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DIGITAL SIGNAL PROCESSING
Unit 1
UNIT I OVERVIEW OF SIGNALS AND SYSTEM S 9
--Basic elements of digital signal Processing
Concept of frequency in continuous time and discrete time signals
Discrete time signals. Discrete time systems
Analysis of Linear t ime invariant systems
Z transform
Convolution and correlation.
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Introduction
W hat is signal processing ?
Signal processing is the action of changing one or more features (parameters)
of a signal according to a predetermined requirement. The parameters that is to
be changed may be , amplitude, frequency, phase etc... of the signal . The signal
which undergoes such a process is know n as input signal. And the entity which
performs this processing is known as "Signal processing system" or simply
system. ie; the input signal is fed to the system and the appropriately processed
signal is coming out of the system. This signal is known as the "Output signal".
This process is illustrated below .
Some of the signal processing tasks are Amplification, Attenuation, Filtering,
Modulation, etc... . To accomplish the required task, the "system" is designed with
appropriate behavior and features. ie; for an amplifier, the "system" is designed
such that any input signal going to the "system" will be amplified. In actual circuit,
such a behavior can be realized using transistors , FET or Op-Amps. This sort of
signal processing is know n as analog signal processing. Well ! , then what is Digital
Signal Processing ....?
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Analog signal processing Vs Digital Signal Processing
In analog processing system the information is represented
as an analog quantity. ie; something that varies continuously with time. This
"something" can be voltage , current etc.. . for example, the signal that comes out
of a microphone is analog in nature. The varying voltage from the microphone is
processed using an analog signal processing system known as "amplifier" , and
reproduced through a speaker.
Where in Digital Signal Processing system , the information is represented in
digital format. The signals that are to be processed is converted into numerical
form before any processing. This conversion is known as sampling. The
information contained in an analog signal is f irst converted to digital samples which
are equally spaced in time. The figure below shows an analog signal and its
sampled version.
Each of these samples are converted in to a numerical value and stored in the
computer' s memory. Processing is then done on these samples. The sampling in a
Digital Signal Processing system is governed by a theorem know n as sampling
theorem.
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Classification of Signals: Continuous-Time verses Discrete-Time Signals
Continuous time or analog signals are signals that are defined for every value of a
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1
DIGITAL SIGNAL PROCESSING
Lecture 1 - Chapter 1
Classification of Signals: Continuous-Time verses Discrete-Time Signals
Continuous time or analog signals are signals that are defined for every value of a < t N
Infinite duration
2. Right-Left sided
( ) 10 Nnnx = left-sided
Some Elementary Discrete-Time Signals
- unit sample sequence ( )
==
00
01
n
nn
- unit step sequence ( )
+
=
+
=
zzz
z
zz
z
nun
Zn
From Table on page 174
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18
( )
+
=
21
11
25.05.01
5.01
zz
z
dz
dzzX
( )4321
543
0625.025.075.01
0625.05.025.0
++
+=
zzzz
zzzzX ; |z| > 0.5
Now, lets use MATLAB to see if weve computed correctly.
b = [0, 0, 0, 0.25, - 0.5, 0.0625];
a = [1, -1, 0.75, -0.25, 0.1625];
n = 0: 20 % checking the fist 21 samples ofx(n)
delta = [n =0]; % creating (n)
x = filter (b, a, delta),
plot (n, x), hold
x = [zeros (1, 2) n.* (0.5.n) * cos(pi * n/3)]; % creating the original signal
n1 = 0:22;
plot (n1, x, r)
RationalZ-Transforms
( )
=
=
=N
k
k
k
M
k
k
k
za
zb
zX
0
0 ifao and bo0, then we can rewrite it as: ( )( )
( )kN
k
k
M
kNM
pz
zzz
a
bzX
=
=
=+
1
1
0
0
It has Mfinite zeros atz1, z2, , zM andNfinite poles atp1, p2, ., pN as well asN Mzeros or
M Npoles at origin and a possible zero/pole at . Depending on the location of the poles, the
signal has different behaviors. Read Section 3.3.2.
The System Function of a LTI System
Y(z) = H(z) X(z) H(z) is called the system-function. A system in general can be presented
by a difference equation:
( ) ( ) ( )
( ) ( ) ( )zXzbzYzazY
knxbknyany
kN
kk
kN
kk
M
kk
N
kk
=
=
==
+=
+=
01
01
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19
( )( )( )
=
=
=
=
=+
==N
k
k
k
M
k
k
k
N
k
k
k
M
ok
k
k
za
zb
za
zb
zX
zYzH
0
0
1
1
, where a0 = 1
Special Cases:
Ifak= 0 for1 < k < N, then ( ) kMM
kkM
kM
kk zb
zzbzH
=
= ==
00
1, which is an all-zeros system.
