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Fourier series, Discrete Time Fourier Transform and Characteristic functions. Fourier series. Fourier proposed in 1807. - PowerPoint PPT Presentation
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Fourier series, Discrete Time Fourier Transform and Characteristic functions
Fourier proposed in 1807
A periodic waveform f(t) could be broken down into an infinite series of simple sinusoids which, when added together, would construct the exact form of the original waveform.
Consider the periodic function
T = Period, the smallest value of T that satisfies the above equation.
Fourier series
To be described by the Fourier Series the waveform f(t) must satisfy the following mathematical properties:
1. f(t) is a single-value function except at possibly a finite number of points.
2. The integral for any t0.
3. f(t) has a finite number of discontinuities within the period T.4. f(t) has a finite number of maxima and minima within the
period T.
0
0
( )t T
tf t dt
Fourier series: conditions
T 2T 3T
t
f(t)
T
ntb
T
nta
atf
N
nn
N
nn
2sin
2cos
2)(
11
0
DC Part Even Part Odd Part
T is a period of all the above signals
Let ω0=2π/T.
)sin()cos(2
)( 01
01
0 tnbtnaa
tfN
nn
N
nn
Fourier series: synthesis
A Fourier Series is an accurate representation of a periodic signal (when N ∞) and consists of the sum of sinusoids at the fundamental and harmonic frequencies.
The waveform f(t) depends on the amplitude and phase of every harmonic components, and we can generate any non-sinusoidal waveform by an appropriate combination of sinusoidal functions.
Fourier series: definition
-0.5
0
0.5
1
1.5
Call a set of functions {ϕk} orthogonal on an interval a < t < b if it satisfies
Example
0
m =1
n = 2
-π π
Orthogonal functions
Define ω0=2π/T.
0 ,0)cos(2/
2/ 0 mdttmT
T0 ,0)sin(
2/
2/ 0 mdttmT
T
nmT
nmdttntm
T
T 2/
0)cos()cos(
2/
2/ 00
nmT
nmdttntm
T
T 2/
0)sin()sin(
2/
2/ 00
We now prove this one
Orthogonal set of sinusoidal functions
dttntmT
T 2/
2/ 00 )cos()cos(
0
)]cos()[cos(2
1coscos
dttnmdttnmT
T
T
T
2/
2/ 0
2/
2/ 0 ])cos[(2
1])cos[(
2
1
2/
2/00
2/
2/00
])sin[()(
1
2
1])sin[(
)(
1
2
1 T
T
T
Ttnm
nmtnm
nm
Case 1: m ≠ n
0
0
Proof or orthogonality
dttntmT
T 2/
2/ 00 )cos()cos(
0
dttmT
T 2/
2/ 02 )(cos
2/
2/
00
2/
2/
]2sin4
1
2
1T
T
T
T
tmm
t
2
T
]2cos1[2
1cos2
dttmT
T 2/
2/ 0 ]2cos1[2
1
nmT
nmdttntm
T
T 2/
0)cos()cos(
2/
2/ 00
Proof or orthogonalityCase 2: m = n
Define ω0=2π/T.
,3sin,2sin,sin
,3cos,2cos,cos
,1
000
000
ttt
ttt
an orthonormal set.an orthonormal set.
Orthogonal set of sinusoidal functions
Decomposition
dttfT
aTt
t
0
0
)(2
0
,2,1 cos)(2
0
0
0
ntdtntfT
aTt
tn
,2,1 sin)(2
0
0
0
ntdtntfT
bTt
tn
)sin()cos(2
)( 01
01
0 tnbtnaa
tfn
nn
n
Example (Square Wave)
112
200
dta
,2,1 0sin1
cos2
200
nntn
ntdtan
,6,4,20
,5,3,1/2)1cos(
1 cos
1sin
2
200
n
nnn
nnt
nntdtbn
π 2π 3π 4π 5π-π-2π-3π-4π-5π-6π
f(t)1
112
200
dta
,2,1 0sin1
cos2
200
nntn
ntdtan
,6,4,20
,5,3,1/2)1cos(
1 cos
1sin
2
100
n
nnn
nnt
nntdtbn
π 2π 3π 4π 5π-π-2π-3π-4π-5π-6π
f(t)1
Example (Square Wave)
ttttf 5sin
5
13sin
3
1sin
2
2
1)(
112
200
dta
,2,1 0sin1
cos2
200
nntn
ntdtan
,6,4,20
,5,3,1/2)1cos(
1 cos
1sin
2
100
n
nnn
nnt
nntdtbn
π 2π 3π 4π 5π-π-2π-3π-4π-5π-6π
f(t)1
Example (Square Wave)
-0.5
0
0.5
1
1.5
ttttf 5sin
5
13sin
3
1sin
2
2
1)(
When series is truncated
Harmonics
T
ntb
T
nta
atf
nn
nn
2sin
2cos
2)(
11
0
DC Part Even Part Odd Part
T is a period of all the above signals
)sin()cos(2
)( 01
01
0 tnbtnaa
tfn
nn
n
Harmonics
tnbtnaa
tfn
nn
n 01
01
0 sincos2
)(
Tf
22 00Define , called the fundamental angular frequency.
