Modelling and Control of Dynamic Systems Linear Systems Sven Laur - - PowerPoint PPT Presentation
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
Formal definition
Let F be a mapping from the state and input signal to an output signal: [x(t0), u(t), t ≥ t0] − →
F
[y(t), t > t0] . Then the corresponding system is linear if for all plausible time moments t0, input signals u1(·), u2(·), states x1(t0), x2(t0), and a constant α ∈ R: [x1(t0), u1(t), t ≥ t0] − →
F
[y1(t), t > t0] , [x2(t0), u2(t), t ≥ t0] − →
F
[y2(t), t > t0] , ⇓ [α · x1(t0), α · u1(t), t ≥ t0] − →
F
[α · y1(t), t > t0] , [α · x2(t0), α · u2(t), t ≥ t0] − →
F
[α · y2(t), t > t0] , [x1(t0) + x2(t0), u1(t) + u2(t), t ≥ t0] − →
F
[y1(t) + y2(t), t > t0] .
Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 1
Canonical decomposition
For linear systems, we can always express y(t) = yzi(t) + yzs(t) , where yzi(·) denotes the zero-input response [x(t0), u(t) ≡ 0, t ≥ t0] − →
F
[yzi(t), t ≥ t0] and yzs(·) denotes the zero-state response [x(t0) = 0, u(t), t ≥ t0] − →
F
[yzs(t), t ≥ t0] .
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 [x(t0), u(t), t ≥ t0] − →
F
[x(t), t > t0]
- r 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) .
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- Examples. Continuous systems
⊲ Low-band filter
R2 C u(t) x2(t) x1(t) x3(t) R3
y(t) = x1(t) ˙ x1(t) = u − x1 R3C − x1 R2C
Zero-input and zero-state responses
u(t) = sin(10t) u(t) = sin(100t) u(t) ≡ 0 x1(0) = 1 Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 4
- Examples. Continuous systems
⊲ Simplest radio receiver
R2 L C u(t) x2(t) x1(t) x3(t) R3
y(t) = x1(t) ˙ x1(t) = u−x1−R3x2
R3C
˙ x2(t) = x1−R2C
L
Zero-input and zero-state responses
x2(t) x1(t) x1(t) x2(t) u(t) = const Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 5
- Examples. Discrete systems
⊲ Smoother (moving average)
Adder
D D D u[k] y[k] y[k] = x1[k] + x2[k] + u[k] x1[k + 1] = u[k] x2[k + 1] = x1[k]
Zero-input and zero-state responses
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x1[0] = 2 x2[0] = 1
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u = [1, 2, 3, 1, 2, 3, . . .] 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]
- y[k] = x[k]
x[k + 1] = 1.02x[k] + u[k] Adder D 1.02 u[k] y[k]
Zero-input and zero-state responses
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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: δ[k] =
- 1,
if k = 0 , 0, if k = 0 . Then we can decompose one-dimensional input signal u[·] into a sum u[k] =
∞
- m=−∞
u[m]δ[k − m] and thus the output is determined by the responses to the pulses δm[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 δm[·] − →
F
g[·, m] captures the behaviour of the system for all inputs. Due to linearity u[·] =
∞
- m=−∞
u[m]δm[·] − →
F ∞
- m=−∞
u[m]g[·, m] = y[·] and thus we can express y[k] =
∞
- m=−∞
g[k, m]u[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[k, m] = g11[k, m] g12[k, m] · · · g1p[k, m] g21[k, m] g22[k, m] · · · g2p[k, m] . . . . . . ... . . . gq1[k, m] gq2[k, m] . . . gqp[k, m] where gij[k, m] measures the influence of an input ui[·] to the output yj[·]. The resulting signal can be computed as a sum y[k] =
∞
- m=−∞
G[k, m]u[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 δ[·] − →
F
g[·] = ⇒ δm[k] = g[k − m] , where g[−k] ≡ 0 for all k = 1, 2, 3, . . .. Consequently, we can express the output signal through a convolution y[k] = yzs[k] + yzi[k] =
k
- m=0
g[k − m]u[m] + yzi[k] , y[k] = yzs[k] + yzi[k] =
k
- m=0
G[k − m]u[m] + yzi[k]
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
- utput signals. For that we use z-transform that maps y[·] to ˆ
y[·]: ˆ y(z) =
∞
- k=0
y[k]z−k, z = 1, 2, 3, . . . .
- Fact. The z-transform is invertible provided that it exists.
