Signals and Systems - ethz.ch · Discretization of CT Systems Systems Signals In this class, we focus on DT signals: CT signals were treated extensively in Control Systems 1 & 2

Signals and SystemsLecture 1: From Continuous Time to Discrete Time

Dr. Guillaume Ducard

Fall 2018

based on materials from: Prof. Dr. Raffaello D’Andrea

Institute for Dynamic Systems and Control

ETH Zurich, Switzerland

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Outline

1 IntroductionSystemsSignals

2 Discretizing a CT Signal by Uniform SamplingSampling a CT Signal

3 CT Signal from a DT SignalUsing a Zero-order HoldUsing a First-order Hold

4 Discretization of CT SystemsArchitectureExact Discretization from Solution of SS Diff. EquationCT, No-input State Equations and SolutionExact Discretization Using the No-input Formulation

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IntroductionDiscretizing a CT Signal by Uniform Sampling

CT Signal from a DT Signal using the Zero-Order HoldDiscretization of CT Systems

SystemsSignals

Outline





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SystemsSignals

What is a system?

A system G operates on the input signal u to produce the outputsignal y, an operation denoted as y = Gu and illustrated as:

Gu

Input

y

Output

Note that, depending on G, the output y could depend on past,current, and future values of the input u.

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SystemsSignals

Definitions: (Dynamic) Systems

Dynamic Systems

systems that are not static, i.e., their state evolves w.r.t. time, dueto:

input signals,

external perturbations,

or naturally.

For example, a dynamic system is a system which changes:

its trajectory → changes in acceleration, orientation, velocity,position.

its temperature, pressure, volume, mass, etc.

its current, voltage, frequency, etc.

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SystemsSignals

Outline





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SystemsSignals

What is a signal?

A signal is a function of time that represents a physical quantitysuch as a force, position, or voltage.

Continuous-time (CT) signals

x(t) : t is continuous, t ∈ R (the set of real numbers)

x(t) takes on continuous values analog signal

t

x(t)

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SystemsSignals

Discrete-time (DT) signals

x[n] : n is discrete, an integer,

n ∈ Z (the set of integer numbers, Z = {. . . ,−1, 0, 1, . . .})

x[n] can take continuous or discrete values

n

x[n]

−1 0 1 2 3 4

1

2

0

1 1

2

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SystemsSignals

In this class, we focus on DT signals:

CT signals were treated extensively in Control Systems 1 & 2.

In DT the math is simpler, which allows us to focus onconcepts.

1 Integrals are replaced by sums,2 differentiation is replaced by finite differences,3 and the Dirac delta function is replaced by the unit impulse.

Algorithms are implemented in discrete time on computers.

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SystemsSignals

We focus on discrete-time signals with the signal x[n] taking oncontinuous values:

Assuming continuous values simplifies the math.

This is a good approximation: On modern computers thedouble data type stores numbers using 8 bytes (= 64 bits)and hence allows for 264 ≈ 1020 different values.

x[n] can be a vector, however we mainly deal with scalarsignals in this class.

x[n] can be a complex number:

x[n] = x1[n] + jx2[n], x1[n], x2[n] ∈ R, j2 = −1

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SystemsSignals

A word on notation

A sequence is a DT signal. The terms are usedinterchangeably.

x denotes a signal. It is a function of time.

x[n] indicates the value of x at the discrete time n.

x(t) indicates the value of x at the continuous time t.

{x[n]} refers to the entire sequence. It is the same as x forDT signals.

Motivated by the above, {x(t)} refers to the entire CT signal.It is the same as x for CT signals.

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SystemsSignals

Three signal representations of DT signals

Graph:

n

x[n]

−2 −1 0 1 2 3

0 0

1 12 1

418

Rule:

x[n] :=

{

(12)n n ≥ 0

0 n < 0

Sequence:

{x[n]} = {. . . , 0, 1↑,1

2,1

4, . . . },

where the arrow indicates the value at n = 0. If an arrow is absent, thefirst value of the sequence is assumed to occur at time n = 0.

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Sampling a CT Signal

Outline





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Sampling a CT Signal

Sampling a CT signal {x(t)} at uniformly-spaced points in timeresults in the DT sequence {x[n]}, where

x[n] = x(nTs)

for all integers n, and where:

Ts is the sampling period.

