pds_modulo5

Digital Signal Processing

Module 5: Linear FiltersDigital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Module Overview:

Module 5.15.3: LTI systems and convolution; examples; stability

Module 5.4: Convolution theorem

Module 5.55.6: Ideal filters and their approximation

Module 5.75.8: Realizable filters; implementation

Module 5.95.10: Filter design

5 1

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


Module 5.1: Linear FiltersDigital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Overview:

Linearity and time invariance

Convolution

5.1 2

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

A generic signal processing device

x [n] H y [n]

y [n] = H{x [n]}

5.1 3

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Linearity

H{ x1[n] + x2[n]} = H{x1[n]}+ H{x2[n]}

5.1 4

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Linearity

5.1 5

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

(Non) Linearity

5.1 6

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Time invariance

y [n] = H{x [n]} H{x [n n0]} = y [n n0]

5.1 7

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Time invariance

5.1 8

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Time (in)variance

5.1 9

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Linear, time-invariant systems

x [n] H y [n]

additionscalarmultiplication

delays

5.1 10

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Linear, time-invariant systems

y [n] = H(x [n], x [n 1], x [n 2], . . . , y [n 1], y [n 2], . . .)

with H() a linear function of its arguments

5.1 11

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Impulse response

h[n] = H{[n]}

Fundamental result: impulse response fully characterizes the LTI system!

5.1 12

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Example

h[n]

b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

3 0 3 6 9 12 150

1

h[n] = n u[n]

x [n]

b b

b

b

b

b b b

2 1 0 1 2 3 4 50

1

2

3

4

x [n] =

2 n = 0

3 n = 1

1 n = 2

0 otherwise

5.1 13

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Example

x [n] = 2[n] + 3[n 1] + [n 2]

we know the impulse response h[n] = H{[n]};

compute y [n] = H{x [n]} exploiting linearity and time-invariance

5.1 14

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Example

y [n] = H{2[n] + 3[n 1] + [n 2]}

= 2H{[n]} + 3H{[n 1]}+H{[n 2]}

= 2h[n] + 3h[n 1] + h[n 2]

5.1 15

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Example

b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b

2h[n]

5 0 5 10 15 20 250

1

2

3

4

5

5.1 16

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Example

b b b b b

b

b

b

b

b

b

b

b

b

b

b

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b b b b b

b b b b b b

b

b

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b

b

b

b

b

b

b b

3h[n 1]

5 0 5 10 15 20 250

1

2

3

4

5

5.1 16

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Example

b b b b b

b

b

b

b

b

b

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b b b b b

b b b b b b

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b b

b b b b b b b

b

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b

b

b

b b b b b b

h[n 2]

5 0 5 10 15 20 250

1

2

3

4

5

5.1 16

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Example

b b b b b

b

b

b

b

b

b

b

b

b

b

b

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b

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b

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b

b

b

y [n]

5 0 5 10 15 20 250

1

2

3

4

5

5.1 16

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution

We can always write (remember Module 3.2):

x [n] =

k=

x [k][n k]

by linearity and time invariance:

y [n] =

k=

x [k]h[n k]

= x [n] h[n]

5.1 17

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Performing the convolution algorithmically

x [n] h[n] =

k=

x [k]h[n k]

Ingredients:

a sequence x [m]

a second sequence h[m]

The recipe:

time-reverse h[m]

at each step n (from to ):

center the time-reversed h[m] in n(i.e. shift by n)

compute the inner product

5.1 18

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Same example, different perspective

h[n]

b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

3 0 3 6 9 12 150

1

h[n] = n u[n]

x [n]

b b

b

b

b

b b b

2 1 0 1 2 3 4 50

1

2

3

4

x [n] =

2 n = 0

3 n = 1

1 n = 2

0 otherwise

5.1 19

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b

b b b b b b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b b b b b b b b b b b b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b b b b b b b b b b b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b b b b b b b b b b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b

b b b b b b b b b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b

b

b b b b b b b b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b

b

b

b


8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b b

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b b b

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b b b

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b b b

b b

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b b b b

b b

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b b b b

b b b

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b b b b b

b b b

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b b b b b b

b b b

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Convolution example

b b b b b b b b b

b

b

b


8 4 0 4 8 12 16 200

2

4x[k]

b b b b b b

b b b

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

8 4 0 4 8 12 16 200

1

h[n

k]

b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b

8 4 0 4 8 12 16 200246

y[n]

