Tutorial 7

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ELE3410 Random Processes and DSP THE CHINESE UNIVERSITY OF HONG KONG Department of Electronic Engineering ELE3410 Random Processes and DSP Tutorial 6 Power Spectral Density 2D Gaussian and n-D Gaussian 1. Power Spectral Density What is Power Spectral Density? Parseval’s Theorem (for aperiodic signal): ∞ ∞2 2y (t) dt = |Y ( f ) | df ∫ ∫− ∞ − ∞For ergodic random process, its average power: T1 2lim x (t)dt∫T → ∞ 2T −T∞1 2= lim | X ( f ) | df∫T → ∞ 2T − ∞∞= G( f ) df∫− ∞1 2 G( f ) ≡ lim | Χ( f ) |Then . T → ∞ 2TNote: (1)G(f) ≥ 0 ∀ f (2)G(f) = G(-f) G(f) power spectral density [watt/Hz] Tutorial 6 [1/5] ELE3410 Random Processes and DSP Power spectral density function(PSD) shows the strength of the variations (energy) as a function of frequency. In other words, it shows at which frequencies variations are strong and at which frequencies variations are weak. The unit of it is energy per frequency and you can obtain energy within a specific frequency range by integrating PSD within that frequency range. How to compute PSD? Computation of PSD is done by computing autocorrelation function and then transforming it using DFT. G(f) & R(τ) constitute a Fourier Transform pair: ∞ ∞1− j2 πf τ j2 πf τG( f ) = R( τ) ⋅e d τ & R( τ) = G( f ) ⋅e df ∫ ∫2 π− ∞ − ∞ Question 1 (Assignment 6) Find the power spectral density G( ω) corresponding to the autocorrelation function 1 for| τ | < T⎧ 0 .⎨R( τ) ...

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• Publié par: Iffe
• Nombre de lectures: 25
• Langue: English

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ELE3410 RandomProcesses and DSP THE CHINESE UNIVERSITY OF HONG KONG Department of Electronic Engineering ELE3410 Random Processes and DSP Tutorial 6 Power Spectral Density 2D Gaussian and nD Gaussian 1.Power Spectral Density What is Power Spectral Density? Parseval’s Theorem (for aperiodic signal): ∞ ∞ 2 2 y(t)dt=|Y(f) |df ∫ ∫ −∞ −∞ For ergodic random process, its average power: T 1 2 limx (t)dt T→∞ ∫ 2T −T ∞ 1 2 =lim|X(f) |df T→∞ ∫ 2T −∞ ∞ =G(f)df ∫ −∞ 1 2 ≡ Χ Then G(f)lim| (f) | . 2T T→∞ Note: (1)G(f)≥0∀f (2)G(f)= G(-f) G(f) power spectral density [watt/Hz] Tutorial 6[1/5]

ELE3410 RandomProcesses and DSP Power spectral density function(PSD) shows the strength of the variations (energy) as a function of frequency. In other words, it shows at which frequencies variations are strong and at which frequencies variations are weak. The unit of it is energy per frequency and you can obtain energy within a specific frequency range by integrating PSD within that frequency range. How to compute PSD? Computation of PSD is done by computing autocorrelation function and then transforming it using DFT. G(f) &R(τa Fourier Transform pair:) constitute ∞ ∞ −j2πfτ 1 j2πfτ G(f)=R(τ)⋅e dτ &R(τ)=G(f)⋅e df ∫ ∫ 2π −∞ −∞ Question 1 (Assignment 6) Find the power spectral densityG(ω) corresponding to the autocorrelation function <T ⎧ 1 for| | 0 ⎨ . R(τ) = 0 elsewhere ⎩

Is there any real random process that may has the aboveR(τwhy or why not.)? Explain Tutorial 6[2/5]

