Time Series Analysis Henrik Madsen Informatics and Mathematical - PowerPoint PPT Presentation

H. Madsen, Time Series Analysis, Chapmann Hall Time Series Analysis Henrik Madsen Informatics and Mathematical Modelling Technical University of Denmark DK-2800 Kgs. Lyngby Henrik Madsen 1

H. Madsen, Time Series Analysis, Chapmann Hall Outline of todays lecture Descriptions of ( deterministic ) linear systems. Chapter 4: Linear Systems Linear (Input-Output) systems Linear systems, Chap. 4, except Sec. 4.7 Cursory material: Sec. 4.6 Henrik Madsen 2

H. Madsen, Time Series Analysis, Chapmann Hall Linear Dynamic Systems Input Output Linear System We are going to study the case where we measure the input and the and the output to/from a system Here we will discuss some theory and descriptions for such systems Later on (in the next lecture) we will consider how we can model the system based on measurements of input and output. Henrik Madsen 3

H. Madsen, Time Series Analysis, Chapmann Hall Dynamic response What would happen to the temperature inside a hollow, insulated, concrete block, which you place it in a controlled temperature environment, wait until everything is settled (all temperatures are equal), and then suddenly raise the temperature by 100 o C outside the block? Sketch the temporal development of the temperature outside and inside the block Henrik Madsen 4

H. Madsen, Time Series Analysis, Chapmann Hall Dynamic response characteristics from data An important aspect of what we aim at later on is to identify the characteristics of the dynamic response based on measurements of input and output signals 4 Input (x) Output (y) 3 2 1 0 −1 −2 0 20 40 60 80 100 Henrik Madsen 5

H. Madsen, Time Series Analysis, Chapmann Hall Dyn. response characteristics from data (cont’nd) Estimated (black) and true (red) step repsponse 0.8 Response 0.4 0.0 0 10 20 30 Lag Henrik Madsen 6

H. Madsen, Time Series Analysis, Chapmann Hall Linear Dynamic Systems – notation x F [ · ] y Input Output Linear System x ( t ) Differential eq., h ( u ) y ( t ) x t Difference eq., h k , h ( B ) y t X ( ω ) H ( ω ) Y ( ω ) X ( z ) H ( z ) Y ( z ) ( X ( s ) H ( s ) Y ( s ) ) Henrik Madsen 7

H. Madsen, Time Series Analysis, Chapmann Hall Dynamic Systems – Some characteristics Def. Linear system: � � � � � � F λ 1 x 1 ( t ) + λ 2 x 2 ( t ) = λ 1 F x 1 ( t ) + λ 2 F x 2 ( t ) Def. Time invariant system: � � � � y ( t ) = F x ( t ) ⇒ y ( t − τ ) = F x ( t − τ ) Def. Stable system: A system is said to be stable if any constrained input implies a constrained output. Def. Causal system: A systems is said to be physically feasible or causal , if the output at time t does not depend on future values of the input. Henrik Madsen 8

H. Madsen, Time Series Analysis, Chapmann Hall Example System: y t − ay t − 1 = bx t Can be written: y t = bx t + ay t − 1 = bx t + a ( bx t − 1 + ay t − 2 ) or ∞ � y t = b ( x t + ax t − 1 + a 2 x t − 2 + a 3 x t − 3 + . . . ) = b a k x t − k k =0 The system is seen to be linear and time invariant The impulse response is h k = ba k , k ≥ 0 (0 otherwise) and the system is seen to be causal � | b | / (1 − | a | ) Since ∞ ∞ ; | a | < 1 | b || a | k = � � | h k | = ∞ ; | a | ≥ 1 k = −∞ k =0 the system is stable for | a | < 1 (stability does not depend on b ) Henrik Madsen 9

H. Madsen, Time Series Analysis, Chapmann Hall Description in the time domain For linear time invariant systems : Continuous time: � ∞ y ( t ) = h ( u ) x ( t − u ) du (1) −∞ ∞ Discrete time: � y t = h k x t − k (2) k = −∞ h ( u ) or h k is called the impulse response j = −∞ h j is called the step response (similar def. in S k = � k continuous time) The impulse response can be determined by “sending a 1 trough the system” Henrik Madsen 10

H. Madsen, Time Series Analysis, Chapmann Hall Example: Calculation of the impulse response fct. The impulse can be determined by ’sending a 1 trough the system’. Consider the linear, time invariant system y t − 0 . 8 y t − 1 = 2 x t − x t − 1 (3) By putting x = δ we see that y k = h k = 0 for k < 0 . For k = 0 we get y 0 = 0 . 8 y − 1 + 2 δ 0 − δ − 1 = 0 . 8 × 0 + 2 × 1 − 0 = 2 i.e. h 0 = 2 . Henrik Madsen 11

H. Madsen, Time Series Analysis, Chapmann Hall Example - Cont. Going on we get y 1 = 0 . 8 y 0 + 2 δ 1 − δ 0 = 0 . 8 × 2 + 2 × 0 − 1 = 0 . 6 y 2 = 0 . 8 y 1 = 0 . 48 . 0 . 8 k − 1 0 . 6 ( k > 0) y k = Hence, the impulse response function is  0 for k < 0  h k = 2 for k = 0 0 . 8 k − 1 0 . 6 for k > 0  which clearly represents a causal system. Furthermore, the system 0 | h k | = 2 + 0 . 6(1 + 0 . 8 + 0 . 8 2 + · · · ) = 5 < ∞ is stable since � ∞ Henrik Madsen 12

