ECT303 DIGITAL SIGNAL
PROCESSING
MODULE 4 - PART II
REALIZATION OF IIR FILTERS
Ms. Neethu Radha Gopan, Asst. Prof., Dept. of ECE, RSET.
Structures for IIR Filters
2
➢ There are mainly 5 different structures for representing IIR Filters :
1. Direct Form I
2. Direct Form II
3. Transposed Form
4. Cascade Form
5. Parallel Form
1. Direct Form I
3
The difference equation for an IIR filter is given by
𝑁 𝑀
𝑦 𝑛 = − 𝑎𝑘 𝑦 𝑛 − 𝑘 + 𝑏𝑘 𝑥(𝑛 − 𝑘)
𝑘=1 𝑘=0
𝑦 𝑛 = − 𝑎1 𝑦 𝑛 − 1 + 𝑎2 𝑦 𝑛 − 2 + ⋯ . . +𝑎𝑁 𝑦 𝑛 − 𝑁 + 𝑏0 𝑥 𝑛 + 𝑏1 𝑥 𝑛 − 1 + ⋯
+𝑏𝑀 𝑥 𝑛 − 𝑀
𝑦 𝑛 = −𝑎1 𝑦 𝑛 − 1 − 𝑎2 𝑦 𝑛 − 2 − ⋯ . . −𝑎𝑁 𝑦 𝑛 − 𝑁 + 𝑏0 𝑥 𝑛 + 𝑏1 𝑥 𝑛 − 1 + ⋯
+𝑏𝑀 𝑥 𝑛 − 𝑀
This structure requires M+N+1 multipliers, M+N adders & M+N delay elements.
x(n) 𝑏0
y(n)
𝑏1
−𝑎1
𝑏2
−𝑎2
𝑏𝑀−1
−𝑎𝑁−1
𝑏𝑀 −𝑎𝑁
4
0.5+2𝑧 −1 +3𝑧 −2
Q) Realize the IIR Filter using direct form I. 𝐻 𝑧 =
1−2𝑧 −1 −3𝑧 −2
𝑦 𝑛 = 2𝑦 𝑛 − 1 + 3𝑦 𝑛 − 2 + 0.5𝑥 𝑛 + 2𝑥 𝑛 − 1 + 3𝑥(𝑛 − 2)
0.5 y(n)
x(n)
2 2
3 3
5
Q) Realize the IIR Filter using direct form I.
y 𝑛 = 0.5 𝑦 𝑛 − 1 − 0.25 𝑦 𝑛 − 2 + 𝑥 𝑛 + 0.4 𝑥 𝑛 − 1
6
2. Direct Form II
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The difference equation for an IIR filter is given by
𝑁 𝑀
𝑦 𝑛 = − 𝑎𝑘 𝑦 𝑛 − 𝑘 + 𝑏𝑘 𝑥(𝑛 − 𝑘)
𝑘=1 𝑘=0
Y z σ𝑀𝑘=0 𝑏𝑘 𝑧
−𝑘
The system function is given by 𝐻 𝑧 = =
𝑋 𝑧 1 + σ𝑁𝑘=1 𝑎 𝑘 𝑧 −𝑘
Y z Y z W z
Let = .
