fourier transform 4: z-transform (part 2) & introduction ...vda.univie.ac.at › teaching ›...

052600 VU Signal and Image Processing

Fourier Transform 4: z-Transform (part 2) & Introduction to

2D Fourier Analysis

Torsten Möller + Hrvoje Bogunović + Raphael Sahann

[email protected] [email protected]

[email protected]

vda.cs.univie.ac.at/Teaching/SIP/17s/

1© Raphael Sahann

mailto:[email protected]



Overview

• Sampling & Impulse Train • Fourier Transform (1D)

– properties – convolution theorem – sampling in the Fourier space

• z-Transform – inverse z-Transform

• 2D Fourier Transform

2© Raphael Sahann

3

Check Yourself

Two signals and two regions of convergence.

≠4≠3≠2≠1 0 1 2 3 4n

x[n] =!7

8"n

u[n]

78

z-planeROC

z

z ≠ 78

≠4 ≠3 ≠2 ≠1 0 1 2n

y[n] = ≠ !78"n

u[≠1 ≠ n]

78

ROC

z-plane

z

z ≠ 78

33© Prof. Dennis Freeman, MIT, 2011

4

ÿ

Properties of Z Transforms

The use of Z Transforms to solve di↵erential equations depends on several important properties.

Property x[n] X(z) ROC

Linearity ax1[n] + bx2[n] aX1(z) + bX2(z) ∏ (R1 fl R2)

Delay x[n ≠ 1] z ≠1X(z) R dX(z)

Multiply by n nx[n] ≠z dz

R

Convolve in n x1[m]x2[n ≠ m] X1(z)X2(z) ∏ (R1 fl R2)

m=≠Œ

45

Œÿ

© Prof. Dennis Freeman, MIT, 2011

Relationship between the z-Transform and the Fourier Transform

5© Raphael Sahann

or


6© Raphael Sahann

This z-Transform can be interpreted as the Fourier transform of the product of the original sequence x[n] and the

exponential sequence r-n

r = 1 reduces to the Fourier transform of x[n].

7


© Raphael Sahann

image source: http://flylib.com/books/2/729/1/html/2/images/0131089897/graphics/06fig13.gif

http://flylib.com/books/2/729/1/html/2/images/0131089897/graphics/06fig13.gif

• Interpreting Fourier transform as the z-transform on the unit circle in the z-plane corresponds to wrapping the frequency axis around the unit circle.

• Inherent periodicity in frequency is captured naturally, since a change of angle of 2π radians in the unit circle corresponds to traversing the unit circle once and returning to the same point.

8


© Raphael Sahann

• If the Region of Convergence (ROC) includes the unit circle, the Fourier transform and all its derivatives with respect to ⍵ must be continuous functions of ⍵.

9


© Raphael Sahann

10

Example 3.5

© Raphael Sahann

Consider the sequence

with a = -1/3 we obtain

and using a = 1/2 yields

11

Example 3.5

© Raphael Sahann

Consider the sequence

with a = -1/3 we obtain

and using a = 1/2 yields

12

Example 3.5

© Raphael Sahann

by the linearity of the z-transform

image source: http://d2r5da613aq50s.cloudfront.net/wp-content/uploads/405300.image2.jpg

http://d2r5da613aq50s.cloudfront.net/wp-content/uploads/405300.image2.jpg

Example 3.7

• Since there is no overlap between |z|>1/2 and |z|<1/3, x[n] has no z-transform (nor Fourier transform) representation.

13

x[n] =

✓1

2

◆n

u[n]�✓�1

3

◆n

u[�n� 1]

X(z) =1

1� 12z

�1

| {z }|z|> 1

2

+1

1 + 13z

�1

| {z }|z|< 1

3

© Raphael Sahann

Example 3.7

• Since there is no overlap between |z|>1/2 and |z|<1/3, x[n] has no z-transform (nor Fourier transform) representation.

14

x[n] =

✓1

2

◆n

u[n]�✓�1

3

◆n

u[�n� 1]

X(z) =1

1� 12z

�1

| {z }|z|> 1

2

+1

1 + 13z

�1

| {z }|z|< 1

3

© Raphael Sahann

Overview

• Sampling & Impulse Train • Fourier Transform (1D) • z-Transform

– inverse z-Transform • 2D Fourier Transform

15© Raphael Sahann

Inverse z-Transform

• The inverse z-transform is a complex contour integral, where C represents a closed contour within the ROC of the z-transform.

