Digital signal processing part2

Books.Books.
1.1. Digital Signal Processing Principles, Algorithms and ApplicationsDigital Signal Processing Principles, Algorithms and Applications
John G.Proakis & Dimitris G.Manolakis
2.2. Digital Signal ProcessingDigital Signal Processing By.
Sen M. Kuo & Woon-Seng Gan
3.3. Digital Signal Processing A Practical Approach.Digital Signal Processing A Practical Approach. By
Emmanuel C. Ifeachor & Barrie W. Jervis
4.4. Digital Signal Processing By Dr. Krishnanaik Vankdoth LAPDigital Signal Processing By Dr. Krishnanaik Vankdoth LAP
LAMBERT Academic Publishing Dnfscland/Germany – 2014LAMBERT Academic Publishing Dnfscland/Germany – 2014
krishnanaik.ece@gmail.com 2

3
Grading Policy
krishnanaik.ece@gmail.com
Assignments 1 & 2
20 Marks
Quiz Test -Mid-Term 30 Marks
Final Exam 50 Marks
Class will be divided different level as per their GPA
Group A- GPA
Group B- GPA
Group C – GPA

z-Transform
• Transform techniques are an important role in the analysis of
signals and LTI system.
• Z- transform plays the same role in the analysis of discrete time
signals and LTI system as Laplace transform does in the
analysis of continuous time signals and LTI system.
• For example, we shall see that in the Z-domain (complex Z-
plan) the convolution of two time domain signals is equivalent
to multiplication of their corresponding Z-transform.
• This property greatly simplifies the analysis of the response of
LTI system to various signals.
Dr. V. Krishnanaik Ph.D 4

1-The Direct Z- Transform
The z-transform of a sequence x[n] is
Where z is a complex variable. For convenience, the z-transform of a
signal x[n] is denoted by X(z) = Z{x[n]}
∑
∞
−∞=
−
=
n
n
znxzX ][)(
ωj
ezX =)(
We may obtain the Fourier transform from the z transform by
making the substitution . This corresponds to
restricting Also with ,1=z
e
j
rz
ω
=
∑
∞
−∞=
−
=
n
njj
ee rnxrX )]([)(
ωω
That is, the z-transform is the Fourier transform of the sequence x[n]r - n
. for r=1
this becomes the Fourier transform of x[n].
The Fourier transform therefore corresponds to the z-transform evaluated on the
unit circle:
5krishnanaik.ece@gmail.com

z-transform(cont:
The inherent periodicity in frequency of the Fourier transform
is captured naturally under this interpretation.
The Fourier transform does not converge for all sequences - the infinite
sum may not always be finite. Similarly, the z-transform does not
converge for all sequences or for all values of z.
For any Given sequence the set of values of z for which the z-transform
converges is called the region of convergence (ROC).

DSP Slide 7
z-transform(cont:
The Fourier transform of x[n] exists if the sum
converges. However, the z-transform of x[n] is just the Fourier
transform of the sequence x[n]r -n
. The z-transform therefore exists
(or converge) if
This leads to the condition
for the existence of the z-transform. The ROC therefore consists of a
ring in the z-plane:
∑
∞
−∞=n
nx ][
∞<=∑
∞
−∞=
−
n
n
rnxzX ][)(
∞<∑
∞
−∞=
−
][n
n
znx
In specific cases the inner radius of this ring may include the origin, and the outer
radius may extend to infinity. If the ROC includes the unit circle , then
the Fourier transform will converge.
1=zkrishnanaik.ece@gmail.com

z-transform(cont:
Most useful z-transforms can be expressed in the form
where P(z) and Q(z) are polynomials in z. The values of z for
which P(z) = 0 are called the zeros of X(z), and the values with
Q(z) = 0 are called the poles. The zeros and poles completely
specify X(z) to within a multiplicative constant.
,
)(
)(
)(
zQ
zP
zX =
In specific cases the inner
radius of this ring may include
the origin, and the outer radius
may extend to infinity. If the
ROC includes the unit circle
, then the Fourier
transform will converge.
1=z

Example: right-sided exponential sequence
∑ ∑
∞
−∞=
∞
=
−−
==
n n
nnn
azznuazX
0
1
)(][)(
∑
∞
∞=
−
∞<
n
az 1
Consider the signal x[n] = an
u[n]. This has the z-transform
Convergence requires that
.az >.11
<−
azwhich is only the case if or equivalently
In the ROC, the series converges to
∑
∞
=
−
−
>
−
=
−
==
0
1
1
,,
1
1
)()(
n
n
az
az
z
az
azzX
since it is just a geometric series.

DSP Slide 10
Example: right-sided exponential sequence
The z-transform has a region of convergence for any finite
value of a.
The Fourier transform of x[n] only exists if the ROC
includes the unit circle, which requires that On
the other hand, if then the ROC does not include
the unit circle, and Fourier transform does not exist. This
is consistent with the fact that for these values of a the
sequence an
u[n] is exponentially growing, and the sum
therefore does not converge.
.1<a
1>a

DSP Slide 11
Example: left-sided exponential sequence
Now consider the sequence ].1[)( −−−= nuanx n
.1−≤nThis sequence is left-sided because it is nonzero only for
The z-transform is
∑ ∑
∑ ∑
∞
=
∞
=
−−
∞
−∞=
−
−∞=
−−
−=−=
−=−−−=
1 0
1
1
)(1
]1[)(
n n
nnn
n n
nnnn
zaza
zaznuazX
For or the series converges to,11
<−
za ,az <
Note that the expression for the
z-transform (and the pole zero
plot) is exactly the same as for
the right-handed exponential
sequence - only the region of
convergence is different.
Specifying the ROC is therefore
critical when dealing with the z-
transform.

Example: Sum of two exponentials
lsexponentiarealtwoofsumtheisu[n]
3
1
u[n]
2
1
x[n]signalThe
nn






−+





=
∑
∞
−∞=
−












−+





=
n
n
nn
znunuzX ][
3
1
][
2
1
)(
∑∑
∞
−∞=
−
∞
−∞=
−






−+





=
n
n
n
n
n
n
znuznu ][
3
1
][
2
1
∑∑
∞
=
−
∞
=
−






−+





=
0
1
0
1
3
1
2
1
n
n
n
n
zz
The z transform is
From the example for the right-handed exponential sequence, the first term in this
sum converges for and the second for The combined
transform X(z) therefore converges in the intersection of these regions, namely when
.
In this case
2/1>z 3/1>z
2/1>z






