Parameters round-off

Ingegneria dell ’Automazione - Sistemi in Tempo Reale

Selected topics on discrete-time and sampled-data systems

Luigi Palopoli

[email protected] - Tel. 050/883444

Outline

Parameters round-off Sample-rate selection

Parameters round-off

So far we have seen the problem of

quantization referred to Input/output data The problem is there also for parametric uncertainties inthe controller

For example suppose you choose a controller given by: y(k + 1) = αy(k) + u(k)

Due to finite precision math the system we will deal with is:

y(k + 1) = (α + δα)y(k) + u(k) + (k)

Parameters round-off

So far we have seen the problem of

quantization referred to Input/output data The problem is there also for parametric uncertainties inthe controller

For example suppose you choose a controller given by: y(k + 1) = αy(k) + u(k)

Due to finite precision math the system we will deal with is:

y(k + 1) = (α + δα)y(k) + u(k) + (k)

Parameters round-off - (continued)

In the example above the Z−transform that we get is z − α + δα

If we want to place a pole at 0.995 we need at least a precision of three digits: δα ≤ 0.005

Robustness analysis can be conducted using the root-locus-Jury critterion methods

We will show a different approach based on linearised sensitivy analysis

Linearised sensitivity analysis

Let the closed loop characteristic polynomial be:

P = zⁿ + α₁zⁿ⁻¹ + . . . + α_n

Let the roots of the system be: λi = rie^jθi

Suppose αk is subject to uncertainity and that we are interested in evaluating variations of λ_j, that we assume to be a single root of P The Polynomial P can be thought of as a function of z and αk: P (z, αk)

Lin. sensitivity an. - (continued)

First order series expansion around z = λj: P (λ_j + δλ_j, α_k + δλ_k) ≈ P (λ^j, α_k) + ^∂P_∂z ˛

˛

˛z=λ_j δλ_j + _∂α^∂P

kδα_k + . . . Observing that P (λ_j, α_k) = 0 we get:

δλ_j ≈ ∂P/∂α_k

∂P/∂z

˛

˛

˛

˛z=λ_j

δα_k

Numerator:

∂P

∂α_k

˛

˛

˛

˛z=λ_j

= λ^n−k_j

Writing P (z, α_k) = (z − λ¹) . . . (z − λⁿ), we get for the denominator:

∂P

∂z

˛

˛

˛

˛z=λ_j

= Y

l6=j

(λ_j − λl)

Lin. sensitivity an. - (continued)

First order series expansion around z = λj: P (λ_j + δλ_j, α_k + δλ_k) ≈ P (λ^j, α_k) + ^∂P_∂z ˛

˛

˛z=λ_j δλ_j + _∂α^∂P

kδα_k + . . . Observing that P (λ_j, α_k) = 0 we get:

δλ_j ≈ ∂P/∂α_k

∂P/∂z

˛

˛

˛

˛z=λ_j

δα_k Numerator:

∂P

∂α_k

˛

˛

˛

˛z=λ_j

= λ^n−k_j

Writing P (z, α_k) = (z − λ¹) . . . (z − λⁿ), we get for the denominator:

∂P

∂z

˛

˛

˛

˛z=λ_j

= Y

l6=j

(λ_j − λl)

Lin. sensitivity an. - (continued)

First order series expansion around z = λj: P (λ_j + δλ_j, α_k + δλ_k) ≈ P (λ^j, α_k) + ^∂P_∂z ˛

˛

˛z=λ_j δλ_j + _∂α^∂P

kδα_k + . . . Observing that P (λ_j, α_k) = 0 we get:

δλ_j ≈ ∂P/∂α_k

∂P/∂z

˛

˛

˛

˛z=λ_j

δα_k

Numerator:

∂P

∂α_k

˛

˛

˛

˛z=λ_j

= λ^n−k_j Writing P (z, α_k) = (z − λ¹) . . . (z − λⁿ), we get for the denominator:

∂P

∂z

˛

˛

˛

˛z=λ_j

= Y

l6=j

(λ_j − λl)

Lin. sensitivity an. - (continued)

First order series expansion around z = λj: P (λ_j + δλ_j, α_k + δλ_k) ≈ P (λ^j, α_k) + ^∂P_∂z ˛

