fourier series and characteristic functions
DESCRIPTION
Fourier Series 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. - PowerPoint PPT PresentationTRANSCRIPT
Fourier Series 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.
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
Synthesis
T 2T 3Tt
f(t)
Tntb
Tntaatf
N
nn
N
nn
2sin2cos2
)(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 tnbtnaatfN
nn
N
nn
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 combinationof sinusoidal functions.
Definition
Orthogonal Functions
Call a set of functions {ϕk} orthogonal on an interval a < t < b if it satisfies
Example
0m =1n = 2
-π π
Orthogonal set of Sinusoidal Functions
Define ω0=2π/T.0 ,0)cos(
2/
2/ 0 mdttmT
T0 ,0)sin(
2/
2/ 0 mdttmT
T
nmTnm
dttntmT
T 2/0
)cos()cos(2/
2/ 00
nmTnm
dttntmT
T 2/0
)sin()sin(2/
2/ 00
nmdttntmT
T and allfor ,0)cos()sin(
2/
2/ 00
We now prove this one
Proof
dttntmT
T 2/
2/ 00 )cos()cos(0
)]cos()[cos(21coscos
dttnmdttnmT
T
T
T
2/
2/ 0
2/
2/ 0 ])cos[(21])cos[(
21
2/
2/00
2/
2/00
])sin[()(
121])sin[(
)(1
21 T
T
T
Ttnm
nmtnm
nm
m ≠ n0
0
Proof
dttntmT
T 2/
2/ 00 )cos()cos(
0
dttmT
T 2/
2/ 02 )(cos
2/
2/0
0
2/
2/
]2sin4
121
T
T
T
T
tmm
t
m = n
2T
]2cos1[21cos2
dttmT
T 2/
2/ 0 ]2cos1[21
nmTnm
dttntmT
T 2/0
)cos()cos(2/
2/ 00
Orthogonal set of Sinusoidal Functions
Define ω0=2π/T.
,3sin,2sin,sin,3cos,2cos,cos
,1
000
000
tttttt
an orthonormal set.
Decomposition
dttfT
aTt
t
0
0
)(20
,2,1 cos)(20
0
0
ntdtntfT
aTt
tn
,2,1 sin)(20
0
0
ntdtntfT
bTt
tn
)sin()cos(2
)( 01
01
0 tnbtnaatfn
nn
n
Example (Square Wave)
1122
00
dta
,2,1 0sin1cos22
00
nntn
ntdtan
,6,4,20
,5,3,1/2)1cos(1 cos1sin
22
00
nnn
nn
ntn
ntdtbn
π 2π 3π 4π 5π-π-2π-3π-4π-5π-6π
f(t)1
1122
00
dta
,2,1 0sin1cos22
00
nntn
ntdtan
,6,4,20
,5,3,1/2)1cos(1 cos1sin
21
00
nnn
nn
ntn
ntdtbn
π 2π 3π 4π 5π-π-2π-3π-4π-5π-6π
f(t)1
Example (Square Wave)
ttttf 5sin
513sin
31sin2
21)(
1122
00
dta
,2,1 0sin1cos22
00
nntn
ntdtan
,6,4,20
,5,3,1/2)1cos(1 cos1sin
21
00
nnn
nn
ntn
ntdtbn
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
513sin
31sin2
21)(
When series is truncated
Harmonics
Tntb
Tntaatf
nn
nn
2sin2cos2
)(11
0
DC Part Even Part Odd Part
T is a period of all the above signals
)sin()cos(2
)( 01
01
0 tnbtnaatfn
nn
n
Harmonics
tnbtnaatfn
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.
tbtaatf nn
nnn
n
sincos2
)(11
0
Harmonics
tbtaatf nn
nnn
n
sincos2
)(11
0
)sincos(2 1
0 tbtaannn
nn
12222
220 sincos2 n
n
nn
nn
nn
nnn t
ba
btba
abaa
1
220 sinsincoscos2 n
nnnnnn ttbaa
)cos(1
0 nn
nn tCC
Amplitudes and Phase Angles
)cos()(1
0 nn
nn tCCtf
20
0aC
22nnn baC
n
nn a
b1tan
harmonic amplitude phase angle
Complex Form of the Fourier Series
Complex Exponentials
Complex Form of the Fourier Series
tnbtnaatfn
nn
n 01
01
0 sincos2
)(
tjntjn
nn
tjntjn
nn eebjeeaa
0000
11
0
221
2
1
0 00 )(21)(
21
2 n
tjnnn
tjnnn ejbaejbaa
1
000
n
tjnn
tjnn ececc
)(21
)(212
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
)(21
)(212
00
nnn
nnn
jbac
jbac
ac
Complex Form of the Fourier Series
2/
2/0
0 )(12
T
Tdttf
Tac
)(21
nnn 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)(
21 T
T
tjnnnn dtetf
Tjbac )(
21
)(212
00
nnn
nnn
jbac
jbac
ac
Complex Form of the Fourier Series
n
tjnnectf 0)(
dtetfT
cT
T
tjnn
2/
2/0)(1
)(21
)(212
00
nnn
nnn
jbac
jbac
ac
If f(t) is real,*nn cc
nn jnnn
jnn ecccecc
|| ,|| *
22
21|||| nnnn bacc
n
nn a
b1tan
,3,2,1 n
00 21 ac
Complex Frequency Spectra
nn jnnn
jnn ecccecc
|| ,|| *
22
21|||| nnnn bacc
n
nn a
b1tan ,3,2,1 n
00 21 ac
|cn|
n
amplitudespectrum
ϕn
n
phasespectrum
Example
2T
2T
TT2d
t
f(t)A
2d
dteTAc
d
d
tjnn
2/
2/0
2/
2/0
01
d
d
tjnejnT
A
2/
0
2/
0
0011 djndjn ejn
ejnT
A
)2/sin2(10
0
dnjjnT
A
2/sin10
021
dnnT
A
TdnT
dn
TAd
sin
TdnT
dn
TAdcn
sin
82
51
T ,
41 ,
201
0
T
dTd
Example
40π 80π 120π-40π 0-120π -80π
A/5
5ω0 10ω0 15ω0-5ω0-10ω0-15ω0
TdnT
dn
TAdcn
sin
Example
40π 80π 120π-40π 0-120π -80π
A/10
10ω0 20ω0 30ω0-10ω0-20ω0-30ω0
Example
dteTAc
d tjnn
00
dtjne
jnTA
00
01
00
110
jne
jnTA djn
)1(10
0
djnejnT
A
2/0
sindjne
TdnT
dn
TAd
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 3Tt
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).
• 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, called 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.
Characteristic functions• CF for finding moments.• Note that we can differentiate the sum “term by term”
• Carrying out the differentiation
• So that
• In fact, repeated differentiation produces the formula for the nth moment as
Moments of geometric RV: example• Since 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 theorem
a 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].
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 φn
X(ω) converge to a given CF, say φX(ω), then the corresponding sequence of PMF, say pn
X[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 problems
1. 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])2] , 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.
Homework
1. 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.