The system has Mtrivial poles at the origin. Such a system has a finite duration impulse response
and therefore is called FIR system.
On the other hand, ifbk= 0 for Mk1 then ( ) 1,1
0
0
0
1
0 ==+
=
=
=
a
za
zb
za
bzH
N
K
kN
k
N
N
k
k
k
.
This system is an all-pole system (has N trivial zeros at origin) and therefore, has an infinite
duration impulse response and thus is called IIR system. A pole-zero system is still IIR because
of the poles.
The Inverse of Z Transform
( ) ( ) kk
zkxzX +
==
By multiplying both sides of the above formula byzn-1
and integrating both sides over a closed
contour within ROC ofX(z), which encloses the origin, we have:
( ) ( ) dzznxdzzzX kn
c k
n
c
+
=
=11
Since the series converges on this contour, we can interchange and . Then
( )43421
==
=
=
kn knj
c
kn
kc
n dzzkxdzzzX
0 2
11)(
Cauchy Integral Theorem
( ) ( ) dzzzxj
nx n
c
1
2
1 =
One of Caushy Theorems states ( )
=
= 1
1
2 n
n
j
odzzz
n
c
o
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20
Let zo = o. Then, f(z) = zn. Ifn is positive the antiderivitive
1
1
+
+
n
znis analytic every where and
therefore, its contour integral is zero. But only forf(z) = z-1
it doesnt have an antiderivitive even
in a punctured plane. For 2n , it is analytic in a punctured plane with origin deleted.
Remember that iff is analytic in a simply connected domain, D, and is any loop (close
contour) in D, then 0)( =
dzzf because in a simply connected domain any loop can be shrunk
to a point. Therefore, the integral of a continuous function over a shrinking loop converges to
zero.
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21
Lecture 5
Calculating the Inverse Z-Transform
( ) ( ) dzzzXj2
1nx 1n=
Three methods to calculate it:
1) Direct Method by contour integration
2) Expansion into a series of termsz/z-1
3) Partial Fraction expansion and look-up table.
Cauchy-Residue Theorem
Let f(z) be a function of the complex variable zand Cbe a closed path in the z-plane. If the
derivative ( )zfdzd
exists on and inside the contour Cand iff(z) has no poles at z = z0, then
( )( )
( )
= else
Cinsideiszifzfdz
zz
zf
j c
00
0 02
1
More generally:
( )( )
( )( )
=
=
else
Cinsidezifdz
zfd
kdzzz
zf
jzzk
k
ck
01
1
0
0
0!1
1
2
1
RHS of the above equation is called residue of the pole zo.
Now suppose the function can be written as( )( )zgzf
wheref(z) has no poles inside Candg(z) is a
polynomial with simple rootsz1, z2, , zninside C. Then,
( ) ( ) ( )( )i
n
i
i
i
in
ic
n
i i
i
c
zAdzzz
zA
jdz
zz
zA
jdz
zg
zf
j ===
=
=
=
111 2
1
2
1
)(2
1
,
where ( ) ( )
( )
( )zgzf
zzzA ii = .
( ) ( ) ( ) ( )izz
nN
i
i
n zzXzzdzzzxj
nx =
=
==1
1
1
2
1
= sum at residue ofx(z)zn-1
at z = zi and N=
number of poles.
Example: Problem 3.56 (C)
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22
( )a
zaz
azzX
1
1>
=
( ) dzzaz
az
ajdzz
az
az
jnx n
c
n
c
11
1
1
2
1
12
1
=
=
arz
1>= This is the contour C.
Three cases:
1) For 1n then ( ) ( ) 11
= nzaz
azf and the only pole inside c is
a
1. Therefore, for 1n
( ) ( )
11
1
1
1
11
1111
+
=
=
=
=
nn
n
az
n
aa
aa
aazaz
anx
2) For ( )( )
dzazz
az
jnxon
==
12
1
then, ( ) ( )aza
zf
=1
and two poles (0 anda
1) inside C. Therefore,
( )( ) ( )
az
az
az
az
anx
az
oz
a
1111
1
=
+
=
==
3) Forn < 0, ( ) ( ) dzzazaz
ajnx
n
c
1
1
1
1
2
1
=
Therefore, it has a pole at zero with order of (n1) and a pole at 1/a. Since
ROC ( )nxa
1z > is right-sided and therefore, it is enough to find where it reaches the zero
on the left side.
( ) ( ) 011
1
1
2
11 122 =
+
=
= == ozaz az
az
dz
d
z
az
adz
azz
az
ajx
With the same method, we can prove thatx(-2)=x(-3)==0 .
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23
The Inverse z-Transform by Power Series Expansion
The method is to use long division. How to divide.
