Post on 13-Jul-2020
Chapter 3 Fourier Representations of Signals
and Linear Time-Invariant Systems
Introduction
Complex Sinusoids and Frequency Response of LTI
Systems.
Fourier Representations for Four Classes of Signals
Discrete-Time Periodic Signals
Continuous-Time Periodic Signals
Discrete-Time Nonperiodic Signals
Continuous-Time Nonperiodic Signals.
Properties of Fourier Representation.
Linearity and Symmetry Properties.
1
Chapter 3 Fourier Representations of Signals
and Linear Time-Invariant Systems
Convolution Property
Differentiation and Integration Properties
Time- and Frequency-Shift Properties
Finding Inverse Fourier Transforms by Using Partial-Fraction
Expansions.
Multiplication Property
Scaling Properties
Parseval Relationships
Time-Bandwidth Product
Duality
2
Introduction
Fourier Analysis
Representing Signals a Superposition of Complex Sinusoids
Four Fourier Representations
DT periodic signals: DT Fourier series (DTFS)
CT periodic signals: CT Fourier series (CTFS)
DT nonperiodic signals: DT Fourier transform (DTFT)
CT nonperiodic signals: CT Fourier transform (CTFT)
3
2. Complex Sinusoids and Frequency
Response of LTI Systems.
The input-output relation of an LTI system is
characterized by convolution with the system’s
impulse response.
By representing the input as a sum of complex
sinusoids, we obtain a more straightforward
understanding of the input-output relation of an LTI
system
4
2. Complex Sinusoids and Frequency
Response of LTI Systems
Discrete-Time LTI System
The output of a complex sinusoidal input to an LTI system is a
complex sinusoid of the same frequency as the input,
multiplied by the frequency response of the system.
5
LTI Systemx(t) or
x[n]
y(t) or
y[n]
k k
knjekhknxkhny
j n j k j j n
k
y n e h k e H e e
( ) [ ]j j
k
H e h k e
Frequency Response
2. Complex Sinusoids and Frequency
Response of LTI Systems
Continuous-Time LTI System
6
( )( ) ( ) ( )
( )
j t j t j
j t
y t h e d e h e d
H j e
( ) ( ) jH j h e d
Frequency response:
Polar form complex number c = a + jb:
arg{ }j cc c e where 2 2 1and arg{ } tan b
ac a b c
Polar form for H (j):
arg{ ( )}( ) ( ) j H jH j H j e
where ( ) Magnitude response and arg ( ) Phase respnseH j H j
argj t H j
y t H j e
2. Complex Sinusoids and Frequency
Response of LTI Systems
RC Circuit: Frequency response
7
/1( ) ( )t RCh t e u t
RC
deueRC
jH jRC
1
deRC
RCj
0
11
1
0
1 1
1
jRCe
RCj
RC
101
11
RCj
RC
RCj
RC1
1
Frequency response:
Magnitude response:
2
2 1
1
RC
RCjH
Phase response:
RCjH arctanarg
2. Complex Sinusoids and Frequency
Response of LTI Systems8
2
4
2
4
arg H j
H j
[ ]nH
( )t
[ ]n
( )t ( ) j tH j e H
j te
H
j ne j j nH e e
2. Complex Sinusoids and Frequency
Response of LTI Systems
Eigenvalues and Eigenfunctions of a LTI System
9
Eigenfunction ( ) j tt e
Eigenvalue ( )H j
If ek is an eigenvector of a matrix A with eigenvalue k, then
k k kAe e
Arbitrary input = weighted superpositions of eigenfunctions
Convolution operation Multiplication
Ex. Input:
M
k
tj
kkeatx
1
Output:
M
k
tj
kkejHaty
1
1
[ ] k
Mj n
k
k
x n a e
1
[ ] k
Mj nj
k
k
y n a H e e
[ ] j nn e
( )jH e
3. Fourier Representations for Four
Classes of Signals
Periodic Signals: Fourier Series Representations
10
0 0is the th harmonic of .jk t j t
e k e
1. x[n] = discrete-time signal with fundamental period N. DTFS of x[n] is
0[ ] [ ]jk n
k
x n A k e (3.4)
o = 2/N Fundamental
frequency of x[n]
2. x(t) = continuous-time signal with fundamental period T. FS of x(t) is
0( ) [ ]jk t
k
x t A k e
(3.5) o = 2/T Fundamental
frequency of x(t)
“^” denotes approximate value. A[k] = the weight applied to the kth harmonic.
3. Fourier Representations for Four
Classes of Signals 11
<pf.> njknjNnKNjeee 000
njNnj ee 02
njke 0
There are only N distinct complex sinusoids of the form exp(jk0n)
should be used in Eq. (3.4).
Symmetries in x[k]:
k = (N 1)/2 to (N 1)/2
3. The complex sinusoids exp(jkon) are N-periodics in the frequency index k.
4. Continuous-time complex sinusoid exp(jk0t) with distinct frequencies k0
are always distinct. FS of continuous-time periodic signal x(t) becomes
0( ) [ ]jk t
k
x t A k e
Mean-square error (MSE) between the signal and its series representation:
Discrete-time case:
1 2
0
1[ ] [ ]
N
n
MSE x n x n dtN
Continuous-time case:
2
0
1( ) ( )
T
MSE x t x t dtT
1
0
0][][ˆN
k
njkekAnx
3. Fourier Representations for Four
Classes of Signals
Nonperiodic Signals: Fourier-Transform
Representations
12
1. FT of continuous-time signal:
X(j)/(2) = the weight applied to a sinusoid
of frequency in the FT representation.
