Radiant. One radian is the measure of a central angle that intercepts an arc s equal in length to the radius r of the circle.

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Lambert Stevenson
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1 Spectral Analysis 1

2 2 Radiant One radian is the measure of a central angle that intercepts an arc s equal in length to the radius r of the circle. Mathematically ( ) θ 2πr = r θ = 1 2π For example, the angle of 90 degree is equivalent to 1 θ = r 2πr/4 θ = π 2 More generally, the radian measure of a central angle is obtained by diving the arc length s by r θ = s r Because the units of measure for arc and radius are the same, the ratio has no units it is simply a real number.

3 3 Trigonometric Functions y t = Acos(ωt ϕ) (1) A is amplitude and y t A ω is frequency, and periodicity is 2π ω = 1 ω/(2π) where ω/(2π) is called natural frequency y t = A when t = ϕ ω, and ϕ ω is phase shift So period is the amount of time it takes the wave to go through a whole cycle. The frequency ω is the angular speed measured in radians per time. The phase is the angular amount ϕ by which the cos wave is shifted. The time shift is ϕ ω.

4 4 Plotting Cos Function A=2, Fre = 0.1, Pri = 62, Pha = 0 A=2, Fre = 0.2, Pri = 31, Pha = 0 g.cos(2, 0.1, 0) g.cos(2, 0.2, 0) time time A=2, Fre = 0.2, Pha = 1, timeshift = 5 A=2, Fre = 0.2, Pha = 2, timeshift = 10 g.cos(2, 0.2, 1) g.cos(2, 0.2, 2) time time

5 5 Complex Numbers e iω = cosω + isinω (2) cosω = eiω + e iω 2 sinω = eiω e iω 2i a + bi = Re iω, R = a 2 + b 2, ω = tan 1 ( b a where the first equation is called Euler s equation; R = a + bi is modulus (amplitude, norm); ω is angle or phase. ) (3) (4) (5)

6 6 Fourier Transformation For any series of number y t, its Fourier transformation (FT) is given by For example, suppose we have a series y 0,y 1,y 2. Its FT is In general, s(ω) is complex-number-valued. s(ω) = e itω y t (6) t y 0 + y 1 e iω + y 2 e 2iω

7 7 Spectral Density The spectral density (SD) of a time series is the FT of its autocovariance For a stationary time series, γ j = γ j. It follows that s(ω) = e i jω γ j (7) j s(ω) = e γ j j (8) = γ 0 + γ 1 (e iω + e iω ) + γ 2 (e 2iω + e 2iω ) +... (9) = γ γ j cos( jω) (10) j=1 which is real-valued. Also note that the spectral density is symmetric: s(ω) = s( ω)

8 8 Examples For white noise, only γ 0 0. So For MA(1), γ 0 = (1 + θ 2 1 )σ 2 e,γ 1 = θ 1 σ 2 e, so s(ω) = γ 0 s(ω) γ 0 = 1 s(ω) = σ 2 e (1 + θ θ 1 cosω) s(ω) γ 0 = 1 + 2θ θ 2 1 cosω

9 9 Autocovariance-Generating Function (AGF) The spectral density is AGF in the sense that γ k = 1 2π Proof: note for j k, π π e ikω γ j e i jω dω = γ j π π π = γ j π = 0 π π e ikω s(ω)dω (11) e i(k j)ω dω cos((k j)ω) + isin((k j)ω)dω

10 10 Variance In particular, let k = 0 in (11) we have γ 0 = 1 2π π π s(ω)dω (12) So s(ω) decomposes the variance into uncorrelated components at each frequency ω

11 11 Filtering A filtered y t series is a moving average of y t = b(l)x t = j b j L j x t = j b j x t j. For example, MA(1) is a filtered series y t = e t + θ 1 e t 1. Another example is AR(1) series y t = 1 1 ϕ 1 L e t = (1 + ϕ 1 L + ϕ 2 1 L2 +...)e t Let b(l) = j b j L j denote the lag polynomial, then s y (ω) = b(e iω )b(e iω )s x (ω) (13) So operations that are difficult in the time domain are just multiplications in the frequency domain.

