Finite Impulse Response of LTI Systems

5/19/2018 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,

2By Assoc.Prof.Dr. Thuong Le-Tien


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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:


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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Finite Impulse Response of LTI Systems