.12.1.

PFE to Real, 2d-Order Sections

When all coefficients of and are existent (implying that is the transfer function of a existent filter), it will e'er happen that the complex i-pole filters will occur in complex conjugate pairs. Let denote any one-pole section in the PFE of Eq. $\,$ (6.7). Then if is circuitous and describes a real filter, we will also detect $(\overline{r},\pc)$ somewhere amidst the terms in the i-pole expansion. These 2 terms tin can be paired to form a real second-lodge section as follows:

$\begin{eqnarray*} H(z) &=& \frac{r}{1-pz^{-1}} + \frac{\overline{r}}{1-\pc z^{-1... ...box{re}\left\{p\right\}z^{-1}+ \left\vert p\right\vert^2 z^{-2}} \end{eqnarray*}$

Expressing the pole in polar form as $p=Re^{j\theta}$ , and the remainder as $r=Ge^{j\phi}$ , the last expression above can be rewritten as

$\displaystyle H(z) \eqsp 2G\frac{\cos(\phi)-\cos(\phi-\theta)z^{-1}}{1-2R\,\cos(\theta)z^{-1}+ R^2 z^{-2}}.$

The utilise of polar-grade coefficients is discussed further in the section on two-pole filters (§B.1.3).

Expanding a transfer part into a sum of second-order terms with real coefficients gives united states the filter coefficients for a parallel banking company of real second-gild filter sections. (Of course, each real pole can be implemented in its own real ane-pole section in parallel with the other sections.) In view of the foregoing, we may conclude that every existent filter with $ Yard<N$ can exist implemented as a parallel bank of biquads.^7.6 However, the full generality of a biquad section (two poles and ii zeros) is not needed considering the PFE requires only one zero per second-order term.

To come across why we must stipulate $ M<N$ in Eq. $\,$ (6.vii), consider the sum of two first-order terms by straight calculation:

$\displaystyle H_2(z) \eqsp \frac{r_1}{1-p_1z^{-1}} + \frac{r_2}{1-p_2z^{-1}} \eqsp \frac{(r_1 + r_2) - (r_1 p_2 + r_2 p_1) z^{-1}}{(1-p_1z^{-1})(1-p_2z^{-1})}$

(7.9)

Notice that the numerator order, viewed as a polynomial in $z^{-1}$ , is one less than the denominator order. In the same way, it is easily shown by mathematical consecration that the sum of one-pole terms

$r_i/(1-p_i z^{-1})$ tin can produce a numerator social club of at about (while the denominator order is if at that place are no pole-goose egg cancellations). Following terminology used for analog filters, nosotros telephone call the case a strictly proper transfer office.^7.7 Thus, every strictly proper transfer function (with distinct poles) can be implemented using a parallel banking concern of 2-pole, one-zippo filter sections.

Inverting the Z Transform

The fractional fraction expansion (PFE) provides a simple ways for inverting the z transform of rational transfer functions. The PFE provides a sum of first-order terms of the form

$\displaystyle H_i(z) \eqsp \frac{r_i}{1-p_iz^{-1}}.$

It is easily verified that such a term is the z transform of

$\displaystyle h_i(n) \eqsp r_i p_i^n, \quad n=0,1,2,\ldots\,.$

Thus, the inverse z transform of is but

$\displaystyle h(n) \eqsp \sum_{i=1}^N h_i(n) \eqsp \sum_{i=1}^N r_i p_i^n, \quad n=0,1,2,\ldots\,.$

Thus, the impulse response of every strictly proper LTI filter (with distinct poles) can be interpreted as a linear combination of sampled complex exponentials. Think that a uniformly sampled exponential is the same thing every bit a geometric sequence. Thus, is a linear combination of geometric sequences. The term ratio of the th geometric sequence is the th pole, , and the coefficient of the thursday sequence is the th residue, .