0 nnDefine , called the n-th harmonic of the periodic function.
tbtaa
tf nn
nnn
n
sincos2
)(11
0
Harmonicstbta
atf n
nnn
nn
sincos2
)(11
0
)sincos(2 1
0 tbtaa
nnnn
n
12222
220 sincos2 n
n
nn
nn
nn
nnn t
ba
bt
ba
aba
a
1
220 sinsincoscos2 n
nnnnnn ttbaa
)cos(1
0 nn
nn tCC
Amplitudes and Phase Angles
)cos()(1
0 nn
nn tCCtf
20
0
aC
22nnn baC
n
nn a
b1tan
harmonic amplitude phase angle
Complex Form of the Fourier Series
Complex Exponentials
Complex Form of the Fourier Series
tnbtnaa
tfn
nn
n 01
01
0 sincos2
)(
tjntjn
nn
tjntjn
nn eeb
jeea
a0000
11
0
22
1
2
1
0 00 )(2
1)(
2
1
2 n
tjnnn
tjnnn ejbaejba
a
1
000
n
tjnn
tjnn ececc
)(2
1
)(2
12
00
nnn
nnn
jbac
jbac
ac
Complex Form of the Fourier Series
1
000)(
n
tjnn
tjnn ececctf
1
10
00
n
tjnn
n
tjnn ececc
n
tjnnec 0
)(2
1
)(2
12
00
nnn
nnn
jbac
jbac
ac
Complex Form of the Fourier Series
2/
2/
00 )(
1
2
T
Tdttf
T
ac
)(2
1nnn jbac
2/
2/ 0
2/
2/ 0 sin)(cos)(1 T
T
T
Ttdtntfjtdtntf
T
2/
2/ 00 )sin)(cos(1 T
Tdttnjtntf
T
2/
2/
0)(1 T
T
tjn dtetfT
2/
2/
0)(1
)(2
1 T
T
tjnnnn dtetf
Tjbac )(
2
1
)(2
12
00
nnn
nnn
jbac
jbac
ac
Complex Form of the Fourier Series
n
tjnnectf 0)(
dtetfT
cT
T
tjnn
2/
2/
0)(1
)(2
1
)(2
12
00
nnn
nnn
jbac
jbac
ac
If f(t) is real,*nn cc
nn jnnn
jnn ecccecc
|| ,|| *
22
2
1|||| nnnn bacc
n
nn a
b1tan
,3,2,1 n
00 2
1ac
Complex Frequency Spectra
nn jnnn
jnn ecccecc
|| ,|| *
22
2
1|||| nnnn bacc
n
nn a
b1tan ,3,2,1 n
00 2
1ac
|cn|
n
amplitudespectrum
ϕn
n
phasespectrum
Example
2
T
2
T TT
2
d
t
f(t)A
2
d
TdnT
dn
T
Adcn
sin
82
5
1
T ,
4
1 ,
20
1
0
T
dTd
Example
40π 80π 120π-40π 0-120π -80π
A/5
5ω0 10ω0 15ω0-5ω0-10ω0-15ω0
TdnT
dn
T
Adcn
sin
Example
40π 80π 120π-40π 0-120π -80π
A/10
10ω0 20ω0 30ω0-10ω0-20ω0-30ω0
Example
dteT
Ac
d tjnn
0
0
d
tjnejnT
A
00
01
00
110
jne
jnT
A djn
)1(1
0
0
djnejnT
A
2/0
sindjne
TdnT
dn
T
Ad
TT d
t
f(t)A
0
)(1 2/2/2/
0
000 djndjndjn eeejnT
A
Discrete-time Fourier transform Until this moment we were talking continuous periodic functions.
However, probability mass function is a discrete aperiodic function. One method to find the bridge is to start with a spectral
representation for periodic discrete function and let the period become infinitely long.