5.1 20

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Convolution properties

linearity and time invariance (by definition)

commutativity: (x h)[n] = (h x)[n]

associativity for absolutely- and square-summable sequences:((x h) w)[n] = (x (h w))[n]

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Convolution properties

linearity and time invariance (by definition)

commutativity: (x h)[n] = (h x)[n]

associativity for absolutely- and square-summable sequences:((x h) w)[n] = (x (h w))[n]

x [n] h[n] w [n] y [n]

x [n] (h w)[n] y [n]

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END OF MODULE 5.1

Digital Sig

nal Proces

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Paolo Pran

doni and M

artin Vett

erli

2013


Module 5.2: Filtering by exampleDigital Sig

nal Proces

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Paolo Pran

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2013

Overview:

Moving average filter

Leaky integrator

5.2 22

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Typical filtering scenario: denoising

0 100 200 300 400 500

6420246

5.2 23

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0 100 200 300 400 500

6420246

0 100 200 300 400 500

6

0

6

5.2 23

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0 100 200 300 400 500

6420246

0 100 200 300 400 500

6

0

6

0 100 200 300 400 500

6420246

5.2 23

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Denoising by Moving Average

idea: replace each sample by the local average

for instance: y [n] = (x [n] + x [n 1])/2)

more generally:

y [n] =1

M

M1k=0

x [n k]

5.2 24

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artin Vett

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M = 2

0 100 200 300 400 500

6

4

2

0

2

4

6

5.2 25

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artin Vett

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M = 4

0 100 200 300 400 500

6

4

2

0

2

4

6

5.2 25

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M = 12

0 100 200 300 400 500

6

4

2

0

2

4

6

5.2 25

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M = 100

0 100 200 300 400 500

6

4

2

0

2

4

6

5.2 25

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MA: impulse response

h[n] =1

M

M1k=0

[n k]

=

{1/M for 0 n < M

0 otherwise

5.2 26

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doni and M

artin Vett

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MA: impulse response


b b b b b b b b b


0 M 10

1/M

5.2 27

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MA: analysis

smoothing effect proportional to M

number of operations and storage also proportional to M

5.2 28

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From the MA to a first-order recursion

yM [n] =1

M(x [n] + x [n 1] + x [n 2] + . . .+ x [n M + 1])

moving average over M points

5.2 29

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yM [n] =1

Mx [n] +

1

M(x [n 1] + x [n 2] + . . .+ x [n M + 1])

5.2 30

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yM [n] =1

Mx [n] +

1

M(x [n 1] + x [n 2] + . . .+ x [n M + 1])

almostyM1[n 1]

i.e., moving average over M 1 points, delayed by one

5.2 30

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Formally:

yM [n] =1

M

M1k=0

x [n k]

yM [n 1] =1

M

M1k=0

x [(n 1) k]

5.2 31

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Formally:

yM [n] =1

M

M1k=0

x [n k]

yM [n 1] =1

M

M1k=0

x [n (k + 1)]

5.2 31

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Paolo Pran

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2013


Formally:

yM [n] =1

M

M1k=0

x [n k]

yM [n 1] =1

M

M1+1k=0+1

x [n k]

5.2 31

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Paolo Pran

doni and M

artin Vett

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2013


Formally:

yM [n] =1

M

M1k=0

x [n k]

yM [n 1] =1

M

Mk=1

x [n k]

5.2 31

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2013


Formally:

yM [n] =1

M

M1k=0

x [n k]

yM [n 1] =1

M

Mk=1

x [n k]

yM1[n] =1

M 1

M2k=0

x [n k]

5.2 31

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doni and M

artin Vett

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2013


Formally:

yM [n] =1

M

M1k=0

x [n k]

yM [n 1] =1

M

Mk=1

x [n k]

yM1[n 1] =1

M 1

M1k=1

x [n k]