In the improper instance, discussed in the next department, we additionally obtain an FIR office in the z transform to be inverted:

$\displaystyle F(z) \eqsp f_0 + f_1z^{-1}+ f_2z^{-2}+ \cdots + f_K z^{-K} \;\longleftrightarrow\; [f_0,f_1,\ldots,f_K,0,0,\ldots].$

The FIR role (a finite-guild polynomial in $z^{-1}$ ) is as well easily inverted by inspection.

The case of repeated poles is addressed in §six.viii.v below.

FIR Part of a PFE

When $M\geq N$ in Eq. $\,$ (6.7), nosotros may perform a stride of long division of to produce an FIR part in parallel with a strictly proper IIR office:

$\displaystyle H(z) \isdefs \frac{B(z)}{A(z)} \eqsp F(z) + \sum_{i=1}^{N} \frac{r_i}{1-p_iz^{-1}} \protect$

(7.10)

where

$\begin{eqnarray*} B(z) &=& b_0 + b_1 z^{-1}+ b_2z^{-2}+ \cdots + b_M z^{-M}\\ A... ...=& f_0 + f_1z^{-1}+ f_2z^{-2}+ \cdots + f_K z^{-K}, \quad K=M-N. \end{eqnarray*}$

When $ K<N$ , we ascertain . This type of decomposition is computed by the residuez function (a matlab part for computing a complete partial fraction expansion, as illustrated in §6.8.8 beneath).

An alternate FIR function is obtained past performing long division on the reversed polynomial coefficients to obtain

$\displaystyle H(z) \eqsp F(z) + z^{-(K+1)}\sum_{i=1}^{N} \frac{r_i}{1-p_iz^{-1}}, \protect$

(vii.eleven)

where

$K=M-N\geq 0$ is again the order of the FIR part. This blazon of decomposition is computed (as function of the PFE) past

residued

, described in §J.vi and illustrated numerically in §half dozen.eight.8 below.

We may compare these two PFE alternatives equally follows: Permit denote , $F_K\isdeftext F(z)$ , and $B_M\isdeftext B(z)$ . (I.e., we use a subscript to betoken polynomial society, and `' is omitted for notational simplicity.) And then for $K=M-N\geq 0$ we have ii cases:

$\begin{eqnarray*} (1) && H(z) \eqsp F_K + \frac{B^\prime_{N-1}}{A_N} \eqsp \frac... ..._N} \eqsp \frac{F_K A_N + z^{-(K+1)}B^{\prime\prime}_{N-1}}{A_N} \end{eqnarray*}$

In the showtime form, the $B^\prime_{N-1}$ coefficients are ``left justified'' in the reconstructed numerator, while in the second form they are ``correct justified''. The second class is generally more than efficient for modeling purposes, since the numerator of the IIR role ( $B^{\prime\prime}_{N-1}(z)$ ) tin can be used to match boosted terms in the impulse response after the FIR part has ``died out''.

In summary, an arbitrary digital filter transfer part with singled-out poles tin always be expressed as a parallel combination of complex i-pole filters, together with a parallel FIR part when $M\geq N$ . When there is an FIR part, the strictly proper IIR part may be delayed such that its impulse response begins where that of the FIR office leaves off.

In artificial reverberation applications, the FIR function may correspond to the early reflections, while the IIR office provides the late reverb, which is typically dense, smooth, and exponentially decaying [86]. The predelay (``pre-delay'') command in some commercial reverberators is the amount of pure delay at the beginning of the reverberator'due south impulse response. Thus, neglecting the early reflections, the guild of the FIR part tin be viewed as the corporeality of predelay for the IIR function.

