Chapter10. Realization of Digital Filter.pptx

Chapter 10
Digital Signal Processing Tarun Rawat Copyright © 2017 by Oxford University Press, Inc.
Realization of Digital Filters
Dr. Tarun Kumar Rawat
Division of Electronics and Communication Engineering
Netaji Subhas Institute of Technology (NSIT), New Delhi, India

FIR and IIR Systems
Methods for realizing the FIR and IIR filters system function H(z) by different
structures.

FIR and IIR Systems
structures.
The difference equation can be obtained from the system function H(z).

FIR and IIR Systems
structures.
The difference equation specifies the actual operations that must be performed by the
digital system on the input data, in the time domain, in order to generate the desired
output.

FIR and IIR Systems
structures.
The difference equation specifies the actual operations that must be performed by the
digital system on the input data, in the time domain, in order to generate the desired
output.
The difference equation, for most practical cases may be written as
NΣ−1 MΣ−1
aky(n −k) = bkx(n −k)
k =0
k=0
NΣ−1
y(n) =−
k=1
aky(n − k) +
MΣ−1
k =0
bkx(n −k); a0 = 1
where x(n) is the input, y(n) is the output, y(n − k) is the previous output and ak , bk
are system coefficients.

FIR Filter (All Zero or Moving Average system)
FIR Filter (All Zero or Moving Average system): The output y(n) ofFIR
system is given by (bk =0)
MΣ−1
y(n) = bkx(n −k)
k=0
Its present output depends only on the present and past inputs.
The present output does not depend on the past outputs, hence, there is no
feedback.
If the filter implementation does not contain feedback, then the implementation is
said to be nonrecursive.
The impulse response h(n) of FIR filter is of finiteduration.
The FIR filter can be viewed as a moving average (MA) system because its output
is just a weighted sum (moving average) of the input terms.

FIR Filter (All Zero or Moving Average system) contd...
y(n) =
The system function H(z) of FIR system is given by
M
Σ− 1
bkx(n −k)
Y (z)=
k=0
M
Σ− 1
k =0
k
b z − k X(z)
Y (z)
X (z)
= H(z) =
MΣ−1
k =0
k
b z − k
All the poles of a FIR filter are located at z = 0.
These poles do not contribute to the magnitude response. Hence, they are said
not to be relevant, and the FIR filter is also called an all-zero filter.
The poles at z = 0, however, do contribute to the phase response, by introducing
linear phase.

IIR Filter (All Pole or Autoregressive system)
y(n) =−
IIR Filter (All Pole or Autoregressive system): The output y(n) ofIIR
system is given by (ak =0)
NΣ−1
k=1
Its present output depends on its own past values y(n − k) and on the present
input.
Since the present output depends on its own past outputs, there is feedback.
If the filter implementation contains feedback, then the implementation is said to
be recursive.
The impulse response h(n) of IIR filter is of infiniteduration.
It is also called an autoregressive (AR) filter, because its output depends
(regresses) on its own previous outputs.
aky(n − k) +b0x(n)

IIR Filter (All Pole or Autoregressive system) contd...
y(n) =−
The system function H(z) of IIR system is given by
N
Σ− 1
aky(n − k) +b0x(n)
Y (z)= −
k=1
N
Σ− 1
k =1
k
a z − k Y (z) +b0X(z)
X (z)
Y (z)
= H(z) =
b0
1+
NΣ−1
k=1
ak z − k
An AR system is an IIR filter whose system function has all its zeros at z =0.
These zeros do not contribute to the magnitude response.
Hence, they are said not to be relevant, and the AR filter is also called an all-pole
(AP) filter.

IIR Filter (Pole-Zero or Autoregressive, Moving average system)
IIR Filter (Pole-Zero or Autoregressive, Moving average system):
The output y(n) of a Pole-Zero IIR system is givenby
NΣ−1 MΣ−1
y(n) = − aky(n −k) + bkx(n −k)
k=1 k=0
Its present output depends on its own past values y(n − k) and on the present
and past values of the input.
Since the present output depends on its own past outputs, there is feedback andit
is realized recursively.
The impulse response h(n) of this IIR filter is of infinite duration.
It is also called an autoregressive, moving average (ARMA) filter, because its
output depends on its own past values y(n − k) and on the present and past
values of the input.