Example:
( ) ( )( ) 232311
211
12
2
2111+=+==
zz
z
zzzzzX
It has two poles: z = 1 andz = 2. Therefore, if the signal is casual, then ROC: |z| > 2 and if the
signal is non-casual, ROC: |z| < 1.
Case 1:
ROC: |z| > 2 signal is casual and therefore ( ) ( ) non
znxzX
=
= has terms with negative power
forz. Therefore, we divide as:
...31715731
3032304515
14150
14217
670
693
230
231
1
231
4321
54
543
43
432
32
321
21
2121
+++++
++
+
+
+
+
++
zzz
zzzzz
zz
zzz
zz
zzz
zz
zzzz
x(n) = {1, 3, 7, 15, 31,}
Case 2: ROC: |z| < 1 signal is non-casual ( ) ( ) nznxzX
=0
has terms with positive
powers ofz. Therefore, we divide in a way to getzwith positive powers.
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24
...8
7
4
3
2
1
4
3
4
70
4
3
4
9
2
3
2
1
2
30
2
1
2
31
1
132
432
32
32
2
21
12
+++
+
+
+
+
+
zzz
zz
zzz
zz
zz
zz
( )
= 0,0,2
1,
4
3,
8
7...,nx
Question: How would you use this method for a case like ROC: 1 < |z| < 2?
The Inverse z-Transform by Partial- Fraction Expansion
( )( )( )zDzN
zX = The goal is to write it in the form
( )( ) ( ) ( ) ( )113
3
1
2
2
1
1
1
1...
111 ++
+
+
=
zp
A
zp
A
zp
A
zp
AzX
N
N
Let ( )N
N
MM
zazaza
zbzbbzX
+++++++=...1
....2
2
1
1
110 . If it is an improper rational function (M > N) then the
first step is to write it as:
( ) ( )( )( )zDzN
zczcczX NMNM11
10 ..... ++++=
, where now the degree ofN1(z) is less thanN.
The inverse z-transformation ofc0 + c1z-1
+ . is very straight-forward. Next step is to write
( )( )zDzN1 or in general a proper function
( )( )zDzN
, where M < N, in terms of positive powers ofz,
factorize denominator and then write( )
z
zXin terms of partial fractions. In general, lets assume
anX(z) hasNsimple poles and L multiple poles at k = j. Then:
( )( ) ( ) ( )N
N
j
j
j
j
j
j
P
A
p
A
p
A
p
A
pz
A
pz
A
z
zX
+
++
+
++
+
=
1...
1....
11....
2
21
2
2
1
1
l
l
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25
Then( ) ( )
kpzk
kz
zXpzA =
= k = 1, 2, ., n excluding k = j and
( ) ( )
=
z
zXpz
zd
dA
k
i
k
k
jk l
l
, k = 1, 2, ..,l
Then
( ) ( )( ) ( )
=
k
N
k
k
n
k
k pzROCnup
pzROCnup
zpZ
:1
:
1
11
1
Example
( )( )( )
2
115.05.11
1 2
21
=+
=
zz
z
zzzX
( )
( )( )2
112
11
21
+
=
=
z
A
z
A
zz
z
z
zX
( )( )
2
21
11 11 === =z
z
zXzA
( ) ( ) 12
12
1
21
212 =
==
=zz
zXzA
( )11
211
1
1
2
21
1
1
2
=
=
zzzzzzX
Now depending on ROC, we get differentx(n).
1) ROC: |z| > 1 casual
( ) ( ) ( ) ( ) ( ) ( ) ( )nunununx nnn 5.022
112 ==
2) ROC: |z| < non-casual and left-sided
( ) ( ) ( ) ( )
( ) ( )15.0215.0112
+=
+=
nu
nununx
n
nn
3) 12
1
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26
( ) ( ) ( ) ( ) ( )nununx nn 5.0112 =
One-Sided Z Transform
One-sided Z Transform is defined as ( ) ( ) ( ) ( ) nnon
znunxznxzX +
=
+
== . It does not contain
information aboutx(n) forn < 0. It is unique only for the causal signals that are zero forn < 0. It
is useful to solve difference equations of the systems that are not relaxed initially, but the input is
not necessarily zero before applying to the system.
Properties
If ( ) ( )zXnx Z ++
then ( ) ( ) ( ) okznxzXzknxk
n
nkZ >
+
=
++ ,1
Proof: ( ){ } ( ) non
zknxknxZ
=
+ = . Let n k = lthen n = 0 l = -kand also -n = -l k.
( )
{( )
( )
( ) ( )
+=
+
=
=
=
+
=
=
=
+
K
1n
nK1
K
zx
on
K
K
K
zXznxzzxzxz
zzx
l l
ll
l
l
43421
ll
l
Time Advance: ( ) ( ) ( ) 0,1
>
+
=
++ kznxzXzknxk
on
nkZ
Final Value Theorem: ( ) ( )zXzimnximzn
+
=1
)1(ll
This limit exists if the ROC of(z-1)X+(z) includes the unit circle.