Example: The General Biquad PFE

The general 2d-order case with (the so-called biquad section) can exist written when $b_0\ne 0$ as

$\displaystyle H(z) \eqsp g\frac{1 + b_1 z^{-1}+ b_2 z^{-2}}{1 + a_1 z^{-1}+ a_2 z^{-2}}.$

To perform a partial fraction expansion, nosotros demand to extract an social club 0 (length 1) FIR part via long division. Let $d=z^{-1}$ and rewrite equally a ratio of polynomials in :

$\displaystyle H(d^{-1}) \eqsp g\frac{b_2 d^2 + b_1 d + 1 }{a_2 d^2 + a_1 d + 1}$

So long division gives $% For typesetting long division --- NEEDED WITHIN THE MAKEIMAGE ENV? % (raw TeX,... ...} & & b_1-\frac{b_2}{a_2}a_1 & 1-\frac{b_2}{a_2} & \end{array}\end{displaymath}$
yielding

$\displaystyle H(d^{-1}) \eqsp g\frac{b_2}{a_2} + g\frac{\left(b_1-\frac{b_2}{a_2}a_1\right)d+ \left(1-\frac{b_2}{a_2}\right)}{a_2d^2 + a_1d + 1}$

$\displaystyle H(z) \eqsp g\frac{b_2}{a_2} + g\frac{\left(1-\frac{b_2}{a_2}\right) +\left(b_1-\frac{b_2}{a_2}a_1\right)z^{-1}}{1 + a_1z^{-1}+ a_2z^{-2}}.$

The delayed grade of the partial fraction expansion is obtained by leaving the coefficients in their original social club. This corresponds to writing as a ratio of polynomials in :

$\displaystyle H(z) \eqsp g\frac{z^2 + b_1 z + b_2 }{z^2 + a_1 z + a_2}$

Long sectionalization now looks like $% For typesetting long division --- NEEDED WITHIN THE MAKEIMAGE ENV?\begin{dis... ...rule width 22\digitwidth}} & & b_1-a_1 & b_2-a_2 & \end{array}\end{displaymath}$
giving

$\displaystyle H(z) \eqsp g + z^{-1}g\frac{(b_1-a_1) + (b_2-a_2)z^{-1}}{1 + a_1 z^{-1}+ a_2 z^{-2}}.$

Numerical examples of partial fraction expansions are given in §6.8.eight beneath. Another worked instance, in which the filter is converted to a set of parallel, second-order sections is given in §3.12. See also §nine.2 regarding conversion to second-club sections in general, and §G.9.1 (specially Eq. $\,$ (Thou.22)) regarding a state-space approach to partial fraction expansion.

Alternate PFE Methods

Some other method for finding the pole residues is to write down the general form of the PFE, obtain a common denominator, expand the numerator terms to obtain a unmarried polynomial, and equate like powers of $z^{-1}$ . This gives a linear system of equations in unknowns , $i=1,\ldots,N$ .

However another method for finding residues is past ways of Taylor serial expansions of the numerator and denominator about each pole , using fifty'Hôpital's rule..

Finally, 1 tin can alternatively construct a country infinite realization of a strictly proper transfer function (using, e.g., tf2ss in matlab) and then diagonalize it via a similarity transformation. (See Appendix K for an introduction to land-space models and diagonalizing them via similarity transformations.) The transfer function of the diagonalized land-space model is trivially obtained every bit a sum of one-pole terms--i.e., the PFE. In other words, diagonalizing a state-infinite filter realization implicitly performs a partial fraction expansion of the filter's transfer office. When the poles are singled-out, the country-infinite model tin be diagonalized; when there are repeated poles, it can be block-diagonalized instead, as discussed further in §Grand.10.