Now observe that z-transform takes convolution into a multiple
ˆ y[z] =
∞
X
k=0
y[k]z−k =
∞
X
k=0 k
X
m=0
g[k − m]u[m] ! z−(k−m)z−m = ∞ X
ℓ=0
g[ℓ]z−ℓ ! ∞ X
m=0
u[m]z−m ! = ˆ g[z]ˆ u[z] .
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Transfer function
Now if we define z-transform for vectors and matrices as an elementwise application, then due to linearity of the z-transform ˆ y[z] = 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
ˆ gij = Nij(z) Dij(z) with deg Nij(z) ≤ deg Dij(z) .
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Examples
⊲ Smoother y[k] = u[k − 2] + u[k − 1] + u[k]: ˆ G[z] = 1 + 1 z + 1 z2 ⊲ Feedback loop y[k] = αy[k + 1] + u[k] ˆ G[z] = 1 + α z + α2 z2 + · · · = z z − α ⊲ Compensator for positive feedback loop y[k] = u[k] − αu[k − 1] ˆ G[z] = 1 − α z = z − α 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 y[z] =
- C(zI − A)−1B + D
ˆ u[z] ⇓ ˆ G[z] = C(zI − A)−1B + 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]: ˆ G[z] = 1 + 1 z + 1 z2 ⊲ Feedback loop y[k] = αy[k + 1] + u[k] ˆ G[z] = 1 + α z + α2 z2 + · · · = z z − α ⊲ Compensator for positive feedback loop y[k] = u[k] − αu[k − 1] ˆ G[z] = 1 − α z = z − α 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 gij(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
Laplace transform
The Laplace transform L allows us to break the convolution into a product ˆ y(s) = ˆ G(s)ˆ y(s) . The Laplace transform is a generalisation of the z-transform ˆ y(s) = ∞ y(t)e−stdt The Laplace transform is also invertible and linear. For lumped systems, the corresponding transfer function is rational. For distributed systems, the transfer function is irrational. For causal lumped systems, the transfer function is proper.
Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 18
From state space to transfer function
As the evolution of the state and output can be expressed ˙ x(t) = A(t)x(t) + B(t)u(t) , y(t) = C(t)x(t) + D(t)u(t) , we can derive the transfer function directly for time-invariant systems ˆ y[z] =
- C(zI − A)−1B + D
ˆ u[z] ⇓ ˆ G[z] = C(zI − A)−1B + D provided that the initial state x(0) is 0.
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Realisations
Basic components of digital electronics
Every linear state equation can be implemented with these components.
Adder ˆ G[z] = [1 1] u1 u2 y y u Unit delay ˆ G[z] = z−1 u Duplicator ˆ G[z] = [1 1]T y1 y2 y u Multiplier ˆ G[z] = α
Fact. Any proper transfer function ˆ G[z] or transfer matrix ˆ G[z] has a corresponding linear state equation.
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Explanation for SISO systems
The following state space description y[k] = β1x1[k] + β2x2[k] + · · · + βrxr[k] x1[k + 1] x2[k + 1] · · · xr[k + 1] = −α1 −α2 · · · −αr 1 · · · 1 · · · . . . . . . ... . . . · · · 1 x1[k] x2[k] · · · xr[k] + u[k] · · · has a transfer function ˆ G[z] = β1zr−1 + β2zr−2 + · · · + βr zr + α1zr−1 + α2zr−2 + · · · + αr .
Modelling and Control of Dynamic Systems, Linear Systems, 29 September, 2008 21
Basic components of analog electronics
Every linear state equation can be implemented with these components. − −
Adder Integrator
R/a R/b R/c R/a R/b R/c R C v1 v2 v3 ˙ y = −(av1 + bv2 + cv3) y = −(ax1 + bx2 + cx3)
x1 x2 x3
Fact. Any proper transfer function ˆ G(s) or transfer matrix ˆ G(s) has a corresponding linear state equation.
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Explanation for SISO systems
The following state space description y(t) = β1x1(t) + β2x2(t) + · · · + βrxr(t) ˙ x1(t) ˙ x2(t) · · · ˙ xr(t) = −α1 −α2 · · · −αr 1 · · · 1 · · · . . . . . . ... . . . · · · 1 x1(t) x2(t) · · · xr(t) + u(t) · · · has a transfer function ˆ G(z) = β1sr−1 + β2sr−2 + · · · + βr sr + α1sr−1 + α2sr−2 + · · · + αr .
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