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Circular shift and Convolution

Spring 2009

© Ammar Abu-Hudrouss -Islamic University Gaza

Slide 2Digital Signal Processing

Circular Shift

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Slide 3Digital Signal Processing

Circular Shift

In previous example, the samples from xp(n-2) 0 to N - 1 result in a circular shifted version of x(n) by 2.

Doing normal shift on xp(n) is equivalent to do circular shift on x(n)

Slide 4Digital Signal Processing

Circular Shift

NknxNknxnx

))(()ulemod,()(

)0())0(()2( 4 xxx

This can be expressed by using the following formula

)1())1(()3( 4 xxx

)3())1(()1( 4 xxx

)2())2(()0( 4 xxx x(1)

x(0)x(2)

x(3)

1

2

4

3

x’(1)

x’(0)x’(2)

x’(3)

3

4

2

1x(n)

x’(n)For our example

++

Green arrow indicates positive direction

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Slide 5Digital Signal Processing

Circular Reflection

)())(( nNxnx N

The time reversal of an N-point sequence is attained by reveresing it samples about the point zero on the circle.

x(0)1

2

4

3 y(0)1

4

2

3x(n) y(n)

Example y(n) = x((-n))N

Slide 6Digital Signal Processing

Circular Symmetry

11)()( NnnxnNx

An N-point sequence is called circularly even if

An N-point sequence is called circularly odd if

11)()( NnnxnNx

ExampleExample : x(n) = {1,2,3,3,2}

Example : x(n) = {0,2,3, -3, 2}

23

3 2

1

23

-3 -2

0

For circularly odd signal x(0) = 0 and if N is even x(N/2) = 0.

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Slide 7Digital Signal Processing

Circular Convolution

The circular convolution is very similar to normal convolution apart from that the signal is shifted using circular shift.

1

021

213

))(()(

)()()(N

nNnmxnx

nxnxmx

If x1(n) and x2(n) is two discrete signals, then the result of the circular convolution x3(n) can be realised by

Slide 8Digital Signal Processing

Example: Perform the circular convolution of the following two sequencesx1(n) = {2,1,2,1} and x2(n) = {1,2,3,4}

↑ ↑

2 1

1

2

1

x1(n)

2

3

4

x2(n)

4

13

2

x2((-n))4 2

4

6

2

x1(n)x2((-n))4

x3(0)=14

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Slide 9Digital Signal Processing

x1(n) = {2,1,2,1} and x2(n) = {1,2,3,4}↑ ↑1

22

1

2x1(n)

4

13

2

x2((-n))4

1

48

3

x1(n)x2((1-n))4

1

24

3

x2((1-n))4

x3(1)=16

Slide 10Digital Signal Processing

x1(n) = {2,1,2,1} and x2(n) = {1,2,3,4}↑ ↑

1

22

1

2x1(n)

4

13

2

x2((-n))4

2

62

4

x1(n)x2((2-n))4

2

31

4

x2((2-n))4

x3(2)=14

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Slide 11Digital Signal Processing

x1(n) = {2,1,2,1} and x2(n) = {1,2,3,4}↑ ↑1

22

1

x1(n)

4

13

2

x2((-n))4

3

84

1

x1(n)x2((3-n))4

3

42

1

x2((3-n))4

x3(3)=16

Slide 12Digital Signal Processing

x1(n) = {2,1,2,1} and x2(n) = {1,2,3,4}↑ ↑

}16,14,16,14{)()()( 213 nxnxmx

The final solution is

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Slide 13Digital Signal Processing

Self Study

Students are encouraged to solve the following questions from the textbook

5.8.