3/22/2011 I. Discrete-Time Signals and Systems 1
M4. Fourier Transform and Its Properties
Reading material: p.40-65
3/22/2011 I. Discrete-Time Signals and Systems 2
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 |||][|
3/22/2011 I. Discrete-Time Signals and Systems 3
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 ][)(
3/22/2011 I. Discrete-Time Signals and Systems 4
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
3/22/2011 I. Discrete-Time Signals and Systems 5
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
3/22/2011 I. Discrete-Time Signals and Systems 6
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
∑∞
+=
−−==
3/22/2011 I. Discrete-Time Signals and Systems 7
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→∞→∞→∞→∞
3/22/2011 I. Discrete-Time Signals and Systems 8
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
3/22/2011 I. Discrete-Time Signals and Systems 9
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
3/22/2011 I. Discrete-Time Signals and Systems 10
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][
3/22/2011 I. Discrete-Time Signals and Systems 11
Using Matlab ...
start matlab
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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
3/22/2011 I. Discrete-Time Signals and Systems 13
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
3/22/2011 I. Discrete-Time Signals and Systems 14
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
3/22/2011 I. Discrete-Time Signals and Systems 15
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 ? …
3/22/2011 I. Discrete-Time Signals and Systems 16
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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])
3/22/2011 I. Discrete-Time Signals and Systems 18
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….
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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)
3/22/2011 I. Discrete-Time Signals and Systems 21
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)( )( −
−∫=
3/22/2011 I. Discrete-Time Signals and Systems 22
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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;