𝑋 𝑧 𝑊 𝑧 𝑋 𝑧
𝑀
Y z W z 1
where = 𝑏𝑘 𝑧 −𝑘 & =
𝑊 𝑧
𝑘=0
𝑋 𝑧 1 + σ𝑁
𝑘=1 𝑎𝑘 𝑧
−𝑘
W z 1 1
Consider = 𝑁 =
𝑋 𝑧 σ
1 + 𝑘=1 𝑎𝑘 𝑧 −𝑘 1 + (𝑎1 𝑧 −1 + 𝑎2 𝑧 −2 + ⋯ +𝑎𝑁 𝑧 −𝑁 )
W z + 𝑎1 𝑧 −1 W z + 𝑎2 𝑧 −2 W z + ⋯ +𝑎𝑁 𝑧 −𝑁 W z = 𝑋 𝑧
W z = 𝑋 𝑧 − 𝑎1 𝑧 −1 W z − 𝑎2 𝑧 −2 W z − ⋯ − 𝑎𝑁 𝑧 −𝑁 W z
Taking Inverse, we get 𝑤(𝑛) = 𝑥(𝑛) − 𝑎1 𝑤(𝑛 − 1) − 𝑎2 𝑤(𝑛 − 2) − ⋯ − 𝑎𝑁 𝑤(𝑛 − 𝑁) ---- (1)
𝑀
Y z
Similarly = 𝑏𝑘 𝑧 −𝑘 = 𝑏0 + 𝑏1 𝑧 −1 + 𝑏2 𝑧 −2 + ⋯ +𝑏𝑀 𝑧 −𝑀
𝑊 𝑧
𝑘=0
𝑌 𝑧 = 𝑏0 𝑊 𝑧 + 𝑏1 𝑧 −1 𝑊 𝑧 + 𝑏2 𝑧 −2 𝑊 𝑧 + ⋯ +𝑏𝑀 𝑧 −𝑀 𝑊 𝑧
Taking Inverse, we get 𝑦 𝑛 = 𝑏0 𝑤 𝑛 + 𝑏1 𝑤 𝑛 − 1 + 𝑏2 𝑤 𝑛 − 2 + ⋯ . .
+𝑏𝑀 𝑤 𝑛 − 𝑀 −− −(2)
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Here equation (1) and (2) contain the same terms w(n) , w(n-1) , w(n-2)… Hence
they can be used common to implement both equations. (less number of delay
elements required)
x(n) w(n) 𝑏0 y(n)
𝑤 𝑛 =𝑥 𝑛 𝑏1
−𝑎1
−𝑎1 𝑤 𝑛 − 1 w(n-1)
−𝑎2 𝑤 𝑛 − 2 − ⋯ … .
−𝑎𝑁 𝑤(𝑛 − 𝑁) −𝑎2 𝑏2
w(n-2)
𝑦 𝑛 = 𝑏0 𝑤 𝑛 +
𝑏𝑀−1 𝑏1 𝑤 𝑛 − 1 +
−𝑎𝑁−1
𝑏2 𝑤 𝑛 − 2 + ⋯ . .
+𝑏𝑀 𝑤 𝑛 − 𝑀
−𝑎𝑁 𝑏𝑀
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Q. Obtain the direct form II for the given IIR filter
𝑦 𝑛 = −0.1 𝑦 𝑛 − 1 + 0.2 𝑦 𝑛 − 2 + 3𝑥 𝑛 + 3.6 𝑥 𝑛 − 1 + 0. 6𝑥(𝑛 − 2)
10
Soln: Taking Z transform on both sides
𝑌 𝑧 = −0.1 𝑧 −1 𝑌 𝑧 + 0.2 𝑧 −2 𝑌 𝑧 + 3𝑋 𝑧 + 3.6 𝑧 −1 𝑋 𝑧 + 0. 6 𝑧 −2 𝑋(𝑧)
𝑌 𝑧 + 0.1 𝑧 −1 𝑌 𝑧 − 0.2 𝑧 −2 𝑌 𝑧 = 3 𝑋 𝑧 + 3.6 𝑧 −1 𝑋 𝑧 + 0. 6 𝑧 −2 𝑋(𝑧)
𝑌 𝑧 (1 + 0.1 𝑧 −1 − 0.2 𝑧 −2 ) = 𝑋 𝑧 (3 + 3.6 𝑧 −1 + 0. 6 𝑧 −2 )
𝑌 𝑧 𝑌 𝑧 𝑊 𝑧 3 + 3.6 𝑧 −1 + 0. 6 𝑧 −2 −1 −2
1
= . = = (3 + 3.6 𝑧 + 0. 6 𝑧 ) ×
𝑋 𝑧 𝑊 𝑧 𝑋 𝑧 1 + 0.1 𝑧 −1 − 0.2 𝑧 −2 1 + 0.1 𝑧 −1 − 0.2 𝑧 −2
𝑌 𝑧 𝑊 𝑧 1
= 3 + 3.6 𝑧 −1 + 0. 6 𝑧 −2 =
𝑊 𝑧 𝑋 𝑧 1 + 0.1 𝑧 −1 − 0.2 𝑧 −2
𝑌 𝑧
= 3 + 3.6 𝑧 −1 + 0. 6 𝑧 −2
𝑊 𝑧
𝑌 𝑧 = 3 𝑊 𝑧 + 3.6 𝑧 −1 𝑊 𝑧 + 0. 6 𝑧 −2 𝑊 𝑧
Taking inverse Z transform on both sides
𝑦 𝑛 = 3 𝑤 𝑛 + 3.6 𝑤 𝑛 − 1 + 0. 6 𝑤 𝑛 − 2 −−−−− −(1)