˛

˛z=λ_j δλ_j + _∂α^∂P

kδα_k + . . . Observing that P (λ_j, α_k) = 0 we get:

δλ_j ≈ ∂P/∂α_k

∂P/∂z

˛

˛

˛

˛z=λ_j

δα_k

Numerator:

∂P

∂α_k

˛

˛

˛

˛z=λ_j

= λ^n−k_j

Writing P (z, α_k) = (z − λ¹) . . . (z − λⁿ), we get for the denominator:

∂P

∂z

˛

˛

˛

˛z=λ_j

= Y

l6=j

(λ_j − λl)

Lin. sensitivity an. - (continued)

...hence,

δλ_j = −

λ^n−k_j

Q

l6=j(λ_j − λl)

δα_k Consideration 1: assuming a stable λj the highest sensitivity is for n = k i.e. for

the variations of αn

Consideration 2: Looking at the denominator roots should be far apart: clusters empahsise senistivity

Lin. sensitivity an. - (continued)

...hence,

δλ_j = −

λ^n−k_j

Q

l6=j(λ_j − λl)

δα_k

Consideration 1: assuming a stable λj the highest sensitivity is for n = k i.e. for the variations of αn

Consideration 2: Looking at the denominator roots should be far apart: clusters empahsise senistivity

Lin. sensitivity an. - (continued)

...hence,

δλ_j = −

λ^n−k_j

Q

l6=j(λ_j − λl)

δα_k

Consideration 1: assuming a stable λj the highest sensitivity is for n = k i.e. for the variations of αn

Consideration 2: Looking at the denominator roots should be far apart: clusters empahsise senistivity

Example

Suppose we need to implement a low pass filter with sharp cutoff

we need a cluster of roots → high

sensitivity for canonical implementations it is much more convenient to use a series or parallel composition of low order filters (sensitivity is on one factor).

Outline

Parameters round-off Sample-rate selection

Factors

A famous rule of the thumb is to choose the sampling period 20-40 times the bandwidth of the system.

Where does it come from?

Shannon theorem limit

Smoothness of time response Immunity to noise

sensitivity to parameters variations

Factors

A famous rule of the thumb is to choose the sampling period 20-40 times the bandwidth of the system.

Where does it come from?

Shannon theorem limit Smoothness of time response

Immunity to noise

sensitivity to parameters variations

Factors

A famous rule of the thumb is to choose the sampling period 20-40 times the bandwidth of the system.

Where does it come from?

Shannon theorem limit

Smoothness of time response Immunity to noise

sensitivity to parameters variations

Factors

A famous rule of the thumb is to choose the sampling period 20-40 times the bandwidth of the system.

Where does it come from?

Shannon theorem limit

Smoothness of time response Immunity to noise

sensitivity to parameters variations

Factors

A famous rule of the thumb is to choose the sampling period 20-40 times the bandwidth of the system.

Where does it come from?

Shannon theorem limit

Smoothness of time response Immunity to noise

sensitivity to parameters variations

Shannon theorem

Consider the block diagram:

Shannon theorem - (cont.)

Recall that setting H^∗(s) = (1 − e^−sT)D^∗(s)n_G(s)

s

o∗

, we get

Y ^∗(s) = H^∗(s)

H^∗(s) + 1R^∗(s) If we want to track a signal r with frequency componente up to ω_b we have to

choose:

ωs

ω_b > 2.

Indeed lower sampling frequencies would determine aliasing of signal r, generating delays and instability

Typically the closed loop roots of _H^H_∗(s)+1^∗(s) should be little lower than the required bnadwidth. Sampling rates lower than the bandwidth would determine aliasing of such roots as well.

In practice Shannon limit is too low!!

Shannon theorem - (cont.)

Recall that setting H^∗(s) = (1 − e^−sT)D^∗(s)n_G(s)

s

o∗

, we get

Y ^∗(s) = H^∗(s)

H^∗(s) + 1R^∗(s)

If we want to track a signal r with frequency componente up to ω_b we have to choose:

ω_s

ω_b > 2.