2. DTFT of discrete-time signal:X(e j )/(2) = the weight applied to the
sinusoid ej n in the DTFT representation.
deeXnx njj )(
2
1][ˆ
dejXnx tj)(2
1][ˆ
3. Fourier Representations for Four
Classes of Signals 13
3. Fourier Representations for Four
Classes of Signals
Periodic Discrete
14
4. Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series15
1. DTFS pair of periodic signal x[n]:
0
1
0
[ ] [ ]N
jk n
k
x n X k e
(3.10)
0
1
0
1[ ] [ ]
Njk n
n
X k x n eN
(3.11)
Fundamental period = N;
Fundamental frequency = o = 2/N
Notation:
0;DTFSx n X k
Fourier coefficients; Frequency domain
representation
4. Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series
Example 3.6
16
MNnM
MnMnx
,0
,1
1. Period of DTFS coefficients = N o = 2/N
2. It is convenient to evaluate Eq.(3.11) over the interval n = M to n = N M 1.
0
0
11[ ] [ ]
1
N Mjk n
n M
Mjk n
n M
X k x n eN
eN
0
0 0
2( )
0
2
0
1[ ]
1
Mjk m M
m
Mjk M jk m
m
X k eN
e eN
3. For k = 0, N, 2N, …, we have
1o ojk jke e
2
0
1 2 11 , 0, , 2 ,
M
m
MX k k N N
N N
4. Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series17
3. For k 0, N, 2N, …, we may sum the geometric series in Eq. (3.15) to
obtain 0 0
0
(2 1)1
[ ] , 0, , 2 , ......1
jk M jk M M
jk
e eX k k N N
N e
0 0 0 0
0 0 0 0
2 1 / 2 2 1 2 1 / 2 2 1 / 2
/ 2 / 2 / 2
1 1 1
1
jk M jk M jk M jk M
jk jk jk jk
e e e eX k
N e e N e e
sin 2 1 /1, 0, , 2 ,
sin /
2 1 / , 0, , 2 ,
k M Nk N N
X k N k N
M N k N N
,2/sin
2/12sin1
0
0
k
Mk
NkX
4. Substituting o = 2/N, yields
0, , 2 ,
sin 2 1 /1 2 1lim
sin /k N N
k M N M
N k N N
L’Hôpital’s
Rule
Nk
NMk
NkX
/sin
/12sin1
The value of X[k] for k = 0, N, 2N, …,
is obtained from the limit as k 0.
L'Hôpital's rule
In its simplest form, l'Hôpital's rule states that for
functions ƒ and g:
18
4. Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series19
Figure 3.12 (p. 211)
The DTFS coefficients for the square
wave shown in Fig. 3.11, assuming a
period N = 50: (a) M = 4. (b) M = 12.
4. Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series20
Symmetry property of DTFS coefficient: X[k] = X[ k].
1. DTFS of Eq. (3.10) can be written as a series involving harmonically related
cosines.2. Assume that N is even and let k range from ( N/2 ) + 1 to N/2. Eq. (3.10)
becomes
2/
12/
0
N
Nk
njkekXnx
12/
1
2/ 0002/0N
m
njknjknjkemXemXeNXXnx
3. Use X[m] = X[ m] and the identity No = 2 to obtain
/ 2 1
0
1
0 / 2 cos 2 cosN
m
X X N n X m m n
0 0/ 2 1
1
0 / 2 22
jm n jm nNj n
m
e ex n X X N e X m
e jn = cos(n)
since sin(n) =
0 for integer n.
4. Define the new set of coefficients
4. Discrete-Time Periodic Signals--
The Discrete-Time Fourier Series21
4. Define the new set of coefficients
, 0, / 2
2 , 1, 2, , / 2 1
X k k NB k
X k k N
and write the DTFS for the square wave in terms of a series of harmonically
related cosines as/ 2
0
0
[ ] [ ]cos( )N
k
x n B k k n
(3.17)
A similar expression may be derived for N odd.
Example 3.7 Building a Square Wave From DTFS Coefficients
0
0
[ ] [ ]cos( )J
J
k
x n B k k n
4. Discrete-Time Periodic Signals--
The Discrete-Time Fourier Series22
(a) J = 1. (b) J = 3. (c) J = 5. (d) J = 23. (e) J = 25.
Example Figure 3.15
Electrocardiograms for two
different heartbeats and the first
60 coefficients of their
magnitude spectra.
(a) Normal heartbeat.
(b) Ventricular tachycardia.
(c) Magnitude spectrum for the
normal heartbeat.
(d) Magnitude spectrum for
ventricular tachycardia.
心動過速
5. Continuous-Time Periodic Signals–
The Fourier Series24
1. Continuous-time periodic signals are represented by the Fourier series (FS).
2. A signal with fundamental period T and fundamental frequency o = 2/T,
the FS pair of x(t) is
0( ) [ ]jk t
k
x t X k e
(3.19)
0
0
1[ ] ( )
Tjk t
X k x t e dtT
(3.20)
3. Notation:
0;FSx t X k
Frequency domain
representation of x(t)
The variable k determines the
frequency of the complex sinusoid
associated with X[k] in Eq. (3.19).
5. Continuous-Time Periodic Signals–
The Fourier Series25
Mean-square error (MSE)
1. Suppose we define
ˆ ojk t
k
x t X k e
2. If x(t) is square integrable,
T
dttxT 0
21
then the MSE between x(t) and x̂ t is zero, or, mathematically,
T
dttxtxT 0
2
01 at all values of t; it simply implies
that there is zero power in their
difference.
ˆ( )x t x t
3. Pointwise convergence of ˆ( )x t x t is guaranteed at all values of t
except those corresponding to discontinuities if the Dirichlet conditions are
satisfied: x(t) is bounded.
x(t) has a finite number of maximum and minima in one period.
x(t) has a finite number of discontinuities in one period.
If a signal x(t) satisfies the Dirichlet conditions and is not continuous, then
x̂ t converges to the mipoint of the left and right limits of x(t) at each
discontinuity.