Similarly,
𝑊 𝑧 1
= 𝑋 𝑧 = 𝑊 𝑧 + 0.1 𝑧 −1 𝑊 𝑧 − 0.2 𝑧 −2 𝑊 𝑧
𝑋 𝑧 1 + 0.1 𝑧 −1 − 0.2 𝑧 −2
𝑊 𝑧 = 𝑋 𝑧 − 0.1 𝑧 −1 𝑊 𝑧 + 0.2 𝑧 −2 𝑊 𝑧
Taking inverse Z transform on both sides
𝑤 𝑛 = 𝑥 𝑛 − 0.1 𝑤 𝑛 − 1 + 0.2 𝑤 𝑛 − 2 −−−− −(2)
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𝑦 𝑛 = 3 𝑤 𝑛 + 3.6 𝑤 𝑛 − 1 + 0. 6 𝑤 𝑛 − 2 −−−−− −(1)
x(n) w(n) 3 y(n)
−0.1 3.6
0.2 0.6
𝑤 𝑛 = 𝑥 𝑛 − 0.1 𝑤 𝑛 − 1 + 0.2 𝑤 𝑛 − 2 −−−− −(2)
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3. Transposition Theorem & Transposed Form
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Transpose of a structure is obtained by the following steps:
➢ Reverse the directions of all branches in the signal flow graph
➢ Interchange the inputs and the outputs
➢ Reverse the roles of all nodes in the flow graph
➢ Summing points become branching points
➢ Branching points become summing points
➢ According to transposition theorem, the system transfer function remains unchanged
by transposition
Signal Flow Graph:
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❑ Graphical representation of relationship between variables
❑ Basic elements are branches and nodes
❑ Contains set of directed branches that connect at nodes
❑ Node – system variable
➢ Source nodes – no entering branches
➢ Sink nodes – have only entering branches
❑ Delay – indicated by 𝑧 −1
Example of signal flow graph
Direct Form II Corresponding Signal Flow Graph
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Q. Determine direct form II and transposed direct form II for the given system
1 1
y 𝑛 = 𝑦 𝑛 − 1 − 𝑦 𝑛 − 2 + 𝑥 𝑛 + 𝑥(𝑛 − 1)
2 4
16
Soln: Taking z transform,
1 −1 1 −2
Y 𝑧 = 𝑧 𝑌 𝑧 − 𝑧 𝑌 𝑧 + 𝑋 𝑧 + 𝑧 −1 𝑋 𝑧
2 4
1 −1 1 −2
Y 𝑧 − 𝑧 𝑌 𝑧 + 𝑧 𝑌 𝑧 = 𝑋 𝑧 + 𝑧 −1 𝑋 𝑧
2 4
1 −1 1 −2
Y 𝑧 (1 − 𝑧 + 𝑧 ) = 𝑋 𝑧 (1 + 𝑧 −1 )
2 4
Y 𝑧 1 + 𝑧 −1 Y 𝑧 W 𝑧
𝐻 𝑧 = = = .