T 2T 3T
t
f(t)
0
continuous
discrete
Discrete-time Fourier transform
We will take a shorter but less direct approach. Recall Fourier series
A spectral representation for the continuous periodic function f(t)
Consider now, a spectral representation for the sequence cn, -∞ < n < ∞
• We are effectively interchanging the time and frequency domains.• We want to express an arbitrary function f(t) in terms of complex
exponents.
Discrete-time Fourier transform
To obtain this, we make the following substitutions in
This is the inverse transform
discrete continuous
Discrete-time Fourier transform
To obtain the forward transform, we make the same substitution in
Discrete-time Fourier transform
Putting everything together
Sufficient conditions of existence
Properties of the Discrete-time Fourier Transform
Initial value
Homework: Prove it
Characteristic functions Determining the moments E[Xn] of a RV can be difficult.
An alternative method that can be easier is based on
characteristic function ϕX(ω).
There are particularly simple results for the characteristic
functions of distributions defined by the weighted sums of
random variables.
37
Characteristic functions
The function g(X) = exp(jωX) is complex but by defining
E[g(X)] = E[cos(ωX) + jsin(ωX)] = E[cos(ωX)] + jE[jsin(ωX)],
we can apply formula for transform RV and obtain
for those integers not included in SX.
Characteristic functions
The definition is slightly different than the usual Fourier
transform(the discrete time Fourier transform), which uses the
function exp(-jωk) in its definition.
As a Fourier transform it has all the usual properties.
The Fourier transform of a sequence is periodic with period of 2π.
Finding moments using CFTo find moments, let’s differentiate the sum “term by term”
Carrying out the differentiation
So that
Repeated differentiation produces the formula for the nth moment as
Moments of geometric RV: exampleSince the PMF for a geometric RV is given by pX[k] = (1 - p)k-1p for k = 1,2,…, we have that
but since |(1-p) exp(jω)| < 1, we can use the result
For z a complex number with |z| < 1 to yield the CF
Note that CF is periodic with period 2π.
Moments of geometric RV: example Let’s find the mean(first moment) using CF.
• Let’s find the second moment using CF and then variance.
Where D = exp(-jω) - (1-p). Since D|ω=0 = p, we have that
Expected value of binomial PMF
Binomial theorema b
By finding second moment we can find variance
Properties of characteristic functions Property 1. CF always exists since
Proof
Property 2. CF is periodic with period 2π. Proof: For m an integer
since exp(j2πmk) = 1 for mk an integer.
Properties of characteristic functions
Property 3. The PMF may be recovered from the CF. Given the CF, we may determine the PMF using
Proof: Since the CF is the Fourier transform of a sequence (although its definition uses a +j instead of the usual -j), it has an inverse Fourier transform. Although any interval of length 2π may be used to perform the integration in the inverse Fourier transform, it is customary to use [-π, π].
Fourier transform
Properties of characteristic functions Property 4. Convergence of characteristic functions guarantees
convergence of PMFs (Continuity theorem of probability).
If we have a sequence of CFs φnX(ω) converge to a given CF, say
φX(ω), then the corresponding sequence of PMF, say pnX[k], must
converge to a given PMF say pX[k].
The theorem allows us to approximate PMFs by simpler ones if we can show that the CFs are approximately equal.
Application example of property 4
Recall the approximation of the binomial PMF by Poisson PMF under
the conditions that p 0 and M ∞ with Mp = λ fixed. To show this using the CF approach we let Xb denote a binomial RV.
And replacing p by λ/M we have
as M ∞.
Application example of property 4
For Poisson RV XP we have that
Since φXb(ω) φXp(ω) as M ∞, by property 4 we must have that
pXb[k] pXp[k] for all k. Thus, under the stated conditions the
binomial PMF becomes the Possion PMF as M ∞.
Practice problems1. Prove that the transformed RV
has an expected value of 0 and a variance of 1.2. If Y = aX + b, what is the variance of Y in terms of the variance of X?3. Find the characteristic function for the PMF pX[k] = 1/5, for k = -2,-1,0,1,2.4. A central moment of a discrete RV is defined as E[(X – E[X])n] , for n positive integer. Derive a formula that relates the central moment to the usual (raw) moments. 5. Determine the variance of a binomial RV by using the properties of the CF. Assume knowledge of CF for binomial RV.
Homework1. Apply Fourier series to the following functions on (0; 2π)
a.
b.
c.
2. Find the second moment for a Poisson random variable by using the characteristic function exp [λ(exp(jω)-1)].3. A symmetric PMF satisfies the relationship pX[ -k] = pX[k] for k = …,-1,0,1,…. Prove that all the odd order moments, E[Xn] for n odd, are zero.