Indeed lower sampling frequencies would determine aliasing of signal r, generating delays and instability

Typically the closed loop roots of _H^H_∗(s)+1^∗(s) should be little lower than the required bnadwidth. Sampling rates lower than the bandwidth would determine aliasing of such roots as well.

In practice Shannon limit is too low!!

Shannon theorem - (cont.)

Recall that setting H^∗(s) = (1 − e^−sT)D^∗(s)n_G(s)

s

o∗

, we get

Y ^∗(s) = H^∗(s)

H^∗(s) + 1R^∗(s)

If we want to track a signal r with frequency componente up to ω_b we have to choose:

ω_s

ω_b > 2.

Indeed lower sampling frequencies would determine aliasing of signal r, generating delays and instability

Typically the closed loop roots of _H^H_∗(s)+1^∗(s) should be little lower than the required bnadwidth. Sampling rates lower than the bandwidth would determine aliasing of such roots as well.

In practice Shannon limit is too low!!

Shannon theorem - (cont.)

Recall that setting H^∗(s) = (1 − e^−sT)D^∗(s)n_G(s)

s

o∗

, we get

Y ^∗(s) = H^∗(s)

H^∗(s) + 1R^∗(s)

If we want to track a signal r with frequency componente up to ω_b we have to choose:

ω_s

ω_b > 2.

Indeed lower sampling frequencies would determine aliasing of signal r, generating delays and instability

Typically the closed loop roots of _H^H_∗(s)+1^∗(s) should be little lower than the required bnadwidth. Sampling rates lower than the bandwidth would determine aliasing of such roots as well.

Smoothness of time response

Attitude control of a satellite System dynamics:

I ¨θ = M_C + M_d, where:

I is the ,oment of inertia of the satellite around its mass center

M_C is the control torque

M_d is the disturbance torque

The system dynamic is approximated by a double integrator

Example - (cont.)

Specifications: tr < 0.5s, Overshoot Mp < 16_% Approximated relations:

t_r ≈ ^1.8_ωn

t_s ≈ _ξω^4.6_n = ^4.6_σ M_p ≈ e^−πξ/

√1−ξ²

...from which we get: ξ = 0.5, σ = 1.8

Example - (cont.)

Try with sample rates equal to 4, 8, 20, 40 times ωn

The root locations in the z−plane are given by z = e^sT = e^{(−xiωn±jωn}√

1−xi²)T = e^−ξωnT(cos(ω_np1 − ξ²T ) ± sin(ωⁿp1 − ξ²T )) Case ωs/ω_n = 4 → ωⁿT = π/2 → T = 0.4363s:

z = e^sT = e^ξπ/2(cos(π/2p

1 − ξ²) ± sin(π/2p

1 − ξ²)) = 0.0952 ± 0.4459i

Closed loop characteristic polynomial:

z² + α₁z + α₂ = z² − 0.1904z + 0.2079

Discrete-time model for the system:

x(k + 1) = Φx(k) + Γu(k) = 2 4

1 T

0 1

3

5x(k) + 2 4

T²/2 T

3

5u(k)

Example - (cont.)

Try with sample rates equal to 4, 8, 20, 40 times ωn

The root locations in the z−plane are given by z = e^sT = e^{(−xiωn±jωn}√

1−xi²)T = e^−ξωnT(cos(ω_np1 − ξ²T ) ± sin(ωⁿp1 − ξ²T )) Case ωs/ω_n = 4 → ωⁿT = π/2 → T = 0.4363s:

z = e^sT = e^ξπ/2(cos(π/2p

1 − ξ²) ± sin(π/2p

1 − ξ²)) = 0.0952 ± 0.4459i

Closed loop characteristic polynomial:

z² + α₁z + α₂ = z² − 0.1904z + 0.2079

Discrete-time model for the system:

x(k + 1) = Φx(k) + Γu(k) = 2 4

1 T

0 1

3

5x(k) + 2 4

T²/2 T

3

5u(k)

Example - (cont.)