𝑋 𝑧 1 −1 1 −2 𝑊 𝑧 𝑋 𝑧
1− 𝑧 + 𝑧
2 4
W 𝑧 1 Y 𝑧
= = 1 + 𝑧 −1
𝑋 𝑧 1 1 𝑊 𝑧
1 − 𝑧 −1 + 𝑧 −2
2 4
W 𝑧 1 1 −1 1 −2
= W 𝑧 − 𝑧 W 𝑧 + 𝑧 W 𝑧 =𝑋 𝑧
𝑋 𝑧 1 1 2 4
1 − 𝑧 −1 + 𝑧 −2
2 4
1 −1 1 −2
W 𝑧 =𝑋 𝑧 + 𝑧 W 𝑧 − 𝑧 W 𝑧
2 4
1 1
𝑤 𝑛 = 𝑥 𝑛 + 𝑤 𝑛 − 1 − 𝑤 𝑛 − 2 −−−− −(1)
2 4
Y 𝑧
Similarly = 1 + 𝑧 −1 Y 𝑧 = 𝑊 𝑧 + 𝑧 −1 𝑊(𝑧)
𝑊 𝑧
y 𝑛 = 𝑤 𝑛 + 𝑤 𝑛 − 1 −−−− −(2)
17
1 1
𝑤 𝑛 = 𝑥 𝑛 + 𝑤 𝑛 − 1 − 𝑤 𝑛 − 2 −−−− −(1)
2 4
x(n) w(n) 1 y(n)
0.5 1
−0.25
y 𝑛 = 𝑤 𝑛 + 𝑤 𝑛 − 1 −−−− −(2)
Corresponding Signal Flow Graph
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➢ Reverse the directions of all branches in the signal flow graph
➢ Interchange the inputs and the outputs
➢ Reverse the roles of all nodes in the flow graph (Summing points become
branching points & branching points become summing points).
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Branching points
Summing points
Signal Flow Graph for transposed structure
20
Transposed Direct form II
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1 + 2𝑧 −1 + 𝑧 −2
Q. Obtain the transposed direct form II for the given IIR filter 𝐻(𝑧) =
3 1
1 − 𝑧 −1 + 𝑧 −2
4 8
Soln: First draw direct form II, & then find the transposed structure.
x(n) w(n) 1 y(n)
3/4 2
−1/8 1
Direct form II Signal flow graph for Direct form II
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Branching points
Summing
points
Signal Flow Graph for transposed structure Transposed Direct form II
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4. Cascade Form
❑ Consider the IIR system with system function expressed as:
𝐻 𝑧 = 𝐻1 𝑧 . 𝐻2 𝑧 . 𝐻3 𝑧 … 𝐻𝑘 𝑧
❑ Realize each 𝐻𝑘 𝑧 in Direct Form II structure and then cascade.
3 1
Q) Realize the system with difference equation 𝑦 𝑛 = 𝑦 𝑛−1 − 𝑦 𝑛−2 +
4 8
1
𝑥 𝑛 + 𝑥(𝑛 − 1) in cascade form.
