n-1

Digital signal processing: Lecture 3
Linear,
Time Invariant Systems - II
Produced by Qiangfu Zhao (Since 1995), All rights reserved ©
DSP-Lec03/1
Review of last lecture
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Linear system
Time-invariant system
Impulse response
Convolution sum
Cascade and parallel
connection
• Causality and stability
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線形システム
時不変システム
インパルス応答
畳み込み和
従続接続と並列接続
因果的システムと安定
的システム
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DSP-Lec03/2
Topics of this lecture
• Systems represented by linear constant-coefficient
difference equations.
• Finding the response (output) of a filter from a given
difference equation.
• Components of digital signal processor.
• Direct-form realization I of digital filters.
• Direct-form realization II of digital filters.
教科書5.5節と5.6節参照
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DSP-Lec03/3
Topics of this lecture
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線形定係数差分方程式で表せるシステム
差分方程式からシステムの応答(出力)を求める方法
デジタルフィルタの構成要素
デジタルフィルタの直接形実現 I
デジタルフィルタの直接形実現 II
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DSP-Lec03/4
Linear constant coefficient difference equation
線形定係数差分方程式
N
N
k =1
k =0
y ( n ) = − ∑ ak y ( n − k ) + ∑ bk x ( n − k ) (5.29)
• The output of an LTI system represented by linear
constant-coefficient different equation is found by
using N input data observed up to now and N output
data already found before.
• N is called the ORDER of the filter.
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DSP-Lec03/5
Implementation (実装方法)
x(n)
y(n)
x(n-1)
y(n-1)
x(n-2)
+
-
+
x(n-N)
y(n-2)
y(n-N)
First-in-first-out memories (Queues)
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DSP-Lec03/6
Why linear constant-coefficient
different equations ?
• Representing an LTI system using its
impulse response is not easy if the length of
h(n) is infinite.
• We can always find the output using finite
number of computations.
• By selecting properly the coefficients, we
can get different kinds of digital filters (or
digital signal processors).
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DSP-Lec03/7
FIR and IIR systems again
• A system is an FIR system if the outputs
are calculated using the input data only.
• If the coefficients ak (k=1,2,…,N) are all
zeros, the impulse response will be h[k]=bk
(see Eq. (5.30) and Eq. (5.31) in p. 76).
• A system is an IIR system if at least one ak
is not zero.
• An IIR filter is also called a recursive filter
(再帰型デジタルフィルタ).
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DSP-Lec03/8
Example 5.4 pp. 76-77
• A system is given as a difference equation.
y ( n ) = −a1 y ( n − 1) + b0 ( n )
(1) Find the impulse response and unit step response of
the filter;
(2) Plot the impulse response and unit step response for
a1=-0.5 and b0=0.5.
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DSP-Lec03/9
Impulse response
y (0) = −a1 y ( −1) + b0δ (0) = b0
y (1) = −a1 y (0) + b0δ (1) = −a1b0
y ( 2) = −a1 y (1) + b0δ ( 2) = ( −a1 ) b0
2

y ( n ) = −a1 y ( n − 1) + b0δ ( n ) = ( −a1 ) n b0 , n ≥ 0
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DSP-Lec03/10
Unit step response
y (0) = −a1 y ( −1) + b0u0 (0) = b0
y (1) = −a1 y (0) + b0u0 (1) = [( −a1 )1 + 1]b0
y ( 2) = −a1 y (1) + b0u0 ( 2) = [( −a1 ) 2 + ( −a1 )1 + 1]b0

y ( n ) = −a1 y ( n − 1) + b0u0 ( n ) = [( −a1 )n + ... + ( −a1 )1 + 1]b0
1 − ( −a1 ) n +1
=
b0 , n ≥ 0
1 − ( −a1 )
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Fig. 5.8 in p. 77
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表5.1
Tはサンプリング周期、1とすることができる
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DSP-Lec03/13
Other symbols for the basic components
x(n)
z-1
y(n)
x1(n)
x2(n)
y(n)
x3(n)
a
x(n)
y(n)
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DSP-Lec03/14
Structure of an IIR system
 The direct-form I
N
N
k =1
r =0
y ( n ) = − ∑ ak y ( n − k ) + ∑ br x ( n − r )
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DSP-Lec03/15
Property of LTI system
The whole system will keep the same even if we
change the positions of the sub-systems
h1(n)
h2(n)
h2(n)
h1(n)
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DSP-Lec03/16
The direct-form II
• The system
contains two subsystems connected
in cascade.
• The positions of the
two sub-systems
can be exchanged
without changing
the property of the
system.
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DSP-Lec03/17
The direct-form II
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DSP-Lec03/18
Homework
• デジタルフィルタが以下の差分方程式で与えられた
とする:
y (n) = x(n) + 0.1 y (n − 1)
• このフィルタのインパルス応答とステップ応答を求
めよ。ただし、y(-1)=0とする。
• pp. 77-78のプログラムを修正し、インパルス応答と
ステップ応答を求め、図で表示せよ。
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DSP-Lec03/19
Quiz and self-evaluation
•
T1
例題5.4の差分方程式で記述され
るデジタルフィルタの直接形実現II
を与えよ。
1.2
1
0.8
0.6
0.4
T5
T2
0.2
0
T4
Name:
Student ID:
T3
.