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Course by: Thad Welch. E-mail the author

# LSI/LTI Systems

Module by: Richard Baraniuk. E-mail the author

Summary: Introduction to LSI/LTI systems.

Definition 1: linear shift invariant system
A linear shift invariant system is one that is both:
• linear
• shift invariant
Also known as LSI, LTI (Figure 1).

## Note:

LSI systems are the bread 'n' budduh of DSP (Figure 2).

## Characterizing LSI Systems

### 1.

Since the system is linear, it can be represented as a matrix multiply (Figure 3).

### 2.

Since is shift invariant, it cannot be just any old matrix. Its values are highly constrained.

In particular we know that

C - m C m = C - m C m
(1)
Let's understand this...

## Understanding Conditions on Matrix ℋ for Shift Invariance

Recall Figure 3.

yn= ( row n of | x ) = ( h n r | x ) y n ( row n of | x ) ( h n r | x )
(2)
y=x y x
·=( ) ·
(3)
where the · is in the nth n th row.

Now shift x x down circularly m m units. If the system is SI then y y will also shift down circularly m m units. i.e.: C m y= C m x C m y C m x

·=( ) ·
(4)
where the · is in the nth n th row, the ▪ is in the n+mth n m th row, and the ↓ has been shifted down circularly mm units.

Key: we want the value ▪ in C m y C m y to equal the · value in yy.

This implies that the rows of must circularly shift as we shift xx and yy. i.e.: row n+m n m of is equal to the circular shift right of row nn of by mm. i.e.: h n + m r = h n r C m T h n + m r h n r C m i.e.: all rows of are circular shifts of each other.

### Example 1

y 0 y 1 y 2 =( 147 258 369 ) x 0 x 1 x 2 y 0 y 1 y 2 1 4 7 2 5 8 3 6 9 x 0 x 1 x 2 LSI also needs m=1 m 1 : y 2 y 0 y 1 =( 147 258 369 ) x 2 x 0 x 1 y 2 y 0 y 1 1 4 7 2 5 8 3 6 9 x 2 x 0 x 1 and m=2 m 2 : y 1 y 2 y 0 =( 147 258 369 ) x 1 x 2 x 0 y 1 y 2 y 0 1 4 7 2 5 8 3 6 9 x 1 x 2 x 0 i.e.: 1=5=9 1 5 9 2=6=7 2 6 7 3=4=8 3 4 8 =( 132 213 321 ) 1 3 2 2 1 3 3 2 1

is called a circulant matrix.

• each row is a circulary shifted version of the row above (right).
• each column is a circularly shifted version of the column to the left (down).
=( 132 213 321 )= h 0 r h 0 r C 1 T h 0 r C N - 1 T=( h 0 c C 1 h 0 c C N - 1 h 0 c ) 1 3 2 2 1 3 3 2 1 h 0 r h 0 r C 1 h 0 r C N - 1 h 0 c C 1 h 0 c C N - 1 h 0 c
(5)
which implies that either the first row or first column are all you need to know to know all of .

### Note:

Circulant matrices are a special case of Toeplitz matrices, which are constant along diagonals. e.g.: T=( )=( 1356 2135 4213 7421 ) T 1 3 5 6 2 1 3 5 4 2 1 3 7 4 2 1 Tn,k=tnk T n k t n k

### Example 2

Also, row nn, column kk element of is

n,k=h(nk)modN n k h n k N
(6)
where 0nN1 0 n N 1 and 0kN1 0 k N 1 and h h is the signal corresponding to the first (i.e. the zeroth!) column of .

### Example 3

N=2 N 2 , h 0 c =-173T h 0 c -1 7 3

=( h(00)modNh(01)modNh(02)modN h(10)modNh(11)modNh(12)modN h(20)modNh(21)modNh(22)modN )=( h0h2h1 h1h0h2 h2h1h0 ) h 0 0 N h 0 1 N h 0 2 N h 1 0 N h 1 1 N h 1 2 N h 2 0 N h 2 1 N h 2 2 N h 0 h 2 h 1 h 1 h 0 h 2 h 2 h 1 h 0
(7)

### Example 4

Apply a 3-point moving average smoother to a signal x=( -2-101210-1 ) x -2 -1 0 1 2 1 0 -1 .

In Figure 4,
yn=1x(n1)modN+2xn+3x(n+1)modN y n 1 x n 1 N 2 x n 3 x n 1 N
(8)
y=( 23000001 12300000 01230000 00123000 00012300 00001230 00000123 30000012 )-2-101210-1=-8-42884-2-8 y 2 3 0 0 0 0 0 1 1 2 3 0 0 0 0 0 0 1 2 3 0 0 0 0 0 0 1 2 3 0 0 0 0 0 0 1 2 3 0 0 0 0 0 0 1 2 3 0 0 0 0 0 0 1 2 3 3 0 0 0 0 0 1 2 -2 -1 0 1 2 1 0 -1 -8 -4 2 8 8 4 -2 -8
(9)

The relationship between rows and columns of :

n,k=h(nk)modN n k h n k N
(10)
where nn is the row, kk is the column, and 0nN1 0 n N 1 and 0kN1 0 k N 1 .

### Note:

Rows and columns run time in reverse order!!!

### Example 5

R4 4

y0y1y2y3=( 2301 1230 0123 3012 )x0x1x2x3 y 0 y 1 y 2 y 3 2 3 0 1 1 2 3 0 0 1 2 3 3 0 1 2 x 0 x 1 x 2 x 3
(11)
where the zeroth column, 2103 2 1 0 3 , is the impulse response, hn h n (Figure 5). In Equation 11, the zeroth row, ( 2301 ) 2 3 0 1 , is the time-reversed impulse response, hk h k .

## Upshot for LSI Systems

i.e.: if we input Figure 6 and measure output h 0 c h 0 c , we can place it in the zeroth column of and then replicate it with circular shifts to build the entire !!

## Summary: LSI Systems and Imuplse Response

Given an LSI system (Figure 3), we can characterize it by the impulse response, hh (Figure 9).

and build up with circular shifts of the zeroth column, hh:
=( h C 1 h C N - 1 h ) h C 1 h C N - 1 h
(12)
Then we can compute y y for any input xCN x N through
y=x y x
(13)

### Exercise 1

How to get the impulse response?

4-point edge detector for 8-point signals in complex space (Figure 10).

h h =

=

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