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

The transfer function is also called the
system function
[
60].
Let

denote the
impulse response of the filter. It turns
out (as we will show) that
the transfer function is equal to the
z transform of the impulse response 
:
Since multiplying the input transform

by the transfer function

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 do
by proving the
convolution theorem for
z transforms.
The Z Transform
The
bilateral z transform of the discrete-time
signal 
is
defined to be
 |
(7.1) |
where

is a
complex variable. Since signals are typically defined
to begin (become nonzero) at time

, and since
filters are often
assumed to be
causal,
7.1 the
lower summation limit given above may be written as 0 rather than

to yield the
unilateral 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

to a
complex
number (

) or
real number (

). The
z transform
of

, on the other hand,

, maps every complex number

to a new complex number

. On a higher
level, the
z transform, viewed as a
linear operator, maps an entire
signal

to its
z transform

. We think of this as a ``function to
function'' mapping. We may say

is the
z transform of

by writing
or, using operator notation,
which can be abbreviated as
One also sees the convenient but possibly misleading notation

, in which

and

must be understood as
standing for the entire domains

and

, as
opposed to denoting particular fixed values.
The
z transform of a signal

can be regarded as a
polynomial in

, with coefficients given by the signal samples. For example,
the signal
has the
z transform

.
Existence of the Z
Transform
The
z transform of a finite-amplitude
signal 
will always
exist provided (1) the signal starts at a finite time and (2) it is
asymptotically exponentially bounded,
i.e., there exists a
finite integer

, and finite
real numbers 
and

,
such that

for all

. The
bounding
exponential may even be growing with

(

). These are
not the most general conditions for existence of the
z transform, but they
suffice for most practical purposes.
For a signal

growing as

, for

, one
would naturally expect the
z transform

to be defined only in the
region

of the
complex plane. This is expected
because the infinite series
requires

to ensure convergence. Since

for a decaying
exponential, we see that
the region of convergence of the
transform of a decaying exponential always includes the unit circle of
the
plane.
More generally, it turns out that, in all cases of practical interest,
the domain of

can be
extended to include the
entire complex plane, except at isolated ``singular''
points
7.2 at which

approaches
infinity (such as at

when

).
The mathematical technique for doing this is called
analytic
continuation, and it is described in §
D.1 as applied to the
Laplace transform (the continuous-time counterpart of the
z transform).
A point to note, however, is that in the extension region (all points

such that

in the above example), the signal
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,
60], and we will derive below only
what we will need.
In this section, we prove the highly useful
shift theorem and
convolution theorem for unilateral
z transforms. We consider the space of
infinitely long,
causal, complex sequences

,

, with

for

.
The
shift theorem says that a
delay of

samples
in the time domain corresponds to a
multiplication by

in the
frequency domain:

S
HIFT
or, using more common notation,
Thus,

, which is the waveform

delayed by

samples, has the
z transform

.
Proof:
where we used the causality assumption

for

.
The
convolution theorem for z transforms states that for any (real or)
complex
causal signals 
and

,
convolution in the time domain is multiplication in the
domain,
i.e.,
or, using operator notation,
where

, and

. (See
[
84] for a development of the convolution theorem for discrete
Fourier transforms.)
Proof:
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.
From Eq.

(
5.5), we have that the output

from a
linear
time-invariant filter with input

and
impulse response 
is given
by the
convolution of

and

,
i.e.,
 |
(7.3) |
where ``

'' means convolution as before. Taking the
z transform of
both sides of Eq.

(
6.3) and applying the
convolution theorem from
the preceding section gives
 |
(7.4) |
where H(z) is the
z transform of the
filter impulse response.
We may divide Eq.

(
6.4) by

to obtain
This shows that, as a direct result of the convolution theorem, the
z transform of an impulse response

is equal to the
transfer
function

of the filter, provided the filter is
linear
and time invariant.
Since
z transforming the
convolution representation for
digital filters was
so fruitful, let's apply it now to the general
difference equation,
Eq.

(
5.1). To do this requires two properties of the
z transform,
linearity (easy to show) and the
shift theorem
(derived in §
6.3 above). Using these two properties, we
can write down the
z transform of any difference equation by inspection, as
we now show. In
§
6.8.2, we'll show how to
invert by inspection as well.
Repeating the general difference equation for
LTI filters, we have
(from Eq.