Repeated Poles

When poles are repeated, an interesting new phenomenon emerges. To see what's going on, let'south consider two identical poles arranged in parallel and in serial. In the parallel instance, we have

$\displaystyle H_1(z) \eqsp \frac{r_1}{1-pz^{-1}} + \frac{r_2}{1-pz^{-1}} \eqsp \frac{r_1+r_2}{1-pz^{-1}} \isdefs \frac{r_3}{1-pz^{-1}}.$

In the series example, we get

$\displaystyle H_2(z) \eqsp \frac{r_1}{1-pz^{-1}} \cdot \frac{r_2}{1-pz^{-1}} \eqsp \frac{r_1r_2}{(1-pz^{-1})^2} \isdefs \frac{r_3}{(1-pz^{-1})^2}.$

Thus, two 1-pole filters in parallel are equivalent to a new ane-pole filter^7.8 (when the poles are identical), while the same two filters in serial give a ii-pole filter with a repeated pole. To accommodate both possibilities, the general partial fraction expansion must include the terms

$\displaystyle \frac{r_{1,1}}{(1-pz^{-1})^2} + \frac{r_{1,2}}{(1-pz^{-1})}$

for a pole having multiplicity two.

Dealing with Repeated Poles Analytically

A pole of multiplicity has residues associated with information technology. For instance,

and the three residues associated with the pole are 1, 2, and 4.

Permit $r_{ij}$ denote the th residual associated with the pole , $j=1,\ldots,m_i$ . Successively differentiating $(1-p_iz^{-1})^{m_i}H(z)$ times with respect to $z^{-1}$ and setting isolates the rest $r_{ik}$ :

$\begin{eqnarray*} r_{i1} &=& \left.(1-p_iz^{-1})^{m_i}H(z)\right\vert _{z=p_i}\\... ...ac{d^3}{d(z^{-1})^3} (1-p_iz^{-1})^{m_i}H(z)\right\vert _{z=p_i} \end{eqnarray*}$

$\displaystyle \zbox {r_{ik} = \left.\frac{1}{(k-1)!(-p_i)^{k-1}}\frac{d^{k-1}}{d(z^{-1})^{k-1}} (1-p_iz^{-1})^{m_i}H(z)\right\vert _{z=p_i}}$

Instance

For the instance of Eq. $\,$ (6.12), we obtain

$\begin{eqnarray*} r_{11} &=& \left.\left(1-\frac{1}{2}z^{-1}\right)^3H(z)\right\... ...}{dz^{-1}} (-5 + 2z^{-1})\right\vert _{z^{-1}=2} = 2\cdot 2 = 4. \end{eqnarray*}$

Impulse Response of Repeated Poles

In the fourth dimension domain, repeated poles give rise to polynomial aamplitude envelopes on the decaying exponentials corresponding to the (stable) poles. For instance, in the case of a single pole repeated twice, nosotros have

$\displaystyle \zbox {\frac{1}{\left(1-pz^{-1}\right)^2} \;\longleftrightarrow\; (n+1) p^n, \quad n=0,1,2,\ldots\,.}$

Proof: First notation that

$\displaystyle \frac{d}{dz^{-1}}\left(\frac{1}{1-pz^{-1}}\right) = (-1)(1-pz^{-1})^{-2}(-p) = \frac{p}{\left(1-pz^{-1}\right)^2}\;.$

Therefore,

Note that is a showtime-order polynomial in . Similarly, a pole repeated three times corresponds to an impulse-response component that is an exponential decay multiplied by a quadratic polynomial in , and and so on. As long every bit $\vert p\vert<1$ , the impulse response will somewhen decay to nil, because exponential decay ever overtakes polynomial growth in the limit equally goes to infinity.

And so What's Up with Repeated Poles?

In the previous section, we found that repeated poles give rise to polynomial amplitude-envelopes multiplying the exponential decay due to the pole. On the other hand, 2 different poles tin just yield a convolution (or sum) of ii different exponential decays, with no polynomial envelope allowed. This is truthful no matter how closely the poles come together; the polynomial envelope tin occur only when the poles merge exactly. This might violate one's intuitive expectation of a continuous modify when passing from two closely spaced poles to a repeated pole.

