Modelling and Control of Dynamic Systems Linear Systems Sven Laur - PowerPoint PPT Presentation

Modelling and Control of Dynamic Systems Linear Systems Sven Laur University of Tartu

Formal definition Let F be a mapping from the state and input signal to an output signal: F [ x ( t 0 ) , u ( t ) , t ≥ t 0 ] �− [ y ( t ) , t > t 0 ] . → Then the corresponding system is linear if for all plausible time moments t 0 , input signals u 1 ( · ) , u 2 ( · ) , states x 1 ( t 0 ) , x 2 ( t 0 ) , and a constant α ∈ R : F [ x 1 ( t 0 ) , u 1 ( t ) , t ≥ t 0 ] �− → [ y 1 ( t ) , t > t 0 ] , F [ x 2 ( t 0 ) , u 2 ( t ) , t ≥ t 0 ] �− → [ y 2 ( t ) , t > t 0 ] , ⇓ F [ α · x 1 ( t 0 ) , α · u 1 ( t ) , t ≥ t 0 ] �− [ α · y 1 ( t ) , t > t 0 ] , → F [ α · x 2 ( t 0 ) , α · u 2 ( t ) , t ≥ t 0 ] �− [ α · y 2 ( t ) , t > t 0 ] , → F [ x 1 ( t 0 ) + x 2 ( t 0 ) , u 1 ( t ) + u 2 ( t ) , t ≥ t 0 ] �− [ y 1 ( t ) + y 2 ( t ) , t > t 0 ] . → Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 1

Canonical decomposition For linear systems, we can always express y ( t ) = y zi ( t ) + y zs ( t ) , where y zi ( · ) denotes the zero-input response F [ x ( t 0 ) , u ( t ) ≡ 0 , t ≥ t 0 ] �− [ y zi ( t ) , t ≥ t 0 ] → and y zs ( · ) denotes the zero-state response F [ x ( t 0 ) = 0 , u ( t ) , t ≥ t 0 ] �− → [ y zs ( t ) , t ≥ t 0 ] . Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 2

What happens with the state? The state of a lumped linear system must also evolve linearly F [ x ( t 0 ) , u ( t ) , t ≥ t 0 ] �− [ x ( t ) , t > t 0 ] → or otherwise non-linearity appears to the output, as well. The latter is true for the state variables that can influence the output. Important consequences: ⊲ A system stays in the zero state if the input is constantly zero. ⊲ Generally, we can describe the evolution through the equations x ( t ) = A ( t ) x ( t ) + B ( t ) u ( t ) , ˙ y ( t ) = C ( t ) x ( t ) + D ( t ) u ( t ) . Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 3

Examples. Continuous systems ⊲ Low-band filter R 3 R 2  y ( t ) = x 1 ( t )  x 1 ( t ) = u − x 1 x 1 x 1 ( t ) u ( t ) ˙ C −  R 3 C R 2 C x 3 ( t ) x 2 ( t ) Zero-input and zero-state responses u ( t ) = sin(10 t ) u ( t ) = sin(100 t ) u ( t ) ≡ 0 x 1 (0) = 1 Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 4

Examples. Continuous systems ⊲ Simplest radio receiver R 3 R 2  y ( t ) = x 1 ( t )   = u − x 1 − R 3 x 2 x 1 ( t ) ˙ R 3 C x 1 ( t ) u ( t ) C L  = x 1 − R 2 C x 2 ( t ) ˙  L x 3 ( t ) x 2 ( t ) Zero-input and zero-state responses u ( t ) = const x 1 ( t ) x 2 ( t ) x 1 ( t ) x 2 ( t ) Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 5

Examples. Discrete systems ⊲ Smoother (moving average) u [ k ] Adder y [ k ]  y [ k ] = x 1 [ k ] + x 2 [ k ] + u [ k ]  D  x 1 [ k + 1] = u [ k ]  x 2 [ k + 1] = x 1 [ k ]  D D Zero-input and zero-state responses &%# ' $%! ! $%# x 1 [0] = 2 & u = [1 , 2 , 3 , 1 , 2 , 3 , . . . ] "%! x 2 [0] = 1 % "%# $ #%! #%# " ! "# "! $# ! "# "! $# Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 6

Examples. Discrete systems ⊲ Bank deposit: y [ k + 1] = 1 . 02 · y [ k ] + u [ k + 1] u [ k ] Adder y [ k ] � y [ k ] = x [ k ] x [ k + 1] = 1 . 02 x [ k ] + u [ k ] D 1.02 Zero-input and zero-state responses (!! ) "'! $ "!! ( x [0] = 1 &'! u [ k ] ≡ 1 # &!! ' '! " ! & ! "! #! $! %! &!! ! "! #! $! %! &!! Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 7

Discrete Linear Systems

Input signal as a sum Let δ [ · ] be the step function with the peak at zero: � 1 , if k = 0 , δ [ k ] = 0 , if k � = 0 . Then we can decompose one-dimensional input signal u [ · ] into a sum ∞ � u [ k ] = u [ m ] δ [ k − m ] m = −∞ and thus the output is determined by the responses to the pulses δ m [ k ] = δ [ k − m ] . Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 8

