m4. fourier transform and its propertieshomes.et.aau.dk/yang/course/ztrans/discrete-time10-4.pdf ·...

3/22/2011 I. Discrete-Time Signals and Systems 1

M4. Fourier Transform and Its Properties

Reading material: p.40-65


Z-Transform

The z-transform of a sequence x[n] is defined as

ROC depends only on |z|, i.e. |X(z)|<∞∞∞∞, if

The z-transform includes a algebraic expression and ROC

e.g., Right-sided sequence anu[n] ⇔⇔⇔⇔ 1/(1-az-1), |z|>|a|

Left-sided sequence -anu[-n-1] ⇔⇔⇔⇔ 1/(1-az-1), |z|<|a|

∑∞

−∞=

−=

n

nznxzX ][)(

∑∞

−∞=

−∞<

n

nznx |||][|


Z-Transform

The z-transform of a sequence x[n] is defined as

ROC depends only on |z|, i.e. |X(z)|<∞∞∞∞, if

The z-transform includes a algebraic expression and ROC

Right-sided exponential sequence (example 3.1 pp.98)

anu[n] ⇔⇔⇔⇔ 1/(1-az-1), |z|>|a|

Left-sided exponential sequence (example 3.2, pp.99)

-anu[-n-1] ⇔⇔⇔⇔ 1/(1-az-1), |z|<|a|

The ROC properties, and

The z-transform properties

Inverse z-transform and its properties

∑∞

−∞=

−=

n

nznxzX ][)(


Laplace Transform & Fourier Transform

Laplace transform for the continuous-time signals

Fourier transform for the continuous-time signals

∫∞

−==

0)())(()( dtetxtxLsX

st

∫∞+

∞−

≥≥=j

j

st tdsesXj

txδ

δδδ

π0,0,)(

2

1)(

∫∞

Ω−=

0)()( dtetxjΩX

tj

∫∞

∞−

ΩΩΩ= ,)(

2

1)( dejXtx tj

π


Motivation for FT

Tuning fork problem

”pure tone signal” and music – a sum of pure tone

signals at different frequencies!

Distinguish two pure tone signals

Signal spectrum analysis


Frequency-Domain Representation

Complex exponential sequences like ejωωωωn are eigenfunctions

for the LTI systems, i.e., y[n] = H(ejωωωω) ejωωωωn

The system’s frequency response

H(ejωωωω) = (∑∑∑∑k=-∞∞∞∞∞∞∞∞ h[k] e-jωωωωk)

describes the change in complex amplitude of a complex

exponential input signal as a function of the frequency ωωωω

H(ejωωωω) is periodic with period 2ππππ

Frequency domain analysis, lowpass filter, highpass filter…

Suddenly Applied Complex Exponential inputs

x[n]=ejωωωωn u[n], there is y[n]=yss[n]+yt[n], where

nj

nk

kj

t

njj

ss eekhnyeeHnyωωωω )][(][,)(][

1

∑∞

+=

−−==


Motivation for Fourier Transform

If we can get the expression for any arbitrary input

sequence as

x[n]= ∑∑∑∑k ααααk ejωωωωkn

Then we could get

y[n]= ∑∑∑∑k ααααkH(ejωωωωk) ejωωωωkn

If the suddenly applied complex exponential sequence

such as x[n]= ∑∑∑∑k ααααk ejωωωωkn u[n] is employed, then we

possibly have

y[n]= ∑∑∑∑k ααααk H(ejωωωωk) ejωωωωkn - ∑∑∑∑k ααααk (∑∑∑∑k=n+1∞∞∞∞ h[k]e-jωωωωkk ejωωωωkn )

If the system is stable, the second part (transient response)

will approach zero as n→∞→∞→∞→∞


The Fourier Transform Many sequences can be represented by a Fourier integral

form as

x[n] can be regarded as a superposition of infinitesimally

small different complex sinusoids

jwk

k

j

jwnj

ekxeX

where

deeXnx

−

∞

−∞=

−

∑

∫

=

=

][)(

)(2

1][

ω

π

π

ωω

π

ωπ

ωω

ω∆= ∑ ∑

∞

−∞=

−

→∆

nj

k m

mj kk eemxnx )][(lim2

1][

0


Expressions of Fourier Transform

X(ejωωωω) can be expressed as the real and imaginary parts

X(ejωωωω)= XR(ejωωωω) + jXI(ejωωωω)

Or in polar form as

X(ejωωωω)= |X(ejωωωω)| e j<<<<X(ejωωωω)

|X(ejωωωω)| is the magnitude (magnitude spectrum, amplitude

spectrum)

<<<<X(ejωωωω) is the phase (phase spectrum)

The principal value of <<<<X(ejωωωω), denoted as ARG[X(ejωωωω) ] is

restricted to the range of -ππππ< ωωωω ≤≤≤≤ ππππ

arg[X(ejωωωω) ] refers to a phase function that is a continuous

function of ωωωω for 0< ωωωω < ππππ


Discussion of Fourier Transform

The frequency response of the LTI system is the Fourier

transform of the impulse response, and vise versa.

Problem 1: Prove that the expression of x[n] and X(ejωωωω)

are inverse of each other (exercise!)

Problem 2: What kind of signals can be expressed as

ωπ

ωπ

π

ω

ωω

deeHnh

enheH

njj

nj

n

j

∫

∑

−

−

∞

−∞=

=

=

)(2

1][

][)(

ωπ

π

π

ωdeeXnx

jwnj

∫−= )(

2

1][


Using Matlab ...

start matlab


Existence of Fourier Transform

Absolute summability is a sufficient Condition for the existence of the Fourier transform (uniform convergence),

why?