To written report this miracle further, consider the convolution of 2 one-pole impulse-responses and :

$\displaystyle h(n) \isdef (h_1\ast h_2)(n) = \sum_{m=0}^n h_1(m)h_2(n-m) = \sum... ...^n p_1^{m}p_2^{n-m} = p_2^n\sum_{m=0}^n \left(\frac{p_1}{p_2}\right)^m \protect$

(seven.fourteen)

The finite limits on the summation upshot from the fact that both and are causal. Remember the closed-form sum of a truncated geometric series:

$\displaystyle \sum_{m=0}^n r^m = \frac{1-r^{n+1}}{1-r}$

Applying this to Eq. $\,$ (half dozen.14) yields

$\displaystyle h(n) = p_2^n \frac{1-(p_1/p_2)^{n+1}}{1-(p_1/p_2)} = \frac{p_2^{n+1}-p_1^{n+1}}{p_2-p_1} = \frac{p_1^{n+1}-p_2^{n+1}}{p_1-p_2}.$

Annotation that the result is symmetric in and . If

$\left\vert p_1\right\vert>\left\vert p_2\right\vert$ , and so becomes proportional to for large , while if

$\left\vert p_2\right\vert>\left\vert p_1\right\vert$ , it becomes instead proportional to .

Going back to Eq. $\,$ (6.xiv), nosotros have

$\displaystyle h(n) = p_2^n\sum_{m=0}^n \left(\frac{p_1}{p_2}\right)^m = p_1^n\sum_{m=0}^n \left(\frac{p_2}{p_1}\right)^m.$

(7.fifteen)

Setting yields

$\displaystyle h(n) = (n+1)p^n$

(7.16)

which is the first-lodge polynomial amplitude-envelope example for a repeated pole. Nosotros can come across that the transition from ``two convolved exponentials'' to ``single exponential with a polynomial amplitude envelope'' is perfectly continuous, equally we would expect.

We besides see that the polynomial amplitude-envelopes fundamentally arise from iterated convolutions. This corresponds to the repeated poles existence arranged in series, rather than in parallel. The simplest case is when the repeated pole is at , in which case its impulse response is a constant:

$\displaystyle \frac{1}{1-z^{-1}} \eqsp 1 + z^{-1}+ z^{-2}+ \cdots \;\longleftrightarrow\; [1,1,1,\ldots]$

The convolution of a constant with itself is a ramp:

$\displaystyle h_1(n)\eqsp \sum_{m=0}^n 1\cdot 1 \eqsp n+1$

The convolution of a constant and a ramp is a quadratic, and so on:^seven.nine

$\begin{eqnarray*} h_2(n)&=&\sum_{m=0}^n (m+1)\cdot 1 \eqsp \frac{(n+1)(n+2)}{2}\... ...+1)(m+2)}{2}\cdot 1\eqsp \frac{(n+1)(n+2)(n+3)}{3!}\\ &\cdots& \end{eqnarray*}$

Alternate Stability Criterion

In §5.half-dozen (folio [*] ), a filter was divers to be stable if its impulse response decays to 0 in magnitude as time goes to infinity. In §6.8.five, nosotros saw that the impulse response of every finite-guild LTI filter can be expressed as a possible FIR role (which is always stable) plus a linear combination of terms of the course , where is some finite-order polynomial in , and is the thursday pole of the filter. In this form, it is clear that the impulse response always decays to nil when each pole is strictly inside the unit circle of the plane, i.e., when $\vert p_i\vert<1$ . Thus, having all poles strictly inside the unit circle is a sufficient criterion for filter stability. If the filter is observable (meaning that in that location are no pole-nil cancellations in the transfer office from input to output), so this is also a necessary criterion.

A transfer part with no pole-cypher cancellations is said to be irreducible. For case, $H(z) = (1+z^{-1})/(1-z^{-1})$ is irreducible, while $H(z) = (1-z^{-2})/(1-2z^{-2}+z^{-2})$ is reducible, since at that place is the common gene of $(1-z^{-1})$ in the numerator and denominator. Using this terminology, nosotros may state the following stability criterion:

$\textstyle \parbox{0.8\textwidth}{\emph{An irreducible transfer function $H(z)$\ is stable if and only if its poles have magnitude less than one.}}$

This characterization of stability is pursued further in §eight.iv, and notwithstanding another stability test (most often used in practice) is given in §8.four.one.