The variable fs =1Ts

is the sampling frequency.

n

x[n]

1 2 3 4

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Using a Zero-order HoldUsing a First-order Hold

Outline





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Zero-order Hold

A CT signal {x(t)} can be obtained from a DT signal {x[n]} by“holding” the value of the DT signal constant for one samplingperiod Ts, such that:

x(t) = x[n] nTs ≤ t < (n+ 1)Ts.

This is known as the zero-order hold.

t

x(t)

Ts 2Ts 3Ts 4Ts

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Outline





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First-order Hold

Alternative methods exist to obtain a CT signal from a DT signal,such as the first-order hold:

t

x(t)

Ts 2Ts 3Ts 4Ts

However, in this class we will exclusively work with zero-order hold,which we will simply call hold.

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ArchitectureExact Discretization from Solution of SS Diff. EquationCT, No-input State Equations and SolutionExact Discretization Using the No-input Formulation

Outline





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Interface between CT and DT systems

To interact with a CT (eg. “real-world”) system Gc from a DTenvironment (eg. a computer), we need to:

convert a DT signal into a CT signal, (ex: Digital to AnalogConverters (DAC)

and vice versa. (ex: Analog to Digital Converters (ADC))

This can be captured formally with the hold and sample operatorsas follows:

H Gc S{u[n]} {u(t)} {y(t)} {y[n]}

By defining the DT system Gd := SGcH, the previous figure can bereplaced by:

Gd

{u[n]} {y[n]}

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In the following, we consider the case where Gc is a CT lineartime-invariant (LTI) system with a state-space description

q(t) = Acq(t) +Bcu(t)

y(t) = Ccq(t) +Dcu(t). (1)

The solution to the differential equation is

q(t) = eActq(0) +

∫

t

0eAc(t−τ)Bcu(τ)dτ .

Demonstration recalled in class on blackboard.

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The goal is to find

a description for the DT LTI system

Gd = SGcH .

The discrete-time equivalent of Equation 1 is of the form:

q[n+ 1] = Adq[n] +Bdu[n]

y[n] = Cdq[n] +Ddu[n]

which is the description of the DT system Gd = SGcH that we arelooking for.

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Outline





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The discrete-time equivalent of Equation 1 is of the form



which is the description of the DT system Gd = SGcH looked for. It can beshown, starting from the solution of the continuous state-space equation

q(t) = eActq(0) +

∫t

0

eAc(t−τ)

Bcu(τ )dτ.

that the corresponding discrete matrices are obtained as:

Ad = eAcTs and Bd =(∫

Ts

0eAcτdτ

)

Bc

with the sampling period Ts.

Demonstration is done on the board during the class. Remarks:

This is the exact solution to the differential equation, there are nodiscretization errors.

While it is exact, information is still lost by the discretization: theinter-sample behavior.

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Outline





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In this section, we will show how the discrete matrices Ad and Bd can beobtained through a slightly different approach. It consists of :

1 simplifying the state equations in (1), by taking into account that theinput is constant during a sampling period (zero-order hold assumption).

2 rewriting the system as a no-input (or homogeneous) system of the type

r(t) = Mr(t). (2)

Such a system has the straightforward solution

r(t) = eMt

r(0), (3)

where eMt is the matrix exponential:

eMt := I +Mt+

(Mt)2

2+ · · ·+

(Mt)k

k!+ · · · =

∞∑

k=0

(Mt)k

k!.

Remarks:

When M is a scalar (matrix of dimension 1× 1), we recover the standarddefinition of the exponential.

It can be shown that the power series defining the matrix exponentialconverges for all real (or complex) square matrices.

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Proof that the solution (3) satisfies the differential equation (2)

Let r(t) = eMtr(0). It follows that r(t) = deMt

dtr(0).

For convergent power series, the derivative can be evaluated as:

deMt

dt=

d

dt

∞∑

k=0

(Mt)k

k!

=∞∑

k=0

kMktk−1

k!=

∞∑

k=1

kMktk−1

k!

=

∞∑

k=1

Mktk−1

(k − 1)!= M

∞∑

k=1

Mk−1tk−1

(k − 1)!

= M

∞∑

l=0

M ltl

l!with the substitution l := k − 1

= MeMt

.

∴ r(t) = MeMt

r(0) = Mr(t).