This can be proved by the following analogy. If the limit )(lim nxn exists, then the function
x(n) can be written as 0)(limand)(limwhere),()( ==+=
nfcnxnfcnxnn
. Thentake the one
sided Z-Transform from both sides and prove the theorem.
Analysis of the Systems in Z-Domain
Without too much restriction, lets assume
( )( )( )zQzN
zx = and ( )( )( )zAzB
zH =
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27
( )( ) ( )( ) ( )zAzQ
zNzBzY
= . If the system is relaxed, the initial condition = o
If the system is not relaxed, then
( ) ( ) ( )( )
( )( )( )
321
43421
zY
o
zY
zi
zs
zA
zN
zxzHzY
+
+= where ( ) ( )n
k
n
kN
kk znyzazN
=
= = 110
H(z) hasp1, ., pNpoles andX(z) has q1,., qL poles.
First lets assume that the poles are distinct and not common. Then
( )
( ) ( ) ( ) ( ) ( )44 344 2144 344 21
responseforce
L
k
n
kk
responsenatural
N
k
n
kk
L
kk
k
kN
kk
k
k
nuqQnuPAny
zq
Q
zP
AzY
==
=
=
+=
+
=
11
11 11
Note that natural response zero-input response. It is in fact the no-input response.
Example
( ) ( ) ( )nxnyny += 12
1Find the output when ( ) ( )nu
nnx
4cos10
=
Solution
( ) ( ) ( )zXzYzzY += 12
1
( )2
1
211
111=
=
P
zzH
However, ( )21
1
2
21
1110
+
=zz
z
zX 4/*124/
1
jj eqqeq ===
( ) ( ) ( )44444 344444 2143421
responseforce
14j
728j
14j
728j
responsenatrual
1 ze1
e786
ze1
e786
z211
36zxzHzY
oo
+
+
==
/
.
/
. ..
/
.
( ) ( )nunyn
nr
=
2
13.6
( ) ( )nunny ofr
= 7.28
4cos5.13
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Testing it with MATLAB:
( )
( ) 321
1
2
1
2
21
2
121
21110
++
=
zzz
z
zY
[ ] ( )
++==
2
1,
2
21,
212,1;2/10,10 ab ;
[R, P, K] = residuez (b, a)
+
=
9074.1
2562.39537.5
2562.39537.5
i
i
R ][
5.0
7071.07071.0
7071.07071.0
=
+
= Ki
i
polesP
norm (R (1)) = 6.78
angle (R (1)) = -0.5005 rad = - 28.7o
zplane (b, a) % plots the zero-poles
Example of a Non-Relaxed System
( ) ( ) ( ) ( ) 022
11
2
3=+ nnxnynyny
( ) ( )nunxn
= 4
1 find y(n) ify (-1) = 4 andy (-2) = 10
Solution
TakingZ+
from both sides:
( ) ( ) ( )[ ] ( ) ( ) ( )[ ]1
211
411
1)(12
2
11
2
3
+++
==++++
zzXzYzyzyzYzyzY
( ) ( )11
21 21
4
11
1
2
1
2
31
+ +
=
+ z
zzzzY
Finally: ( )21
21
21
231
21
492
+
+
+=
zz
zzzY
or ( )111
411
31
1
32
211
1
+
+
+
=
zzzzY
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( ) ( )nu4
1
3
1
3
2
2
1ny
responseforce
n
responsenatural
n
++
=
32143421
Note that natural response is due to system poles and force response is due to the input poles.
Transient response is due to the poles inside the unit circle and steady-state response is due to
poles on the unit circle. In this case,
Transient Response )(2
1
4
1
3
1nu
nn
+
and Steady-State Response )(3
2nu .
Zero-Input (or Initial condition) Response is: )().()( zXzHzY ICZI = andZero-State Response is:
)().()( zXzHzYZS = . In this case,
)(3
8
2
12
4
1
3
1)( nuny
nn
zs
+
=
and )(22
13)( nuny
n
zi
= .
Note that complete response is eitherTransient Response+Steady-State Response, orNatural Response + Force Response, or Zero-Input Response + Zero-State Response. Each response
emphasizes a different aspect of system analysis.
Checking with MatLab:
n = [0: 7]; % just checking the first 8 samples
; n4
1x
= xic = [1, -2] % terms due to initial conditions (1 2z-1)
b = [1, 0];
21
231=a ;
y1 = filter (b, a, x, xic);
( ) ( ) ( )8,1*3
22
141*
31 onesnny ++= ;
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Since y andy1are the same, then our solution is correct!