• Too complex for a typical task in this domain, so we use less formal procedures

16

x[n] =1

2⇡j

I

CX(z)zn�1

dz

© Raphael Sahann

Inspection Method

• We “inspect” a z-transform by looking up its transform in a table of common z-transforms.

• Find the inverse z-transform for:

17© Raphael Sahann

X(z) =1

1� 12z

�1, |z| > 1

2

Common z-transform pairs

18

image source: http://www.dip.ee.uct.ac.za/~nicolls/lectures/eee401f/03_ztrans.pdf© Raphael Sahann

http://www.dip.ee.uct.ac.za/~nicolls/lectures/eee401f/03_ztrans.pdf

Inspection Method• Find the inverse z-transform for:

19© Raphael Sahann

X(z) =1

1� 12z

�1, |z| > 1

2

right-sided

x[n] =

✓1

2

◆n

u[n]

Inverse z-transform

• Inspection Method

• Partial Fractions —> build partial fractions until they can be interpreted by the inspection method

• Power Series Expansion —> Taylor series expansion of z-transform, interpret result with inspection method

20

21

ÿ

Properties of Z Transforms

The use of Z Transforms to solve di↵erential equations depends on several important properties.

Property x[n] X(z) ROC

Linearity ax1[n] + bx2[n] aX1(z) + bX2(z) ∏ (R1 fl R2)

Delay x[n ≠ 1] z ≠1X(z) R dX(z)

Multiply by n nx[n] ≠z dz

R

Convolve in n x1[m]x2[n ≠ m] X1(z)X2(z) ∏ (R1 fl R2)

m=≠Œ

45

Œÿ

© Prof. Dennis Freeman, MIT, 2011

Overview

• Sampling & Impulse Train • Fourier Transform (1D) • z-Transform • 2D Fourier Transform

22© Raphael Sahann

© Laurent Condat / Torsten Möller

How to represent an image?

• An image is made of pixels (=picture elements)

• the coordinate values are discretized

23

f(t) ⇡1X

n=�1cne

j 2⇡nT t

cn =1

T

Z T/2

�T/2f(t)e�j 2⇡n

T tdt

© Torsten Möller

What is a Fourier Transform? (1D)

• Let’s go back to (spatial) representation of functions:

• Fourier series — into Frequency Domain:

24

f(t) ⇡1X

n=�1

✓Z 1

�1f(t) (t� n�T )dt

◆�(t� n�T )

f(t) ⇡1X

n=�1c[n]�(t� n�T )

• Let’s go back to (spatial) representation of functions:

• Fourier series — into Frequency Domain:

f(x) ⇡1X

n=�1cn

ej2⇡nT ·x

cn

=1

Tx

Ty

ZT/2

�T/2f(x)e�j

2⇡nT ·xdx

© Torsten Möller

What is a Fourier Transform? (2D)

25

f(x) ⇡1X

n=�1

✓Z 1

�1f(s) (s� n�X)ds

◆�(x� n�X)

f(x) ⇡1X

n=�1c[n]�(x� n�X) �X =

✓�x 00 �y

◆

© Torsten Möller

There are 4 Fourier Transforms! (1D)

• Recall Fourier series:

• f(t) is periodic with period T! • General Fourier Transform requires no

periodicity:

26

cn =1

T

Z T/2


T t

f(t) ⇡1X

n=�1cne

j 2⇡nT t

F (!) =

Z 1

�1f(t)e�j2⇡!tdt

f(t) =

Z 1

�1F (!)ej2⇡!tdt

© Torsten Möller

There are 4 Fourier Transforms! (2D)

• Recall Fourier series:

• f(x) is periodic with period T! • General Fourier Transform requires no

periodicity:

27

f(x) ⇡1X

n=�1cn

ej2⇡nT ·x

cn

=1

Tx

Ty

ZT/2

�T/2f(x)e�j

2⇡nT ·xdx

F (!) =

Z 1

�1f(x)e�j2⇡!·xdx

f(x) =

Z 1

�1F (!)ej2⇡!·xdx

© Torsten Möller

DFT — the most important one (1D)