+





−






−
=
+
+
−
=
−−
3
1
2
1
12
1
2
3
1
1
1
2
1
1
1
)(
11
zz
zz
zz
zX

13
Example: Sum of two exponentials
The pole-zero plot and region of convergence of the signal is

DSP Slide 14
Example: finite length sequence
The pole-zero plot and region of convergence of the signal is
The signal
1
11
0
1
1
0 1
)(1
)()( −
−−
=
−
−
=
−
−
−
=== ∑∑ az
az
azzazX
nN
n
n
N
n
nn
.
1
1
az
az
z
NN
N
−
−
= −
has z transform
Since there are only a finite number of nonzero terms the sum always converges when
is finite. There are no restrictions on ,and the ROC is the entire z-
plane with the exception of the origin z = 0 (where the terms in the sum are infinite). The N roots of
the numerator polynomial are at
1−
az )( ∞<a
1,......1,0,)/2(
−== NkaeZ Nkj
k
π
*since these values satisfy the equation ZN
= aN
The zero at k = 0 cancels the pole at z = a, so there are no poles except
at the origin, and the zeros are at zk = aej(2k/N)
k = 1; : : : ;N -1 The zero at k = 0 cancels the pole at z = a, so there
are no poles except at the origin, and the zeros are at zk = aej(2k/N)
k = 1; : : : ;N -1krishnanaik.ece@gmail.com

2-Properties of the region of convergence
∞
).0( ∞≤<<≤ LR
z ττ
).( 21
∞<≤≤<∞ NnN
The ROC is a ring or disk in the z-plane, centered on the origin
The Fourier transform of x[n] converges absolutely if and only if the ROC of
the z-transform includes the unit circle.
The ROC cannot contain any poles.
If x[n] is finite duration (ie. zero except on finite interval
), then the ROC is the entire Z-plan except perhaps at z=0 or
z= .
If x[n] is a right-sided sequence then the ROC extends outward from the
outermost finite pole to infinity.
 If x[n] is left-sided then the ROC extends inward from the innermost nonzero
pole to z = 0.
A two-sided sequence (neither left nor right-sided) has a ROC consisting of a
ring in the z-plane, bounded on the interior and exterior by a pole (and not
containing any poles).
 The ROC is a connected region.
The properties of the ROC depend on the nature of the signal. Assuming that the
signal has a finite amplitude and that the z-transform is a rational function:

DSP Slide 16
3 - The inverse z-transform
,2
1
,
2
1
1
1
)(
1
>












−
=
−
z
z
zX
Formally, the inverse z-transform can be performed by evaluating a
Cauchy integral. However, for discrete LTI systems simpler
methods are often sufficient.
A-Inspection method: If one is familiar with (or has a table
of) common z-transform pairs, the inverse can be found by
inspection. For example, one can invert the z-transform
],[
2
1
x[n]
thatrecogniseweinspectionBy
.zfor,........
1
1
][a 1
n
nu
a
az
nu
n
z






=
>
−
→← −
Also, if X(z) is a sum of terms then one may be able to do a term-by-
term inversion by inspection, yielding x[n] as a sum of terms.
Using Z-transform pair

3 - The inverse z-transform
( )
( )
X(z).ofpolesandnonzerothearesck'thewhere
1
1b
X(z)
asfactorX(z)topossiblealwaysisIt
,)(
1
1
1
1
0
0
0
0
∏
∏
∑
∑
=
−
=
−
=
−
=
−
−
−
=
=
N
k k
M
k k
N
k
k
k
M
k
k
k
zd
zc
a
za
zb
zX
B-Partial fraction expansion:
For any rational function we can obtain a partial fraction expansion,
and identify the z-transform of each term. Assume that X(z) is
expressed as a ratio of polynomials in z-1
:

The inverse z-transform
( ) kdzk
k
k
zXzd
zd
=
−
=
−
−=
−
= ∑
)(1A
bygivenareAtscoefficienthecasein this
,
1
A
X(z)
1
k
N
1k
1
k
Partial fraction expansion (Continue:)
NMusingobtainedbecans'AThe
1
k
0 1
1
<
−
+= ∑ ∑= =
−
−
M-N
r
N
k k
kr
r
,
zd
A
zBX(z)
If M>N and the poles are first order, then an expression of the form
cab be used, and Br’s be obtained by long division of the numerator.
If M<N and the poles are all first order, then X(z) can be expressed
as

3 - The inverse z-transform Partial fraction expansion
( )∑ ∑∑= =
−
≠=
−
−
−
+
−
+=
M-N
r
s
m
m
i
m
N
ikk k
kr
r
zd
C
zd
A
zBX(z)
0 1
1
,1
1
.
11
sided.-rightorsided-leftare
sequencesthewhetherdecidetousedbemustpropertiesROC
thetermsFor thesesequences.lexponentiatocorrespond
ispoles,order-multiplewithdealalsocanwhich
expansion,fractionpartialforformgeneralmostThe
stermfractionalTher].-[nBform
theoftermstoinvertandsequences,impulsescaled
andshiftedtocorrespondBtermsThetexts.DSP
standardmostinfoundbecans'CthefindingofWays
r
r
m
δ
r
z−
1
1 −
− zd
A
k
k

DSP Slide 20
Example: inverse by Partial fractions
( )
( )
.1z,
1
2
1
1
1
2
1
2
3
1
1
transform-zwithx[n]sequencetheConsider
11
21
21
21
2 >
−−
+
=
+−
++
=
−−
−
−−
−−
zz
z
zz
zzX(z)
,
1
2
1
1
asexpressedbecanthis2NMSince
1
2
1
1
0
z
A
z
A
BX(z) −
− −
+
−
+=
==
( )zz
zzzz
11
1
1
12
1212
0
1
2
1
1
51-
2X(z)
15
23
2
12)1
2
3
2
1
:divisionlongbyfoundbecanBvalueThe
z
z
zz
−−
−
−
−−
−−−−
−





−
+
+=
−
+−
+++−

21
Example: inverse by Partial fractions
( ) .)(1
usingfoundbecanAandAcoecientsThe
1
21
d
zX
k
zkk zdA =
−
−=
zz
z
zzA
z
zzA
z
z
1
1
1
1
21
2
1
1
21
1
1
8
2
1
1
9
-2X(z)foreThere
9
2/1
121
2
1
1
21
and
9
21
441
1
21
So
1
1
−
−
=
−
−−
=
−
−−
−
+
−
=
=
++
=
−
++
=
−=
−
++
=
−
++
=
−
−
Using the fact that the ROC , the terms can be inverted one at a time
by inspection to give
.1z >
[ ] [ ] ].[)2/1(92 nunnx n
−= δ

DSP Slide 22
C- Power Series Expansion
( )
......]2[]1[]0[]1[]2[..................
][z
forminseriespowerasgivenistransformZIf
2112
zz
z
zxxzx
nxX
n
n
+++−+−+=
= ∑
∞
−∞=
−
then any value in the sequence can be found by identifying the
coefficient of the appropriate power of z-1
.

