discrete-time systems - tu berlin · tu berlin discrete-time control systems 2 sampling a...

TU Berlin Discrete-Time Control Systems 1

Discrete-Time Systems

Overview

• Sampling a Continuous-Time State-Space Model

• Input-Output-Models: The Pulse Response

• Shift Operator

• The Pulse-Transfer Operator

• The z -Transform

• Computation of the Pulse-Transfer Function

• Poles and Zeros

21st April 2014


Sampling a Continuous-Time State-Space Model

• Assuming a continuous-time system given in the following state-space form

dx(t)

dt= Ax(t) +Bu(t)

y(t) = Cx(t) +Du(t)

The system has r inputs, p outputs, and is of order n.

• General solution x(t) of the state over a time interval starting at t0

x(t) = eA(t−t0)x(t0) +∫ t

t0

eA(t−s′)Bu(s′)ds′

• Assuming the time interval now goes from one sampling instant t0 = k∆ to the next sampling

instant t = k∆+∆.

• The control signal u(t) is kept constant over the sampling interval:

u(s′) = u(k∆), k∆ ≤ s′ < k∆+∆

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• With the additional time transformation s = k∆+∆− s′ we can obtain:

x(k∆+∆) = eA∆x(k∆) +

∫ ∆

0

eAsdsBu(k∆)

Let us define

Φ(∆) = eA∆

Γ (∆) =

∫ ∆

0

eAsdsB

in order to obtain the following discrete-time system for the sampled continuous-time system:

x[k + 1] = Φx[k] + Γu[k]

y[k] = Cx[k] +Du[k]

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How to compute Γ andΦ

Can be computed by

• Numerical calculation (Scilab, Matlab, ...)

• Series expansion of the matrix exponential

• Laplace transform (L{eAt} = (sI −A)−1)

• Transformation of the state-space model to diagonal or Jordan forms

Calculation by hand are feasible for low order systems, n ≤ 2, and for high-order systems with

special structures.

Using the definition of the matrix exponential yields

Φ(∆) = eA∆ = I +A∆+A2∆2

2!+A3∆3

3!+ · · ·+

Ai∆i

(i)!+ · · · =

∞∑

k=0

Ak∆k

(k!),

which can be rewritten into

Φ(∆) = I +AΨ(∆)

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with

Ψ(∆) = I∆+A∆2

2!+A2∆3

3!+ · · ·+

Ai∆i+1

(i+ 1)!+ · · · =

∞∑

k=0

Ak∆k+1

(k + 1)!.

Γ (∆) =

∫ ∆

0

eAsdsB =

∫ ∆

0

( ∞∑

k=0

Aksk

(k!)

)dsB =

[ ∞∑

k=0

Aksk+1

(k + 1)!

]s=∆

s=0

B

=

∞∑

k=0

Ak∆k+1

(k + 1)!B = Ψ(∆)B

Examples (blackboard) ...

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Solution of the System Equation

Assume that the initial condition x[k0] and the input signals u[k0],u[k0 + 1], . . . for the

discrete-time system in state-space form are given.

x[k0 + 1] = Φx[k0] + Γu[k0]

x[k0 + 2] = Φx[k0 + 1] + Γu[k0 + 1]

= Φ2x[k0] +ΦΓu[k0] + Γu[k0 + 1]

...

x[k] = Φk−k0x[k0] +Φk−k0−1Γu[k0] + · · ·+ Γu[k − 1]

= Φk−k0x[k0] +k−1∑

j=k0

Φk−j−1Γu[j]

The solution consists of two parts: One depends on the initial condition and the other is the weighted

sum of the input signals.

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• The Eigenvalues ofΦ will determine the properties of the solution.

• The Eigenvalues are obtained from the characteristic equation:

det (λI −Φ) = 0

Example (blackboard) ...

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Changing Coordinates in State-Space Models

Coordinate transformation

z[k] = Tx[k]

where T is a nonsingular matrix. It follows:

z[k + 1] = Tx[k + 1] = TΦx[k] + TΓu[k] = TΦT−1z[k] + TΓu[k]

= Φz(k) + Γ u[k]

and

y[k] = Cx[k] +Du[k] = CT−1z[k] +Du[k]

= Cz[k] +Du[k]

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The characteristic equation

det(λI −Φ) = 0

is invariant when new states are introduced through the nonsingular transformation matrix T .