5.2 31

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doni and M

artin Vett

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2013


yM [n] =1

M

M1k=0

x [n k] yM1[n 1] =1

M 1

M1k=1

x [n k]

M1k=0

x [n k] = x [n] +

M1k=1

x [n k]

MyM [n] = x [n] + (M 1)yM1[n 1]

5.2 32

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doni and M

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2013


yM [n] =M 1

MyM1[n 1] +

1

Mx [n]

yM [n] = yM1[n 1] + (1 )x [n], =M 1

M

5.2 33

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doni and M

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2013

The Leaky Integrator

when M is large, yM1[n] yM [n] (and 1)

try the filtery [n] = y [n 1] + (1 )x [n]

filter is now recursive, since it uses its previous output value

5.2 34

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artin Vett

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2013

Denoising recursively with the Leaky Integrator

= 0.2

0 100 200 300 400 500

6

4

2

0

2

4

6

5.2 35

Digital Sig

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Paolo Pran

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artin Vett

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2013


= 0.5

0 100 200 300 400 500

6

4

2

0

2

4

6

5.2 35

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Paolo Pran

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2013


= 0.8

0 100 200 300 400 500

6

4

2

0

2

4

6

5.2 35

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Paolo Pran

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artin Vett

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2013


= 0.98

0 100 200 300 400 500

6

4

2

0

2

4

6

5.2 35

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doni and M

artin Vett

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2013

What about the impulse response?

y [n] = y [n 1] + (1 )[n]

y [n] = 0 for all n < 0

y [0] = y [1] + (1 )[0] = (1 )

y [1] = y [0] + (1 )[1] = (1 )

y [2] = y [1] + (1 )[2] = 2(1 )

y [3] = y [2] + (1 )[3] = 3(1 )

. . .

5.2 36

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Impulse response

h[n] = (1 )n u[n]

b b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

1

0 15 300

5.2 37

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Paolo Pran

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Leaky Integrator: why the name

Discrete-time integrator is a boundless accumulator:

y [n] =

nk=

x [k]

We can rewrite the integrator as

y [n] = y [n 1] + x [n]

5.2 38

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Paolo Pran

doni and M

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2013

Leaky Integrator: why the name

To prevent explosion pick < 1

y [n] = y [n 1] + (1 )x [n]

keep only a fraction ofthe accumulated valueso far and forget(leak) a fraction 1

add only a fraction 1 of the current value tothe accumulator

5.2 39

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2013

END OF MODULE 5.2

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


Module 5.3: Filter StabilityDigital Sig

nal Proces

sing

Paolo Pran

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2013

Overview:

Filter classification in the time domain

Filter stability

5.3 40

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Filter types according to impulse response

Finite Impulse Response (FIR)

Infinite Impulse Response (IIR)

causal

noncausal

5.3 41

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FIR

impulse response has finite support

only a finite number of samples are involved in the computation of each output sample

5.3 42

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FIR (example)

Moving Average filter


b b b b b b b b b


M 1

1/M

15 0 150

5.3 43

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Paolo Pran

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2013

IIR

impulse response has infinite support

a potentially infinite number of samples are involved in the computation of each outputsample

surprisingly, in many cases the computation can still be performed in a finite amount ofsteps

5.3 44

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IIR (example)

Leaky Integrator


b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b b

15 0 150

5.3 45

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Causal vs Noncausal

causal:

impulse response is zero for n < 0

only past samples (with respect to the present) are involved in the computation of eachoutput sample

causal filters can work on line since they only need the past

noncausal:

impulse response is nonzero for some (or all) n < 0

can still be implemented in a oine fashion (when all input data is available on storage, e.g.in Image Processing)

5.3 46

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Causal example

Moving Average filter


b b b b b b


18 12 6 0 6 12 180

5.3 47

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Noncausal example

Zero-centered Moving Average filter


b b b b b


18 12 6 0 6 12 180

5.3 48

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Stability

key concept: avoid explosions if the input is nice

a nice signal is a bounded signal: |x [n]| < M for all n

Bounded-Input Bounded-Output (BIBO) stability: if the input is nice the output shouldbe nice

5.3 49

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Fundamental Stability Theorem