However, for large order difference equations, it is tedious to determine xic(n) analytically.
MatLab command filtic does findxic as well.
Xic = filtic (b, a, Y, x) where Yandx are initial conditions.
Y = [y(-1), y(-2), , y(-N)]
x = [x(-1), x(-2), , x(-M)]
Ifx(n) = 0 forn < -1, then x need not to be defined. In our example:
Y = [4, 10];
Xic = 1, -2
Causality and Stability
Ifh(n) = 0, forn < 0, then the system is causal. Then its ROC is the exterior of a circle. The
stability of a system is quarantined by the condition that the ROC includes the unit circle.
Because the necessary and sufficient condition for a BIBO system is that
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31
Lecture 6- Chapter 4
Frequency Analysis of Signals and Systems
Continuous Signals and Discrete-Time Signals
Periodic Aperiodic
Starting with periodic CT signals:
Recall that a linear combination of harmonically related complex exponentials of the form
( ) tkFjk
keCtx02
+
=
= is a periodic signal with fundamental periodo
pF
T1
= . In order to find Ck,
multiply both sides byFot2je l and integrate over one period:
dteCdteCedtetx
lkTlk
T
tFlkj
k
k
tkFj
k
k
T
tlFj
T
tlFj
P
PPP
43421
=
+
=
+
=
==
0
0
)(22
0
2
0
2 0000)(
( ) plFotj
Tp
oTCetx = l
2
( ) FotjT
p
etxT
Cp
l
l
21 = Fourier Series
An important issue is that whether kFotj
k
keC2
0+
=
representation is equal to x(t) for every moment
oft. The Dirichlet conditions guarantee that this series is equal to x(t) except at the values oftfor
which x(t) is discontinuous. At those values oft, the series converges to the midpoint (average
value) of the discontinuity.
Dirichlet conditions are:
1) x(t) has a finite number of discontinuity in any period.
2) x(t) has a finite number of maxima and minima during each period sufficient but not
3) x(t) is absolutely integrable in any period: necessary
( )
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Power Density Spectrum of Periodic Signals
A periodic signal has infinite energy but finite average power.
Parsevals Theorem:
( ) ( ) ( )
( ) ( )
2
K
CT
TpKFot2j
Kp
KFot2jKTp
p
Tpp
2
Tpp
x
C
dtetxCT
1dteCtx
T
1
dttxtxT
1
dttxT
1
P
Kp
+
+
+
=
==
==
444 3444 21
**
*
Ifx(t) is real then PSDCCCC kkkk == 2
*2* is an even function in frequency and the
phase is an odd function.Example:
( )
( )
0
0
0
0
2
20
22/
2/
2
sin
sin
2
2
1
00
0
kFcT
A
kF
kF
T
Aee
kTF
A
kFj
e
T
AdtAe
TC
p
p
kFjtkFj
p
kFotj
p
tkFj
p
k
=
=
=
==
x(t)
t
-Tp Tp-/2 /2
po
T
1F =
pTA
F
kC
A
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33
Now ifpT
decreases ,pT then Ck 0, which means the signal becomes aperiodic
average power becomes zero.
CT Aperiodic Signals
We can say ( ) ( ) = txtx pTplim
( ) ( ) ( )
( ) ( ) ( ) dtetxdtetxFX
anddeXdFeFxtx
tjtFj
tjFtj
+
+
+
+
==
==
2
2
2
1
Aperiodic signals are energy signals.
( )
( ) dFFX
dttxEx
2
2
+
+
=
=
Energy Density Spectrum: Sxx(F) = |X(F)|2
A couple of points:
1) Remember that from only ESD or PSD we cannot reconstruct x(t)because phase information
is lost.
2) Ckforxp(t) is just samples ofX(F)
( )op
k kFXT
C1
=
DT Frequency Analysis
First consider a periodic DT signalx(n) = x(n + N)
( )n
N
kjN
ok
keCnx
=
=21
Multiply both sides byn
N2j
e
l
and sum over one period.
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34
( )( )
=
=
=
=
=
=
elseo
N2NoKifN
eCenx1N
on
1N
oK
nN
K2j
K
nN
2j1N
on
,,l
ll
( )l
l
CNenxn
NjN
on
.21
=
=
( )
( )
( )2121
1/2
/21
1
1,....,1
=
=
=
=
==
=
==
N
on
N
ok
kx
N
oK
Nknj
k
NknjN
onk
nxN
CP
eCnx
NokenxN
C
*DTFS is periodic like Periodic DT*
Ck+N= Ck. Therefore, the spectrum of a periodic DT,x(n), is also periodic with periodN.