12 12 Examples For MA(1) s ma1 (ω) = (1 + θ 1 e iω )(1 + θ 1 e iω )σ 2 e = (1 + 2θ 1 cosω + θ 2 1 )σ 2 e When ω = 0, s ma1 (ω) = (1 + θ 1 ) 2 σ 2 e When ω = π, s ma1 (ω) = (1 θ 1 ) 2 σ 2 e For AR(1) When ω = 0, s ar1 (ω) = σ 2 e (1 ϕ 1 ) 2 When ω = π, s ar1 (ω) = σ 2 e (1+ϕ 1 ) 2 s ar1 (ω) = σ 2 e 1 2ϕ 1 cosω + ϕ 2 1 (14)

13 13 Periodogram MA1 MA1, theta=1 Series: g.ma1(1, 1000) Smoothed Periodogram s spectrum 5e 04 5e 02 5e omega frequency bandwidth = MA1, theta= 1 Series: g.ma1( 1, 1000) Smoothed Periodogram s spectrum 1e 03 1e 01 1e omega frequency bandwidth =

14 14 Periodogram AR1 AR1, x=0.9 Series: g.ar1(0.9, 1000, 0) Smoothed Periodogram spectrum 5e 02 5e omega frequency bandwidth = AR1, x= 0.5 Series: g.ar1( 0.5, 1000, 0) Smoothed Periodogram spectrum omega frequency bandwidth =

15 15 AR2 For AR(2) s ar2 (ω) = σ 2 e 1 2ϕ 1 cosω + ϕ 2 1 2ϕ 2 cos(2ω) + 2ϕ 1 ϕ 2 cosω + ϕ 2 2 We can show this spectral density is not monotonic with ω (15)

16 16 Periodogram AR2 AR2, x=0.5,0.7 AR2, x= 0.5,0.7 s s omega omega AR2, x=0.5, 0.7 AR2, x= 0.5, 0.7 s s omega omega

17 17 Periodogram Simulated AR2 Series: g.ar2(c(1.2, 0.35), 1000, c(0, 0)) Smoothed Periodogram Series: g.ar2(c(0.2, 0.35), 1000, c(0, 0)) Smoothed Periodogram spectrum 5e 02 5e 01 5e+00 5e spectrum frequency bandwidth = frequency bandwidth = Series: g.ar2(c( 0.2, 0.35), 1000, c(0, 0)) Smoothed Periodogram Series: g.ar2(c( 1.2, 0.35), 1000, c(0, 0)) Smoothed Periodogram spectrum spectrum 5e 02 1e+00 5e frequency bandwidth = frequency bandwidth =

18 18 Seasonal AR For a seasonal AR model y t = ϕ 1 y t 4 + e t we can show In general s seasonal AR4 = σ 2 e 1 + ϕ 2 1 2ϕ 1 cos(4ω) s seasonal ARm = σ 2 e 1 + ϕ 2 1 2ϕ 1 cos(mω) The seasonality can easily be seen in the periodogram, but not in the time series plot.

19 19 Periodogram Seasonal AR4 and AR8 Seasonal AR4 Seasonal AR8 s s omega omega Series: g.sar14(0.8, 1000) Smoothed Periodogram Series: g.sar18(0.8, 1000) Smoothed Periodogram spectrum spectrum frequency bandwidth = frequency bandwidth =

20 20 Random Walk When ϕ 1 = 1, we can show that at the frequency ω = 0, s random walk (0) = So the smallest frequency dominates the spectral density for the random walk. When ϕ 1 = 0.5, s ar(0.5) (0) = 4σ 2 e

21 21 Bandpass Filter Consider the bandpass filter given by b(e iω ) = 1, if ω (ω 1,ω 2 ) 0, otherwise. (16) Applying the filter leads to b(e iω )b(e iω )s(ω) = s(ω), if ω (ω 1,ω 2 ) 0, otherwise. (17) In words, only part of original density is passed after the filter being applied.

22 22 Constructing Bandpass Filter b j = 1 π 2π = 1 2π π ω2 e i jω b(e iω )dω ω 1 e i jω dω + 1 2π = sin( jω 2) sin( jω 1 ) jπ ω1 ω 2 e i jω dω For example, at j = 0, L Hopital s rule indicates that b 0 = ω 2 ω 1 π

23 Plotting Bandpass Filter Bandpass Filter, Omega2 = 1, Omega1= j

24 24 First Difference Filter For the first difference 1 L we can show that (1 e iω )(1 e iω ) = 2(1 cosω) So the frequency response is zero at the zero frequency, but it is four at the frequency ω = π.

Digital Signal Processing

COMP ENG 4TL4: Digital Signal Processing Notes for Lecture #29 Wednesday, November 19, 2003 Correlation-based methods of spectral estimation: In the periodogram methods of spectral estimation, a direct