H. Madsen, Time Series Analysis, Chapmann Hall Description in the frequency domain The Fourier transform is a way of representing a signal y ( t ) or y t by it’s distribution over frequencies: � ∞ ∞ y ( t )e − iωt dt � y t e − iωt Y ( ω ) = or Y ( ω ) = −∞ t = −∞ If the time unit is seconds, ω is the angular frequency in radians per second. In discrete time − π ≤ ω < π For a linear time invariant system it holds that Y ( ω ) = H ( ω ) X ( ω ) where H ( ω ) is the Fourier transform of the impulse response function. H ( ω ) = |H ( ω ) | e i arg {H ( ω ) } = G ( ω )e iφ ( ω ) Henrik Madsen 13

H. Madsen, Time Series Analysis, Chapmann Hall Description in the frequency domain (cont.) The function H ( ω ) is called the Frequency response function , and it is the Fourier transformation of the impulse response function, ie. ∞ � h k e − iωk H ( ω ) = ( − π ≤ ω < π ) (4) k = −∞ The frequency response function is complex. Thus, it is possible to split H ( ω ) into a real and a complex part: H ( ω ) = |H ( ω ) | e i arg {H ( ω ) } = G ( ω )e iφ ( ω ) (5) where G ( ω ) is the amplitude ( amplitude function ) and φ ( ω ) is the phase ( phase function ). Henrik Madsen 14

H. Madsen, Time Series Analysis, Chapmann Hall Single harmonic input Lets consider the a single harmonic signal as input: x ( t ) = A e iωt = A cos ωt + iA sin ωt (6) Then the output becomes also single harmonic, cf.: � ∞ y ( t ) = h ( u ) x ( t − u ) du −∞ � ∞ h ( u ) A e iω ( t − u ) du = −∞ � ∞ h ( u )e − iωu du A e iωt = −∞ � � H ( ω ) A e iωt = G ( ω ) A e i ωt + φ ( ω ) = (7) Henrik Madsen 15

H. Madsen, Time Series Analysis, Chapmann Hall Single harmonic input (cont.) A single harmonic input to a linear, time invariant system will give an output having the same frequency ω . The amplitude of the output signal equals the amplitude of the input signal multiplied by G ( ω ) . The change in phase from input to output is φ ( ω ) . Henrik Madsen 16

H. Madsen, Time Series Analysis, Chapmann Hall Sampling From continuous time to discrete time – what is lost? | | | | | | | x(t) | | | T xt T is the sampling time ω 0 = 2 π/T is the sampling frequency Henrik Madsen 17

H. Madsen, Time Series Analysis, Chapmann Hall Sampling (cont’nd) If we work out the mathematical theory of sampling it turns out that the Fourier transform of the sampled signal X s ( ω ) is composed by the Fourier transform of the original signal X ( ω ) at the correct frequency ω and at the frequencies ω ± ω 0 , ω ± 2 ω 0 , ω ± 3 ω 0 , . . . If X ( ω ) is zero outside the interval [ − ω 0 / 2 , ω 0 / 2] = [ − π/T, π/T ] then X s ( ω ) = X ( ω ) If not the values outside the interval cannot be distinguished from values inside the interval ( aliasing ) Henrik Madsen 18

H. Madsen, Time Series Analysis, Chapmann Hall Sampling (cont’nd) T = 1 , w0 = 6.2832 −w0/2 w0/2 0.6 G(w) 0.0 −3.1416 0.0 3.1416 w T = 0.6 , w0 = 10.472 −w0/2 w0/2 0.6 G(w) 0.0 −5.2360 0.0 5.2360 w Henrik Madsen 19

H. Madsen, Time Series Analysis, Chapmann Hall The z -transform A way to describe dynamical systems in discrete time ∞ � x t z − t Z ( { x t } ) = X ( z ) = ( z complex) t = −∞ The z -transform of a time delay: Z ( { x t − τ } ) = z − τ X ( z ) ∞ The transfer function of the system is called H ( z ) = � h t z − t t = −∞ ∞ � y t = h k x t − k ⇔ Y ( z ) = H ( z ) X ( z ) k = −∞ Relation to the frequency response function : H ( ω ) = H (e iω ) Henrik Madsen 20

H. Madsen, Time Series Analysis, Chapmann Hall Linear Difference Equation y t + a 1 y t − 1 + · · · + a p y t − p = b 0 x t − τ + b 1 x t − τ − 1 + · · · + b q x t − τ − q (1 + a 1 z − 1 + · · · + a p z − p ) Y ( z ) = z − τ ( b 0 + b 1 z − 1 + · · · + b q z − q ) X ( z ) Transfer function: z − τ ( b 0 + b 1 z − 1 + · · · + b q z − q ) H ( z ) = (1 + a 1 z − 1 + · · · + a p z − p ) z − τ (1 − n 1 z − 1 )(1 − n 2 z − 1 ) · · · (1 − n q z − 1 ) b 0 = (1 − λ 1 z − 1 )(1 − λ 2 z − 1 ) · · · (1 − λ p z − 1 ) Where the roots n 1 , n 2 , . . . , n q is called the zeros of the system and λ 1 , λ 2 , . . . , λ p is called the poles of the system The system is stable if all poles lie within the unit circle Henrik Madsen 21