3
Soln: Taking Z Transform on both sides,
3 −1 1 1
𝑌 𝑧 = 𝑧 𝑌 𝑧 − 𝑧 −2 𝑌 𝑧 + 𝑋 𝑧 + 𝑧 −1 𝑋 𝑧
4 8 3 1
3 −1 1 −2 1 −1 1 + 𝑧 −1
∴ 𝐻1 𝑧 = 3
𝑌 𝑧 1− 𝑧 + 𝑧 =𝑋 𝑧 1+ 𝑧 1
4 8 3 1 − 𝑧 −1
1 −1 2
1 + 𝑧 1
𝑌(𝑧) 3 𝐻2 𝑧 =
𝐻 𝑧 = = 1
𝑋(𝑧) 3 −1 1 −2 1 − 𝑧 −1
1− 𝑧 + 𝑧 4
4 8
1
1 + 𝑧 −1
3
𝐻 𝑧 = = 𝐻1 𝑧 . 𝐻2 𝑧
1 −1 1 −1
1− 𝑧 1− 𝑧
2 4
𝑌1 𝑧 𝑌1 𝑧 𝑊1 𝑧 1 −1 1
𝐻1 𝑧 = = . =1+ 𝑧 .
𝑋1 (𝑧) 𝑊1 (𝑧) 𝑋1 (𝑧) 3 1
1 − 𝑧 −1
2
𝑊1 𝑧 1
= 𝑌2 𝑧 𝑌2 𝑧 𝑊2 𝑧 1
𝑋1 (𝑧) 1 − 1 𝑧 −1 𝐻2 𝑧 = = . = 1.
1
2 𝑋2 (𝑧) 𝑊2 (𝑧) 𝑋2 (𝑧) 1 − 𝑧 −1
4
1 −1 𝑊2 𝑧 1
𝑊1 𝑧 − 𝑧 𝑊1 𝑧 = 𝑋1 (𝑧) =
2 𝑋2 (𝑧) 1 − 1 𝑧 −1
1 4
𝑤1 𝑛 = 𝑥1 𝑛 + 𝑤1 𝑛 − 1 −− −(1) 1
2 𝑊2 𝑧 − 𝑧 −1 𝑊2 𝑧 = 𝑋2 (𝑧)
4
𝑌1 𝑧 1 −1
=1+ 𝑧 1
𝑊1 (𝑧) 3 𝑤2 𝑛 = 𝑥2 𝑛 + 𝑤2 𝑛 − 1 −− −(3)
4
1 𝑌2 𝑧
𝑌1 𝑧 = 𝑊1 𝑧 + 𝑧 −1 . 𝑊1 (𝑧) =1 𝑌2 𝑧 = 𝑊2 𝑧
3 𝑊2 (𝑧)
1
𝑦1 𝑛 = 𝑤1 𝑛 + . 𝑤1 𝑛 − 1 −−− −(2) 𝑦2 𝑛 = 𝑤2 𝑛 −−− −(4)
3
26
1 1
𝑤1 𝑛 = 𝑥1 𝑛 + 𝑤1 𝑛 − 1 𝑤2 𝑛 = 𝑥2 𝑛 + 𝑤2 𝑛 − 1
2 4
27
Direct II form of 𝑯𝟏 𝒛 Direct II form of 𝑯𝟐 𝒛
𝒙 𝒏 = 𝒙𝟏 𝒏 𝒘𝟏 𝒏 𝒚𝟏 𝒏 𝒙𝟐 𝒏 𝒘𝟐 𝒏 𝒚 𝒏 = 𝒚𝟐 𝒏
𝟏
𝟏/2 𝟏/3 𝟏/4
1
𝑦1 𝑛 = 𝑤1 𝑛 + . 𝑤1 (𝑛 − 1) 𝑦2 𝑛 = 𝑤2 𝑛
3
Q) Obtain the cascade realization for the following system:
3 1 3
1 + 𝑧 −1 + 𝑧 −2 1 − 𝑧 −1 + 𝑧 −2
2 2 2
𝐻 𝑧 =
1 1 1
(1 + 𝑧 −1 + 𝑧 −2 ) 1 + 𝑧 −1 + 𝑧 −2
4 4 2
3 1 3
1 + 𝑧 −1 + 𝑧 −2 1 − 𝑧 −1 + 𝑧 −2
2 2 2
Soln: Given 𝐻 𝑧 = = 𝐻1 𝑧 . 𝐻2 (𝑧)
1 1 1
(1 + 𝑧 −1 + 𝑧 −2 ) 1 + 𝑧 −1 + 𝑧 −2
4 4 2
3 1 3 −1
1 + 𝑧 −1 + 𝑧 −2 1 − 𝑧 + 𝑧 −2
Where 2 2 2
𝐻1 𝑧 = & 𝐻2 𝑧 =
1 1 1
(1 + 𝑧 −1 + 𝑧 −2 ) 1 + 𝑧 −1 + 𝑧 −2
4 4 2
3 1 3 −1
1 + 𝑧 −1 + 𝑧 −2 1 − 𝑧 + 𝑧 −2