5. Continuous-Time Periodic Signals–
The Fourier Series26
1. Fundamental frequency: o = 2/T.
2. X[k]: by integrating over T/2 t T/2
00 0
0
0
0
0
0 0 0 0
/ 2
/ 2
0
0
0 0
0
1 1
1, 0
2, 0
2
2sin, 0
T Tjk t jk t
T T
T
jk t
T
jk T jk T
X k x t e dt e dtT T
e kTjk
e ek
Tk j
k Tk
Tk
For k = 0, we have
0
0
0210
T
T T
Tdt
TX
3. By means of L’Hôpital’s rule, we have
5. Continuous-Time Periodic Signals–
The Fourier Series27
T
T
Tk
Tk
k
0
0
00
0
2sin2lim
0
00sin2
Tk
TkkX
4. X[k] is real valued. Using o = 2/T gives X[k] as a function of the ratio To/T:
02sin( 2 / )[ ]
2
k T TX k
k
(3.23)
5. Fig. 3.22 (a)-(c) depict X[k], 50 k 50, for To/T =1/4, To/T =1/16, and To/T =
1/64, respectively.
Definition of Sinc function:
sin( )sinc( )
uu
u
(3.24) Fig. 3.23.
1) Maximum of sinc function is unity at u = 0, the zero crossing occur at
integer values of u, and the amplitude dies off as 1/u.
2) The portion of this function between the zero crossings at u = 1 is
manilobe of the sinc function.
5. Continuous-Time Periodic Signals–
The Fourier Series28
The FS coefficients, X[k], –50 k 50, for three square waves. (see Fig. 3.21.) (a) To/T
= 1/4 . (b) To/T = 1/16. (c) To/T = 1/64.
5. Continuous-Time Periodic Signals–
The Fourier Series
Example 3.15
29
J
k
J tkkBtx0
0cos
Figure 3.25b-3 (p. 226) (b) J = 3. (c) J = 7. (d) J = 29.
Gibbs phenomenon
Ripple near discontinuiitiies
5. Continuous-Time Periodic Signals–
The Fourier Series
Trigonometric FS for Real Signal
30
0 0
1
( ) [0] [ ]cos( ) [ ]sin( )k
x t B B k k t A k k t
(3.25)
FS coefficients:
0
1[0] ( )
T
B x t dtT
00
2[ ] ( )cos( )
T
B k x t k t dtT
00
2[ ] ( )sin( )
T
A k x t k t dtT
(3.26)
B[0] = X[0] represents the time-
averaged value of the signal.
[ ] [ ] [ ]B k X k X k
[ ] ( [ ] [ ])A k j X k X k for k 0
Relation between exponential FS and trigonometric FS coefficients
6. Discrete-Time Nonperiodic Signals–
The Discrete-Time Fourier Transform31
1. The DTFT is used to represent a discrete-time nonperiodic signal as a
superposition of complex sinusoids.
2. DTFT representation of time-domain signal x[n]:
(3.31) 1
[ ] ( )2
j j nx n X e e d
( ) [ ]j j n
n
X e x n e
(3.32)
DTFT
pair
3. Notation:
Frequency-domain representation x[n]
DTFT jx n X e
6. Discrete-Time Nonperiodic Signals–
The Discrete-Time Fourier Transform
Condition for convergence of DTFT:
32
If x[n] is of infinite duration, then the sum converges only for certain classes
of signals. If x[n] is absolutely summable, i.e.,
n
x n
The sum in Eq. (3.32) converges uniformly to a
continuous function of .
If x[n] is not absolutely summable, but does satisfy (i.e., if x[n] has finite
energy),
n
nx2
It can be shown that the sum in Eq. (3.32)
converges in a mean-square error sense,
but does not converge pointwise.
6. Discrete-Time Nonperiodic Signals–
The Discrete-Time Fourier Transform33
Let 1,
0,
n Mx n
n M
1. DTFT of x[n]: 1 .
Mj j n
n M
X e e
2 2
( )
0 0
2( 1)1, 0, 2 , 4 ,
1
2 1, =0, 2 , 4 ,
M Mj j m M j M j m
m m
j Mj M
j
X e e e e
ee
e
M
Change of variable
m = n + M
0, 2 , 4 , The expression for X(e j ) when
2 1 / 2 2 1 / 2 2 1 / 2
/ 2 / 2 / 2
2 1 / 2 2 1 / 2
/ 2 / 2
( )
j M j M j M
j j M
j j j
j M j M
j j
e e eX e e
e e e
e e
e e
sin 2 1 / 2( )
sin 2
jM
X e
0, 2 , 4 , ,
sin 2 1 / 2lim 2 1;
sin 2
MM
L’Hôptital’s Rule
With understanding that X(e j ) for
is obtained as limit.0, 2 , 4 ,
6. Discrete-Time Nonperiodic Signals–
The Discrete-Time Fourier Transform
Example 3.19 Inverse DTFT of A Rectangular Pulse
34
1,
,0,
jW
X eW n
1
2
W
j n
W
x n e d
sin / ,W
x n c Wn
6. Discrete-Time Nonperiodic Signals–
The Discrete-Time Fourier Transform
Example 3.20 DTFT of The Unit Impulse
35
Find the DTFT of x[n] = [n].
<Sol.>1. DTFT of x[n]:
1j j n
n
X e n e
2. This DTFT pair is depicted in Fig. 3. 32.
1.DTFTn
Figure 3.32 (p. 235)
Example 3.20. (a) Unit impulse
in the time domain. (b) DTFT
of unit impulse in the
frequency domain.
6. Discrete-Time Nonperiodic Signals–
The Discrete-Time Fourier Transform36
Example 3.21 Inverse DTFT of A Unit Impulse Spectrum
1. Inverse DTFT of X(e j ): 1
.2
j nx n e d
1
,2
DTFT
.
2. This DTFT pair is depicted in Fig. 3. 33.
We can define X(e j ) over all by writing it as an infinite sum of delta
functions shifted by integer multiples of 2.
Dilemma: The DTFT of x[n] = 1/(2) does not converge, since it is not a
square summable signal, yet x[n] is a valid inverse DTFT!