Proof.

det(λTIT−1 − TΦT−1) = det(T ) det(λI −Φ) detT−1 = det(λI −Φ)

To find a transformation it is the same as solving for the n2 elements of T from a linear set of

equations:

TΦ = ΦT (TΦT−1 = Φ)

Coordinates can be chosen to give simple forms of the system equations.

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Diagonal Form

AssumeΦ has distinct eigenvalues. Then there exists a T such that

TφT−1 =

λ1 0

. . .

0 λn

where λi are the eigenvalues ofΦ. Assuming single input u[k], with the transformation a set of

decoupled first order difference equations is obtained:

z1[k + 1] = λ1z1[k] + β1u[k]

...

zn[k + 1] = λnzn[k] + βnu[k]

y[k] = γ1z1[k] + · · ·+ γnzn[k]

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The solution is now simple. Each mode will have the solution:

zi[k] = λki zi[0] +

k−1∑

j=0

λk−j−1i βiu[j]

Jordan Form

IfΦ has multiple Eigenvalues, then it is generally not possible to diagonalizeΦ. Introduce the

notation

Lk(λ) =

λ 1 0 · · · 0

0 λ 1 0...

. . .. . .

. . .

0 · · · 0 λ 1

0 · · · 0 0 λ

where Lk is a k × k matrix.

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Then there exists a matrix T such that

TφT−1 =

Lk1(λ1) 0

. . .

0 Lkn(λn)

with k1 + k2 · · ·+ kr = n. The transform matrix, Φ, has the eigenvalues in the diagonal and

some 1’s in the superdiagonal.

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Input-Output-Models: The Pulse Response

Consider a system with one input and one output only. The input and output vectors over a finite

interval can be represented as finite-dimensional vectors

U =(u[0] · · ·u[N − 1]T

)

Y =(y[0] · · · y[N − 1]T

)

A general linear model that relatesU and Y can be expressed as

Y =HU + Y p

whereH is a N ×N matrix and Y p accounts for initial conditions. If the relation betweenU and

Y is causal, the matrixH must be lower triangular. The element h(k,m) ofH is thus zero if

m > k.

The input-output relation can also be written as

y[k] =

k∑

m=0

h(k,m)u[m] + yp[k]

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where yp accounts for initial conditions of the system.

• The function h(k,m) is called the pulse-response function.

• The pulse-response function can be easily measured by injecting a pulse of unit magnitude and

the width of the sampling interval and recording the output. For zero initial conditions the value

h(k,m) is the output at sampling instant k for a unit pulse at instant m.

• For time-invariant systems the pulse response becomes

h(k,m) = h(k −m).

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• It is easy to compute the pulse response of the system defined by the state-space model

(k0 = 0):

y[k] = CΦk−k0x[k0] +k−1∑

j=k0

CΦk−j−1Γu[j] +Du[k]

h(k) =

0 k < 0

D k = 0

CΦk−1Γ k ≥ 1

(1)

• The pulse response is invariant with respect to the coordinate transf. of the state-space model.

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Shift Operator

• In operator calculus, all signals are considered as doubly infinite sequences

{f [k] : k = · · · − 1, 0, 1, · · · }.

• The forward-shift operator is denoted as q. It has the property

qf [k] = f [k + 1]

• The shift operator has unit norm. It does not change norm of signal it operates on.

• The inverse of the forward shift operator is called backward-shift operator or delay operator.

• Operator calculus gives compact description of systems and makes it easy to derive

relationships among system variables.

• Difference equations can be reduced to a purely algebraic problem.