Input-output description An impulse response sequence g [ · , · ] defined through the mapping F F δ m [ · ] �− g [ · , m ] → captures the behaviour of the system for all inputs. Due to linearity ∞ ∞ � � F u [ · ] = u [ m ] δ m [ · ] �− u [ m ] g [ · , m ] = y [ · ] → m = −∞ m = −∞ and thus we can express ∞ � y [ k ] = g [ k, m ] u [ m ] . m = −∞ Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 9

Extension to multi-dimensional case If an output signal consists of q components, then we must define a different impulse response sequence for every output component. If an input signal is p dimensional, then we must consider excitements in every dimension. As a result, we get an impulse response matrix   g 11 [ k, m ] g 12 [ k, m ] g 1 p [ k, m ] · · · g 21 [ k, m ] g 22 [ k, m ] · · · g 2 p [ k, m ]   G [ k, m ] = . . ... . . . .   . . .   g q 1 [ k, m ] g q 2 [ k, m ] . . . g qp [ k, m ] where g ij [ k, m ] measures the influence of an input u i [ · ] to the output y j [ · ] . The resulting signal can be computed as a sum ∞ � y [ k ] = G [ k, m ] u [ m ] . m = −∞ Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 10

Extension to time-invariant systems If the input state is zero and then the it stays there until input is excited. Consequently, it is irrelevant when the first peak arrives F δ [ · ] �− g [ · ] = δ m [ k ] = g [ k − m ] , → ⇒ where g [ − k ] ≡ 0 for all k = 1 , 2 , 3 , . . . . Consequently, we can express the output signal through a convolution k � y [ k ] = y zs [ k ] + y zi [ k ] = g [ k − m ] u [ m ] + y zi [ k ] , m =0 k � y [ k ] = y zs [ k ] + y zi [ k ] = G [ k − m ] u [ m ] + y zi [ k ] m =0 Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 11

Z-transform To get rid of the convolution, we must change the domain for input and output signals. For that we use z -transform that maps y [ · ] to ˆ y [ · ] : ∞ � y [ k ] z − k , y ( z ) = ˆ z = 1 , 2 , 3 , . . . . k =0 Fact. The z -transform is invertible provided that it exists. Now observe that z -transform takes convolution into a multiple k ! ∞ ∞ y [ k ] z − k = X X X z − ( k − m ) z − m y [ z ] = ˆ g [ k − m ] u [ m ] m =0 k =0 k =0 ∞ ! ∞ ! X g [ ℓ ] z − ℓ X u [ m ] z − m = = ˆ g [ z ]ˆ u [ z ] . m =0 ℓ =0 Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 12

Transfer function Now if we define z -transform for vectors and matrices as an elementwise application, then due to linearity of the z -transform y [ z ] = ˆ ˆ G [ z ]ˆ u [ z ] , where ˆ G is referenced as a transfer function . Fact. For lumped systems, the transfer function ˆ G [ z ] has always rational matrix elements. For distributed systems, the transfer function is irrational. Fact. For causal lumped systems, the transfer function is always proper g ij = N ij ( z ) ˆ with deg N ij ( z ) ≤ deg D ij ( z ) . D ij ( z ) Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 13

Examples ⊲ Smoother y [ k ] = u [ k − 2] + u [ k − 1] + u [ k ] : G [ z ] = 1 + 1 z + 1 ˆ z 2 ⊲ Feedback loop y [ k ] = αy [ k + 1] + u [ k ] z + α 2 G [ z ] = 1 + α z ˆ z 2 + · · · = z − α ⊲ Compensator for positive feedback loop y [ k ] = u [ k ] − αu [ k − 1] G [ z ] = 1 − α z = z − α ˆ z Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 14

From state space to transfer function As the evolution of the state and output can be expressed x [ k + 1] = A [ k ] x [ k ] + B [ k ] u [ k ] , y [ k ] = C [ k ] x [ k ] + D [ k ] u [ k ] , we can derive the transfer function directly for time-invariant systems � ˆ C ( z I − A ) − 1 B + D � y [ z ] = ˆ u [ z ] ⇓ ˆ G [ z ] = C ( z I − A ) − 1 B + D provided that the initial state x (0) is 0 . Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 15

Examples ⊲ Smoother y [ k ] = u [ k − 2] + u [ k − 1] + u [ k ] : G [ z ] = 1 + 1 z + 1 ˆ z 2 ⊲ Feedback loop y [ k ] = αy [ k + 1] + u [ k ] z + α 2 G [ z ] = 1 + α z ˆ z 2 + · · · = z − α ⊲ Compensator for positive feedback loop y [ k ] = u [ k ] − αu [ k − 1] G [ z ] = 1 − α z = z − α ˆ z Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 16

Continuous Linear Systems

What happens if the sampling rate grows? If the sampling rate grows infinitely, then the effect of δ ∆ [ · ] decreases to zero. Hence, we must consider pulses with constant area δ ∆ [ · ] / ∆ . As a result, we obtain the following equality � ∞ y ( t ) = G ( t, τ ) u ( τ ) dτ , −∞ where g ij ( t, τ ) corresponds to excitements with Dirac delta function. For causal time-invariant systems, we again get a convolution � ∞ y ( t ) = G ( t − τ ) u ( τ ) dτ , −∞ Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 17