Any stable system will have a finite and continuous frequency response, and any FIR system …

∞≤≤

≤

=

∑

∑

∑

∞

−∞=

−

∞

−∞=

−

∞

−∞=

|][|

|||][|

|][||)(|

k

jwk

k

jwk

k

j

kh

ekh

ekheXω


Mean-Square Convergent Sequences

If some sequences are not absolutely summable, but are

square summable, i.e.,

∑∑∑∑k=-∞∞∞∞∞∞∞∞ |h[k]|2 < ∞∞∞∞

then its Fourier transform expression

X(ejωωωω)= ∑∑∑∑k=-∞∞∞∞∞∞∞∞ h[k] e-jωωωωk

has the property – mean square convergence

where

XM(ejωωωω)= ∑∑∑∑k=-MM h[k] e-jωωωωk

0|)()(|lim 2=−∫−∞→

ωω

π

π

ω deXeX j

M

j

M


Example 2.22 Square-summable ideal lowpass filter

Frequency response is

Impulse response is

Noncausal

hlp[n] is not absolutely

summable, but square

summable

≤<

≤=

πωω

ωωω

||,0

||,1)(

c

cjeH lp

∞<<∞−= nn

nwnh c

lp ,sin

][π


Some Useful Fourier Transforms

(Example 2.23 p.53) The Fourier transform of the constant sequence, i.e., x[n]=1 for all n, is the periodic impulse train, i.e.,

X(ejωωωω)= ∑∑∑∑r=-∞∞∞∞∞∞∞∞ 2 πδπδπδπδ(ωωωω+2ππππr)

(Example 2.24 p.54) Fourier transform of the complex exponential sequence i.e., x[n]= ejωωωω0n for all n, is

X(ejωωωω) = ∑∑∑∑r=-∞∞∞∞∞∞∞∞ 2 πδπδπδπδ(ωωωω- ωωωω0+2ππππr)

The Fourier transform of the step sequence x[n]=u[n] is represented by (p.54)

U(ejωωωω) = 1/(1- ejωωωω)+∑∑∑∑r=-∞∞∞∞∞∞∞∞ πδπδπδπδ(ωωωω+2ππππr)

Can you prove it ? …


Symmetry Properties

Motivation: Symmetry properties are often very useful for

simplifying the solution of the problems

Some definitions:

If xe[n]=xe*[-n], xe[n] is called conjugate-symmetric sequence

If xo[n]= -xo*[-n], xe[n] is called conjugate-antisymmetric

sequence

Any sequence can be expressed as a sum of a conjugate-symmetric and conjugate-antisymmetric sequence, i.e.,

X[n]= xe[n]+ xo[n], where

xe[n]=1/2(x[n]+x*[-n] )

xo[n]=1/2(x[n]-x*[-n] )

A real sequence that is conjugate-symmetric is called even sequence, while which is conjugate-antisymmetric is called odd sequence (! an error in p.55 xo[n]= -xo[-n])


Symmetry Properties of FT

A Fourier transform X(ejωωωω) can be decomposed into a sum

of conjugate-symmetric and conjugate-antisymmetric

functions, i.e., X(ejωωωω)= Xe(ejωωωω) + Xo(e

jωωωω), where

Xe(ejωωωω)=1/2(X(ejωωωω)+ X*(e-jωωωω) )

Xo(ejωωωω)=1/2(X(ejωωωω)- X*(e-jωωωω) )

If x[n] is a real sequence, the real part XR(ejωωωω) of the

Fourier transform X(ejωωωω) is an even function, and the

imaginary part XI(ejωωωω) is an odd function

If x[n] is a real sequence, the magnitude |X(ejωωωω)| of the

Fourier transform X(ejωωωω) in polar form is an even

function, and the phase <<<<X(ejωωωω) is an odd function

See example 2.25 page57….


Fourier Transform Theorems

Linearity

ax1[n]+bx2[n] ⇔⇔⇔⇔ aX1(ejωωωω)+bX2(ejωωωω)

Time shifting

x[n-nd] ⇔⇔⇔⇔ e-jωωωωnd X(ejωωωω)

frequency shifting

ejωωωω0n x[n] ⇔⇔⇔⇔ X(ej(ωωωω-ωωωω0) )

Time reversal

x[-n] ⇔⇔⇔⇔ X*(ejωωωω)

Differentiation in frequency

nx[n] ⇔⇔⇔⇔ j (dX(ejωωωω)/dωωωω)


Fourier Transform Theorems (continue)

Parseval’s Theorem

|X(ejωωωω)|2 is the energy density spectrum, i.e., it determines

how the energy is distributed in the frequency domain

The convolution Theorem: if y[n]=x[n]*h[n], then

Y(ejωωωω)=X(ejωωωω)H(ejωωωω)

The modulation (windowing) theorem: if y[n]=x[n]w[n],

then

ωπ

π

π

ω deXnxE j

n

22 |)(|2

1|][| ∫∑

−

∞

−∞=

==

θπ

θωπ

π

θωdeWeXeY

jjj )()(2

1)( )( −

−∫=


Exercise Four

Problem 2.32 on page76 of the textbook.

(optional) Prove that The Fourier transform –

equation (2.134) in page 48 of the textbook is the

inverse of the (synthesis) equation (2.133) in page

48 of the textbook;

m4. fourier transform and its propertieshomes.et.aau.dk/yang/course/ztrans/discrete-time10-4.pdf ·...

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