2 .j
k
X e k
7. Continuous-Time Nonperiodic
Signals– The Fourier Transform37
1. The Fourier transform (FT) is used to represent a continuous-time non-
periodic signal as a superposition of complex sinusoids.
2. FT representation of time-domain signal x(t):
1( ) ( )
2
j tx t X j e d
(3.35)
( ) ( ) j tX j x t e dt
(3.36)
Frequency-domain representation of
the signal x(t)
3. Notation for FT pair:
.FTx t X j
7. Continuous-Time Nonperiodic
Signals– The Fourier Transform38
Convergence condition for FT:
1) Approximation: 1
ˆ ,2
j tx t X j e d
Squared error between x(t) and : the error energy, given by x̂ t
2
ˆ ,x t x t dt
is zero if x(t) is square
integrable, i.e., if
2
.x t dt
Zero squared error does not imply pointwise convergence [i.e., ] ˆx t x tat all values of t.
Pointwise convergence of ˆ( )x t x t is guaranteed at all values of t
except those corresponding to discontinuities if x(t) satisfies the Dirichlet
conditions for nonperiodic signals:
x(t) has a finite number of maximum, minima, and discontinuities in any
finite interval.
The size of each discontinuity is finite.
There is zero energy in the difference of ˆ( ) andx t x t
x(t) is absolutely integrable: .x t dt
7. Continuous-Time Nonperiodic
Signals– The Fourier Transform39
Example 3.25 FT of A Rectangular Pulse
Find the FT of x(t).
Consider the rectangular pulse depicted in Fig. 3.40 (a) and defined as
0 0
0
1,
0,
T t Tx t
t T
1. The rectangular pulse x(t) is absolutely integrable, provided that To < .
2. FT of x(t):
0
0
0
0 0
1 2sin , 0
Tj t j t
T
Tj t
T
X j x t e dt e dt
e Tj
3. For = 0, the integral simplifies to 2To.
0 00
2lim sin 2 .T T
0
2sin ,X j T
With understanding that
the value at = 0 is
obtained by evaluating
a limit.
7. Continuous-Time Nonperiodic
Signals– The Fourier Transform40
0sin
2 ,T
X j
5. Phase spectrum:
0
0
sin 00,arg
sin 0,
TX j
T
6. X(j) in terms of sinc function:
0 02 sinc .X j T T
Fig. 3. 40 (b).
v ; p
As To increases, the nonzero time
extent of x(t) increases, while X(j)
becomes more concentrated about
the frequency origin.
4. Magnitude spectrum:
7. Continuous-Time Nonperiodic
Signals– The Fourier Transform41
Example 3.26 Inverse FT of A Rectangular Pulse
Find the inverse FT of the rectangular spectrum depicted in Fig. 3. 42 (a) and
given by
1,
0,
W WX j
W
<Sol.>
Figure 3.42 (p. 246)
Example 3.26. (a) Rectangular spectrum in
the frequency domain.
(b) Inverse FT in the time domain.
v ; p
1. Inverse FT of x(t):
1 1 1
sin( ), 02
Wj t j t W
WW
x t e d e Wt tj t t
7. Continuous-Time Nonperiodic
Signals– The Fourier Transform42
0
1lim sin ,t
Wt Wt
3. Inverse FT is usually written as
1
sin ,x t Wtt
or sinc ,W Wt
x t
Fig. 3. 42 (b).
With understanding that the value
at t = 0 is obtained as a limit.
As W increases, the frequency-domain reprsentation becomes less
concentrated about = 0, while the time-domain representation X(j)
becomes more concentrated about t = 0.
2. When t = 0, the integral simplifies to W/, i.e.,
7. Continuous-Time Nonperiodic
Signals– The Fourier Transform43
Notation for FT pair: 1,FTt
Example 3.28 Inverse FT of An Impulse Spectrum
Find the inverse FT of X(j) = 2().
<Sol.>
1. Inverse FT of X(j):
1
2 12
j tx t e d
2. Notation of FT pair:
1 2FT
Duality between
Example 3.27 and 3.28
Example 3.27 FT of The Unit Impulse
Find the FT of x(t) = (t).
<Sol.> 1. x(t) does not satisfy the Dirichlet conditions, since the discontinuity at the
origin is infinite.
2. FT of x(t): 1j tX j t e dt
8. Properties of Fourier Representation44
8. Properties of Fourier Representation45
8. Properties of Fourier Representation
Periodic Discrete
46
9. Linearity and Symmetry Properties
Linearity property for four Fourier representations
47
;
;
o
o
FT
FS
DTFT j j j
DTFS
z t ax t by t Z j aX j bY j
z t ax t by t Z k aX k bY k
z n ax n by n Z e aX e bY e
z n ax n by n Z k aX k bY k
9. Linearity and Symmetry Properties48
1. Signal z(t):
3 1
2 2z t x t y t
where x(t) and y(t)
2. X[k] and Y[k]:
;2
;2
1 sin 4
1 sin 2
FS
FS
x t X k k k
y t Y k k k
3. FS of z(t):
;2 3 2 sin 2 1 2 sin 2FSz t Z k k k k k
9. Linearity and Symmetry Properties
Symmetry Properties: Real and Imaginary Signals
49
Symmetry property for real-valued signal x(t):
1. FT of x(t):
** *( ) ( ) ( )j t j tX j x t e dt x t e dt
(3.37)
2. Since x(t) is real valued, x(t) = x(t). Eq. (3.37) becomes
* ( )j tX j x t e dt
*( ) ( )X j X j (3.38)
X(j) is complex-conjugate symmetric
Re ( ) Re ( ) and Im ( ) Im ( )X j X j X j X j
9. Linearity and Symmetry Properties50
The conjugate symmetry property for DTFS:
*[ ] [ ]X k X N k
Because the DTFS coefficients are N periodic, and thus
[ ] [ ]X k X N k
The symmetry conditions in all four Fourier representations of real-valued
signals are indicated in Table 3.4.