(
5.1))

Let's take the
z transform of both sides, denoting the transform by

. Because

is a
linear operator,
it may be distributed through the terms on the right-hand side as
follows:
7.3
where we used the superposition and scaling properties of linearity
given on page
![[*]](../icons/crossref.png)
, followed by use of the shift
theorem, in that order. The terms in

may be grouped together
on the left-hand side to get
Factoring out the common terms

and

gives
Defining the polynomials
the
z transform of the difference equation yields
Finally, solving for

, which is by definition
the
transfer function 
, gives
 |
(7.5) |
Thus, taking the
z transform of the general difference equation led to a new
formula for the transfer function in terms of the difference equation
coefficients. (Now the minus signs for the feedback coefficients in
the difference equation Eq.

(
5.1) are explained.)
Factored Form
By the
fundamental theorem of algebra, every

th order polynomial
can be
factored into a product of

first-order polynomials.
Therefore, Eq.

(
6.5) above can be written in
factored form as
 |
(7.6) |
The numerator roots

are called the
zeros
of the
transfer function, and the denominator roots

are called the
poles of the
filter.
Poles and zeros are
discussed further in Chapter
8.
The transfer function conveniently captures the
algebraic structure of a
filtering operation with respect to
series or parallel combination. Specifically, we have the
following cases:
- Transfer functions of filters in series multiply together.
- Transfer functions of filters in parallel sum together.
Figure
6.1 illustrates the
series connection of two
filters

and

.
The output

from filter 1 is used as the input to filter 2.
Therefore, the overall transfer function is
In summary, if the output of filter

is given as input to
filter

(a series combination), as shown in Fig.
6.1, the
overall transfer function is

--transfer
functions of filters connected in series
multiply together.
Figure
6.2 illustrates the
parallel combination of two
filters. The filters

and

are driven by the
same input
signal 
, and their respective outputs

and

are
summed. The transfer function of the parallel
combination is therefore
where we needed only linearity of the
z transform to have that

.
Since multiplication of
complex numbers is commutative, we have
which implies that any ordering of
filters in series results in the
same overall
transfer function. Note, however, that the
numerical performance of the overall filter is usually affected
by the ordering of filter stages in a series 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:
Partial 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 first and/or
second-order terms. The case of first-order terms is the simplest and
most fundamental:
 |
(7.7) |
where
and

. (The case

is addressed in the next 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 general only if the poles

are
distinct. (Repeated poles are addressed in
§
6.8.5 below.) Both the poles and their residues may be
complex. The poles may be found by factoring the polynomial

into first-order terms,
7.4e.g., using the
roots function in
matlab.
The residue

corresponding to pole

may be found
analytically as
 |
(7.8) |
when the poles

are distinct. The matlab function
residuez7.5 will find poles and residues
computationally, given the
difference-equation (transfer-function)
coefficients.
Note that in Eq.

(
6.8), there is always a
pole-zero cancellation at

. That is, the term

is always cancelled by an
identical term in the denominator of

, which must exist because

has a pole at

. The residue

is simply the
coefficient of the one-pole term

in the partial
fraction expansion of

at

. The transfer function
is

, in the limit, as

.
Consider the
two-pole filter
The
poles are

and

. The corresponding residues are then
We thus conclude that
As a check, we can add the two one-pole terms above to get
as expected.
To illustrate an example involving complex
poles, consider the
filter
where

can be 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 form can be written as
Using Eq.

(
6.8), the residues are found to be
Thus,
A more elaborate example of a
partial fraction expansion into complex
one-pole sections is given in §
3.12.1.
PFE to Real, Second-Order Sections
When all coefficients of

and

are real (implying that

is the
transfer function of
a
real filter), it will
always happen that the complex one-
pole filters will occur in
complex conjugate pairs. Let

denote any one-pole
section in the PFE of Eq.

(
6.7). Then if

is complex and

describes a
real filter, we will also find

somewhere among
the terms in the one-pole expansion. These two terms can be paired to
form a
real second-order section as follows:
Expressing the pole

in
polar form as

,
and the residue as

,
the last expression above can be rewritten as
The use of polar-form coefficients is discussed further in the section
on
two-pole filters (§
B.1.3).
Expanding a transfer function into a sum of second-order terms with
real coefficients gives us the filter coefficients for a parallel bank
of real second-order filter sections. (Of course, each real pole can
be implemented in its own real one-pole section in parallel with the
other sections.) In view of the foregoing, we may conclude that every
real filter with

can be implemented as a parallel bank
of
biquads.
7.6 However, the full generality of a biquad
section (two poles and two zeros) is not needed because the PFE
requires only one zero per second-order term.
To see why we must stipulate

in Eq.

