EE385_Chapter_6.pdf

EE385: Introduction to Digital Signal Processing
Chapter 6: Design of FIR and IIR Filters
Dr. Mohammed W. Baidas
Professor
Department of Electrical Engineering
College of Engineering and Petroleum
Kuwait University
EE385: Introduction to Digital Signal Processing - Chapter 6 Prof. M. W. Baidas 1/19

Introduction (1)
• Given an LTI system with a rational system function H[z], the input and output are
related by an LCCDE.
I For example, with a system function
H[z] =
b[0] + b[1]z−1
1 + a[1]z−1
, (1)
the input x[n] and output y[n] are related by the LCCDE
y[n] = −a[1]y[n − 1] + b[0]x[n] + b[1]x[n − 1]. (2)
I Note that this system may also be implemented with the following pair of coupled
diﬀerence equations
w[n] = −a[1]w[n − 1] + x[n]
y[n] = b[0]w[n] + b[1]w[n − 1].
(3)
With this implementation, it is only necessary to provide one memory
location to store w[n − 1]; whereas Eq. (2) requires two memory locations,
one to store y[n − 1] and one to store x[n − 1].
I This suggests that the amount of computation and/or memory required will de-
pend on the implementation.
• For a linear time-invariant system with a rational system function, the input x[n] and
the output y[n] are related by an LCCDE
y[n] =
q
∑
k=0
b[k]x[n − k] −
p
∑
k=1
a[k]y[n − k]. (4)

Introduction (2)
I Basic required computational elements at time n: adders, multipliers and delays.
I A network is often represented pictorially using a signal flowgraph.
Consists of directed branches that are connected at nodes.
Each branch has input and output, with direction indicated by an arrowhead.
I There are two special types of nodes:
Source Nodes: nodes with no incoming branches and used as inputs to filters.
Sink Nodes: nodes with only entering branches and used as outputs to filters.
Figure 1: Notation Used for an Adder, Multiplier and Delay
Figure 2: Signal Flowgraph Consisting of Nodes, Branches, and Node Variables - Node j
is an Adder, and Node k is a Branch Point

Structures for FIR Systems (1)
• A causal finite impulse response (FIR) filter has a system function that is polynomial in
z−1, as
H[z] =
N
∑
n=0
h[n]z−n. (5)
I For an input x[n], the output is
y[n] =
N
∑
k=0
h[k]x[n − k].
For each value of n, evaluating this sum requires N + 1 multiplications and N
additions.
• Direct Form:
I Implement an FIR filter is in direct form using a tapped delay line.
This structure requires N + 1 multiplications, N additions, N delays.
Figure 3: Direct Form Implementation of FIR Filters

• Cascade Form:
I For a causal FIR filter, the system function may be factored into a product of first-
order factors,
H[z] =
N
∑
n=0
h[n]z−n = A
N
∏
k=1
(
1 − αk z−1
)
, (6)
where αk for k = 1,2,...,N are the zeros of H[z].
If h[n] is real, the complex roots of H[z] occur in complex conjugate pairs.
These conjugate pairs may be combined to form second-order factors with
real coefficients,
H[z] = A
Ns
∏
k=1
[
1 + bk [1]z−1 + bk [2]z−2
]
. (7)
Written in this form, H[z] may be implemented as a cascade of second-order
FIR filters.
Figure 4: An FIR Filter Implemented as a Cascade of Second-Order Systems

• Linear Phase Filters:
I Linear phase ﬁlters have a unit sample response that is either symmetric,
h[n] = h[N − n], (8)
or antisymmetric,
h[n] = −h[N − n]. (9)
I This symmetry may be exploited to simplify the network structure.
I For example, if N is even and h[n] is symmetric (type I ﬁlter),
y[n] =
N
∑
n=0
h[k]x[n − k] =
N/2−1
∑
k=0
h[k][x[n − k] + x[n − N + k]] + h
[
N
2
]
x
[
n −
N
2
]
.
(10)
Therefore, forming the sums [x[n − k] + x[n − N + k]] prior to multiplying by
h[k] reduces the number of multiplications.
Figure 5: Direct Form Implementation of Type I Linear Phase Filter

I If N is odd and h[n] is symmetric, then a type II linear phase ﬁlter is obtained.
Figure 6: Direct Form Implementation of Type II Linear Phase Filter
I Example 6.1: An LTI system has a unit sample response given by h[n] =
{−0.01,0.02,−0.10,0.40,−0.10,0.02,−0.01}. Draw a signal ﬂowgraph for this sys-
tem that requires the minimum number of multiplications.