IIR Filter (Pole-Zero or Autoregressive, Moving average system) contd...
k =1
k
Y (z) = − a z − k Y (z)+
The system function H(z) of Pole-Zero IIR system is given by
NΣ−1 MΣ−1
k=1 k=0
NΣ−1 MΣ−1
k =0
k
b z − k X(z)
X (z)
Y (z)
= H(z) =
MΣ−1
k=0
bkz− k
1+
NΣ−1
k=1
ak z − k
This general IIR filter, having both poles and zeros, is also known as Pole-zero
(PZ) system.

Basic Building Blocks
A transfer function H(z) can be conveniently represented in block diagram form using
the basic building blocks representing the unit delay, the multiplier, the adder and the
pick-off node as shown below.
(a)
z1
x(n) ax(n)
x(n) x(n)
x2(n)
x1(n) x1(n)  x2(n)
x(n)
(b)
a

x(n) x(n1) x(n) z1 x(n1)
x(n) ax(n)
x(n) x(n)
x(n)
a
x2(n)
x1(n) x1(n)  x2(n)
Unit delay:
Multiplier:
Adder:
Pick-off node:

Canonic and Transposed Structure
Canonic Structure: A realization is canonic if the number of delay units used in
the realization is equal to the order of the transfer function realized. Thus canonic
realization has no redundant delay units.

Canonic and Transposed Structure
Canonic Structure: A realization is canonic if the number of delay units used in
the realization is equal to the order of the transfer function realized. Thus canonic
realization has no redundant delay units.
Transposed Structure: Two realizations are said to be equivalent if they have
the same transfer function. A simple way to generate an equivalent structure from a
given realization is via the transpose operation, which is as follows:
1. Interchange the input and output nodes.
2. Reverse all paths.
3. Replace pick-off nodes by adders, and vice versa.

FIR Filter Structures
H(z) =
A finite-duration impulse response filter has a system function of the form
MΣ−1
k =0
b z − k = b0 + b1z−1 + b2z − 2 + ··· + b z − ( M −1 )
k M − 1
Hence, the impulse response h(n) is
b
n
h(n) =
0
0 ≤ n ≤ M − 1
otherwise
and the difference equation representation is
y(n) = b0x(n) + b1x(n − 1)+ b2x(n − 2) + ··· + bM − 1 x(n − M + 1)
The length of the filter (which is equal to the number of coefficients) is M . The order of
the filter is M −1.

Direct Form (Transversal or Tapped-Delay Line) Structure
Direct Form Structure: Structures in which the multiplier coefficients are directly
available as the coefficients of H(z) are called direct form structures.

A direct form structure of 4th order FIR filter requires 4 memory locations,5
multiplications and 4 additions per outputsample.

A direct form structure of 4th order FIR filter requires 4 memory locations,5
multiplications and 4 additions per outputsample.
For the direct form structure of a FIR filter of order M −1, the computation of each
output sample, y(n), requires
M − 1 memory locations to store the M − 1 previous input samples.
M memory locations to store the M coefficients.
M multiplications, and
M − 1 additions.

Direct Form (Transversal or Tapped-Delay Line) Structure contd...
Let M = 5(i.e., a fourth order FIR filter), then
y(n) = b0x(n) + b1x(n − 1)+ b2x(n − 2) + b3x(n − 3)+ b4x(n − 4)
z1 z1 z1 z1
x(n)
y(n)
x(n1)

x(n2)

x(n3)

x(n4)

b0 b1 b2 b3 b4
z1
x(n)
x(n1)
x(n2)
x(n3)
x(n4)
b0
b1
b2
b3
b4
 y(n)



z1
z1
z1
(a)
x(n)
y(n) b0
b4

z1
z1
b1

z1
b2

z1
b3


Cascade-Form Structure
Cascade-Form Structure: If we have an FIR filter of higher order, it may be
realized as a cascade of FIR filters of lower order, preferably as second-order filters
with real coefficients when the order is even.