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Consider the following higher-order difference equation

y[k + na] + a1y[k + na − 1] + · · · anay[k] = b0u[k + nb] + · · ·+ bnbu[k]

where na ≥ nb. Use of shift operator gives

(qna + a1qna−1 + · · ·+ ana

)y[k] = (b0qnb + b1q

nb−1 + · · ·+ bnb)u[k]

With the polynomials

A(q) = qna + a1qna−1 + · · ·+ ana

B(q) = b0qnb + b1q

nb−1 + · · ·+ bnb

it follows

A(q)y[k] = B(q)u[k]

The degree of a polynomial can be expressed by a subscript, for example, Ana(q).

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The before introduced higher-order difference equation can be expressed in terms of back-shift

operator

y[k] + a1y[k − 1] + · · · anay[k − nna

] = b0u[k − d] + · · ·+ bnbu[k − d− nb]

where d = na − nb is the pole excess (time-delay) of the system.

The reciprocal polynomial

A∗(q) = 1 + a1q + · · ·+ anaqna = qnaA(q−1)

is obtained from the polynomial A by reversion the order of the coefficients. With reciprocal

polynomials the system we can write

A∗(q−1)y[k] = B∗(q−1)u[k − d]

Attention: A∗∗ is not necessary the same as A (see for example A(q) = q). A polynomial is

self-reciprocal if

A∗(q) = A(q)

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A difficulty with the shift operator

• Difference equations can be multiplied by powers of q.

• Equations for shifted times can be multiplied by real numbers and added (corresponds to

multiplying by a polynomial in q.

• However, it is not possible to divide by a polynomial in q unless special assumptions are made.

• Division by an arbitrary polynomial are only allowed if it is assumed that there is some k0such that all sequences are zero for k ≤ k0!

Example (see blackboard) ...

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The Pulse-Transfer Operator

• Expression of input-output relations as rational functions in either the forward- or the

backward-shift operator pulse-transfer operator

• Obtained from any description by eliminating internal variables using purely algebraic

manipulations

x[k + 1] = qx[k] = Φx[k] + Γu[k]

Hence

(qI −Φ)x[k] = Γu[k]

This gives

y[k] = Cx[k] +Du[k] =(C(qI −Φ)−1Γ +D

)u[k]

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The pulse transfer-operator is thus given by

H(q) =(C(qI −Φ)−1Γ +D

)=

B(q)

A(q)

The pulse transfer-operator can also be expressed in terms of the backward-shift operator

H∗(q−1) =(C(I − q−1Φ)−1q−1Γ +D

)

If the state vector is of dimension n and if the polynomials A and B do not have common factors,

then the polynomial A is of degree n.


• The pulse-transfer operator H(q) for a state-space model is independent of the state

representation.

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Poles and Zeros - Pulse-Transfer Operator

• The poles are the zeros of the denominator of H(q), the characteristic polynomial A(q).

• The zeros are obtained from B(q) = 0 and are the poles of the inverse of the system.

• Time delay (pole excess) gives rise to poles at the origin.

The Order of the System - Pulse-Transfer Operator

• Order of the system = dimension of the state-space representation = number of poles

• Important: Always use the forward-shift form to determine the order because of time delays.


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The z-Transform

• The discrete-time analogy of the Laplace transform is the z-transform – a convenient tool to

study linear difference equations without and with initial conditions (in contrast to the calculus

operator).

• The z-Transform maps a semi-infinite time sequence (in contrast to the operator calculus) into a

function of a complex variable.

• The z-Transform is defined as

Z{f(k∆} = F (z) =

∞∑

k=0

f(k∆)z−k

where z is a complex variable (in contrast to q).

• The inverse transform is given by

f(k∆) =1

2πj

∮F (z)zk−1dz

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Some properties of the z-Transform (h ≡ ∆):

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The z-Transform can be used to solve difference equations, for instance

x[k + 1] = Φx[k] + Γu[k]

y[k] = Cx[k] +Du[k]

If the z-transform of both sided is taken

∞∑

k=0

z−kx[k + 1] = z

( ∞∑

k=0

z−kx[k]− x[0])

=

∞∑

k=0

Φz−kx[k] +∞∑

k=0

Γ z−ku[k]

Hence

z (X(z)− x[0]) = ΦX(z) + ΓU(z)

X(z) = (zI −Φ)−1 (zx[0] + ΓU(z))

Y (z) = C(zI −Φ)−1zx[0] +(C(zI −Φ)−1Γ +D

)U(z)

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• Pulse-transfer function:

H(z) =(C(zI −Φ)−1Γ +D

)

• The time sequence y[k] can now be obtained using the inverse transform.