Summary of the Partial Fraction Expansion

In summary, the partial fraction expansion can be used to expand any rational z transform

$\displaystyle H(z) \eqsp \frac{B(z)}{A(z)} \eqsp \frac{b_0 + b_1 z^{-1}+ \cdots + b_M z^{-M}}{1 + a_1 z^{-1}+ \cdots + a_N z^{-N}}$

as a sum of first-order terms

$\displaystyle H(z) \isdefs \frac{B(z)}{A(z)} \eqsp \sum_{i=1}^{N} \frac{r_i}{1-p_iz^{-1}} \protect$

(vii.17)

for $ M<N$ , and

$\displaystyle H(z) \eqsp F(z) + z^{-(K+1)}\sum_{i=1}^{N}\frac{r_i}{1-p_iz^{-1}} \protect$

(vii.18)

for $M\geq N$ , where the term

$z^{-(K+1)}$ is optional, but frequently preferred. For existent filters, the complex i-pole terms may be paired up to obtain 2d-social club terms with real coefficients. The PFE procedure occurs in 2 or three steps:

When $M\geq N$ , perform a step of long segmentation to obtain an FIR part and a strictly proper IIR part $B^\prime(z)/A(z)$ .
Find the poles , $i=1,\ldots,N$ (roots of ).
If the poles are distinct, find the residues , $i=1,\ldots,N$ from
$\displaystyle r_i = \left.(1-p_iz^{-1})\frac{B(z)}{A(z)}\right\vert _{z=p_i}$
If there are repeated poles, find the boosted residues via the method of §half dozen.viii.five, and the general form of the PFE is

$\displaystyle H(z) \eqsp F(z) + z^{-(K+1)}\sum_{i=1}^{N_p}\sum_{k=1}^{m_i}\frac{r_{i,k}}{(1-p_iz^{-1})^k} \protect$ (7.nineteen)

where denotes the number of distinct poles, and $m_i\ge 1$ denotes the multiplicity of the thursday pole.

In pace two, the poles are typically found by factoring the denominator polynomial . This is a unsafe pace numerically which may fail when in that location are many poles, specially when many poles are clustered close together in the plane.

The post-obit matlab code illustrates factoring $A(z) = 1 - z^{-3}$ to obtain the three roots, $p_k=e^{jk2\pi/3}$ , :

A = [one 0 0 -1];  % Filter denominator polynomial poles = roots(A) % Filter poles

See Chapter ix for additional discussion regarding digital filters implemented every bit parallel sections (especially §9.2.2).

Software for Partial Fraction Expansion

Figure 6.three illustrates the use of residuez (§J.v) for performing a partial fraction expansion on the transfer part

$\displaystyle H(z) \eqsp \frac{1 + 0.5^3 z^{-3}}{1 + 0.9^5z^{-5}}$

The complex-conjugate terms can be combined to obtain 2 real second-lodge sections, giving a total of one real beginning-order section in parallel with 2 real second-order sections, every bit discussed and depicted in §three.12.

**Figure 6.3:** Use of `residuez` to perform a fractional fraction expansion of an IIR filter transfer function .
B = [1 0 0 0.125]; A = [ane 0 0 0 0 0.9^5]; [r,p,f] = residuez(B,A) % r = % 0.16571 % 0.22774 - 0.02016i % 0.22774 + 0.02016i % 0.18940 + 0.03262i % 0.18940 - 0.03262i % % p = % -0.90000 % -0.27812 - 0.85595i % -0.27812 + 0.85595i % 0.72812 - 0.52901i % 0.72812 + 0.52901i % % f = [](0x0)