23
Example; Power Series Expansion
( ) ( ) aazX >+= −
z,1logz
transformZheConsider t
1
Using the power series expansion for log(1 + x), with /x/< 1, gives
( ) ∑
∞
=
−+
−
=
1
1
,
)1(
z
n
nnn
n
za
X

24
Example; Power Series Expansion by long division
( ) a
az
X >
−
= −
z,
1
1
z
transformtheConsider
1
Since the ROC is the exterior of a circle, the sequence is right-sided. We therefore
divide to get a power series in powers of z-1
:
( )
].[][......................zaaz1
1
1
.....
1
zaaz1
11z
2-21
1
22
221
1
1
-221
1
nuanxTherefore
az
za
zaaz
az
az
azX
n
=+++=
−
+
−
−
++
−=
−
−
−
−−
−
−
−
−

25
Example; Power Series Expansion for left-side Sequence
( ) a
az
X <
−
= −
z,
1
1
z
ormtransf-ZtheConsider
1
Because of the ROC, the sequence is now a left-sided one. Thus we
divide to obtain a series in powers of z:
].1[][..............
zaa-
z
1
21
2..-21
−−−=
−
−
+−
−
−
−
nuanxThus
az
zaz
z
za
n

26
4- Properties of the z-transform
RROC),(][ x
zXnx z
→←
if X(z) denotes the z-transform of a sequence x[n] and the ROC of X(z) is
indicated by Rx, then this relationship is indicated as
Furthermore, with regard to nomenclature, we have two sequences such
that
RROC),(][
RROC),(][
x222
x111
zXnx
zXnx
z
z
→←
→←
A—Linearity: The linearity property is as follows:
.RRcontainsROC),(][)(][ x1x12121
∩+→←+ zbXzaXnbXnax z
B—Time Shifting: The time shifting property is as follows:
x0
RROC),(][ 0
zXznnx
nz −
→←−
(The ROC may change by the possible addition or deletion of z =0 or z = ∞.)
This is easily shown:
).(][
][][)(
00
0 )(
0
zXzzmxz
zmxznnxzY
n
n
mn
n
nm
n
n
−
∞
−∞=
−−
∞
−∞=
+−
∞
−∞=
−
∑
∑∑
==
=−=

27
Example: shifted exponential sequence
4
1
z,
4
1
1
)( >
−
=
z
zX
Consider the z-transform
From the ROC, this is a right-sided sequence. Rewriting,
.]1[)4/1(][ 1
−= −
nunx n
The term in brackets corresponds to an exponential sequence (1/4)n
u[n]. The
factor z-1
shifts this sequence one sample to the right.
The inverse z-transform is therefore
4
1
z
4
1
-1
1
,
4
1
1
)(
1
1
1
1
>












=
−
=
−
−
−
−
z
z
z
z
zX

28
C - Multiplication by an exponential sequence
The exponential multiplication property is
where the notation indicates that the ROC is scaled by (that is,
inner and outer radii of the ROC scale by ). All pole-zero locations are
similarly scaled by a factor z0: if X(z) had a pole at then X(z/z0)
will have a pole at z=z0z1.
( ).][
)( 00
ee
jFnj
Xnx
ωωω −
 →←
•If z0 is positive and real, this operation can be interpreted as a shrinking or
expanding of the z-plane | poles and zeros change along radial lines in the z-
plane.
If z0 is complex with unit magnitude (z0 = ejw0
) then the scaling operation
corresponds to a rotation in the z-plane by and angle w0, That is, the poles and
zeros rotate along circles centered on the origin. This can be interpreted as a
shift in the frequency domain, associated with modulation in the time domain
by ejw0n
. If the Fourier transform exists, this becomes
1
zz =
,RROC],/[][ x000 zzzXnx zn
z →←
,0 x
Rz 0
z
0
z

29
Example: exponential multiplication
The z-transform pair
can be used to determine the z-transform of x[n] = rn
cos(w0n)u[n].
Since cos (w0n) = 1/2ejw0n
+ 1/2e –jw0n
. The signal can be written as
( ) ( ) ].[
2
1
][
2
1
][ 00
nurenurnx
njnj
e
ωω −
+=
1z,
1
1
][ 1
>
−
→← −
z
nu z
From the exponential multiplication property,
( ) .z,
1
2/1
][
2
1
10
0
r
zr
nur
e
e j
znj
>
−
→← −ω
ω
( ) .z,
1
2/1
][
2
1
10
0
r
zr
nur
e
e j
znj
>
−
→← −−
−
ω
ω
.z
1
2/1
1
2/1
X(z)
So
11 00
r
zrzr ee
jj
>
−
+
−
= −−− ωω
.z,
cos21
cos1
221
0
1
0
r
zrzr
zr
>
+−
−
−−
−
ω
ω

30
D- Differentiation
The differentiation property states that
This can be seen as follows: since
.RROC,
)(
][ x
=−→←
dz
zdX
znnx z
.]}[{][][)(
)( 1
∑∑
∞
−∞=
−−−
==−−=−
n
nn
nnxzznnxznxnz
dz
zdX
z
haveWe
,][)(
-n
∑
∞
∞=
−
= n
znxzX
Example: second order pole
The z-transform of the sequence ][][ nunanx n
=
Can be found
( )
.z,
11
1
dz
d
-X(z)
beto
,z
1
1
][
21
1
1
1
a
az
az
az
a
z
nua zn
>
−
=





−
=
>
−
→←
−
−
−
−

31
E- Conjugation
This property is
F- Time reversal.
.RROC*),(*][* x
=→← zXnx z
Example: Time-reversed exponential sequence
The Signal is a time-reversed version of an
u[n]. The
z-transform is therefore
][][ nuanx n
−= −
.
R
1
ROC*),/1(*][*Here
x
=→←− zXnx z
,z LR
rr <<
The notation 1/Rx means that the ROC is inverted, so if Rx is the set
of values such that then the ROC is the set of values of z su
that .1/r/1 R
<zrl
.Rz,
11
1
)( x
1
11
11
=<
−
−
=
−
= −
−−
−−
a
za
za
az
zX

32
G- Convolution
This property state that
.RRcontainsROC),()(][*][ x2x12121 zXzXnxnx z
→←
Example: evaluating a convolution using the z-transform
The z-transforms of the signal x1[n] =an
u[n] and x2[n] = u[n] are
.
R
1
ROC*),/1(*][*Here
x
=→←− zXnx z
,1a <
1z,
1
1
)(.
and
z,
1
1
)(
1
0
2
1
0
1
>
−
==
>
−
==
−
∞
=
−
−
∞
=
−
∑
∑
az
zzX
a
az
zazX
n
n
n
nn
For The z-transforms of the convolution y[n] = x1[n] *x2[n] is
( )( ) ( )( )
1z
111
1
)(
2
11
>
−−
=
−−
= −−
zaz
z
azaz
zY
( )( ) ( )( )
1z
111
1
)(
2
11
>
−−
=
−−
= −−
zaz
z
azaz
zY krishnanaik.ece@gmail.com

33
Using a partial fraction expansion,
( )
( ).][][
1
1
)(
1z,
1
-
1
1
1
1
)(
1
11
nuanu
a
ny
So
az
a
za
zY
n+
−−
−
−
=
>