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Example : Double integrator with zero input

”Standard solution approach”Consider the system described by the differential equation

y(t) = 0

with initial conditions y(0) = v0, y(0) = y0.

We can immediately write the solution:

y(t)= v0,

y(t) = y0 + v0t.

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Example : Double integrator with zero input

The solution can also be derived using the matrix exponential: let

r(t) =

[r1(t)r2(t)

]

:=

[y(t)y(t)

]

r(0) =

[y0v0

]

.

Then[r1(t)r2(t)

]

=

[r2(t)0

]

=

[0 10 0

]

︸︷︷︸

= M

[r1(t)r2(t)

]

.

Note that the matrix M is nilpotent because M2 = 0, and therefore

eMt = I +Mt

[r1(t)r2(t)

]

=

[1 t

0 1

] [y0v0

]

.

which is identical solution as the one found before, i.e.

y(t) = v0, y(t) = y0 + v0t. (4)29 / 38




Relation to the Euler forward method

Matrix exponential approximation

When t is very small, the matrix exponential can be approximated by

eMt

≈ I +Mt

and therefore r(t) = eMtr(0) ≈ (I +Mt)r(0) ≈ r(0) + tMr(0).

Euler forward method

The Euler forward method approximates a function’s derivative as

r(0) ≈r(t)− r(0)

t.

For small t, and with r(t) = Mr(t) we have r(t)−r(0)t

≈ Mr(0), which yieldsr(t) ≈ r(0) + tMr(0).

⇒ The Euler forward method gives the same result as a first-orderapproximation to the matrix exponential.

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Outline





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The original problem was, for the first time step

q(t) = Acq(t) +Bcu(t), q(0) = q[0],

u(t) = u[0] 0 ≤ t < Ts

where

1 u is constant over the sampling interval due to the zero-order hold device

2 and we set the initial state q(0) to its discrete-time counter-part q[0].

We can rewrite these equations as the zero-input system (2)

[q(t)u(t)

]

=

[Ac Bc

0 0

]

︸︷︷︸

=: M

[q(t)u(t)

]

by including the input in the state of this adapted system.

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Therefore, the solution at time T−

s , just before the sampling time, is

[q(T−

s )u(T−

s )

]

= eMTs

[q(0)u(0)

]

with

F = eMTs =

[F11 F12

F21 F22

]

.

It is easy to show that

F21 = 0 and F22 = I,

since

Mk =

[∗ ∗

0 0

]

for any integer k ≥ 1.

This result is expected, since u is constant over the sampling interval and thusu(T−

s ) = u(0).

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Ad= F11, Bd = F12

Cd= Cc, Dd = Dc.

Then

q[1] = Adq[0] +Bdu[0]

y[0] = Cdq[0] +Ddu[0].

We can do this for any time period, since the CT system is time invariant, andthus obtain:



which is the description of the DT system Gd = SGcH that we were looking for.

This is the exact solution to the differential equation, there are nodiscretization errors.

While it is exact, information is still lost by the discretization: Theinter-sample behavior.

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Mass-spring-damper system

Consider a mass-spring-damper system with the differential equation

p(t) + p(t) + p(t) = f(t)

and compare:

1 the Euler discretization,

2 to the exact discretization,

with sampling times of 0.1s and 0.5s.

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The input to the continuous and discretized models is a pulse with a width of0.5 seconds, identical to the longest sampling time used.

0 2 4 6 8 10

0

1

2

Time (s)

Input(N

)continuous

Ts = 0.1s

Ts = 0.5s

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0 2 4 6 8 10

0

0.5

Time (s)

Position(m

)Euler Discretization

continuous

Ts = 0.1s

Ts = 0.5s

When discretizing using the Euler discretization, the output strongly dependson the discretization time, and differs from the continuous-time output even forsmall sampling times (remember that the Euler discretization is identical to afirst-order approximation of the matrix exponential – the errors seen here stemfrom this approximation)

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The exact discretization has no discretization errors. The discrete-time outputcoincides with the continuous-time output at the sampling times:

0 2 4 6 8 10

0

0.2

0.4

0.6

Time (s)

Position(m

)

Exact Discretization

continuous

Ts = 0.1s

Ts = 0.5s

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Documents

Signals and Systems - ethz.ch · Discretization of CT Systems Systems Signals In this class, we focus on DT signals: CT signals were treated extensively in Control Systems 1 & 2