(
6.7), consider the sum of two
first-order terms by direct calculation:
 |
(7.9) |
Notice that the numerator order, viewed as a polynomial in

, is
one less than the denominator order. In the same way, it is easily
shown by
mathematical induction that the sum of

one-pole terms

can produce a numerator order of at most

(while the denominator order is

if there are no
pole-zero
cancellations). Following terminology used for analog filters, we
call the case

a
strictly proper transfer
function.
7.7 Thus, every strictly
proper transfer function (with distinct poles) can be implemented
using a parallel bank of two-pole, one-zero filter sections.
The
partial fraction expansion (
PFE) provides a simple means for
inverting the
z transform of
rational transfer functions. The PFE
provides a sum of first-order terms of the form
It is easily verified that such a term is the
z transform of
Thus, the inverse
z transform of

is simply
Thus, the
impulse response of every strictly proper
LTI filter (with distinct
poles) can be interpreted as a
linear combination of sampled
complex exponentials.
Recall that a
uniformly sampled
exponential is the same thing as 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

th
sequence is the

th residue,

.
In the
improper case, discussed in the next section, we
additionally obtain an
FIR part in the
z transform to be inverted:
The FIR part (a
finite-order polynomial in

) is also easily
inverted by inspection.
The case of repeated poles is addressed in §
6.8.5 below.
FIR Part of a PFE
When

in Eq.

(
6.7), we may perform a step of
long division
of

to produce an
FIR part in parallel with a
strictly proper IIR part:
 |
(7.10) |
where
When

, we define

. This type of decomposition is
computed by the
residuez function (a
matlab function for
computing a complete
partial fraction expansion, as illustrated in
§
6.8.8 below).
An alternate FIR part is obtained by performing long division on the
reversed polynomial coefficients to obtain
 |
(7.11) |
where

is again the order of the FIR part.
This type of decomposition is computed (as part of the PFE)
by
residued, described in §
J.6 and illustrated
numerically in §
6.8.8 below.
We may compare these two PFE alternatives as follows:
Let

denote

,

, and

.
(
I.e., we use a subscript to indicate polynomial order, and `

' is
omitted for notational simplicity.) Then for

we have two cases:
In the first form, the

coefficients are ``left
justified'' in the reconstructed numerator, while in the second form
they are ``right justified''. The second form is generally more
efficient for
modeling purposes, since the numerator of the IIR
part (

) can be used to match additional
terms in the
impulse response after the FIR part

has
``died out''.
In summary, an arbitrary
digital filter transfer function 
with

distinct
poles can always be expressed as a
parallel combination
of
complex one-pole filters, together with a parallel FIR part
when

. 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 part leaves off.
In
artificial reverberation applications, the FIR part may correspond
to the
early reflections, while the IIR part provides the
late reverb, which is typically dense, smooth, and
exponentially decaying [
86]. The
predelay (``pre-delay'') control in some commercial reverberators
is the amount of pure delay at the beginning of the reverberator's
impulse response. Thus, neglecting the early reflections, the order of
the FIR part can be viewed as the amount of predelay for the IIR part.
The general second-order case with

(the so-called
biquad section) can be written when

as
To perform a
partial fraction expansion, we need to extract an order 0
(length 1) FIR part via
long division. Let

and rewrite

as a ratio of polynomials in

:
Then long division gives
yielding
or
The delayed form of the partial fraction expansion is obtained by
leaving the coefficients in their original order. This corresponds
to writing

as a ratio of polynomials in

:
Long division now looks like
giving
Numerical examples of partial fraction expansions are given in §
6.8.8
below. Another worked example, in which the
filter

is converted to a set of parallel, second-order
sections is given in §
3.12. See also §
9.2 regarding
conversion to second-order sections in general, and §
G.9.1 (especially
Eq.