( ) ( ) ( ) kN
on
NknjN
on
NnNkj
Nk CenxN
enxN
C ===
=
=
++
1/2
1/2 11
Example
Find DTFS of the following signals:
( )
( ) ( )
155
2sin
3
2cos
53 22
1
1
=+=
==
Nnnnx
Nnx
Nnx
4342143421
For this case, we can directly write it is a sum of exponentials.
( )( )
( )
+=
+=
43421
3
n22j
e
3
n232j
3
n2j
3
n2j
3
n2j
1 ee2
1ee
2
1nx
2
1C
2
1C 21 == , forx1(n)
Interchange the
sums
=
=
= 1if,1
1
1if,1
aa
a
aN
a NN
on
n
Power
The smallest common denominator
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35
( )( )
=
==
5
n254j
5
n2j
5
n2j
5
n2j
2 eej2
1ee
j2
1n
5
2nx
sin
j2
1C
j2
1C
41
== , forx2(n) and 0 else where.
Ckforx(n) is like 21 35x
k
x
k CC +
=
=
=
=
else
122
1
1052
1
32
1
o
kj
,k
kj
Ck
Fourier Transform for Aperiodic D.T. Signals
( ) ( ) ( )
( ) ( )
=
=
+
=
deXnx
enxXeX
nj
n
njj
22
1
(recall that for C.T. signals it was over+
and here is over 2 which means that X() is
periodic).
Two Basic Differences Between CTFT and DTFT:
1) X() X(ej) is periodic with period 2( ) ( ) ( ) ( ) ( ) XenxenxkX njnkj
n
===+ +
++
= 22
2) SinceX() is periodic, (in fact a it is a periodic C.T. signal), then it has a Fourier Seriesand in factx(n) are the coefficients of that Fourier Series.
Before visiting a famous example, lets review the concept of convergence.
If we have a limited observation, we will have the truncation effect, and the famous theory of the
Gibbs. Let ( ) ( ) njN
Nn
N enxX
== if ( ) ( ) 0lim
44 344 21
N
N XX , then XN() converges uniformly
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36
toX() as N . This convergence is guaranteed ifx(n) is absolutely summable (3rd Dirichlet
condition).
( )
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37
( ) njN
N
cN e
n
nX
=sin
Matlab definition: ( )x
xxc
sinsin =
Section 4.2.12
There are two time-domain characteristics that determine the type of signal spectrum and they
are: Periodicity and Continuity
Signals
Continuous Discrete-Time
** Periodicity with period in one domain automatically implies discretion with spacing
1 in
the other domain.
Properties of the Fourier Transform for Discrete-Time Signals
1. Real signals if x(n) is real, theX*() = X(-)
Spectrum magnitude: ( ) ( ) = XX even function
Spectrum phase ( ) ( )= XX pp odd function.
2. Real and evenx(n) Real and EvenX()
3. Real and oddx(n) Imaginary and oddX()
4. Imaginary and oddx(n) Real and oddX()
5. Imaginary and evenx(n) Imaginary and evenX()
6. Linearity a1x1(n) + bx2(n) F aX1() +bX2()
7. Time-Shiftingx(n) F X()
x(n-k) F e-jk
X()
Aperiodic
&
Continuous
Spectrum
Periodic PeriodicAperiodic Aperiodic
Aperiodic
&
Discrete
Spectrum
Periodic
&
Continuous
Spectrum
Periodic
&
Discrete
Spectrum
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38
8. Time-Reversalx(n) F X()
x(-n) F X(-) Therefore, FT of an even function is an even function too.
9. Convolution: x1(n) * x2(n)F
X1(
).X2(
)Proof:
( ) ( ) ( )knxkxnxk
= +
=
21
( ) ( ) ( ){
( ) ( )
)().( 21
)(
21
.