When the order is odd, it may be realized as a cascade of second-order filters and one
first-order filter.

first-order filter.
We can factorize the given FIR transfer function H(z) in the form
H(z) =H1(z)H2(z).

first-order filter.
We can factorize the given FIR transfer function H(z) in the form H(z) = H1(z)H2(z).
Example: Obtain the cascade form realization of the given FIR filter system function
6 7 26 1
H(z) =1+ z − 1 + z − 2 + z − 3 + z − 4
5 5 25 5
Solution: The given FIR filter is of length M = 5. Factorizing the givensystem
function, we obtain
5 5
Σ Σ Σ Σ
1 2
1 1
H(z) = 1 + z − 1 + z − 2 1 +z − 1 + z − 2 = H (z)H (z)

Cascade-Form Structure contd...
A cascade realization of the given system function H(z) is shown below.


z 1
1
1 1
5
z 1
x(n)

y(n)

z 1
z 1
15

Linear-Phase Structure
Linear-Phase Structure: It is generally desirable for an FIR digital filter to have
linear phase response; that is, we want
∠H(ejω ) = β − αω − π < ω < π
2
where β = 0 or ± π and α is a constant.

Linear-Phase Structure
Linear-Phase Structure: It is generally desirable for an FIR digital filter to have
linear phase response; that is, we want
∠H(ejω ) = β − αω − π < ω < π
where β = 0 or ± π and α is a constant.
2
For an FIR digital filter to have linear phase response [Eq. (10.12)], its impulse
response [h(n), 0 ≤ n ≤ M − 1] must be either symmetric or antisymmetric about
some point in time.
Symmetric: h(n) = h(M − 1 − n); β = 0, 0 ≤ n ≤ M − 1
Antisymmetric: h(n) = −h(M − 1− n);
π
β = ± , 0 ≤ n ≤ M − 1
2

Linear-Phase Structure contd...
Example:Obtain the linear phase structure for the linear phase FIR system given by
1 1 1 1 1
H(z) =1 + z − 1 + z − 2 + z − 3 + z − 4 + z − 5 +z − 6
2 3 6 3 2
Solution: Consider the given system function [length= M = 7,
order= M − 1 = 6],
Y (z) 1 1 1 1 1
H(z) = = 1 + z − 1 + z − 2 + z − 3 + z − 4 + z − 5 + z − 6
X(z) 2 3 6 3 2
2 3 6
Y (z) = [1 + z − 6 ]X(z) +
1
[z−1 + z− 5 ]X(z) +
1
[z−2 + z− 4 ]X(z) +
1
z − 3 X(z)
1
2
1
3
y(n) = [x(n) + x(n − 6)] + [x(n − 1) + x(n − 5)] + [x(n − 2)+ x(n − 4)]+
1
6
x(n −3)

Linear-Phase Structure contd...
A linear phase structure of the given system function H(z) is shown below.
x(n)
 
 
z1
z1

z1
z1
z1
z1
 y(n)
13 16
12
1

IIR Filter Structures
The system function of an IIR filter is given by
MΣ− 1
bk z− k
H(z) = k=0
1+
NΣ−1
k=1
ak z − k
where bk and ak are the coefficients of the filter and N ≥ M . The order of the filter is
N − 1 if aN − 1 ƒ= 0. The difference equation of an IIR filter is given by
NΣ−1 MΣ−1
k=1 k=0

Direct-Form I Structure
Direct-Form I Structure: Structures in which the multipliers coefficients are
precisely the coefficients of the transfer function are called direct form structures.

In direct form I, the system function H(z) is divided into two parts, the first part H1(z)
containing only the zeros, followed by the part H2(z) containing only the poles. Or, we
can say that H(z) is the product of an all-zero function H1(z), and an all-pole function
H2(z).

