finite impulse response of lti systems
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By Assoc.Prof.Dr. Thuong Le-Tien 1
DIGITAL SIGNAL PROCESSING
FINITE IMPULSE RESPONSE
OF LTI SYSTEMS
Lectured by: Assoc. Prof. Dr. Thuong Le-Tien
National Distinguished Lecturer
HCMC September, 20111
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Practical DSP methods fall in two basis classes:
1. Block Processing Methods
2. Sample Processing Methods
In block processing methods, data are collected andprocessed in blocks (DFT/FFT spectrumcomputation, speech analysis and synthesis, imageprocessing.
In sample processing methods, data are processedone at a time (real-time applications, audio effectsprocessing , digital controls,
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In this chapter, block processing and sampleprocessing methods applied to FIR filteringand Convolution. Several computationalaspects of convolution equations areconsidered:
* Direct form* Convolution table
* LTI form
* Matrix form
* Flip-and-slide form* Overlap-add block convolution form.
* Z-Transform (discussed in the Z-transformChapter)
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1. Block processing methods
1.1. Convolution
Sampling time interval, T=1/fs.
Number of time samples: L = TLfs
x(n) = [x0, x1, , xL-1]where n = 0, 1, , L 1:
The direct and convolution forms
Convolution table form
mm
mnhmxmnxmhny )()()()()(
)()()()(.
njijxihnyji
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1.2 DIRECT FORMConsider a causal FIR filter of order M withimpulse response h(n),
h = [h0, h1, , hM]
where n = 0, 1, , M
the length of the filter or the number of filtercoefficients LH = M + 1
The output of the filter:
m
mnxmhny )()()(
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with conditions 0 m Mand 0 n m L 1
m n L 1 + m
The limit of output index n:
0 m n L 1 + m L 1 + M 0 n L 1 + M
y = [y0, y1, y2, , yL 1 + M]
The length of output: Ly = L + M
Ly = Lx + Lh1
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M must satisfy simultaneously the inequalities
0 m Mn L + 1 m n
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It follows:
max(0, n L + 1 ) m min(n,M)
The direct form of convolution
Example: an order 3 filter and a length 5-inputsignal: h = [h0, h1, h2, h3]
x = [x0, x1, x2, x3, x4]
y = h * x = [y0, y1, y2, y3, y4, y5, y6, y7]
),min(
)1,0max()()()(
Mn
Lnmmnxmhny
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Equation applied:
for n= 0, 1, 2,.7
max (0, 0 4 ) m min(0, 3) => m = 0
max (0, 1 4 ) m min(1, 3) => m = 0, 1max (0, 2 4 ) m min(2, 3) => m = 0,1 ,2
max (0, 3 4 ) m min(3, 3) => m = 0, 1, 2, 3
max (0, 4 4 ) m min(4, 3) => m = 0, 1, 2, 3
max (0, 5 4 ) m min(5, 3) => m = 1, 2, 3
max (0, 6 4 ) m min(6, 3) => m = 2, 3
max (0, 7 4 ) m min(7, 3) => m = 3
i.e. n = 5, y5 = h1x4 + h2x3 + h3x2
7...,,1,0)()()()3,min(
)4,0max(
nmnxmhny
n
nm
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All the output samples:y0 = h0x0y1 = h0x1 + h1x0y
2= h
0x
2+ h
1x
1+ h
2x
0y3 = h0x3 + h1x2 + h2x1 + h3x0y4 = h0x4 + h1x3 + h2x2 + h3x1y5 = h1x4 + h2x3 + h3x2y6 = h2x4 + h3x3y7 = h3x4
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1.3. Convolution table:
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Example: Find the convolution of the following
filter and input signalh = [1, 2, -1, 1]
x = [1, 1, 2, 1, 2, 2, 1, 1]
y = [1, 3, 3, 5, 3, 7, 4, 3, 3, 0, 1]
Ly = L + M = 8 + 3 = 11
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1.4. LTI form:
h = [h0, h0, h2, h3]
x = [x0, x1, x2, x3, x4]
Input x can be rewritten as a linear combination of delayedimpulses
x = x0[1, 0, 0, 0, 0] + x1[0, 1, 0, 0, 0] + x2[0, 0, 1, 0, 0] +
x3[0, 0, 0, 1, 0] + x4[0, 0, 0, 0, 1]
x(n)=x0(n)+x1(n1)+x2(n2)+x3(n3)+x4(n4)
Then:
y(n)=x0h(n)+x1h(n1)+x2h(n2)+x3h(n3)+x4h(n4)
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We can present the input and output signals as blocks:
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LTI form of convolution
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Example: h = [1, 2, -1, 1] and
x = [1, 1, 2, 1, 2, 2, 1, 1]
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The LTI form can also be written in a formsimilar by determine the proper limits of
For n = 0, 1, , L + M 1
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1.5. Matrix formLinear matrix form: y = Hx
The filter matrix H must be rectangular withdimensions:Ly * Lx = (L + M) * L
H is also called the TOEPLITZ MATRIX in the sense of thatit has the same entry a long each diagonal
Hx
x
x
x
x
x
h
hhhhh
hhhh
hhhh
hhh
hh
h
y
yy
y
y
y
y
y
y
4
3
2
1
0
3
23
123
0123
0123
012
01
0
7
6
5
4
3
2
1
0
0000
00000
0
0
00
000
0000
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Example:
There is also a matrix
Form written as follows
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1.6. Flip and slide
yn = h0xn + h1xn-1 + h2xn-2 + + hMxn-Mthe first M outputs correspond to the input-on
transient behavior of the filter, and the last Moutputs beyond the end of the input data arethe input-off transients. The remains are thesteady-state outputs.
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1.7. Transient and Steady-State Behavior
Length of input signal is L, filter order M, the outputcan be separated into 3 parts
0 n < M (input on transient)
M n L 1 (steady state)
L 1 < n L 1 + M (input off transient)
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Example: An IIR filter has xung h(n) =(0.75)nu(n). Using convolution to find y(n)when the inputs are:
a) x(n) = u(n)
b) x(n) = (-1)nu(n)
c) x(n) = u(n) u(n 25)
Find the steady state response for each case.
Solve:
a)
n
m
n
m
ny00
)( m)-u(m)u(n(0.75)m)-h(m)x(n n
n
m
nn
0
1
)75.0(3475.01
)75.0(1n(0.75)
475.01
1)(
nylim
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b)
Steady state response:
n
m 0
( mn (0.75)-1)
n
m
n
m
ny00
)( m-nm (-1)(0.75)m)-h(m)x(n
(0.75)73+
74(-1)=
75.01)75.0(1(-1) nn
1
n
n
7
4)1(
75.01
11)(-y(n) n n
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n
Lnm
mn
Lnm
mnmn xhy
)1,0max()1,0max(
)75.0(c)
Two cases:
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nnn
m
ny )75.0(34
75.01
)75.0(1(0.75)
1
0
m
n
nm
ny24
2524-nm
0.75-1
(0.75)-1(0.75)=(0.75)
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1.8. Overlap-Add Block Convolution Method
Overlap-add convolution method25By Assoc.Prof.Dr. Thuong Le-Tien
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Example:
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2. Sample processing method
Adder
Multiplier
Delay
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FIR Filtering in Direct Form
Direct form realization of third-order filter.
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