−−−
=
Example: evaluating a convolution using the z-transform
H-H- Initial Value theoremInitial Value theorem
If x[n] is zero for n<0, then
).(]0[ lim zXx
z ∞→
=

34
Some common z-transform pairs are:

I- Relationship with the Laplace transform:
.
havewe
,
)( TjdTTjd
eeez
jds
ωω
ω
==
==
+
Continuous-time systems and signals are usually described by the Laplace
transform. Letting z = esT
, where s is the complex Laplace variable
,/2f/f2Tzand
Therefore
s s
dT
ez ωπωπω ==== 
where ws is the sampling frequency. As varies from to , the s-plane is
mapped to the z-plane:
 The j axis in the s-plane is mapped to the unit circle in the z-plane.
 The left-hand s-plane is mapped to the inside of the unit circle.
The right-hand s-plane maps to the outside of the unit circle.
ω ∞ ∞
ω

36
Frequency Analysis
Voltage Vs Time Representation That
become
Magnitude Vs Frequency
,
Phase Vs Frequency Representation
And
Vice Versa

37
Frequency Analysis of Signals
•Fourier transform and Fourier series basically involve the
decomposition of the signal in terms of sinusoidal components.
With such a decomposition ,a signal is said to be represented in
the frequency domain.
•These decompositions are very important in the analysis of LTI
systems because response of a system to a sinusoidal input
signal is a sinusoid of the same frequency but of different
amplitude and phase.
•Many other decompositions of signals are possible, only the
class of sinusoidal signals possess this desirable property in
passing through a LTI system.

38
The Fourier Series for Continuous-Time Periodic Signals
• The Fourier Series of a periodic analogue signal x(t) is given by
Tptdt
Tpto
to
Tpto
to
==
++
∫
1...............................)( 02
∑
∞
−∞=
=
k
tkFj
k
ectx
π
is a periodic signal with fundamental period Tp=1/Fo and k = 0,±1, ±2…,
We can construct periodic signals of various types by proper choice of
fundamental frequency and the coefficients CK.FO determines the fundamental
period of x(t) and coefficient Ck specify the shape of waveform. We determine the
expression for Ck
2........)(
1 02
∫
−
=
pT
tkFj
p
k
dtetx
T
c
π
dtecedtetx
k
kFotj
k
lFotj
Tpto
to
lFotj
Tpto
to






= ∑∫∫
∞
−∞=
+−
+
−
+
πππ 222
)(
pl
lFotj
Tpto
to
TCdtetx =−
+
∫
π2
)(

39
• In general, the Fourier Coefficients ck are complex valued.
• If the periodic signal is real, ck and c-k are complex conjugates. As a result,
if
k
k
j
kk
j
kk
ecc
then
ecc
θ−
−
θ
=
=

40
Other forms of Fourier Series
Representation
As we have just mentioned
∑
∞
−∞=
π
=
k
tkF2j
k
0
ec)t(x
The above equation can be re-written as
[ ]∑
∞
=
−
−++=
1
)(22
0
00
)(
k
tFkj
k
tkFj
k ececctx ππ
since kj
kk ecc θ
= and kj
kk ecc θ−
− =
( ) ( )
[ ]kk tkFjtkFj
k
k eecctx θπθπ +−+
∞
=
++=∴ ∑ 00 22
1
0)(
( )k
k
k tkFcc θπ ++= ∑
∞
=
0
1
0 2cos2
This is called the Cosine
Fourier Series.

41
Other forms of Fourier Series
Representation
Yet another form for the Fourier Series can be obtained by
expanding the cosine Fourier series as
[ ]∑
∞
=
θπ−θπ+=
1k
k0k0k0 sintkF2sincostkF2cosc2c)t(x
Consequently, we may rewrite the above equation in
the form
( )∑
∞
=
−+=∴
1
000
2sin2cos)(
k
kk
tkFbtkFaatx ππ
This is called the Trigonometric form of the FS,
where a0 = co, ak = 2|ck|cosθk and bk = 2|ck|sinθk.krishnanaik.ece@gmail.com

42
Power Density Spectrum of Periodic
Signals
A periodic signal has infinite energy and a
finite average power, which is given as
dt)t(x
T
1
P
2
Tp
x
p
∫=
If we take the complex conjugate of (1) and
substitute for x*
(t), we obtain
∑ ∫∫ ∑
∞
−∞=
π−
∞
−∞=
π−








==
k
tkF2j
T
p
*
k
T
k
tkF2j*
k
p
x dte)t(x
T
1
cdtec)t(x
T
1
P 0
pp
0
2
k
kc∑
∞
−∞=
=

43
Therefore, we have established the relation
2
k
k
2
Tp
x cdt)t(x
T
1
P
p
∑∫
∞
−∞=
==
Which is called Parse Val's relation for power signals.
This relation states that the total average power in the periodic
signal is simply the sum of the average powers in all the
harmonics.
If we plot the |ck| as a function of the frequencies kFo ,k=0,±1,±2,….
the diagram we obtain shows how the power of the periodic signal
is distributed among the various frequency components. This diagram
is called the Power Density Spectrum of the periodic signal x(t). A
typical PSD is shown in the next slide.

44
-2F0 -F0 0 F0 2F0
|ck|2
F
Power density spectrum of a continuous time periodic signalkrishnanaik.ece@gmail.com

45
Example1: Determine the Fourier Series and the Power
Density Spectrum of the rectangular pulse train signal
illustrated in the following figure.
Solution:
-τ/2 τ/2
Tp
Tp
A
x(t)
∫
τ
τ−
τ
==
2/
2/
pp
0
T
A
Adt
T
1
c
τπ
τπτ
== ∫
τ
τ−
π−
0
0
p
2/
2/
tkF2j
p
k
kF
kFsin
T
A
dtAe
T
1
c 0
and
where k = ±1, ±2, …..
Figure (a), (b) and (c) illustrate the Fourier
coefficients when Tp is fixed and the pulse width
τ is allowed to vary.krishnanaik.ece@gmail.com

46
-60 -40 -20 0 20 40 60-0.05
0
0.05
0.1
0.15
0.2
τ = 0.2Tp
-60 -40 -20 0 20 40 60-0.04
-0.02
0
0.02
0.04
0.06
0.08
0.1
τ = 0.1Tp
Fig.(a)
Fig. (b)

47
-60 -40 -20 0 20 40 60-0.02
-0.01
0
0.01
0.02
0.03
0.04
0.05
τ = 0.05Tp
Fig. (c)
From these three figures we observe that the
effect of decreasing τ while keeping Tp fixed is to
spread out the signal power over the frequency
range. The Spacing between the adjacent lines is
independent of the value of the width τ.krishnanaik.ece@gmail.com

48
The following figures demonstrate the effect
of varying Tp when τ is fixed.

49
The figures on the previous slide ()
show that the spacing between adjacent spectral lines
decreases as Tp increases. In the limit as Tp → ∞, the
Fourier coefficients ck approach zero. This behavior is
consistent with the fact that as Tp → ∞ and τ remains
fixed, the resulting signal is no longer a power signal.
Indeed it becomes an energy signal and its average
power is zero.
The Power Density Spectrum for the rectangular pulse train is