In direct form I, the system function H(z) is divided into two parts, the first part H1(z)
containing only the zeros, followed by the part H2(z) containing only the poles. Or, we
can say that H(z) is the product of an all-zero function H1(z), and an all-pole function
H2(z).
Consider the system function
H(z) =
Y (z)
X (z)
MΣ−1
= k=0
bkz− k
1+
NΣ−1
k=1
a k z − k
=
ΣMΣ−1
bk z − k
Σ Σ
k=0
v ˛¸
zeros
cv
1
1+
Σ N − 1
a k z − k
k
=
˛¸1
Σ
c
poles
H(z) =
Y (z)
X (z)
=
Σ
W (z)
X (z)
v ˛¸
zeros
Y (z)
Σ Σ Σ
W (z)
cv ˛¸ c
poles
= H1(z)H2(z)

Direct-Form I Structure contd...
where
H1(z) =
W (z)
X (z)
=
MΣ−1
k =0
k
b z − k =
Σ3
k =0
k
b z − k
− k
bkz X(z)
Σ3
W (z) =
k=0
Σ3
w(n) =
k=0
bkx(n −k)
w(n) = b0x(n) + b1x(n − 1) + b2x(n − 2)+ b3x(n − 3)
and
Y (z) 1
k =1
H2(z) = =
W (z) 1+
Σ N − 1
a k z − k
1
= Σ 3
1+
k =1 k
a z − k
Σ3
Σ
Y (z) 1 + a
k =1
k z − k
Σ
= W (z)

Direct-Form I Structure contd...
− k
ak z Y (z) = W (z)
Σ3
Y (z)+
k=1
Σ3
y(n) +
k=1
ak y(n − k) = w(n)
y(n) = −a1y(n − 1)− a2y(n − 2)− a3y(n − 3)+ w(n)
y(n)
x(n) b0 w(n)
b1
b2
b3

Z1

(b)

Z1

Z1
a3
a2
a1
Z1
Z1
Z1

Direct-Form II Structure
Direct-Form II Structure: In direct form II, the system function H(z) is divided
into two parts, the first part H1(z)containing only the poles, followed by the part H2(z)
containing only the zers. Or, we can say that H(z) is the product of an all-polefunction
H1(z), and an all-zero function H2(z).

Direct-Form II Structure
Direct-Form II Structure: In direct form II, the system function H(z) is divided
into two parts, the first part H1(z)containing only the poles, followed by the part H2(z)
containing only the zers. Or, we can say that H(z) is the product of an all-polefunction
H1(z), and an all-zero function H2(z).
Consider the system function
H(z) =
Y (z)
X (z)
MΣ−1
= k=0
bkz− k
1+
NΣ−1
k=1
a k z − k
=
Σ
1
1+
Σ N − 1
a k z − k
k
=
˛¸1
Σ Σ
v
poles
MΣ−1
bk z − k
Σ
c v
k=0
˛¸ c
zeros
H(z) =
Y (z)
X (z)
=
Σ Σ Σ Σ
V (z) Y (z)
poles
X(z) V (z)
v ˛¸ cv ˛¸ c
zeros
= H1(z)H2(z)

Direct-Form II Structure contd...
where
V (z) 1
k =1
H1(z) = =
X(z) 1+
Σ N − 1
a k z − k
1
k =1
= Σ 3
1 + ak z − k
Σ3
V (z) 1 + a
k =1
k z − k
Σ Σ
= X(z)
− k
ak z V (z) =X(z)
Σ3
V (z) +
k=1
Σ3
v(n) +
k=1
akv(n − k) = x(n)
Σ3
v(n) = − akv(n − k) +x(n)
k=1
v(n) = −a1v(n − 1) − a2v(n − 2)− a3v(n − 3) + x(n)

and
H2(z) =
Y (z)
V (z)
=
MΣ−1
k =0
k
b z − k =
Σ3
k =0
k
b z − k
− k
bkz V (z)
Σ3
Y (z) =
k=0
Σ3
y(n) =
k=0
bkv(n −k)
y(n) = b0v(n) + b1v(n − 1)+ b2v(n − 2)+ b3v(n − 3)