A filter is BIBO stable if and only if its impulse response is absolutely summable

5.3 50

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Proof ()

Hypotheses:

|x [n]| < M

n |h[n]| = L

Proof ()

Hypotheses:

|x [n]| < M

|y [n]| < P

Thesis:

h[n] absolutelysummable

Proof (by contradiction):

assume

n |h[n]| =

build x [n] =

{+1 if h[n] 0

1 if h[n] < 0

clearly, x [n] is bounded

however

y [0] = (xh)[0] =

k=

h[k]x [k] =

k=

|h[k]| =

5.3 52

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The good news

FIR filters are always stable

5.3 53

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Checking the stability of IIRs

Lets check the Leaky Integrator:

n=

|h[n]| = |1 |

n=0

||n

= limn

|1 |1 ||n+1

1 ||

Checking the stability of IIRs

We will study indirect methods for filter stability later in this Module.

5.3 55

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END OF MODULE 5.3

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


Module 5.4: Frequency ResponseDigital Sig

nal Proces

sing

Paolo Pran

doni and M

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2013

Overview:

Eigensequences

Convolution theorem

Frequency and phase response

5.4 56

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A remarkable result

e j0n H ?

5.4 57

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A remarkable result

y [n] = e j0n h[n]

= h[n] e j0n

=

k=

h[k]e j0(nk)

= e j0n

k=

h[k]ej0k

= H(e j0) e j0n

5.4 58

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doni and M

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2013

A remarkable result

e j0n H H(e j0) e j0n

complex exponentials are eigensequences of LTI systems, i.e., linear filters cannot changethe frequency of sinusoids

DTFT of impulse response determines the frequency characteristic of a filter

5.4 59

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2013

Magnitude and phase

If H(e j0) = Ae j, then

H{e j0n} = Ae j(0n+)

amplitude:amplification (A > 1)or attenuation (0 A < 1)

phase shift:delay ( < 0)or advancement ( > 0)

5.4 60

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doni and M

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2013

The convolution theorem

In general:

DTFT {x [n] h[n]} =?

Intuition: the DTFT reconstruction formula tells us how to build x [n]from a set of complex exponential basis functions. By linearity...

5.4 61

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The convolution theorem

DTFT {x [n] h[n]} =

n=

(x h)[n]ejn

=

n=

k=

x [k]h[n k]ejn

=

n=

k=

x [k]h[n k]ej(nk)ejk

=

k=

x [k]ejk

n=

h[n k]ej(nk)

= H(e j)X (e j)

5.4 62

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doni and M

artin Vett

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2013

Frequency response

H(e j) = DTFT {h[n]}

Two effects:

magnitude: amplification (|H(e j)| > 1) or attenuation (|H(e j)| < 1) of inputfrequencies

phase: overall delay and shape changes

5.4 63

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2013

Moving Average revisited

h[n] = (u[n] u[nM])/M


b b b b b b b b b b

b b b b b b b b b b b

0 M 10

1/M

5.4 64

Digital Sig

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sing

Paolo Pran

doni and M

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2013

Moving Average, magnitude response

|H(e j)| = 1M

sin(2 M)sin(2 ) (see Module 4.7)

M = 9

pi pi/2 0 pi/2 pi0

1

|H(e

j)|

5.4 65

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doni and M

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2013


|H(e j)| = 1M


M = 20

pi pi/2 0 pi/2 pi0

1

|H(e

j)|

5.4 65

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Paolo Pran

doni and M

artin Vett

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2013


|H(e j)| = 1M


M = 100

pi pi/2 0 pi/2 pi0

1

|H(e

j)|

5.4 65

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Paolo Pran

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2013

Denoising revisited

0 100 200 300 400 500

6420246

0 100 200 300 400 500

6

0

6

5.4 66

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Paolo Pran

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2013

Denoising revisited

0 100 200 300 400 500

6420246

0 100 200 300 400 500

6

0

6

pi pi/2 0 pi/2 pi0

1

pi pi/2 0 pi/2 pi0

1

5.4 66

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Paolo Pran

doni and M

artin Vett

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2013

Denoising revisited

pi pi/2 0 pi/2 pi0

1

5.4 67

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By the way, remember the time-domain analysis...