Try with sample rates equal to 4, 8, 20, 40 times ωn

The root locations in the z−plane are given by z = e^sT = e^{(−xiωn±jωn}√

1−xi²)T = e^−ξωnT(cos(ω_np1 − ξ²T ) ± sin(ωⁿp1 − ξ²T )) Case ωs/ω_n = 4 → ωⁿT = π/2 → T = 0.4363s:

z = e^sT = e^ξπ/2(cos(π/2p

1 − ξ²) ± sin(π/2p

1 − ξ²)) = 0.0952 ± 0.4459i Closed loop characteristic polynomial:

z² + α₁z + α₂ = z² − 0.1904z + 0.2079

Discrete-time model for the system:

x(k + 1) = Φx(k) + Γu(k) = 2 4

1 T

0 1

3

5x(k) + 2 4

T²/2 T

3

5u(k)

Example - (cont.)

Try with sample rates equal to 4, 8, 20, 40 times ωn

The root locations in the z−plane are given by z = e^sT = e^{(−xiωn±jωn}√

1−xi²)T = e^−ξωnT(cos(ω_np1 − ξ²T ) ± sin(ωⁿp1 − ξ²T )) Case ωs/ω_n = 4 → ωⁿT = π/2 → T = 0.4363s:

z = e^sT = e^ξπ/2(cos(π/2p

1 − ξ²) ± sin(π/2p

1 − ξ²)) = 0.0952 ± 0.4459i Closed loop characteristic polynomial:

z² + α₁z + α₂ = z² − 0.1904z + 0.2079 Discrete-time model for the system:

x(k + 1) = Φx(k) + Γu(k) = 2 4

1 T

0 1

3

5x(k) + 2 4

T²/2 T

3

5u(k)

Example - (cont.)

Try with sample rates equal to 4, 8, 20, 40 times ωn

The root locations in the z−plane are given by z = e^sT = e^{(−xiωn±jωn}√

1−xi²)T = e^−ξωnT(cos(ω_np1 − ξ²T ) ± sin(ωⁿp1 − ξ²T )) Case ωs/ω_n = 4 → ωⁿT = π/2 → T = 0.4363s:

z = e^sT = e^ξπ/2(cos(π/2p

1 − ξ²) ± sin(π/2p

1 − ξ²)) = 0.0952 ± 0.4459i Closed loop characteristic polynomial:

z² + α₁z + α₂ = z² − 0.1904z + 0.2079 Discrete-time model for the system:

x(k + 1) = Φx(k) + Γu(k) = 2 4

1 T

0 1

3

5x(k) + 2 4

T²/2 T

3

5u(k)

Example - (cont.)

State feedback: u(k) = −k¹x₁(k) − k²x₂(k) Closed loop evolution x(k + 1) =

2 4

1 − k1T²/2 T − k2T²/2

−k1T 1 − k²T 3

5x(k) The characteristic polynomial is:

z² + (T k₂ + (T²/2)k₁ − 2)z + (T²/2)k₁ − T k2 + 1

The gains are found as follows:

2 4

k₁ k₂

3 5 =

2 4

T²/2 T T²/2 −T

3 5

−1 2 4

2 + α₁

−1 + α² 3 5

Example - (cont.)

State feedback: u(k) = −k¹x₁(k) − k²x₂(k) Closed loop evolution x(k + 1) =

2 4

1 − k1T²/2 T − k2T²/2

−k1T 1 − k²T 3

5x(k) The characteristic polynomial is:

z² + (T k₂ + (T²/2)k₁ − 2)z + (T²/2)k₁ − T k2 + 1

The gains are found as follows:

2 4

k₁ k₂

3 5 =

2 4

T²/2 T T²/2 −T

3 5

−1 2 4

2 + α₁

−1 + α² 3 5

Example - (cont.)

State feedback: u(k) = −k¹x₁(k) − k²x₂(k) Closed loop evolution x(k + 1) =

2 4

1 − k1T²/2 T − k2T²/2

−k1T 1 − k²T 3

5x(k) The characteristic polynomial is:

z² + (T k₂ + (T²/2)k₁ − 2)z + (T²/2)k₁ − T k2 + 1 The gains are found as follows:

2 4

k₁ k₂

3 5 =

2 4

T²/2 T T²/2 −T

3 5

−1 2 4

2 + α₁

−1 + α² 3 5

Example - (cont.)