±±=





τπ
τπ







 τ
=







 τ
=
,...2,1k,
kF
kFsin
T
A
0k,
T
A
c 2
0
0
2
p
2
p2
k

50
Frequency Analysis of Discrete-Time
Signals

51
Frequency Analysis of Discrete-Time Signals
•We have already discussed the Fourier series representation for continuous-
time periodic (power) signals and the Fourier transform for finite energy
aperiodic signals.
•The frequency range for continuous-time periodic signals
extends from -∞ to ∞,that contain infinite number of frequency
components with frequency spacing (1/Tp).
•The frequency range for discrete-time signals is unique over the
interval (-π,π) or (0,2π).
•A discrete-time signal of fundamental period N can consist of
frequency components separated by 2π/N radians or f= 1/N
cycles.
•Consequently, the Fourier series representation of the discrete-
time periodic signal will contain N frequency components (the
basic difference b/w Fourier series representation for continuous-
time and discrete-time periodic signals).

52
The Fourier Series for Discrete-Time Signals
Suppose that we are given a periodic sequence with
period N. The Fourier series representation for x[n]
consists of N harmonically related exponential
functions
ej2πkn/N
, k = 0, 1,2,…….,N-1
and is expressed as
∑
−
=
π
=
1N
0k
N/kn2j
kec]n[x
where the coefficients ck can be computed as:
∑
∞
=
π−
=
0n
N/kn2j
k e]n[x
N
1
c

53
Example: Determine the spectra of the following signals:
(a) x[n] = [1, 1, 0, 0], x[n] is periodic with period 4 (b) x[n] = cosπn/3
(c) x[n] = cos(√2)πn
Solution: (a) x[n] = [1, 1, 0, 0]
Now
[ ] [ ]
3
0
0
1 1 1 1
[ ] [0] [1] [2] [3] 1 1 0 0
4 4 4 2n
c x n x x x x
=
= = + + + = + + + =∑
3 3
2 / 4 / 2 / 2
1
0 0
1 1 1
[ ] [ ] [0] [1] 0 0
4 4 4
j n j n j
n n
c x n e x n e x x eπ π π− − −
= =
 = = = + + + ∑ ∑
( ) ( ) ( )2 2
1 1 1
1 1 cos sin 1 0 1
4 4 4
j j jπ π
 = + − = + − = −   
∑∑ =
π−
−
=
π−
==
3
0n
N/kn2j
1N
0n
N/kn2j
k e]n[x
4
1
e]n[x
N
1
c
↑

54
The magnitude spectra are:
and the phase spectra are:
00 =Φ 4
1
π−
=Φ undefined=Φ2 4
3
π
=Φ
[ ] 0sinjcos1
4
1
=π−π+=
( )[ ] [ ] [ ]j1
4
1
j01
4
1
)2/3sin(j2/3cos1
4
1
e]n[x
4
1
c
3
0n
4/3n2j
3 +=++=π−π+== ∑=
π−
[ ]π
=
π
=
π
+=== ∑∑ j
3
0n
nj
3
0n
4/n22j
2 e.11
4
1
e]n[x
4
1
e]n[x
4
1
c
4
2
c1 = 0c2 = 4
2
c3 =
2
1
c0 =

55
(b) x[n] = cosπn/3
Solution: In this case, f0 = 1/6 and hence x[n] is periodic with
fundamental period N = 6.
Now
∑∑∑ =
−−
==
−
===
5
0
3/6/2
5
0
5
0
/2
3
cos
6
1
3
cos
6
1
][
6
1
n
knjknj
nn
Nknj
k e
n
e
n
enxc πππ ππ
[ ] ( ) ( )
[ ]∑∑ =
+−−−−
=
+=+=
5
0
113/3/3/
5
0
33
12
1
2
1
6
1
n
kjkjknjnjnj
n
nn
eeeee
ππ
πππ
[ ] 0coscoscoscoscos0cos
6
1
3
cos
6
1
3
cos2
12
1
3
5
3
4
3
3
3
2
3
5
0
5
0
0
=+++++=
==∴ ∑∑ ==
πππππ
ππ
nn
nn
c
Similarly, c2 = c3 = c4 = 0, c1 = c5 = ½.krishnanaik.ece@gmail.com

56
(c) Cos(√2)πn
Solution: The frequency f0 of the signal is 1/√2 Hz.
Since f0 is not a rational number, the signal is
not periodic. Consequently, this signal cannot be
expanded in a Fourier series.

57
Power density Spectrum of Periodic Signals
The average power of a discrete time periodic signal with period N is
∑
−
=
=
1
0
2
)(
1 N
n
x nx
N
P
The above relation may also be written as
∑ ∑∑
−
=
−
=
π−
−
=






==
1N
0n
1N
0n
N/kn2*
k
1N
0n
*
x ec]n[x
N
1
]n[x]n[x
N
1
P
or
∑∑
∑∑
−
=
−
=
−
=
π−
−
=
==






=
1N
0n
2
21N
0k
k
1N
0n
N/kn2j
1N
0n
*
kx
]n[x
N
1
c
e]n[x
N
1
cP
This is Parse Val's Theorem for Discrete-Time Power Signals.

58
The Fourier Transform of Discrete-Time Aperiodic
Signals
The Fourier Transform of a finite energy discrete time signal x[n] is defined as
∑
∞
−∞=
−
=
n
jwn
e]n[x)w(X
X(w) may be regarded as a decomposition of x[n] into its Frequency
components. It is not difficult to verify that X(w) is periodic with frequency
2π.Indeed,X(ω) is periodic with period 2π,that is,
( ) ( ) ( )
( )
( ) )(.....................
...................
2
2
2
ω
πω
ω
πω
πω
Xenx
eenx
enxkX
nj
n
knjnj
n
n
nkj
==
=
=+
−
∞
−∞=
−−
∞
−∞=
∞
−∞=
+−
∑
∑
∑

59
• We observe two basic differences b/w the Fourier transform of a discrete-
time finite-energy signal and the Fourier transform of a finite-energy
analog signal .
• First, for continuous time signals, the spectrum of the signal have a
frequency range of (-∞,∞). In contrast, the frequency range for a discrete -
time signal is unique over the frequency interval of (-π,π).
• The second one is also a consequence of the discrete-time nature of the
signal. Since the signal is discrete in time , the Fourier transform of the
signal involves the summation of terms instead of an integral, as in the
case of continuous – time signals.
• Let us evaluate the sequence x(n) from X(ω).we multiply both sides of
X(ω) by ejωm
and integrate over the interval (-π,π).
ωωω ω
π
π
ωω
π
π
deenxdeX mjnj
n
mj
∫ ∑∫ −
−
∞
−∞=−