2 2 2
𝐻1 𝑧 = & 𝐻2 𝑧 =
1 1 1
(1 + 𝑧 −1 + 𝑧 −2 ) 1 + 𝑧 −1 + 𝑧 −2
4 4 2
𝑥1 (n) 𝑤1 (n) 1 𝑦1 (n)
-1 3/2
−1/4 1/2
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Parallel form
❑ The parallel form realization of an IIR system can be obtained by performing a partial
fraction expansion of:
𝑁
𝑐𝑘
𝐻 𝑧 =𝑐+
1 − 𝑝𝑘 𝑧 −1
𝑘=1
where 𝑝𝑘 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑝𝑜𝑙𝑒𝑠.
𝑐1 𝑐2 𝑐𝑁
𝐻 𝑧 =𝑐+ + +⋯
1 − 𝑝1 𝑧 −1 1 − 𝑝2 𝑧 −1 1 − 𝑝𝑁 𝑧 −1
𝑌(𝑧)
𝐻 𝑧 = = 𝑐 + 𝐻1 𝑧 + 𝐻2 𝑧 + . . . 𝐻𝑁 𝑧
𝑋(𝑧)
Parallel Form Realization
c
x(n) 𝑯𝟏 (𝒛)
𝑯𝟐 (𝒛)
𝑯𝟑 (𝒛)
𝑯𝑵 (𝒛) y(n)
Q) Realize the system with difference equation 𝑦 𝑛 = −0.1 𝑦 𝑛 − 1 +
0.72 𝑦 𝑛 − 2 + 0.7𝑥 𝑛 − 0.252 𝑥(𝑛 − 2) in parallel form.
Soln: Taking Z Transform on both sides,
𝑌 𝑧 = −0.1𝑧 −1 𝑌 𝑧 + 0.72𝑧 −2 𝑌 𝑧 + 0.7𝑋 𝑧 − 0.252𝑧 −2 𝑋 𝑧
𝑌 𝑧 1 + 0.1𝑧 −1 − 0.72𝑧 −2 = 𝑋 𝑧 0.7 − 0.252𝑧 −2
𝑌(𝑧) 0.7−0.252𝑧 −2 0.35 − 0.035𝑧 −1
𝐻 𝑧 = = Slide 34 = 0.35 +
1 + 0.1𝑧 −1 − 0.72𝑧 −2
𝑋(𝑧) 1+0.1𝑧 −1 −0.72𝑧 −2
0.35 − 0.035𝑧 −1
= 0.35 +
(1 + 0.9𝑧 −1 )(1 − 0.8𝑧 −1 )
Solving partial fraction,
0.35 − 0.035𝑧 −1 𝐴 𝐵
= +
(1 + 0.9𝑧 −1 )(1 − 0.8𝑧 −1 ) (1 + 0.9𝑧 −1 ) (1 − 0.8𝑧 −1 )
Solving for A and B, A=0.206, B=0.144
0.206 0.144 𝒙 𝒏 0.35 𝒚𝟏 𝒏
𝐻 𝑧 = 0.35 + +
(1 + 0.9𝑧 ) (1 − 0.8𝑧 −1 )
−1
𝐻1 𝑧 = 0.35 𝒙 𝒏 0.206
0.206 𝒙 𝒏 𝒚𝟐 𝒏
𝐻2 𝑧 =
1 + 0.9𝑧 −1
0.144 −0.9
𝐻3 𝑧 =
1 − 0.8𝑧 −1 𝒙 𝒏 0.144 y 𝒏
Now 𝐻1 𝑧 , 𝐻2 𝑧 & 𝐻3 𝑧 can be 𝒚𝟑 𝒏
realized using direct form II and the
outputs can be added.