0 100 200 300 400 500

6

4

2

0

2

4

6

5.4 68

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By the way, remember the time-domain analysis...

M = 12

0 100 200 300 400 500

6

4

2

0

2

4

6

5.4 68

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What about the phase?

Assume |H(e j)| = 1

zero phase: H(e j) = 0

linear phase: H(e j) = d

nonlinear phase

5.4 69

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Phase and signal shape

x [n] =1

2sin(0n) + cos(20n) 0 =

2

40

0 28 56 84 112 140 168

1

0

1

5.4 70

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2013

Phase and signal shape: linear phase

x [n] =1

2sin(0n + 0) + cos(20n+ 20) 0 =

8

5

0 28 56 84 112 140 168

1

0

1

5.4 71

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Phase and signal shape: nonlinear phase

x [n] =1

2sin(0n) + cos(20n + 20)

0 28 56 84 112 140 168

1

0

1

5.4 72

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Linear phase

x [n] D x [n d ]

y [n] = x [n d ]

Y (e j) = ejd X (e j)

H(e j) = ejd

linear phase term

5.4 73

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Linear phase

In general, if H(e j) = A(e j)ejd , with A(e j) R

x [n] A(e j) delay by d y [n]

5.4 74

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Paolo Pran

doni and M

artin Vett

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2013

Moving Average is linear phase

H(e j) =1

M

sin(2M

)sin(2

) ej M12

5.4 75

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Paolo Pran

doni and M

artin Vett

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2013

Leaky integrator revisited

h[n] = (1 )n u[n]

b b b b b b b b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

1

0 15 300

5.4 76

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H(e j) =1

1 ej(Module 4.4)

Finding magnitude and phase require a little algebra...

5.4 77

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Recall from complex algebra:1

a + jb=

a jb

a2 + b2

so that if x = 1/(a + jb),

|x |2 =1

a2 + b2

x = tan1[b

a

]

5.4 78

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doni and M

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2013


H(e j) =1

(1 cos) j sin

so that:

|H(e j)|2 =(1 )2

1 2 cos + 2

H(e j) = tan1[

sin

1 cos

]

5.4 79

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Paolo Pran

doni and M

artin Vett

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2013

Leaky integrator, magnitude response

= 0.9

pi pi/2 0 pi/2 pi0

1

|H(e

j)|

5.4 80

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


= 0.95

pi pi/2 0 pi/2 pi0

1

|H(e

j)|

5.4 80

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


= 0.98

pi pi/2 0 pi/2 pi0

1

|H(e

j)|

5.4 80

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Leaky integrator, phase response

= 0.9

pi/2

pi/2

pi pi/2 0 pi/2 pi

H(e

j)

5.4 81

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


= 0.95

pi/2

pi/2

pi pi/2 0 pi/2 pi

H(e

j)

5.4 81

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


= 0.98

pi/2

pi/2

pi pi/2 0 pi/2 pi

H(e

j)

5.4 81

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Phase is sufficiently linear where it matters

pi pi/2 0 pi/2 pi0

1|H

(ej)|

pi/2

pi/2

pi pi/2 0 pi/2 pi

H(e

j)

5.4 82

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Karplus-Strong revisited, again!

x [n] + b y [n]

zM

y [n] = y [n M] + x [n]

5.4 83

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

KS revisited

y [n] = n/Mx [n mod M] u[n]

0 166 332 498 664 830 996

1

0

1

5.4 84

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Karplus-Strong revisited

y [n] = x [0], x [1], . . . , x [M 1], x [0], x [1], . . . , x [M 1], 2x [0], 2x [1], . . .

5.4 85

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Karplus-Strong revisited

y [n] = x [0], x [1], . . . , x [M 1], x [0], x [1], . . . , x [M 1], 2x [0], 2x [1], . . .

1st period 2nd period . . .