B = [1 0 0 0.125]; A = [ane 0 0 0 0 0.9^5]; [r,p,f] = residuez(B,A) % r = %   0.16571 %   0.22774 - 0.02016i %   0.22774 + 0.02016i %   0.18940 + 0.03262i %   0.18940 - 0.03262i % % p = %   -0.90000 %   -0.27812 - 0.85595i %   -0.27812 + 0.85595i %    0.72812 - 0.52901i %    0.72812 + 0.52901i % % f = [](0x0)

Case ii

For the filter

we obtain the output of

residued

(§J.6) shown in Fig.6.4. In dissimilarity to

residuez

residued

delays the IIR part until after the FIR function. In dissimilarity to this consequence,

residuez

returns

r=[-24;xvi]

and

f=[10;2]

, respective to the PFE

$\displaystyle H(z) = 10+2z^{-1}-\frac{24}{1-z^{-1}} + \frac{16}{(1-z^{-1})^2},$

(7.22)

in which the FIR and IIR parts accept overlapping impulse responses.

Encounter Sections J.five and J.half-dozen starting on folio [*] for listings of residuez, residued and related word.

Polynomial Multiplication in Matlab

The matlab role conv (convolution) can be used to perform polynomial multiplication. For example:

B1 = [1 ane];   % 1st row of Pascal's triangle B2 = [1 2 1]; % 2nd row of Pascal's triangle B3 = conv(B1,B2) % 3rd row % B3 = 1  3  iii  1 B4 = conv(B1,B3) % 4th row % B4 = 1  4  half dozen  four  1 % ...

The matlab

conv(B1,B2)

is identical to

filter(B1,one,B2)

, except that

conv

returns the complete convolution of its two input vectors, while

filter

truncates the result to the length of the ``input signal''

B2

.^7.10 Thus, if

B2

is nada-padded with

length(B1)-1

zeros, it will render the complete convolution:

B1 = [1 ii 3]; B2 = [4 5 6 7]; conv(B1,B2) % ans = iv  13  28  34  32  21 filter(B1,1,B2) % ans = 4  13  28  34 filter(B1,1,[B2,zeros(one,length(B1)-1)]) % ans = 4  thirteen  28  34  32  21

Polynomial Division in Matlab

The matlab function deconv (deconvolution) can be used to perform polynomial long division in order to dissever an improper transfer function into its FIR and strictly proper parts:

B = [ 2 half dozen 6 2]; % 2*(i+i/z)^iii A = [ 1 -ii i];  % (ane-ane/z)^2 [firpart,rest] = deconv(B,A) % firpart = %   two  10 % remainder = %    0    0   24   -8

Thus, this example finds that is as written in Eq. $\,$ (vi.21). This outcome can be checked past obtaining a common denominator in social club to recalculate the straight-class numerator:

Bh = residuum + conv(firpart,A) %  = two six 6 ii

The operation deconv(B,A) can exist implemented using filter in a manner analogous to the polynomial multiplication example (see §6.viii.8 higher up):

firpart = filter(B,A,[1,zeros(1,length(B)-length(A))]) %       = 2 ten remainder = B - conv(firpart,A) %         =  0 0 24 -eight

That this must work can exist seen by looking at Eq. $\,$ (six.21) and noting that the impulse-response of the residual (the strictly proper part) does not begin until time , then that the offset two samples of the impulse-response come simply from the FIR part.

In summary, we may conveniently utilise convolution and deconvolution to perform polynomial multiplication and sectionalisation, respectively, such as when converting transfer functions to various alternate forms.

When carrying out a fractional fraction expansion on a transfer part having a numerator club which equals or exceeds the denominator order, a necessary preliminary step is to perform long division to obtain an FIR filter in parallel with a strictly proper transfer function. This section describes how an FIR function of whatever length can exist extracted from an IIR filter, and this can exist used for PFEs too every bit for more avant-garde applications [].

Issues

Come across http://ccrma.stanford.edu/~jos/filtersp/Transfer_Function_Analysis_Problems.html.

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