9. Linearity and Symmetry Properties51
1. Input signal of LTI system:
cosx t A t
2. Real-valued impulse response of LTI system is denoted by h(t).
/ 2 / 2
j t j tx t A e A e
Euler’s formula
3. Output signal of LTI system:
cos argy t H j A t H j
( arg ( ) ) ( arg ( ) )( ) ( ) ( / 2) ( ) ( / 2)
j t H j j t H jy t H j A e H j A e
Applying Eq. (3.2)
and linearity
Exploiting the symmetry conditions:
( ) ( )H j H j
arg ( ) arg ( )H j H j
Continuous-time case:
9. Linearity and Symmetry Properties52
Discrete-time case: x[n] = Acos(n )
2. Real-valued impulse response of LTI system is denoted by h[n].
3. Output signal of LTI system:
cos argj jy n H e A n H e
The LTI system modifies the amplitude of the input sinusoid by H(e j)and
the phase by arg{H(e j)}.
x(t) is purely imaginary:
1. x(t) = x(t).
2. Eq.(3.37) becomes
* ( )j tX j x t e dt
*( ) ( )X j X j (3.39)
Re ( ) Re ( ) and Im ( ) Im ( )X j X j X j X j
9. Linearity and Symmetry Properties
Symmetry Properties: Even and Odd Signals
53
1. x(t) is real valued and has even symmetry.
x(t) = x(t) and x ( t) = x(t) x(t) = x( t)
2. Eq.(3.37) becomes
* ( )j tX j x t e dt
* jX j x e d X j
3. Conclusion:
1) The imaginary part of X(j) = 0: ( ) ( )X j X j
If x(t) is real and even, then X(j) is real.
2) If x(t) is real and odd, then X(j) = X(j) and X(j) is imaginary.
Change of variable = t
10. Convolution Property
Nonperiodic Signals
54
1. Convolution of two nonperiodic continuous-time signals x(t) and h(t):
y t h t x t h x t d
2. FT of x(t ):
( )1
2
j tx t X j e d
3. Substituting this expression into the convolution integral yields
1
2
j ty t H j X j e d
and we identify H(j)X(j) as the FT of y(t).
( ) ( )* ( ) ( ) ( ) ( )FTy t h t x t Y j X j H j (3.40)
1
2
1
2
j t j
j j t
y t h X j e e d d
h e d X j e d
10. Convolution Property
Example
55
2
2
4sinFTx t X j
x t z t z t
1, 1
0, 1
FTt
z t Z jt
10. Convolution Property56
• Convolution of Nonperiodic Signals — Discrete-time case
If DTFT jx n X e DTFT jh n H e and
[ ] [ ]* [ ] ( ) ( ) ( )DTFT j j jy n x n h n Y e X e H e (3.41)
10.2 Filtering57
1. Multiplication in frequency domain Filtering.
3. System Types of filtering:
1) Low-pass filter
2) High-pass filter
3) Band-pass filter
Fig. 3.53 (a), (b), and (c)
The characterization of DT filter is based
on its behavior in the frequency range
< because its frequency
response is 2 -periodic.
4. Realistic filter:
1) Gradual transition band
2) Nonzero gain of stop band
5. Magnitude response of filter: 20log H j 20log jH e or [dB]
2. The terms “filtering” implies that some frequency components of the input
are eliminated while others are passed by the system unchanged.
10.2 Filtering58
10. Convolution Property59
Example 3.33 RC Circuit: Filtering
For the RC circuit depicted in Fig. 3.54, the impulse response for the case where
yC(t) is the output is given by
1 t RC
Ch t e u tRC
Since yR(t) = x(t) yC(t), the impulse response for
the case where yR(t) is the output is given by
1 t RC
Rh t t e u tRC
Figure 3.54 (p. 264)
RC circuit with input x(t) and outputs yc(t) and
yR(t).
1
1CH j
i RC
1
R
i RCH j
i RC
60
10. Convolution Property
Frequency response of a system
The ratio of the FT or DTFT of the output to the input .
61
For CT system:( )
( )( )
Y jH j
X j
(3.42)
For DT system:( )
( )( )
jj
j
Y eH e
X e
(3.43)
Frequency response
10. Convolution Property
Recover the input of the system from the output
Equalizer
The process of recovering the input from the output is known
as equalization
62
invX j H j Y j and invj j jX e H e Y e
CT case DT case
where inv 1/ ( )H j H j and inv 1/ ( )j jH e H e
10. Convolution Property
Convolution of Periodic Signals
63
Define the periodic convolution of two CT signals x(t) and z(t), each having
period T, as
0
T
y t x t z t x z t d #
where the symbol denotes that integration is performed over a single period
of the signals involved.
2;
( ) ( ) ( ) [ ] [ ] [ ]FS
Ty t x t z t Y k TX k Z k
#
10. Convolution Property
Discrete-Time Convolution of Periodic Sequences
Define the discrete-time convolution of two N-periodic
sequences x[n] and z[n]
64
1
0
N
k
y n x n z n x k z n k
#
2;
[ ] [ ] [ ] [ ] [ ] [ ]DTFS
Ny n x n z n Y k NX k Z k
#
DTFS of y[n]
( ) ( ) ( ) ( )FT
x t z t X j Z j
0;( ) ( ) [ ] [ ]
FSx t z t TX k Z k
#
[ ] [ ] ( ) ( )FT j jx n z n X e Z e
0;[ ] [ ] [ ] [ ]
DTFSx n z n NX k Z k
#
Table 3.5 Convolution Properties
11. Differentiation and Integration
Properties
Differentiation in Time
65
1. A nonperiodic signal x(t) and its FT, X(j), is related by
dejXtx tj
2
1 Differentiating both
sides with respect to
t
dejjXtxdt
d tj
2
1
◆ Differentiation of x(t) in Time-Domain (j) X(j) in Frequency-Domain
( ) ( )FTd
x t j X jdt
00
( ) ( ) ( ) 0d
x t j X jdt
F
11. Differentiation and Integration
Properties
Differentiation in Frequency
66
1. FT of signal x(t):
dtetxjX tj
Differentiating both
sides with respect to
dtetjtxjXd
d tj
2. Differentiation property in frequency domain:
FT djtx t X j
d
◆ Differentiation of X(j) in Frequency-Domain ( jt) x(t) in Time-Domain
11. Differentiation and Integration
Properties
Integration
Integrating nonperiodic signals
67
1. Define dxty
t
( ) ( )
dy t x t
dt (3.51)
2. By differentiation property, we have
1( ) ( )Y j X j
j
(3.52)
Indeterminate at = 0
3. When the average value of x(t) is not zero, then it is possible that y(t) is not
square integrable.
