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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.
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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 formof 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.
Synthesis f(t) t T 2T 3T T is a period of all the above signals Odd Part Even Part DC Part Let ω0=2π/T.
Definition A Fourier Seriesis 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 theamplitudeand phaseof every harmonic components, and we can generate any non-sinusoidal waveform by an appropriate combination of sinusoidal functions.
Orthogonal Functions Call a set of functions {ϕk}orthogonal on an interval a < t < b if it satisfies Example m =1 0 n = 2 π -π
Orthogonal set of Sinusoidal Functions Define ω0=2π/T. We now prove this one
Proof m ≠n 0 0
Proof m = n 0
Define ω0=2π/T. Orthogonal set of Sinusoidal Functions an orthonormal set.
f(t) 1 -6π -5π -4π -3π -2π -π π 2π 3π 4π 5π Example (Square Wave)
f(t) 1 -6π -5π -4π -3π -2π -π π 2π 3π 4π 5π Example (Square Wave)
f(t) 1 -6π -5π -4π -3π -2π -π π 2π 3π 4π 5π Example (Square Wave) When series is truncated
Harmonics T is a period of all the above signals Even Part Odd Part DC Part
Define , called the fundamental angular frequency. Define , called the n-th harmonicof the periodic function. Harmonics
harmonic amplitude phase angle Amplitudes and Phase Angles
Complex Form of the Fourier Series If f(t) is real,
amplitude spectrum |cn| n ϕn phase spectrum n Complex Frequency Spectra
f(t) A t Example
A/5 0 -120π -80π -40π 40π 80π 120π -15ω0 -10ω0 -5ω0 5ω0 10ω0 15ω0 Example
A/10 0 -120π -80π -40π 40π 80π 120π -30ω0 -20ω0 -10ω0 10ω0 20ω0 30ω0 Example
f(t) A t 0 Example
0 Discrete-time Fourier transform f(t) continuous • 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 T 2T 3T 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 continuous discrete
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 Homework: Prove it Initial value
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-1pfor 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 • By finding second moment we can find variance a b binomial theorem
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 • Property3. 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 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 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.