State feedback: u(k) = −k¹x₁(k) − k²x₂(k) Closed loop evolution x(k + 1) =

2 4

1 − k1T²/2 T − k2T²/2

−k1T 1 − k²T 3

5x(k) The characteristic polynomial is:

z² + (T k₂ + (T²/2)k₁ − 2)z + (T²/2)k₁ − T k2 + 1 The gains are found as follows:

2 4

k₁ k₂

3 5 =

2 4

T²/2 T T²/2 −T

3 5

−1 2 4

2 + α₁

−1 + α2

3 5

Example - (cont.)

An alternative way: construct the matrix T = [Γ ΦΓ]

Consider the coordinate transformation x = M ˜x, where M = T 2 4

1 α^(ol)₁

0 1

3 5 In the new coordinates we find

˜

x(k + 1) = 2 4

2 −1

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k)

With u(k) = −˜k˜x we get the close loop evolution

˜

x(k + 1) = 2 4

2 − ˜k¹ −1 − ˜k₂

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k)

, where the top row are the coeeficient of the characteristic polynomial.

Find solving:2 − ˜k¹ = −α¹, −1 − ˜k² = −α² k = −˜kM⁻¹

Example - (cont.)

An alternative way: construct the matrix T = [Γ ΦΓ]

Consider the coordinate transformation x = M ˜x, where M = T 2 4

1 α^(ol)₁

0 1

3 5 In the new coordinates we find

˜

x(k + 1) = 2 4

2 −1

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k)

With u(k) = −˜k˜x we get the close loop evolution

˜

x(k + 1) = 2 4

2 − ˜k¹ −1 − ˜k₂

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k)

, where the top row are the coeeficient of the characteristic polynomial.

Find solving:2 − ˜k¹ = −α¹, −1 − ˜k² = −α² k = −˜kM⁻¹

Example - (cont.)

An alternative way: construct the matrix T = [Γ ΦΓ]

Consider the coordinate transformation x = M ˜x, where M = T 2 4

1 α^(ol)₁

0 1

3 5 In the new coordinates we find

˜

x(k + 1) = 2 4

2 −1

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k) With u(k) = −˜k˜x we get the close loop evolution

˜

x(k + 1) = 2 4

2 − ˜k¹ −1 − ˜k₂

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k)

, where the top row are the coeeficient of the characteristic polynomial.

Find solving:2 − ˜k¹ = −α¹, −1 − ˜k² = −α² k = −˜kM⁻¹

Example - (cont.)

An alternative way: construct the matrix T = [Γ ΦΓ]

Consider the coordinate transformation x = M ˜x, where M = T 2 4

1 α^(ol)₁

0 1

3 5 In the new coordinates we find

˜

x(k + 1) = 2 4

2 −1

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k)

With u(k) = −˜k˜x we get the close loop evolution

˜

x(k + 1) = 2 4

2 − ˜k¹ −1 − ˜k₂

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k)

, where the top row are the coeeficient of the characteristic polynomial.

Find solving:2 − ˜k¹ = −α¹, −1 − ˜k² = −α² k = −˜kM⁻¹

Example - (cont.)

An alternative way: construct the matrix T = [Γ ΦΓ]

Consider the coordinate transformation x = M ˜x, where M = T 2 4

1 α^(ol)₁

0 1

3 5 In the new coordinates we find

˜

x(k + 1) = 2 4

2 −1

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k)

With u(k) = −˜k˜x we get the close loop evolution

˜

x(k + 1) = 2 4

2 − ˜k¹ −1 − ˜k₂

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k) k = −˜kM⁻¹

Example - (cont.)

An alternative way: construct the matrix T = [Γ ΦΓ]

Consider the coordinate transformation x = M ˜x, where M = T 2 4

1 α^(ol)₁

0 1

3 5 In the new coordinates we find

˜

x(k + 1) = 2 4

2 −1

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k)

With u(k) = −˜k˜x we get the close loop evolution

˜

x(k + 1) = 2 4

2 − ˜k¹ −1 − ˜k₂

1 0

3

5x(k) +˜ 2 4 1 0

3

5u(k)

, where the top row are the coeeficient of the characteristic polynomial.