FT of y(t) may not converge!
W can get around this problem by including impulse in the transform, i.e.
1( ) ( ) ( )Y j X j c
j
c can be determined by
the average value of x(t)
C = X(j0)
11. Differentiation and Integration
Properties68
68
Table 3.6 Commonly Used Differentiation and Integration Properties.
( ) ( )FTd
x t j X jdt
0;0
FSdx t jk X k
dt
FT djtx t X j
d
DTFT jdjnx n X e
d
1( ) ( ) ( 0) ( )
tFTx d X j X j
j
12. Time- and Frequency-Shift Properties
Time-Shift Property
69
1. Let z(t) = x(t t0) be a time-shifted version of x(t).
2. FT of z(t):
dtetzjZ tj
dtettx tj
0
3. Change variable by = t t0:
dexjZtj 0
dexe jtj 0 jXe
tj 0
◆ Time-shifting of x(t) by t0 in Time-Domain
(e j t0) X(j) in Frequency-Domain
and( ) ( )Z j X j 0arg{ ( )} arg{ ( )}Z j X j t
12. Time- and Frequency-Shift Properties70
0
0
j tFTx t t e X j
0 0 0;
0
FT jk tx t t e X k
0
0
j nDTFT jx n n e X e
0 0 0;
0
DTFS jk nx n n e X k
12. Time- and Frequency-Shift Properties
Frequency Shift
71
1. Suppose that:
( ) ( )FTx t X j ( ) ( )
FTz t Z jand
2. By the definition of the inverse FT, we have
dejZtz tj
2
1
dejX tj
2
1
Substituting variables = into above Eq. gives
dejXtz tj
2
1
dejXe tjtj
2
1 txe tj
◆ Frequency-shifting of X(j) by in Frequency-Domain [i.e. X(j( ))]
(e t) x(t) in Time-Domain
12. Time- and Frequency-Shift Properties72
FTj te x t X j
0 0 0;
0
jk t FSe x t x k k
[ ]
jDTFTj ne x n X e
0 0 0;
0[ ]jk n FS
e x n X k k
13. Finding Inverse Fourier Transforms
by Using Partial-Fraction Expansions. 73
1. FT X(j) in terms of a ratio of polynomials in j:
jA
jB
ajajaj
bjbjbjX
N
N
N
M
M
01
1
1
01
Assume that M < N. If M N, then we may use long division to express X(j)
in the form
jA
jBjfjX
NM
k
k
k
0
Partial-fraction expansion
is applied to this term
Applying the differentiation property
and the pair to these
terms
( ) 1FTt
2. Let the roots of the denominator
A(j) be dk, k = 1, 2, …, N. These
roots are found by replacing j
with a generic variable v and determining the roots of the polynomial
001
1
1
avavav N
N
N
13. Finding Inverse Fourier Transforms
by Using Partial-Fraction Expansions74
3. For M < N, we may the write
N
k
k
M
k
k
k
dj
jb
jX
1
0
Assuming that all the roots dk, k = 1, 2, …, N, are distinct, we may write
N
k k
k
dj
CjX
1
Ck, k = 1, 2, …, N are determined
by the method of residues
Appendix B
In Example 3.24, we derived the transform pair
1FTdte u t
j d
This pair is valid even if d is
complex, provided that Re{d} < 0.
Assuming that the real part of each dk, k = 1, 2, …, N, is negative, we use
linearity to write
1 1
k
N Nd t FT k
k
k k k
Cx t C e u t X j
j d
13. Finding Inverse Fourier Transforms
by Using Partial-Fraction Expansions
Inverse Discrete-Time Fourier Transform
75
1. Suppose X(e j) is given by a ratio of polynomial in e j, i.e.
11
1
1
01
jNj
N
Nj
N
jMj
Mj
eee
eeeX
Normalized to 1
2. Factor the denominator polynomial as
N
k
j
k
jNj
N
Nj
N edeee1
1
1
1 11
Replace e j with the generic variable v:
01
2
2
1
1
NN
NNN vvvv Find dk, the roots of
this polynomial
3. Partial-fraction expansion:
Assuming that M < N and all the dk are distinct, we may express X(e j)as
N
kj
k
kj
ed
CeX
1 1
13. Finding Inverse Fourier Transforms
by Using Partial-Fraction Expansions
Since
76
1
1
n DTFT
k j
k
d u nd e
The linearity property implies that
N
k
n
kk nudCnx1
Expansions for
repeated roots are
treated in Appendix B.
14. Multiplication Property77
Non-periodic continuous-time signals
1. Non-periodic signals: x(t), z(t), and y(t) = x(t)z(t). Find the FT of y(t).
2. FT of x(t) and z(t):
dvejvXtx jvt
2
1
dejZtz tj
2
1and
dvdejZjvXty tvj
22
1
Change variable: = vInner Part: Z(j) X(j)
Outer Part: FT of y(t)3. FT of y(t):
1 1
2 2
j ty t X jv Z j v dv e d
1( ) ( ) ( ) ( ) ( )* ( )
2
FTy t x t z t Y j X j Z j
(3.56)
Scaled by 1/2
14. Multiplication Property78
where
dvvjZjvXjZjX *
◆ Multiplication of two signals in Time-Domain
Convolution in Frequency-Domain (1/2)
Non-periodic continuous-time signals
1. Non-periodic DT signals: x[n], z[n], and y[n] = x[n]z[n].
Find the
DTFT of
y[n].2. DTFT of y[n]:
1[ ] [ ] [ ] ( ) ( ) ( )
2
DTFT j j jy n x n z n Y e X e Z e
# (3.57)
◆ Multiplication of two signals in Time-Domain
Convolution in Frequency-Domain (1/2)
where the symbol denotes periodic convolution.