(
G.22)) regarding
a
state-space approach to partial fraction expansion.
Alternate PFE Methods
Another 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 single polynomial, and equate like powers of

.
This gives a
linear system of

equations in

unknowns

,

.
Yet another method for finding residues is by means of
Taylor series
expansions of the numerator

and denominator

about each
pole

, using
l'Hôpital's
rule..
Finally, one can alternatively construct a
state space
realization of a
strictly proper transfer function (using,
e.g.,
tf2ss in
matlab) and then
diagonalize it via a
similarity transformation. (See Appendix
G for an
introduction to
state-space models and diagonalizing them via
similarity transformations.)
The
transfer function of the diagonalized state-space model is
trivially obtained as a sum of one-pole terms--
i.e., the PFE. In other
words, diagonalizing a state-space
filter realization implicitly
performs a
partial fraction expansion of the filter's transfer
function. When the poles are distinct, the state-space model can be
diagonalized; when there are repeated poles, it can be
block-diagonalized instead, as discussed further in §
G.10.
When poles are repeated, an interesting new phenomenon emerges. To
see what's going on, let's consider two identical poles arranged in
parallel and in series. In the parallel case, we have
In the series case, we get
Thus, two one-pole
filters in
parallel are equivalent to a new
one-pole filter
7.8 (when the poles are identical), while the same two
filters in
series give a
two-pole filter with a repeated
pole. To accommodate both possibilities, the general
partial fraction
expansion must include the terms
for a pole

having multiplicity 2.
Dealing with Repeated Poles Analytically
A pole of
multiplicity 
has

residues associated with it. For example,
and the three residues associated with the pole

are 1, 2, and 4.
Let

denote the

th residue associated with the pole

,

.
Successively differentiating

times with
respect to

and setting

isolates the residue

:
or
For the example of Eq.

(
6.12), we obtain
In the time domain, repeated poles give rise to
polynomial
amplitude envelopes on the decaying
exponentials corresponding to the
(stable) poles. For example, in the case of a single pole repeated
twice, we have
Proof:
First note that
Therefore,
Note that

is a first-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 so on. As long as

, the impulse response will
eventually decay to zero, because
exponential decay always overtakes
polynomial growth in the limit as

goes to infinity.
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, two
different poles can only
yield a
convolution (or sum) of two different
exponential decays, with
no polynomial envelope allowed. This is true no matter how closely
the poles come together; the polynomial envelope can occur only when
the poles merge exactly. This might violate one's intuitive
expectation of a continuous change when passing from two closely
spaced poles to a repeated pole.
To study this phenomenon further, consider the convolution of two
one-pole
impulse-responses

and

:
 |
(7.14) |
The finite limits on the summation result from the fact that both

and

are
causal. Recall the closed-form sum of a truncated
geometric series:
Applying this to Eq.

(
6.14) yields
Note that the result is symmetric in

and

. If

, then

becomes proportional to

for large

, while if

, it becomes
instead proportional to

.
Going back to Eq.

(
6.14), we have
 |
(7.15) |
Setting

yields
 |
(7.16) |
which is the first-order polynomial
amplitude-envelope case for a
repeated pole. We can see that the transition from ``two convolved
exponentials'' to ``single exponential with a polynomial amplitude
envelope'' is perfectly continuous, as we would expect.
We also see that the polynomial amplitude-envelopes fundamentally
arise from
iterated convolutions. This corresponds to the
repeated poles being 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:
The convolution of a constant with itself is a ramp:
The convolution of a constant and a ramp is a quadratic, and so
on:
7.9
Alternate
Stability Criterion
In §
5.6 (page
![[*]](../icons/crossref.png)
), a
filter was defined to
be
stable if its
impulse response 
decays to 0 in
magnitude as time

goes to infinity. In §
6.8.5, we saw that
the
impulse response of every
finite-order LTI filter can be expressed
as a possible FIR part (which is always stable) plus a
linear
combination of terms of the form

, where

is some
finite-order polynomial in

, and

is the

th
pole of the
filter. In this form, it is clear that the impulse response always
decays to zero when each pole is strictly inside the unit circle of
the

plane,
i.e., when

. Thus, having all poles strictly
inside the unit circle is a
sufficient criterion for
filter
stability. If the filter is
observable (meaning that there are
no
pole-zero cancellations in the
transfer function from input to
output), then this is also a
necessary criterion.
A transfer function with no pole-zero cancellations is said to be
irreducible. For example,

is
irreducible, while

is reducible, since
there is the common factor of

in the numerator and
denominator. Using this terminology, we may state the following
stability criterion:
This characterization of stability is pursued further in §
8.4,
and yet another stability test (most often used in practice)
is given in §
8.4.1.
In summary, the partial fraction expansion can be used to expand
any rational
z transform
as a sum of first-order terms
 |
(7.17) |
for