• Discrete Fourier Transform (DFT) requires periodicity in both transform pairs

28

Fm =M�1X

n=0

fne�j2⇡mn/M

fn =1

M

M�1X

m=0

Fmej2⇡mn/M

© Torsten Möller

DFT — the most important one (2D)

• Discrete Fourier Transform (DFT) requires periodicity in both transform pairs

29

Fab =M�1X

m=0

N�1X

n=0

fmne�j2⇡(am/M+bn/N)

fmn =1

MN

M�1X

a=0

N�1X

b=0

Fabej2⇡(am/M+bn/N)

F (!) =

Z 1

�1f(t)e�j2⇡!tdtf(t) =

Z 1

�1F (!)ej2⇡!tdt

f(t) =1X

n=�1cne

j 2⇡nT t

Fm =M�1X

n=0

fne�j2⇡mn/Mfn =

1

M

M�1X

m=0

Fmej2⇡mn/M

fn =1

2⇡

Z ⇡

�⇡F (!)ej!nd! F (!) =

1X

n=�1fne

�j!nd!

cn =1

T

Z T/2


T tdt

© Torsten Möller

All Fourier Transforms (1D)

30

FT

FS —Fourier Series

DFT —Discrete FT

DTFT —Discrete Time FT

Spatial Domain Frequency Domain

continuous continuous

discretecontinuous + periodic

discrete + periodic discrete + periodic

continuous + periodicdiscrete

© Torsten Möller

All Fourier Transforms (nD)

31

FT

FS —Fourier Series

DFT —Discrete FT

DTFT —Discrete Time FT

Spatial Domain Frequency Domain

continuous continuous

discretecontinuous + periodic

discrete + periodic discrete + periodic

continuous + periodicdiscrete

© Torsten Möller

What happens to an impulse? (1D)

• it is basically a constant!

32

cn =1

T

Z T/2

�T/2�(t)e�j 2⇡n

T t

cn =1

Te0

cn =1

T

© Torsten Möller

What happens to an impulse? (2D)

• it is basically a constant!

33

cn

=1

Tx

Ty

ZT/2

�T/2�(x)e�j

2⇡nT ·xdx

cn

=1

Tx

Ty

e0

cn

=1

Tx

Ty

© Torsten Möller

What about a shifted impulse? (1D)

• the shifts remain as frequencies

34

cn =1

T

Z T/2

�T/2�(t� t0)e

�j 2⇡nT t

cn =1

Te�j 2⇡n

T t0

© Torsten Möller

What about a shifted impulse? (2D)

• the shifts remain as frequencies

35

cn

=1

Tx

Ty

ZT/2

�T/2�(x� x

0

)e�j

2⇡nT ·xdx

cn

=1

Tx

Ty

e�j

2⇡nT ·x0

© Torsten Möller

What happens to an impulse train?• Impulse train is periodic — apply Fourier

series, will not do the math here, see book:

• distance between impulses grows inversely

36

s�T (t) =1X

�1�(t� n�T )

S�T (!) =1

�T

1X

n=�1�(! � n

�T)

• Impulse train is periodic — apply Fourier series, will not do the math here, see book:

• distance between impulses grows inversely

© Torsten Möller

What happens to an impulse train?

37

S�X(!) =1

|�X|

1X

n=�1�(! ��X�1n)

s�X(x) =1X

n=�1�(x��Xn)

�X =

✓�x 00 �y

◆

© Torsten Möller

What happens to a box?

• it is the well-known sinc function

39

sinc(Tt, Zz) =sin(T⇡t)

⇡t

sin(Z⇡z)

⇡z

© Torsten Möller

What is the Fourier Transform of a convolution?

• convolution <==> multiplication:

• multiplication <==> convolution:

40

f ⇤ h(x) () F (!)H(!)

f(x)h(x) () F (!) ⇤H(!)

© Torsten Möller

What is sampling in 2D? (mathematically speaking)

41

�(x, y) =

⇢1 if x = y = 00 if t 6= 0

Z 1

�1

Z 1

�1�(x, y)dxdy = 1

• modeled through an ‘impulse’

• not really a function, but a distribution:

© Torsten Möller

The sifting property

• picking a value off from f:

• more general:

42

Z 1

�1f(x)�(x)dt = f(0)

Z 1

�1f(x)�(x� x0)dx = f(x0)

fourier transform 4: z-transform (part 2) & introduction ...vda.univie.ac.at › teaching ›...

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