Here, X(e j ) and X(e j ) are 2-periodic, so we evaluate the convolution over
a 2 interval:
jjX j Z j X e Z e d
#
14. Multiplication Property79
Multiplication property
can
be used to study the
effects
of truncating a time-domain
signal on its frequency-
domain.
Windowing !
Truncate signal x(t) by a
window function w(t) is
represented by
y(t) = x(t)w(t)
Fig. 3. 65 (a)
0 0
Time Interval
T t T
14. Multiplication Property80
Figure 3.65b (p. 293)
(b) Convolution of the signal
and window FT’s resulting
from truncation in time.
1. FT of y(t):
1
*2
FTy t Y j X j W j
p ; v 2. If w(t) is the rectangular window depicted
in Fig. 3. 65 (b), we have
0sin2
TjW
Smoothing the details
in X(j) and
introducing
oscillation near
discontinuities in X(j).
14. Multiplication Property81
Example 3.46 Truncating the Impulse Response
The frequency response H(e j) of an ideal discrete-time system is depicted in
Fig. 3. 66 (a). Describe the frequency response of a system whose impulse
response is the ideal system impulse response truncated to the interval M
n M.
<Sol.>
1. The ideal impulse response is the inverse DTFT of H(e j ).
Using the result of Example 3.19, we write
1
sin2
nh n
n
2. Let ht[n] be the truncated impulse response:
,0, otherwise
t
h n n Mh n
[ ] [ ] [ ]th n h n w n
1,
0, otherwise
n Mw n
14. Multiplication Property82
3. Let [ ] ( )DTFT j
t th n H e
Using the multiplication property in Eq. (3.57), we have
1
2
jj j
tH e H e W e d
4. Since
2/,0
2/,1jeH
sin 2 1 / 2
sin / 2
jM
W e
On the basis of
Example 3.18
and
2/
2/2
1
dFeH j
t
where we have defined
, / 2
0, / 2
j
jjW e
F H e W e
Discussion: refer to
p. 295 in textbook.
14. Multiplication Property83
Figure 3.66 (p. 294)
The effect of truncating the impulse
response of a discrete-time system.
(a) Frequency response of ideal
system. (b) F() for near zero.
v , u
14. Multiplication Property84
Figure 3.66 (p. 294)
The effect of truncating the impulse response of a discrete-
time system. (c) F() for slightly greater than /2.(d)
Frequency response of system with truncated impulse
response.
v , u
Ht(ej ) is the area under
F( ) between = /2
and = /2.
14. Multiplication Property85
1( ) ( ) ( ) ( )
2
FTx t z t X j Z j
;( ) ( ) [ ] [ ]oFS
x t z t X k Z k
1[ ] [ ] ( ) ( )
2
DTFT j jx n z n X e Z e
#
;[ ] [ ] [ ] [ ]oDTFT
x n z n X k Z k
#
Table 3.9 Multiplication Properties of Fourier Representations
15. Scaling Properties86
1. Let z(t) = x(at).
2. FT of z(t) :
( ) ( ) ( )j t j tZ j z t e dt x at e dt
Changing
variable: = at
( / )
( / )
(1/ ) ( ) , 0( )
(1/ ) ( ) , 0
j a
j a
a x e d aZ j
a x e d a
( / )( ) (1/ ) ( ) ,j aZ j a x e d
( ) ( ) (1/ ) ( / ).FTz t x at a X j a
◆ Scaling in Time-Domain Inverse Scaling in Frequency-Domain
Signal expansion or compression!
Refer to Fig. 3.70.
(3.60)
15. Scaling Properties87
Figure 3.70 (p. 300)
The FT scaling property. The figure assumes that 0 < a < 1.
v
15. Scaling Properties88
Example 3.48 Scaling a Rectangular Pulse
<Sol.>
Let the rectangular pulse
1, 1( )
0, 1
tx t
t
Use the FT of x(t) and the scaling property to
find the FT of the scaled rectangular pulse
1, 2( )
0, 2
ty t
t
1. Substituting T0 = 1 into the result of Example 3. 25 gives
2( ) sin( )X j
2. Note that y(t) = x(t/2).
2 2( ) 2 ( 2 ) 2 sin(2 ) sin(2 )
2Y j X j
Scaling property and a = 1/2
Fig. 3.71
Substituting T0 = 2 into the result of Example 3.25 can also give the answer!
15. Scaling Properties89
Figure 3.71
(p. 301)
Application of
the FT scaling
property in Example 3.48.
(a) Original time
signal. (b)
Original FT. (c)
Scaled time signal y(t) =
x(t/2). (d) Scaled
FT Y(j) =
2X(j2).
v
16. Parseval Relationships90
The energy or power in the time-domain representation of a signal is equal to
the energy or power in the frequency-domain representation.
◆ Case for CT nonperiodic signal: x(t)
1. Energy in x(t):
2( )xW x t dt
2. Note that 2( ) ( ) ( )x t x t x t
Express x*(t) in terms of its FT X(j):1
( ) ( )2
j tx t X j e d
1( ) ( )
2
j t
xW x t X j e d dt
The integral inside
the braces is the FT
of x(t).