5.4 85

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

DTFT of KS signal, using the convolution theorem

key observation:

y [n] = x [n] w [n], w [n] =

{k for n = kM

0 otherwise

Y (e j) = X (e j)W (e j)

5.4 86

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

DTFT of KS signal, using the convolution theorem

X (e j) = ej(M + 1

M 1

)1 ej(M1)

(1 ej)2

1 ej(M+1)

(1 ej)2

W (e j) =1

1 ejM

5.4 87

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

DTFT of KS

|X (e j)|

pi pi/2 0 pi/2 pi

5.4 88

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

DTFT of KS

|W (e j)|

pi pi/2 0 pi/2 pi

5.4 88

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

DTFT of KS

|Y (e j)|

pi pi/2 0 pi/2 pi

5.4 88

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

END OF MODULE 5.4

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


Module 5.5: Ideal FiltersDigital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Overview:

Filter classification in the frequency domain

Ideal filters

Demodulation revisited

5.5 89

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Filter types according to magnitude response

Lowpass

Highpass

Bandpass

Allpass

Moving Average and Leaky Integrator are lowpass filters

5.5 90

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Filter types according to phase response

Linear phase

Nonlinear phase

5.5 91

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

What is the best lowpass we can think of?

ccpi 0 pi0

1

H(e

j)

5.5 92

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

What is the best lowpass we can think of?

cc

b = 2c

pi 0 pi0

1

H(e

j)

5.5 92

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Ideal lowpass filter

H(e j) =

{1 for || c

0 otherwise(2-periodicity implicit)

perfectly flat passband

infinite attenuation in stopband

zero-phase (no delay)

5.5 93

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Ideal lowpass filter: impulse response

h[n] = IDTFT{H(e j)

}=

1

2

pipi

H(e j)e jnd

=1

2

cc

e jnd

=1

n

e jcn ejcn

2j

=sincn

n

5.5 94

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Ideal lowpass filter: impulse response

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

0

c/pi

20 10 0 10 20

5.5 95

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

The bad news

impulse response is infinite support, two-sided= cannot compute the output in a finite amount of time= thats why its called ideal

impulse response decays slowly in time= we need a lot of samples for a good approximation

5.5 96

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Nevertheless...

The sinc-rect pair:

rect(x) =

{1 |x | 1/20 |x | > 1/2

sinc(x) =

sin(x)

xx 6= 0

1 x = 0

(note that sinc(x) = 0 when x is a nonzero integer)

5.5 97

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

The ideal lowpass in canonical form

rect

(

2c

)DTFT

cpi

sinc(cpin

)

5.5 98

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Example

c = /3: H(ej) = rect(3/2), h[n] = (1/3)sinc(n/3)

pi 2pi/3 pi/3 0 pi/3 2pi/3 pi0

1

H(e

j)

b

b b

b

b b

b

b b

b

b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b b

b

b b

b

b b

b

b b

b

1/3

30 20 10 0 10 20 30

0

h[n]

5.5 99

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Little-known fact

the sinc is not absolutely summable

the ideal lowpass is not BIBO stable!

example for c = /3: h[n] = (1/3) sinc(n/3)

take x [n] = sign{sinc(n/3)} and

y [0] = (x h)[0] =1

3

k=

| sinc(k/3)| =

5.5 100

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Divergence is however very slow...

s(n) = (1/3)n

k=n

| sinc(k/3)|

1 5000 100000

1

2

3

4

5.5 101

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Ideal highpass filter

ccpi 0 pi0

1

H(e

j)

5.5 102

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Ideal highpass filter

Hhp(ej) =

{1 for || c


Hhp(ej) = 1 Hlp(e

j)

hhp[n] = [n]csinc(

cn)

5.5 103

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013

Ideal bandpass filter

0 c 0 0 + c0pi 0 pi0

1

H(e

j)

5.5 104

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


ccpi 0 pi0

1

5.5 105

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


0pi 0 pi0

1

5.5 105

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


00pi 0 pi0

1

5.5 105

Digital Sig

nal Proces

sing

Paolo Pran

doni and M

artin Vett

erli

2013


Hbp(ej) =

{1 for | 0| c


hbp[n] = 2 cos(0n)csinc(

cn)

5.5 106

Digital Sig

nal Proces

sing

pds_modulo5

Documents

exampley n

example x n

x n n0

time invariancey n

compute y n

y n n05

timeinvariant systemsy

martin vetterli 2013linearity5