H1(z) H2(z)
y(n)
x(n)
Y(z)
X(z)
v(n)
V(z)
Poles Zeros
(a) (b)
y(n)
x(n) v(n)
a1
a2
a3
z1

z1

z1
b3
b2
b1
b0
z 1
z1
z1
(c)
Fig (d) is the transpose of Fig. (c).
(d)
y(n)
x(n) v(z)
a1
a2

z1

z1
b0
b1
b2
a3 z1
b3
x(n)
y(n)
a1
a2
a3
z1
z1
z1
b0
b1
b2
b3

Cascade-Form Structure: In cascade form, the output of the first system is the
input to the second system.

The Cascade form of H(z) is obtained by factorizing the numerator and denominator
polynomials of H(z).

The system function is decomposed as the product of transfer functions in the form
(assuming M = N )
N1(z)N2(z)···N K (z)
H(z) =
D1(z)D2(z)···D K (z)
H(z) =
Σ
N1(z)
D1(z)
Σ Σ
N2(z)
D2(z)
Σ Σ
···
N K (z)
K
D (z)
Σ
= H1(z)H2(z) ···H K (z)

The system function is decomposed as the product of transfer functions in the form
(assuming M = N )
N1(z)N2(z)···N K (z)
H(z) =
D1(z)D2(z)···D K (z)
H(z) =
Σ
N1(z)
Σ Σ
N2(z)
Σ Σ
···
N K (z)
D1(z) D2(z) DK (z)
Σ
= H1(z)H2(z) ···H K (z)
Each of the transfer functions H1(z)H2(z) ···H K (z) is realized by the direct form II or
its transpose structure and then connected in cascade.

Cascade-Form Structure contd...
4
Example: Obtain a cascade realization using second-order sections where possible
for
1 + 1z − 1 + 1z − 2
H(z) = . Σ. 2
Σ.
2 2 2 4
1 + 1z − 1 1 + z − 1 + 1z − 2 1+ 1z − 1 + 1z − 2
Σ
Solution: The given system function can be writtenas
H(z) =
1
1+ 1z − 1
Σ Σ Σ
4 2
1 + 1z − 1 + 1z − 2
2 2
1+ z − 1 + 1z − 2
Σ Σ
1
1 + 1z − 1 + 1z − 2
2 4
Σ
= H1(z)H2(z)H3(z)
where H1(z) is a first order section and H2(z) and H3(z) are second-order (biquad)
sections. Biquads are realized using direct form II.
1
1
z1


z1
z1
z1

1 y(n)
   
x(n)
2

 1 z1 1
2 2
2
 1
4
 1
 1
4

Parallel-Form Structure
Parallel-Form Structure: In parallel form, the input signal is processed
separately by different subsystems. The system output is a sum of the outputs of the
subsystems.

subsystems.
An IIR system can be realized in parallel form by using partial fraction expansion of
H(z).

subsystems.
An IIR system can be realized in parallel form by using partial fraction expansion of
H(z).
Example: Obtain the parallel form realization structures for the system function
4
1+ 1z − 1
H(z) = . Σ.
2 2 4
1 + 1z − 1 1 + 1z − 1 + 1z − 2
Σ
Solution: Partial fraction expansion of H(z) yields
4
1+ 1z − 1
H(z) = .
Σ.
2 2 4
1 + 1z − 1 1 + 1z − 1 + 1z − 2
Σ =
1
2
1+ 1z − 1
+
Σ Σ Σ
2
= H1(z) + H2(z)
2 4
1 − 1 z − 1
2 4
1+ 1z − 1 + 1z − 2
Σ

Parallel-Form Structure contd...
where H1(z) is a first-order section and H2(z) is a second-order (biquad) section
which are realized using direct form II. The output of each section is summed to form
the filter output.





1
2
1
2 1
4
1
4
1
2
1
2
Z1
Z1
Z1
x(n) y(n)
(b)

Realization of Digital Filters
Thank You!

Chapter10. Realization of Digital Filter.pptx

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