0.8
34
Q) Obtain the parallel form Realization of the system with difference equation
𝑦 𝑛 = −0.1 𝑦 𝑛 − 1 + 0.2 𝑦 𝑛 − 2 + 3𝑥 𝑛 + 3.6 𝑥 𝑛 − 1 + 0. 6𝑥(𝑛 − 2)
Soln: Taking Z Transform on both sides,
𝑌 𝑧 = −0.1𝑧 −1 𝑌 𝑧 + 0. 2𝑧 −2 𝑌 𝑧 + 3𝑋 𝑧 + 3. 6𝑧 −1 𝑋 𝑧 + 0.6𝑧 −2 𝑋 𝑧
𝑌 𝑧 1 + 0.1𝑧 −1 − 0. 2𝑧 −2 = 𝑋 𝑧 3 + 3.6 𝑧 −1 + 0.6 𝑧 −2
𝑌(𝑧) 3 + 3.6𝑧 −1 + 0.6𝑧 −2
𝐻 𝑧 = =
𝑋(𝑧) 1 + 0.1𝑧 −1 − 0. 2𝑧 −2
6 + 3.9𝑧 −1 6 + 3.9𝑧 −1
= −3 + = −3 +
1 + 0.1𝑧 −1 − 0.2𝑧 −2 (1 + 0.5𝑧 −1 )(1 − 0.4𝑧 −1 )
𝐴 𝐵
H z = −3 + +
1 + 0.5𝑧 −1 1 − 0.4𝑧 −1
Solving for A and B, A = 7, B = -1
7 −1
H z = −3 + + 𝒙 𝒏 −3 𝒚𝟏 𝒏
1 − 0.4𝑧 −1 1 + 0.5𝑧 −1
𝐻1 𝑧 = −3 𝒙 𝒏 7
7 𝒙 𝒏 𝒚𝟐 𝒏
𝐻2 𝑧 =
1 − 0.4𝑧 −1
−1 0.4
𝐻3 𝑧 =
1 + 0.5𝑧 −1 𝒙 𝒏 −1 y 𝒏
𝒚𝟑 𝒏
−0.5
36
FIR Filter Vs IIR Filter
37
FIR Filter IIR Filter
Finite Impulse Response Infinite Impulse Response
Non Recursive Recursive
Present output depends on present input and Present output depends on present input, past
past input. input and past outputs.
𝑁−1 𝑁 𝑀
𝑦 𝑛 = 𝑏𝑘 𝑥(𝑛 − 𝑘) 𝑦 𝑛 = − 𝑎𝑘 𝑦 𝑛 − 𝑘 + 𝑏𝑘 𝑥(𝑛 − 𝑘)
𝑘=0 𝑘=1 𝑘=0
𝑁−1
σ𝑀𝑘=0 𝑏𝑘 𝑧 −𝑘
𝐻(𝑧) = 𝑏𝑘 𝑧 −𝑘 𝐻 𝑧 =
1 + σ𝑁𝑘=1 𝑎𝑘 𝑧
−𝑘
𝑘=0
Transfer function consists of only zeros Transfer function consists of both poles &
zeros.
More Stable Less Stable
References
38
1. Proakis J. G. and Manolakis D. G., Digital Signal Processing, 4/e, Pearson Education,
2007.
2. P. Ramesh Babu, Digital Signal Processing, Scitech Publications (India) Pvt Ltd.
39 END of PART -II
THANK YOU!