= )()(
∫−
−



≠
=
=
π
π
ω π
ω
nm
nm
de nmj
.........0
........2)(

60
Energy Density Spectrum of Aperiodic Signals
Energy of a discrete time signal x[n] is defined as
2
n
x ]n[xE ∑
∞
−∞=
=
Let us now express the energy Ex in terms of the spectral characteristic
X(w). First we have
∑ ∑ ∫
∞
−∞=
∞
−∞=
π
π−
−∗






π
==
n n
jwn*
x dwe)w(X
2
1
]n[x]n[x]n[xE
If we interchange the order of integration and summation in the above
equation, we obtain
dw)w(X
2
1
dwe]n[x)w(X
2
1
E
2
n
jwn
x ∫∫ ∑
π
π−
π
π−
∞
−∞=
−∗
π
=





π
=
ωω
π
π
ω
ω
π
π
π
π
ω
deXnx
nm
nmnx
denx
nj
nmj
n
∫
∫∑
−
−
−
∞
−∞=
=



≠
=
=
)(
2
1
)(
.........0
)........(2
)( )(

61
Therefore, the energy relation between x[n] and X(w) is
dw)w(X
2
1
]n[xE
22
n
x ∫∑
π
π−
∞
−∞= π
==
This is Parse Val's relation for discrete-time aperiodic signals.

62
Example: Determine and sketch the energy density
spectrum of the signal x[n] = an
u[n],
-1<a<1
Solution:
( )∑∑∑
∞
=
−
−−
∞
=
∞
−∞=
−
−
====
0n
jw
njwjwn
0n
n
n
jwn
ae1
1
aeeae]n[x)w(X
The energy density spectrum (ESD) is given by
( )( )jwjw
2
xx
ae1ae1
1
)w(X)w(X)w(X)w(S
+−
=== −
∗
2
awcosa21
1
+−
=
0 ππ
w
X(w)
a = 0.5 a= -0.5

63
Example: Determine the Fourier Transform and the energy
density spectrum of the sequence
Solution:


 −≤≤
=
otherwise,0
1Ln0,A
]n[x
)2/wsin(
)2/wLsin(
Ae
e1
e1
AAee]n[x)w(X )1L)(2/w(j
jw
jwL1L
0
jwn
n
jwn −−
−
−−
−
∞
−∞=
−
=
−
−
=== ∑∑
The magnitude of x[n] is



 =
=
otherwise,A
0w,LA
)w(X
)2/wsin(
)2/wLsin(
and the phase spectrum is
)2/wsin(
)2/wLsin(
2
w
)2L(A)w(X ∠+−∠−∠=∠
The signal x[n] and its magnitude is plotted on the next slide. The
Phase spectrum is left as an exercise.

64
x[n]
|X(w)|

65
Properties of DTFT
Symmetry Properties:
Suppose that both the signal x[n] and its transform X(w) are complex
valued. Then they can be expressed as
x[n] = xR[n] + j xI[n] (1)
X(ω) = XR(ω) + j XI(ω) (2)
The DTFT of the signal x[n] is defined as
∑
∞
−∞=
−
=
n
nj
enxX ω
ω ][)( (3)
Substituting (1) and (2) in (3) we get
[ ] nj
n
IRIR enxnxjXX ω
ωω −
∞
−∞=
∑ +=+ ][][)()(
but
njne nj
ωωω
sincos −=−

66
[ ][ ]njnnxnxjXX
n
IRIR
ωωωω sincos.][][)()( −+=+∴ ∑
∞
−∞=
separating the real and imaginary parts, we have
[ ]∑
∞
−∞=
+=
n
IRR nnxnnxX ωωω sin][cos][)( (4)
[ ]∑
∞
−∞=
−−=
n
IRI nnxnxX ωωωω cos][sin)()( (5)
In a similar manner, one can easily prove that
[ ]
[ ] ωωωωω
π
ωωωωω
π
π
π
dnXnXnx
dnXnXnx
IRI
IRR
∫
∫
+=
−=
2
2
cos)(sin)(
2
1
][
sin)(cos)(
2
1
][

67
DTFT Theorems and Properties
• Linearity
If x1[n] ↔ X1(w) and x2[n] ↔ X2(w), then
a1x1[n] + a2x2[n] ↔ a1X1(w) + a2X2(w)
Example 1: Determine the DTFT of the signal
x[n] = a|n|, -1< a<1
Solution: First, we observe that x[n] can be expressed as
x[n] = x1[n] + x2[n]
where

68



<
≥
=
0n,0
0n,a
]n[x
n
1 and



≥
<
=
−
0n,0
0n,a
]n[x
n
2
Now ( )∑∑∑
∞
=
−
∞
=
−−
∞
−∞=
===
00
11 ][)(
n
nj
n
njnnj
n
aeeaenxX ωωω
ω
( ) ( ) ω
ωωω
j
jjj
ae
aeaeae −
−−−
−
=++++=
1
1
....1
32
and ( ) ( )∑∑∑
−
−∞=
−−
−
−∞=
−
∞
−∞=
−
===
11
22 ][
n
njnj
n
n
n
nj
aeeaenxX ωωω
ω
( ) ω
ω
ωωω
j
j
jj
k
kj
ae
ae
aeaeae
−
=++==∑
∞
= 1
...)( 2
0
Now
2
2
21
cos21
1
11
1
)()()(
aa
a
ae
ae
ae
XXX j
j
j
+−
−
=
−
+
−
=+= −
ω
ωωω ω
ω
ω

69
• Time Shifting
If x[n] ↔ X(ω) then x[n-k] = e-jωk
X(ω)
Proof: [ ] ∑
∞
−∞=
−
−=−
n
nj
eknxknxF ω
][][
Let n – k = m or n = m+k
[ ] )(][][][ )(
ωωωω
XeemxeemxknxF kj
m
mjjwk
m
kmj −
∞
−∞=
−−
∞
−∞=
+−
===−∴ ∑∑
• Time Reversal property
If x[n] ↔ X(ω) then x[-n] ↔ X(-ω)
Proof:
[ ] ∑ ∑∑
−∞
∞=
∞
−∞=
−−
∞
−∞=
−
−===−=−
m m
mjmj
n
nj
XemxemxenxnxF )(][][][][ )(
ωωωω

70
• Convolution Theorem
If x1[n] ↔ X1(ω) and x2[n] ↔ X2(ω)
then
x[n] = x1[n]*x2[n] ↔ X (ω) = X1(ω)X2(ω)
Proof: As we know[convolution formula]
]kn[x]k[x]n[x*]n[x]n[x 2
k
121 −== ∑
∞
−∞=
nj
n kn
nj
eknxkxenxX ωω
ω −
∞
−∞=
∞
−∞=
∞
−∞=
−
∑ ∑∑ 





−== ][][][)( 21
Therefore
Interchanging the order of summation and making a substitution
n-k = m, we get
)()(][][
][][)(
2121
)(
21
ωω
ω
ωω
ω
XXemxekx
emxkxX
m
mjkj
k
kmj
k m
=











=






=
∑∑
∑ ∑
∞
−∞=
−−
∞
−∞=
+−
∞
−∞=
∞
−∞=
If we convolve two signal in time domain, then this is equivalent to
multiplying their spectra in frequency domain.