ELE3410 RandomProcesses and DSP 2. 2D Gaussian and nD Gaussian Definitions: 2 2 1 (x−µ) p(x;µ,σ)=exp(−) ¾1D Gaussian: 2 2 2σ 2πσ ¾ 2D Gaussian: Ifx,yare independent : 2 2 ⎡ ⎤ ⎛ − (x−µ) ⎞ 2 2 ( 1 1 xµ) y x ⎜ ⎟ p(x,y;µ,µ,σ,σ) = exp ⎥ ⎢− + x y xy 2 2 ⎜ ⎟ 2πσ σ2 σ σ x y ⎢ x y ⎥ ⎣ ⎝⎠⎦ Ifx,yare dependent with correlationρ: 2 2 ⎡ ⎤ ⎛ y−µ ρx−µy−µ ⎞ 2 2 (1 1x−µ x ) ( y () 2 x )( y ) ⎜ ⎟ p(x,y; µ , µ , σ , σ )=exp− ⎥⎢− + x y x y 2 22 2 ⎜ ⎟ 2(1−ρ)σ σσ σ 2πσ σ1−ρ ⎢ x yx y ⎥ x y ⎠⎦⎣ ⎝ ¾nD Gaussian: Fornrandom variablesx 1 ,x 2 , …x n , they can form arandom vectorx= [x 1 ,x 2 , …x n ]. Its T correlation matrixR:R= E[x⋅x], T T and its covariance matrixC : C = E[( x η)( x ηwhere) ]η= E[ x ]. Ifx 1 ,x 2 , …x n are ALL zero mean, i.e. E[x 1 ] = E[x 2 ] =…= E[x n ] = 0, their jointp.d.f. is given by: 1 T p(x,x,...,x)=exp[−xCx] 1 2n n (2π) det(C) It isnD Gaussian. Properties: ¾ Ifxandyare jointly Gaussian and uncorrelated, thenxandyare independent. 2 ⎡ ⎛ 2 ⎞⎤ (y−µ) 1 1 (x−µ) y ⎜ ⎟ x ⎢ ⎥ p(x,y)=exp− + 2 2 ⎜ ⎟ 2πσ σ ⎢ 2 σ σ ⎥ x yx y ⎣ ⎝⎠⎦ ¾ Ifxandyare jointly Gaussian, thenxandyare marginally Gaussian. ∞ p(x)=p(x,y)dy ∫ −∞ ¾ x 1 ,x 2 , …x n are jointly Gaussian, iff the suma 1 x 1 +a 2 x 2 + … +a n x n is a Gaussian r.v. for alla 1 ,a 2 , …a n . ¾ A real random processx(t) is said to be Gaussian (or normal), if the r.v.s.x(t 1 ), x(t 2 ), …,x(t n ) are jointly Gaussian for anyn,t 1 ,t 2 , …,t n . Q1: −2τ τ Given a Gaussian processx(t) with meanη x = 0 an ( )4 e form the random d R x =e , w variablesz=x(t+1) andw=x(t1). 2 (a)Find E[zw] and E[(z+w) ]; Tutorial 6[3/5]

ELE3410 RandomProcesses and DSP (b)Findf z (z), the p.d.f. of the random variablez, andP{z≤1}; (c)Findf zw (z,w), the joint p.d.f. of the random variableszandw. Jointly Gaussian: Definition: Two real random processesf 1 (t) andf 2 (t) are said to be jointly Gaussian, or jointly normal, if the r.v.sf 1 (t 1,1 ),f 1 (t 1, 2 ), …,f 1 (t 1,m ),f 2 (t 2,1 ),f 2 (t 2, 2 ), …,f 2 (t 2,n ), for anym,n,t 1,1 ,t 1,2 , …,t 1,m ,t 2,1 , t 2,2 , …,t 2,n ; that is the joint distribution of the random variables is Gaussian. If the random process f1(t), f2(t) f3(t) and f4(t) are jointly Gaussian with E[f 1 (t)] = E[f 2 (t)] = E[f 3 (t)] = E[f 4 (t)], then E[f 1 (t)f 2 (t)f 3 (t)f 4 (t)] =E[f 1 (t)f 2 (t)]E[f 3 (t)f 4 (t)] + E[f 1 (t)f 3 (t)]E[f 2 (t)f 4 (t)] + E[f 1 (t)f 4 (t)]E[f 2 (t)f 3 (t)]. Q2: 2 4 Ifx(t) is a Gaussian process with zero mean and varianceσ, determine E[x]. 4 22 4 E[x]=3E[x]E[x]=3σ Q1: −2τ τ Given a Gaussian processx(t) with meanη x = 0 and R( )=4e we form x random, the variablesz=x(t+1) andw=x(t1). 2 (a) Find E[zw] and E[(z+w) ]; −4 E[zw]=E[x(t+1)x(t−1)]=E[x(t)x(t+2)]=R(2)=4e x 2 22 E[(z+w) ]=E[x(t+1) ]+E[x(t−1) ]+2E[x(t+1)x(t−1)] −4 =2R(0)+2R(2)=8+8e x x (b) Findf z (z), the p.d.f. m variablez, andP{z≤1}; 2 2 1 (x−µ) p(x;µ,σ)=exp(−) 2 2 2σ 2 πσ 2 2 u=E x t+ = Z [ (1)] 0; σ Z =E[x(t+1) ]=4; 2 1z f(z)=exp(−)  So Z 2 2π 8 Tutorial 6[4/5]

ELE3410 RandomProcesses and DSP 1 11 11 1 (1er( )erf erf P{z≤1}= −f)+( )= +( ) 2 22 22 2 (c) Findf zw (z,w), the joint p.d.f. of the random variableszandw. 2 2 1x 2ρ(τ)x x+x 1 12 2 p(x,x;τ)=exp { − } 1 2 2 2 ×ρ τ 2π×4 1ρ(τ ) 42[1 ()] − − −4 µE[(z−z)(w−w)] 4e − 11 4 ρ(τ)= =.= =e 2 2 µ µ 4 20 02 σ σ Z W 2 42 1z 2ezw+w f(z,w)=exp(−) ZW  So −8 8 8(1e) 8π1e

Tutorial 6

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