Structures for IIR Systems (1)
• The input x[n] and output y[n] of a causal IIR filter with a rational system function
H[z] =
B[z]
A[z]
=
∑q
k=0 b[k]z−k
1 +
∑p
k=1 a[k]z−k
(11)
is described by the LCCDE
y[n] =
q
∑
k=0
b[k]x[n − k] −
p
∑
k=1
a[k]y[n − k]. (12)
• Direct Form:
I Two direct form filter structures: direct form I and direct form II.
I Direct Form I:
An implementation that results when Eq. (12) is written as
w[n] =
q
∑
k=0
b[k]x[n − k]
y[n] = w[n] −
p
∑
k=1
a[k]y[n − k].
(13)
First equation corresponds to an FIR filter with input x[n], and output w[n];
while second equation corresponds to an all-pole filter with input w[n] and
output y[n].

Figure 7: Direct Form I Realization of an IIR Filter
• Direct Form: (Continued)
I Direct Form I: (Continued)
Therefore, this pair of equations represents a cascade of two systems,
Y[z] =
1
A[z]
[B[z]X[z]]. (14)
Computational requirements for a Direct Form I structure are as follows:
Number of multiplications: p + q + 1 per output sample.
Number of additions: p + q per output sample.
Number of delays: p + q.

• Direct Form: (Continued)
I Direct Form II:
Structure is obtained by reversing order of cascade of B[z] and 1/A[z], such
that x[n] is first filtered with the all-pole filter 1/A[z] and then with B[z], as
Y[z] = B[z]
[
1
A[z]
X[z]
]
. (15)
If we denote the output of the all-pole filter 1/A[z] by w[n], this structure is
given by
w[n] = x[n] −
p
∑
k=1
a[k]w[n − k]
y[n] =
q
∑
k=0
b[k]w[n − k].
(16)
This structure may be simplified by noting that the two sets of delays are
delaying the same sequence. Therefore, they may be combined.
Computational requirements for a Direct Form II structure are as follows:
Number of multiplications: p + q + 1 per output sample.
Number of additions: p + q per output sample.
Number of delays: max(p,q).
Direct form II structure is said to be canonic because it uses the minimum
number of delays for a given H[z].

Figure 8: Direct Form II Realization of an IIR Filter
Figure 9: Direct Form II Realization of an IIR Filter with p = q

• Cascade Structure:
I The cascade structure is derived by factoring the numerator and denominator
polynomials of H[z]:
H[z] =
∑q
k=0 b[k]z−k
1 +
∑p
k=1 a[k]z−k
= A
max(p,q)
∏
k=1
1 − βk z−1
1 − αk z−1
. (17)
This factorization corresponds to a cascade of first-order filters, each having
one pole and one zero.
In general, the coefficients αk and βk will be complex.
However, if h[n] is real, roots of H[z] will occur in complex conjugate pairs,
and these complex conjugate factors may be combined to form second-order
factors with real coefficients, such that
Hk [z] =
1 + β1k z−1 + β2k z−2
1 + α1k z−1 + α2k z−2
. (18)
I A sixth-order IIR filter implemented as a cascade of three second-order systems
in direct form II is shown below.