• The pulse response and the pulse-transfer function are a z-transform pair, that is,

Z{h(k)} = H(z).

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Computation of the Pulse-Transfer Function

• Determine the pulse-transfer function H(z) directly from the cont.-time transfer-function G(s).

• Let us assume the following system:

holdZero-order

G(s)u[k] = u(k∆) u(t) y(t) y[k] = y(k∆)

H(z)

• H(z) is uniquely determined by the response to a given signal.

Consider for example a unit step input: {u[k]} is a sequence of ones and u(t) is a unit step. Let

Y (s) denote the Laplace transform of y(t), that is,

Y (s) =G(s)

s(2)

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Let the output sequence {y[k]} of the sampled system have the z-transform

Y (z) = Z{L−1{Y (s)}}.

Division of Y (z) by the z-transform of the input, which is z/(z − 1) gives

H(z) = (1− z−1)Y (z)

The pulse-transfer function H(z) is obtained as follows:

1. Determine the step response of the system with the transfer-function G(s).

2. Determine the corresponding z-transform of the step response.

3. Divide by the z-transform of the step function.


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Some time functions and corresponding Laplace and z-transforms (h ≡ ∆).

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Poles and Zeros

• Poles and Zeros can be obtained from the denominator and numerator of the pulse-transfer

function.

• A pole z = a corresponds to a free mode of the system associated with the sequence

z[k] = ak.

• Poles are also the eigenvalues of the system matrixΦ.

• The zeros are related to how inputs and outputs are coupled to the states.

• Zeros can also be characterised by their signal blocking properties. A zero z = a means that a

transmission u(k) = ak is blocked by the system.

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Poles

Consider the continuous-time system described by the nth-order state-space model

dx(t)

dt= Ax(t) +Bu(t)

y(t) = Cx(t)

The poles of the system are the eigenvalues ofA: λi(A), i = 1, . . . , n. The zero-order-hold

sampling gives the discrete-time system

x[k + 1] = Φx[k] + Γu[k]

y[k] = Cx[k].

The poles of this system are the eigenvalues ofΦ: λi(Φ), i = 1, . . . , n.

• BecauseΦ = eA∆ is follows from the properties of matrix functions that

λi(Φ) = eλi(A)∆

This defines the mapping of the complex s-plane into the z-plane, when z = es∆.

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The conformal map z = es∆ (aus Astroem & Wittenmark, Computer-Controlled Systems, 1997)

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• This map is not bijective - several points in the s-plane are mapped into the same point in the

z-plane (aliasing effect) (∆ ≡ h).

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Complex Poles:

Consider the continuous-time system

ω20

s2 + 2ζω0s+ ω0.

The poles of the corresponding discrete-time system are given by the characteristic equation

z2 + a1z + a2 = 0

where

a1 = −2ζe−ζω0∆ cos(√

1− ζ2ω0∆)

a2 = e−2ζω0∆

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Step responses of the second order discrete-time system for different values of ∆ when ζ = 0.5

and ω0 = 1.83, which gives a rise time tr = 1 s. (a) ∆ = 0.125s, (b) ∆ = 0.25s, (c) ∆ = 0.5s,

and (d) ∆ = 1s.

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Loci of constant ζ (solid) and ω0 (dashed) when a second order continuous-time system is sampled.

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Zeros

• It is not possible to give a simple formula for the mapping of zeros.

• The discrete-time system has in general n− 1 zeros! The sampling procedure gives extra zeros.

• For fast sampling periods, a discrete-time system will have zeros in

zi ≈ esi∆

where the si’s are the zeros of the continuous-time system.

• A continuous time system with a stable inverse may become a discrete-time system with a

unstable inverse (poles outside the unit circle). Also continuous-time nonminimum-phase system

will not always become a discrete-time system with a unstable inverse.

• Large effect of sampling period on zeros.

21st April 2014

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