, and
 |
(7.18) |
for

, where the term

is optional, but often
preferred. For
real filters, the complex one-
pole terms may be paired
up to obtain second-order terms with real coefficients.
The
PFE procedure occurs in two or three steps:
- When
, perform a step of long division to obtain
an FIR part
and a strictly proper IIR part
.
- Find the
poles
,
(roots of
).
- If the poles are distinct, find the
residues
,
from
- If there are repeated poles, find the additional residues via
the method of §6.8.5, and the general form of the PFE is
 |
(7.19) |
where
denotes the number of distinct poles, and
denotes the multiplicity of the
th pole.
In step 2, the poles are typically found by
factoring the
denominator polynomial

. This is a dangerous step numerically
which may fail when there are many poles, especially when many poles
are clustered close together in the

plane.
The following
matlab code illustrates factoring

to
obtain the three roots,

,

:
A = [1 0 0 -1]; % Filter denominator polynomial
poles = roots(A) % Filter poles
See Chapter
9 for additional discussion regarding
digital filters
implemented as parallel sections (especially §
9.2.2).
Figure
6.3 illustrates the use of
residuez (§
J.5)
for performing a partial fraction expansion on the
transfer function
The complex-conjugate terms can be combined to obtain two real
second-order sections, giving a total of one real first-order section
in parallel with two real second-order sections, as discussed and
depicted in §
3.12.
Figure 6.3:
Use of residuez to perform
a partial fraction expansion of an IIR filter transfer function
.
B = [1 0 0 0.125];
A = [1 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)
|
For the
filter
we obtain the output of
residued (§
J.6) shown in
Fig.
6.4. In contrast to
residuez,
residued delays the
IIR part until after the FIR part. In contrast to this result,
residuez returns
r=[-24;16]
and
f=[10;2], corresponding to the
PFE
 |
(7.22) |
in which the FIR and IIR parts have overlapping
impulse responses.
See Sections
J.5 and
J.6 starting on page
![[*]](../icons/crossref.png)
for
listings of
residuez,
residued and related
discussion.
The matlab function
conv (
convolution) can be used to
perform
polynomial multiplication. For example:
B1 = [1 1]; % 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 3 1
B4 = conv(B1,B3) % 4th row
% B4 = 1 4 6 4 1
% ...
The matlab
conv(B1,B2) is identical to
filter(B1,1,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
zero-padded with
length(B1)-1 zeros, it will return the complete convolution:
B1 = [1 2 3];
B2 = [4 5 6 7];
conv(B1,B2)
% ans = 4 13 28 34 32 21
filter(B1,1,B2)
% ans = 4 13 28 34
filter(B1,1,[B2,zeros(1,length(B1)-1)])
% ans = 4 13 28 34 32 21
Polynomial Division in Matlab
The matlab function
deconv (
deconvolution) can be used
to perform
polynomial long division in order to split an improper
transfer function into its FIR and strictly proper parts:
B = [ 2 6 6 2]; % 2*(1+1/z)^3
A = [ 1 -2 1]; % (1-1/z)^2
[firpart,remainder] = deconv(B,A)
% firpart =
% 2 10
% remainder =
% 0 0 24 -8
Thus, this example finds that

is as written in Eq.

(
6.21).
This result can be checked by obtaining a common denominator in order
to recalculate the
direct-form numerator:
Bh = remainder + conv(firpart,A)
% = 2 6 6 2
The operation
deconv(B,A) can be implemented using
filter in a manner analogous to the
polynomial
multiplication case (see §
6.8.8 above):
firpart = filter(B,A,[1,zeros(1,length(B)-length(A))])
% = 2 10
remainder = B - conv(firpart,A)
% = 0 0 24 -8
That this must work can be seen by looking at Eq.

(
6.21) and
noting that the
impulse-response of the remainder (the strictly proper
part) does not begin until time

, so that the first two samples
of the
impulse-response come only from the FIR part.
In summary, we may conveniently use
convolution and deconvolution to
perform polynomial multiplication and division, respectively, such as
when converting transfer functions to various alternate forms.
When carrying out a
partial fraction expansion on a transfer function
having a numerator order 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 part of any length can be
extracted from an
IIR filter, and this can be used for
PFEs as well as
for more advanced applications [].
See
http://ccrma.stanford.edu/~jos/filtersp/Transfer_Function_Analysis_Problems.html.
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