1( ) ( )
2xW X j X j d
1( ) ( )
2
j t
xW X j x t e dt d
2 21( ) ( )
2x t dt X j d
16. Parseval Relationships91
Representations Parseval Relationships
FT 2 21
( ) ( )2
x t dt X j d
FS 2 2
0
1( ) [ ]
T
kx t dt X k
T
DTFT 22 1
[ ] ( )2
j
nx n X e d
DTFS 1 12 2
0 0
1[ ] [ ]
N N
n kx n X k
N
16. Parseval Relationships92
Example 3.50 Calculating Energy in a Signal
Letsin( )
[ ]Wn
x nn
Use Parseval’s theorem to evaluate
21( )
2
jX e d
22
2 2
sin ( )[ ]
n n
Wnx n
n
<Sol.>
1. Using the DTFT Parseval relationship in Table 3.10, we have
2. Since 1,[ ] ( )
0,
DTFT jW
x n X eW
11 /
2
W
Wd W
Direct calculation in time-
domain is very difficult!
17. Time-Bandwidth Product93
1. Preview: 1,( ) ( ) 2sin( ) /
0,
o FT
o
o
t Tx t X j T
t T
Figure 3.72 (p. 305)
Rectangular pulse illustrating the inverse relationship between the time
and frequency extent of a signal.
Fig. 3.72
2. x(t) has time extent 2T0. X(j) is actually of infinite extent frequency.
v , p
3. The mainlobe of sinc function is fallen in the interval of < /T0, in which
contains the majority of its energy.
17. Time-Bandwidth Product94
4. The product of the time extent T0 and mainlobe width 2/T0 is a constant.
5. Compressing a signal in time leads expansion in the frequency domain and
vice versa.
Signla’s time-bandwidth product !
6. Bandwidth: The extent of the signal’s significant frequency content.
1) A mainlobe bounded by nulls.
Ex. Lowpass filter One half the width of mainlobe.
2) The frequency at which the magnitude spectrum is 1/2 times its peak values.
7. Effective duration of signal x(t):1/ 2
22
2
( )
( )d
t x t dtT
x t dt
(3.63)
8. Effective bandwidth of signal x(t):1/ 2
22
2
( )
( )w
X j dB
X j d
(3.64)
17. Time-Bandwidth Product95
9. The time-bandwidth product for any signal is lower bounded according to
the relationship
1/ 2d wT B (3.65)
We cannot simultaneously decrease the duration and bandwidth of
a signal.
Uncertainty principle !
Example 3.51 Bounding the Bandwidth of a Rectangular Pulse
<Sol.>
Let 1,( )
0,
o
o
t Tx t
t T
Use the uncertainty principle to place a lower
bound on the effective bandwidth of x(t).
1. Td of x(t): 1/ 22o
o
o
o
T
T
d T
T
t dtT
dt
17. Time-Bandwidth Product96
1/ 22
3 1/ 2[(1/(2 ))(1/ 3) ] / 3
o
o o
oo
o
T
T T
d o T dT
T
t dtT T t T
dt
2. The uncertainty principle given by Eq. (3.65) states that
1/(2 )w dB T 3 /(2 )w oB T
18. Duality
Duality of the FT
97
1. FT pair:
1( ) ( )
2
j tx t X j e d
( ) ( ) j tX j x t e dt
and
2. General equation:
1( ) ( )
2
jvy v z e dv
(3.66)
Choose = t and = the Eq. (3.66) implies that
3. Interchange the role of time and frequency by letting = and = t, then
Eq. (3.66) implies that
1( ) ( )
2
j ty t z e d
( ) ( )FTy t z (3.67)
1( ) ( )
2
j ty z t e dt
( ) 2 ( )FTz t y (3.68)
18. Duality98
Example99
Example 3.52 Applying Duality
Find the FT of 1( )
1x t
jt
<Sol.>
1. Note that:1
( ) ( ) ( )1
FTtf t e u t F jj
2. Replacing by t, we obtain
1( )
1F jt
jt
x(t) has been expressed as F(jt).
3. Duality property:
( ) 2 ( ) 2 ( )X j f e u
Duality100
Figure 3.74 (p. 309)
The FT duality property.
v ; p
18. Duality101
1. DTFS Pair for x[n]:
1
0
[ ] [ ] O
Njk n
k
x n X k e
1
0
1[ ] [ ] O
Njk n
n
X k x n eN
and
2. The DTFS duality is stated as follows: if
2;
[ ] [ ]DTFS
Nx n X k
(3.71)
2; 1
[ ] [ ]DTFS
NX n x kN
(3.72)
N = time index; k =
frequency index
18. Duality
DTFT
102
Table 3.11 Duality Properties of Fourier Representations
FT ( ) ( )FTf t F j ( ) 2 ( )FTF jt f
DTFS ; 2 /[ ] [ ]DTFS Nx n X k ; 2 /[ ] (1/ ) [ ]DTFS NX n N x k
FS-DTFT [ ] ( )DTFS jx n X e ;1( ) [ ]FSjtX e x k
( ) [ ] ojk t
k
z t Z k e
( ) [ ]j j n
k
X e x n e
3. Duality relationship between z(t) and X(e j)
Require
T = 2
In the DTFT t in the FS
n In the DTFT k in the FSAssumption: 0 = 1
[ ] ( )DTFS jx n X e
;1( ) [ ]FSjtX e x k
A.1 Wind Instruments
Fipples, pressure-excited reeds, or buzzing lips are
coupled with resonating air column.
A.2 String Instruments
One or more stretched stringed strings are made to vibrate,
typically by bowing, plunking or striking, in conjunction with
resonating boxes or sound boards.
A.3 Percussion Instruments
Include virtually any object that may be struck, typically
consisting of bars, plates, or stretched membrances and
associated resonators
A.4 Cello & Clarinet
A.5 Trumpet & Bass Drum
A.6 Speech
Models
y n a y n k G b x n rk r
r
q
k
p
( ) ( ) ( )
01
0 ( ) z
Pitch period P
Voiced
Unvoiced
V
U
voiced/unvoiced
switch
signal speech
Gain estimate
All-Pole
Filtery(n)