Fitting Sine Waves and Computing the Periodogram

Question 1

What is spectral analysis?

Answer 1

-in spectral analysis, a time series is analysed by decomposing it into sine waves of different amplitude and frequency

Question 2

Spectral Analysis

Low Frequency

Answer 2

-slowly changing trends

Question 3

Spectral Analysis

High Frequency

Answer 3

-rapid changes (noise)

Question 4

Spectral Analysis Model

Answer 4

-if the data have a periodic component of known frequency, we can fit the following model:
Xt = B sin[2πf (t-to)] + μ + εt
-where:
μ is the baseline/mean
f is the frequency (in cycles per unit time)
to is the start time of one period
B is the amplitude

Question 5

Spectral Analysis Model in Terms of Sine and Cosine

Answer 5

Xt = Acos(2πft) + Bsin(2πft) + μ + εt

-therefore only two parameters (A,B) are required per frequency

Question 6

Aliasing

Definition

Answer 6

-two frequencies that are indistinguishable from each other are aliases of each other

Question 7

Nyquist Frequency

Definition

Answer 7

• assume that the observations are given at times 0, Δ, 2Δ 3Δ, …
• we have a sampling frequency of 1/Δ
• the frequency f=1/2Δ is called the Nyquist frequency for the sampling frequency 1/Δ
• this is the fastest possible oscillation that can be recognised when we have a sampling frequency of 1/Δ

Question 8

Nyquist Frequency

Explaination

Answer 8

-assume that the observations are given at times 0, Δ, 2Δ 3Δ, …
-we have a sampling frequency of 1/Δ
-a component with frequency f will be observed as:
Xk = Acos(2πfkΔ) + Bsin(2πfkΔ)
-the bigger f, the faster Xk oscillates until when f=1/2Δ, then:
Xk = A (-1)^k
-this is the fastest possible oscillation that can be recognised when we have a sampling frequency of 1/Δ
-for f>1/2Δ we only have aliases

Question 9

What are Discrete Fourier Transforms?

Answer 9

-a method for decomposing a time series into periodic components

Question 10

Fourier Frequencies

Definition

Answer 10

• if we have observations Xt at t=0,1,2,…,n-1

- frequencies fj = j/n are called Fourier frequencies

Question 11

The Discrete Fourier Transform

Definition

Answer 11

-the DFT of Xo,…,Xn-1 is given by:
Xj^ = 1/√n Σ Xt exp(-2πifjt)
-sum from t=0 to t=n-1, for j=0,…,n-1
-here ^ indicates a DFT, not an estimate

Xj^ = 1/√n Σ Xt[cos(2πifjt) - isin(2πifjt)]

Question 12

Inverse DFT Lemma

Answer 12

Σ exp(2πifjt) exp(-2πifkt)
= n, j=k or 0, else
-sum from t=0 to t=n-1

Question 13

The Inverse DFT

Answer 13

Xt = 1/√n Σ Xj^ exp(2πifjt)

-sum from j=0 to j=n-1

Question 14

Properties of the DFT

Answer 14

1) linearity
2) conservation of energy
3) time shifts (cyclic shift)
4) time shifts (linear shift)

Question 15

Linearity of the DFT

Answer 15

-let Yt=cXt :
Yj^ = cXj^
-let Zt = Xt + Yt :
Zj^ = Xj^ + Yj^

Question 16

Conservation of Energy with the DFT

Answer 16

-energy in the frequency domain equals energy in the time domain:
Σ |Xj^|² = Σ |Xt|²
-sum from j=0 to j=n-1 and t=0 to t=n-1

Question 17

Cyclic Time Shifts with DFT

Answer 17

-can take the set of measurements {X0,X1,…,Xn-2,Xn-1} can be viewed as a cycle of measurements so you can pick any measurement as the first one
-let Y = (Xn-1,X0,…,Xn-2) :
Yj^ = exp(-2πifjt) Xj^
=>
|Yj^| = |Xj^|

Question 18

Linear Time Shifts with DFT

Answer 18

• view the measurements as a linear line, and shift the window e.g. miss one measurement off the end and pick one up at the start
• let Z = (X-1,X0,X1,…,Xn-2) with the new measurement X-1∈R

Question 19

Compare Cyclic and Linear Shift

Answer 19

||Y^-Z^|| = |Xn-1 - X-1|

-therefore the relative difference between Y^ and Z^ goes to zero as n->∞

Question 20

Spectral Density

Definition

Answer 20

-the function I : [0,1] -> [0,∞) with:
I(fj) = |Xj|²
-is called the spectral density of X
-the spectral density I(fj) is a measurement of how strongly the frequency fj is represented in the data

Question 21

Periodogram

Definition

Answer 21

-plot of the spectral density I(fj) against fj

Question 22

How many frequencies are needed?

Answer 22

-for real data {Xt}, we have:
Xn-j^ = Xj^*
-where * indicates a complex conjugate
-so only half the frequencies are needed
-this is expected since fj fj=j/n is greater than the Nyquist frequency for j>n/2

Question 23

Xo^ in terms of X~

Answer 23

-let X~ be the mean of {Xt}, then:

Xo^ = X~√n

Question 24

Xn/2^ when n is even

Answer 24

Xn/2^ = Xn-n/2^* = Xn/2^* so they must be real numbers