21)(
XX
eknxekx
eknxkxX
n
knj
k
kj
ee
nj
n k kjknj
=
=
=
+
=
+
=
+
=
+
=
10. Correlation Theorem: ( )nr21xx
FX1()X2(-)
Proof:
( ) ( ) ( )nkxkxnrk
xx = +
=
2121
FT
( ) ( ) ( ) ( )( )
321KjKnj
2121
ee
nj2
n K
1nj
n
xxxx enKxKxenrS
+
=
+
=
+
=
==
( )
( )
( )[ ] ( )
( )44444 344444 21444 3444 21
+
=
+
=
=
21 x
n
Knj2
x
K
Kj1 eKnxeKxRHS
Now ifx(n) is real, thenX*() =X(-)
( ) ( ) ( ) ( )( )
( ) 2
x
xxxx xxxSr
==321
l*
Energy Spectral Density
11. Frequency Shifting
ejon
F X( -o)
12 Modulation Theorem
Cross-Energy
Density Spectrum
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39
( ) ( ) ( ) ( )[ ]
[ ]njnj
oo
een
XXnnx
00
2
1)cos(
2
1cos
0
0
+=
++
13. Parseval Theorem
( ) ( ) ( ) ( )
dXXnxnxn
*
21
*
212
1
+
=
=
Proof:
( )
( ) )()(2
1)(
)(2
1
*
21
)(
*
21
*
21
*2
nxnxdeXnx
dXenxRHS
n
nx
nj
n
nj
n
=
=
=
==
=
444 3444 21
Special case: ( ) ( ) ( ) ( )
dx2
1nxnxnx
22
2
12 ==
( ) ( ) ( )( )
===+
=
2
dS
22
nxxx
xx
dx2
1nx0rE
43421
14. Windowing
( ) ( )nxnx 21 F ( ) ( )[ ]
21 *2
1XX
( ) ( ) ( ) ( ) nj21nj enxnxenxx
+
+
==
( ) ( )
( ) ( )
( ) ( ) ( ) ( )[ ]
2122
1
22
1
22
1
*
2
1
2
1
2
1
2
1
XXdXX
eenxdX
enxdeXRHS
njnj
n
njnj
=
=
=
+
=
+
15. Differentiation in Frequency Domain
n(x)n F j( )d
dx
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40
Skipping to Section 4.4.8
Correlation Function and Power spectra for Random Input Signals
When the input signal is random, then we have to consider statistical moments of input and
output. So here is a bit of introduction about Stationary Random Process. Starting with the
Definition of Stationary Signals:
If X(t) is a random process with a point Probability Density Function (PDF),
( ) nxxxxPxP tttt ,...,,, 31 2= forn random variables. ( ) ( ) nIii itxtX ,...2,1,
If the joint probability of ( ) ( ) += 11 ,.........,,, 111t
n
t
n xxPxxxP for all t1 and then the random
process X(t) is stationary in strict sense. In other words, statistical properties of a stationary
random process is time-invariant, meaning that its mean and variance and other moments are
time invariant.
One Observation
Set 1
Set 2
Set 3
t
t
t
t1
t1
t1
t1+
t1+
t1+
x(t,S1)
x(t,S2)
x(t,S3)
x1(t1)
x2(t
1)
x3(t1)
x3(t+)
x1(t1+)
x2(t+)
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41
Statistical (ensemble) Average
( ) ( )iiii tttt
dxxPxxE +
=
If we dont haveP(xti) but have many observations, then ( ) ( )1111 ...
121
t
N
ttt
xxxNxE +++= , which of
a stationary process it is equal toE(xti) for any ti.
Also autocorrelation function:
( ) ( )
[ ]21
21212121 ,,
tt
tttttttt
xx
xxE
dxdxxxPxxxx
=
= +
If this xx depends only on the time difference t1 t2 = , then ( ) [ ],, 11 += ttxx xxE which is the
case for stationary process. If a process has two features:
1) Its xx depends only on time difference, , and
2) E(xt1) = E(x
t2) = E(x
ti)
then the process is said to bestationary in wide sense.
Now if all statistical averages can be obtained by one single realization (or one sample set), then
the process is also Ergodic meaningEnsemble average time average.
( )nx xE= and ( )
=
=1
2
1 N
on
nx
N
)
x)
is an estimate ofx. It will be said it is an unbiased estimate if ( ) xxE =)
. Also, it is a
good estimator if
( ) ( ) 022 = xxx EVar ))
as N
Therefore, time average ensemble average.
Autocorrelation: ( ) ( ) ( )mnxnxN
mN
on
xx +=
=
1*1
( )[ ] ( )mrmE xxxx = the true autocorrelation.
Now back to systems:
x(n) y(n)h(n)
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42
( ){ } ( ) ( )
( ) ( ){ } ( ) ( )0HkhknxEkh
knxkhEnyE
xx
n
y
===
==
+
+
+
=
The autocorrelation sequence:
( ) ( ) ( ){ } ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ){ }
( ) ( ) ( )kmhkh
mnxknxEhkh
mnxhknxkhEmnynyEm
xx
k
k
k
yy
+=
+=
+=+=
+
=
+
=
ll
ll
ll
l
l
l
.*
**
Special Form: when x(n) is a white noise, then ( ) ( )mm xxx 2= and ( )02 xxx = . Then
( ) ( )mm hhxyy 2=
( ) ( ) ( )
dHxhhxyy2
2
22
2
100 ==
by getting the Fourier transform in general:
2
)()(realissignaltheIf
)().().(
)()()()(
..let
)()()(
)()(
H
HH
euelhekh
eeeeklmu
eklmlhkh
em
xx
xx
u
uj
xx
l
ljkj
k
yy
kjljujmj
mj
m k l
xx
m
mj
yyyy
=
=
=
=+=
+=
=
=
=
=
=
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1. The Concept and Representation of Periodic
Sampling of a CT Signal
2. Analysis of Sampling in the Frequency Domain
3. The Sampling Theorem the Nyquist Rate
4. In the Time Domain: Interpolation
5. Undersampling and Aliasing
Signals and SystemFall 2003
Lecture #13
21 October 2003
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We live in a continuous-time world: most of the signa
encounter are CT signals, e.g.x(t). How do we conversignalsx[n]?