71
• Example 2: Determine the convolution of the sequences
x1[n] = x2[n] = [1, 1, 1]
Solution:
∑∑ −=
−
∞
−∞=
−
===
1
1
1121 ][][)()(
n
nj
n
nj
enxenxXX ωω
ωω
[ ] ωωωωω
cos21]1[]1[]0[]1[ 211 +=++=++−= −− jjjj
eeexxex
Then X(ω) = X1(ω)X2(ω) = (1 + 2cosω)2
=1 + 4cosω+ 4(cosω)2
.
= 1 + 4cosω+ 4(1+cos2ω/2) = 1 + 4cosω+ 2(1+cos2ω).
= 1 + 4cosω+ 2+2cos2ω).
= 3 + 4cos ω + 2cos2ω
= 3 + 2(ejω
+ e-jω
) + (ej2ω
+ e-j2ω
)
Hence the convolution of x1[n] and x2[n] is
x[n] = [1 2 3 2 1]
As known
1 2( ) ( ) 1 2cos( )X Xω ω ω= = +

72
• The Wiener-Khintchin Theorem:
Let x[n] be a real signal. Then rxx[k] ↔Sxx(w)
In other words, the DTFT of autocorrelation function is
equal to its energy density function.*
Proof: The autocorrelation of x[n] is defined as
∑
∞
−∞=
−=
k
xx nkxkxnr ][][][
Now jwn
kn
xx enkxkxnrF −
∞
−∞=
∞
−∞=






−= ∑∑ ][][]][[
Re-arranging the order of summations and making
Substitution m = k-n we get
)(
][][]][[ mkjw
mk
xx emxkxnrF −−
∞
−∞=
∞
−∞=






= ∑∑

73
( )ωωωωωω
xx
m
mj
k
kj
SXXXemxekx ==−=











−= ∑∑
∞
−∞=
∞
−∞=
2
|)(|)()(][][
• Frequency Shifting:
)(][
)(][
0
0
ωω
ω
−↔
↔
Xnxe
then
Xnx
If
njw
As from above property, multiplication of a sequence x(n) by
is equivalent to a frequency translation of the spectrum X(w)
by wo. So it be periodic, The shift ωo applies to the spectrum of
the signal in every period
oj n
e ω
Displacement in frequency multiplies the time/space function by a
unit phasor which has angle proportional to time/space and to the
amount of displacement.

74
• The Modulation Theorem:
If x[n] ↔ X(w) then
x[n] cosω0[n] ↔ ½X(ω + ω0) + ½X(ω - ω0)
Proof: Multiplication of a time/space function by a
cosine wave splits the frequency spectrum of the
function.
Half of the spectrum shifts left and half shifts right.
This is simply a variant of the shift theorem
which makes use of Euler's relationship
[ ] nj
n
njnj
n
nj
e
ee
nxnenxnnxF ω
ωω
ω
ωω −
∞
−∞=
−∞
−∞=
−
∑∑ 




 +
==
2
][cos][cos][
00
00
Use
frequenc
y shift
property
2
)cos(
jxjx
ee
x
−
+
=∴

75
• The Modulation Theorem: If x[n] ↔ X(w) then
x[n] cosω0[n] ↔ ½X(ω + ω0) + ½X(ω - ω0)
Proof:
[ ])()( 00
][
2
1 ωωωω +−−−
∞
−∞=
+= ∑
jnj
n
eenx
[ ] nj
n
njnj
n
nj
e
ee
nxnenxnnxF ω
ωω
ω
ωω −
∞
−∞=
−∞
−∞=
−
∑∑ 




 +
==
2
][cos][cos][
00
00
( ) ( )00
)()(
2
1
2
1
][
2
1
][
2
1 00
ωωωω
ωωωω
−++=+= ∑∑
∞
−∞=
−−
∞
−∞=
+−
Xenxenx
n
nj
n
nj
Use frequency shift property

76
• Parseval’s Theorem:
If x1[n] ↔ X1(w) and x2[n] ↔ X2(w) then
dw)w(X)w(X
2
1
]n[x]n[x *
21
n
*
21 ∫∑
π
π−
∞
−∞= π
=
Proof: ( ) ( ) [ ] ωω
π
ωωω
π
π
π
ω
π
π
dXenxdXXSHR nj
n
)(
2
1
2
1
... *
21
*
21 ∫ ∑∫ −
−
∞
−∞=−






==
SHLnxnxdeXnx
n
nj
n
..][][)(
2
1
][ *
21
*
21 === ∑∫∑
∞
−∞=
−
−
∞
−∞=
ωω
π
ω
π
π
In the special case where x1[n] = x2[n] = x[n], the Parseval’s
Theorem reduces to
ωω
π
π
π
dXnx
n
22
)(
2
1
)( ∫∑
−
∞
−∞=
=
We observe that the LHS of the above equation is energy Ex
of the Signal and the R.H.S is equal to the energy
density spectrum.

77
Thus we can re-write the above equation as
∫∫∑
−−
∞
−∞=
===
π
π
π
π
ωω
π
ωω
π
dSdXnxE xx
n
x )(
2
1
)(
2
1
][
2
2
• Multiplication of two sequences: [Windowing
Theorem]
If x1[n] ↔ X1(ω) and x2[n] ↔ X2(ω) then
( ) ( ) λλωλ
π
π
π
dXXnxnx −↔ ∫−
2121
2
1
][][
Windowing is the process of taking a small subset of a larger
dataset, for processing and analysis. A naive approach, the
rectangular window, involves simply truncating the dataset before
and after the window, while not modifying the contents of the
window at all. However, as we will see, this is a poor method of
windowing and causes power leakage.
Application of a window to a dataset will alter the spectral
properties of that dataset. In a rectangular window, for instance, all
the data points outside the window are truncated and therefore
assumed to be zero. The cut-off points at the ends of the sample will
introduce high-frequency components

78
Multiplication of two sequences: [Windowing Theorem](cont:)
If x1[n] ↔ X1(ω) and x2[n] ↔ X2(ω) then
( ) ( ) λλωλ
π
π
π
dXXnxnx −↔ ∫−
2121
2
1
][][
Proof:
[ ] ( ) nj
n
njnj
n
enxdeXenxnxnxnxF ωλ
π
π
ω
λλ
π
−
∞
−∞= −
−
∞
−∞=
∑ ∫∑ 





== ][
2
1
][][][][ 212121
( ) ( ) λλωλ
π
λλ
π
π
π
λω
π
π
dXXenxdX
n
nj
)(
2
1
][
2
1
21
)(
21 −=





= ∫∑∫ −
∞
−∞=
−−
−
Show periodic Convolution
Technique use for FIR filter design

79
• Differentiation in the Frequency Domain:
If x[n] ↔ X(w) then Fnx[n] ↔ jdX(w)/dw
Proof: nj
nn
nj
e
d
d
nxenx
d
d
d
dX ωω
ωωω
ω −
∞
−∞=
∞
−∞=
−
∑∑ =





= ][][
)(
∑
∞
−∞=
−
−=∴
n
nj
ennxj
d
dX ω
ω
ω
][
)(
Multiplying both sides by j we have
∑
∞
−∞=
−
=
n
nj
ennx
d
dX
j ω
ω
ω
][
)(
OR ]][[
)(
nnxF
d
dX
j =
ω
ω
Differentiation of a function induces a 90° phase shift in the
spectrum and scales the magnitude of the spectrum in proportion
to frequency. Repeated differentiation leads to the general result:
This theorem explains why differentiation of a signal has the reputation
for being a noisy operation. Even if the signal is band-limited, noise will
introduce high frequency signals which are greatly amplified by
differentiation.