• Parallel Structure:
I An alternative to factoring H[z] is to expand the system function using a partial
fraction expansion. For example, with
H[z] =
∑q
k=0 b[k]z−k
1 +
∑p
k=1 a[k]z−k
= A
∏q
k=1
(
1 − βk z−1
)
∏p
k=1
(
1 − αk z−1
) , (19)
if p q and αi , αj (i.e. roots of denominator polynomial are distinct), H[z] may
be expanded as a sum of p first-order factors as follows:
H[z] =
p
∑
k=1
Ak
1 − αk z−1
, (20)
where the coefficients Ak , and αk are, in general, complex.
This expansion corresponds to a sum of p first-order system functions and
may be realized by connecting these systems in parallel.
If h[n] is real, the poles of H[z] will occur in complex conjugate pairs, and
these complex roots in the partial fraction expansion may be combined to
form second-order systems with real coefficients, as
H[z] =
Ns
∑
k=1
γ0k + γ1k z−1
1 + α1k z−1 + α2k z−2
. (21)
I If p ≤ q, the partial fraction expansion will also contain a term of the form
c0 + c1z−1 + ··· + cq−pz−(q−p), (22)
which is an FIR filter in parallel with the other terms in the expansion of H[z].

Figure 10: A Sixth-Order IIR Filter Implemented as a Parallel Connection of Three
Second-Order Direct Form II Structures

• Example 6.2: Consider the causal LTI filter with system function
H[z] =
1 + 0.875z−1
(
1 + 0.2z−1 + 0.9z−2
)(
1 − 0.7z−1
) .
Draw a signal flowgraph for this system using:
(a) Direct form I.
(b) Direct form II.
(c) Cascade of first- and second-order systems realized in direct form II.
(d) Parallel connection of first- and second-order systems realized in direct form II.
Solution:
(a) Writing the system function as a ratio of polynomials in z−1, yields
H[z] =
1 + 0.875z−1
1 − 0.5z−1 + 0.76z−2 − 0.63z−3
,
it follows that the direct form I realization of H[z] is as follows.

• Example 6.2: (Continued)
(b) For a direct form II realization of H[z], we have
(c) The realization of H[z] is as follows:

• Example 6.2: (Continued)
(d) For a parallel structure, H[z] must be expanded using a partial fraction expan-
sion,
H[z] =
1 + 0.875z−1
(
1 + 0.2z−1 + 0.9z−2
)(
1 − 0.7z−1
) =
A + Bz−1
1 + 0.2z−1 + 0.9z−2
+
C
1 − 0.7z−1
.
(23)
Solving for A, B, and C, we ﬁnd A = 0.2794, B = 0.9265, and C = 0.7206. There-
fore, the partial fraction expansion is
H[z] =
0.2794 + 0.9265z−1
1 + 0.2z−1 + 0.9z−2
+
0.7206
1 − 0.7z−1
. (24)
Thus, a parallel structure for H[z] is shown below.

Solved Problems (1)
• S-Problem 7.1: Draw the structure of the FIR filter for
y[n] =
3
∑
k=0
b[k]x[n − k] = 3x[n] + 3x[n − 1] + 2x[n − 2] − 2x[n − 3].
Figure 11: FIR Filter Structure of S-Problem 7.1
• S-Problem 7.2: Consider the filter structure shown in the figure below. Find the system
function and the unit sample response of this system.

Solved Problems (2)
• S-Problem 7.2:
Solution: For the three nodes labeled in the flowgraph below, we have the following
node equations:
w[n] = x[n] + 0.2w[n − 1]
v[n] = x[n] + w[n]
y[n] = v[n] + 2w[n − 1].
Using the z−transforms, the first equation becomes W[z] = 1
1−0.2z−1 X[z]. Taking the
z−transform of the second equation, and substituting the expression above for W[z],
we have
V[z] = X[z] + W[z] = X[z] +
1
1 − 0.2z−1
X[z] =
2 − 0.2z−1
1 − 0.2z−1
X[z].
Finally, taking the z−transform of the last equation, we find
Y[z] = V[z] + 2z−1W[z] =
[
2 − 0.2z−1
1 − 0.2z−1
+ 2z−1 1
1 − 0.2z−1
]
X[z] =
2 + 1.8z−1
1 − 0.2z−1
X[z].
Therefore, the system function is H[z] = 2+1.8z−1
1−0.2z−1 , and the unit sample response is
h[n] = 2(0.2)nu[n] + 1.8(0.2)n−1u[n − 1].

EE385_Chapter_6.pdf

Recommended

Recommended

More Related Content

Similar to EE385_Chapter_6.pdf

Similar to EE385_Chapter_6.pdf (20)

Recently uploaded

Recently uploaded (20)

EE385_Chapter_6.pdf