SAMPLING
How do we perform sampling
Sampling, taking snap shots ofx(t) every Tse
T sampling period
x[n] x(nT), n = ..., -1, 0, 1, 2, ... regularly spaced
Applications and Examples
Digital Processing of Signals
Strobe
Images in Newspapers
Sampling Oscilloscope
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Why/When Would a Set of Samples Be A
Observation: Lots of signals have the same sample
By sampling we throw out lots of information
all values ofx(t) between sampling points are l
Key Question for Sampling:Under what conditions can we reconstruct the orix(t) from its samples?
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Impulse Sampling Multiplying x(t) by the sam
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Analysis of Sampling in the Frequency D
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Illustration of sampling in the frequency-do
band-limited (X(j)=0 for | |> M) s
No overlap between shifted spectra
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Reconstruction ofx(t) from sampled s
If there is no overlap between
shifted spectra, a LPF canreproduce x(t) from xp(t)
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The Sampling Theorem
Suppose x(t) is bandlimited, so t
Then x(t) is uniquely determined by its samp
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Observations on Sampling
(1) In practice, we obviously
dont sample with impulses
or implement ideal lowpass
filters.
One practical example:
The Zero-Order Hold
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Observations (Continued)
(2) Sampling is fundamentally a time-varyingoperati
multiply x(t) with a time-varying function p(t). H
is the identity system (which is TI) for bandlimi
the sampling theorem (s > 2M).
(3) What ifs 2M? Something different: more la
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Time-Domain Interpretation of Reconst
Sampled Signals Band-Limited Inter
The lowpass filter interpolates the samples assuming
no energy at frequencies c
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Graphic Illustration of Time-Domain Inte
Original
CT signal
After sampling
After passing the LPF
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Interpolation Methods
Bandlimited Interpolation
Zero-Order Hold First-Order Hold Linear interpolation
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Undersampling and Aliasing
When s 2 M Undersampling
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Undersampling and Aliasing (contin
Higher frequencies ofx(t) are folded back and
aliases of lower frequencies
Note that at the sample times, xr(nT) = x(nT)
Xr(j
Dis
ofa
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Demo: Sampling and reconstruction of cosot
A Simple Example
Picture would beModified
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M ultiple Choice
1. The system y(t) = x(t) + 2x(t+3) is(a) causal system (b) non-causal system
(c) part ly (a) and (b) (d) none of these
2. The systemdt
tdy )(+3y(t) = x( t) is a
(a) time invariant system (b) time variant system
(c) partly (a) and (b) (d) none of these
3. The system y(t) = 3x(t) + 4 is(a) linear system (b) non linear system
(c) partly (a) and (b) (d) none of these
4. A power signal has inf ini te energy whereas an energy signal has zeroaverage power.
(a) true (b) false
5. Find dynamic system of the follow ing(a) y(n) = 3x(n) (b) y(n) = x(n) + x(n-1)
6. z transform of the signal {0,1,3,2,0,0,0.} is(a) 1+
z
2+
2
3
z
(b) 1+
z
3
(c) 1+z
3+
2
2
z(d) 1+
2
2
z
7. The ROC of Question no.6 is(a) whole of z plane (b) whole of z plane except z =
(c) whole of z plane except z = 0 (d) none of these
8.
If x(n) and y(n) are two finite sequences, then x(n) * y(n) is(a) X(z)/ Y(z) (b) Y(z)/ X(z)
(c) X 2 (z) Y(z) (d) X(z) Y(z)
9. If X(z) =)45.095.0)(1(
5.02
2
zzz
z, the initial value of x(n) is
(a) 0.5 (b) 1
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(c) 0 (d) 0.45
10.z transform of x(nk) is(a) kz X(z) (b) nz X(z)
(c) 1z X(z) (d) none of these
11. Find correct representation for the following(a) x1(n) * x2(n) = x2(n) * x1(n)
(b) x1(n) * [x2(n) * x3(n)] = [x1(n)* x2(n)] * x3(n)
(c) x1(n)* [x2(n)+x3(n) ] = x1(n) * x2(n) + x1(n) * x3(n)
(d) all are valid
12.The ROC of the z transform of the discrete time sequencex(n) = n)
3
1( u(n) n)
2
1( u(-n-1) is
(a)3
1
2
1
(c) |z|