80
The response of any LTI system to an arbitrary input
signal x[n] is given by convolution sum Formula
The Frequency Response Function:
∑
∞
−∞=
−=
k
]kn[x]k[h]n[y (6)
In this I/O relationship, the system is characterized
in the time domain by its unit impulse response
h[k]. To develop a frequency domain
characterization of the system, let us excite the
system with the complex exponential
x[n] = Aejwn
. -∞ < n < ∞ (7)
where A is the amplitude and w is an arbitrary frequency
confined to the frequency interval [-π, π]. By substituting
(7) into (6), we obtain the responsekrishnanaik.ece@gmail.com

81
[ ]∑
∞
−∞=
−
=
k
)kn(jw
Ae]k[h]n[y
jwn
k
jwk
ee]k[hA 





= ∑
∞
−∞=
−
or jwn
e)w(AH]n[y =
where
∑
∞
−∞=
−
=
k
jwk
e]k[h)w(H
(8)
(9)
The exponential Aejwn
is called an Eigen-function of
the system. An Eigen function of a system is an input
signal that produces an output that differs from the
input by a constant multiplicative factor. The
multiplicative factor is called an Eigen-value of the
System.
Response is also in form of complex exponential with same frequency as input, but altered by the multiplicative factor H(w).

82
Example: Determine the magnitude and phase of H(w)
for the three point moving average(MA) system
y[n] = 1/3[x[n+1] + x[n] + x[n-1]]
Solution: since h[n] = [1/3, 1/3, 1/3]
It follows that
H(w) = 1/3(ejw
+1 + e-jw
) = 1/3(1 + 2cosw)
Hence
|H(w)| = 1/3|1+2cosw| and



π<≤ππ
π≤≤
=Φ
w3/2,
3/2w0,0
)w(

83
1
|H(w)|
2π/3
w
0
1
0 2π/3
π
Φ(w)
w

84
Example:An LTI system is described by
the following difference equation:
y[n] = ay[n-1] + bx[n], 0 < a < 1
(a) Determine the magnitude and phase of
the frequency response H(w) of the system.
(b) Choose the parameter b so that the
maximum value of |H(w)| is unity.
(c) Determine the output of the system to
the input signal
x[n] = 5 + 12sin(π/2)n – 20cos (πn + π/4)

85
Solution:
(a) The frequency response is
)w(bX)w(Yae)w(Y jw
+= −
)w(bX)w(Y)ae1( jw
=− −
( ) wsinjawcosa1
b
)wsinjw(cosa1
b
ae1
b
)w(X
)w(Y
)w(H jw
−−
=
−−
=
−
== −
Now
( ) ( ) wcosa2a1
b
wsinawcosa1
b
)w(H
222
−+
=
+−
=
and 





−
−=Φ −
wcosa1
wsina
tan)w( 1
These responses are sketched on the next slide.

86
(b) It is easy to find that |H(w)| attains its
maximum value at w = 0. At this frequency,
we have
1
a1
b
)0(H =
−
= Which implies that b = ±(1-
a).
At b = (1-a), we have
wcosa2a1
a1
)w(H
2
−+
−
=
and wcosa1
wsina
tan)w( 1
−
−=Φ −
(c) The input signal consists of components of frequencies
w = 0, π/2 and π radians.
For w = 0, |H(0)| = 1 and θ(0) = 0.
For w = π/2,
.074.0
)9.0(1
9.01
a1
a1
2
H
22
=
+
−
=
+
−
=




 π
011
2
42)9.0(tanatan −=−=−=




 π
Θ −−krishnanaik.ece@gmail.com

87
For w = π,
0)(
053.0
a1
a1
|)w(H|
=πΘ
=
+
−
=
Therefore, the output of the system is
( )





πΘ+
π
+ππ−










 π
Θ+
π





 π
+=
4
ncos)(H20
2
n
2
sin
2
H12)0(H5]n[y
)ncos(06.1)42nsin(888.05 4
0
2
ππ
+π−−+=

88
Response to A-periodic input signals
Consider the LTI system of the following figure
where x[n] is the input, and y[n] is the output.
LTI System
h[n], H(w)
x[n] y[n]
If h[n] is the impulse response of the system, then
y[n] = h[n]*x[n]
The corresponding frequency domain representation is
Y(w) = H(w)X(w) [Corresponding Fourier transform of the y(n),x(n), & h(n) respectively]
Now the squared magnitude of both sides is given by
|Y(w)|2
= |H(w)|2
|X(w)|2
Or
Syy(w) = |H(w)|2
Sxx(w)
where Sxx(w) and Syy(w) are the energy density spectra of the
input and output signals, respectively.krishnanaik.ece@gmail.com

89
The energy of the output signal is
∫∫
π
π−
π
π−
π
=
π
= dw)w(S|)w(H|
2
1
dw)w(S
2
1
E xx
2
yyy
Example: An LTI system is characterized by its
impulse response h[n] = (1/2)n
u[n]. Determine the
spectrum and the energy density spectrum of
the output signal when the system is excited by the
signal x[n] = (1/4)n
u[n].
Solution:
( ) jw
2
1
jwn
n
0n
2
1
e1
1
e)w(H −
−
∞
= −
== ∑
Similarly, jw
4
1
e1
1
)w(X −
−
=
Hence the spectrum of the signal at the output of the system
is krishnanaik.ece@gmail.com

90
( )( )jw
4
1jw
2
1
e1e1
1
)w(X)w(H)w(Y −−
−−
==
The corresponding energy density spectrum is
222
yy )w(X)w(H)w(Y)w(S ==
( )( )wcoswcos
1
2
1
16
17
4
5
−−
=

Dr. Krishnanaik VankdothDr. Krishnanaik Vankdoth
B.EB.E(ECE),(ECE), M.TechM.Tech (ECE),(ECE), Ph.DPh.D (ECE)(ECE)
Professor in ECE Dept
Aksum University, Ethiopia– 1010
Krishnanaik.ece@gmail.com
Krishnanaik_ece@yahoo.com
Phone : +919441629162
krishnanaik.ece@gmail.com 92

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