Discrete Time Systems: Impulse responses and …maapc/static/files...Systems and Control Theory...

Systems and Control Theory

STADIUS - Center for Dynamical Systems,

Signal Processing and Data Analytics

Discrete Time Systems:Impulse responses and convolution; An introduction to the Z-transform

Lecture 5




Impulse response and system output

Using impulse response, output can be calculated as:y[k] = h[k] * u[k]Proof:

ConclusionThe impulse of a system describes the input/output behavior completely.

Definition of impulse response

Time-invariance

Linearity

Definition of convolution

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Impulse response and system outputVisually:

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Z-transform

Discrete equivalent to the Laplace-transformConverts time dependent descriptions of systems to the time-independent Z-domain.Simplifies many calculations

Convolution theorem → convolution becomes multiplicationLinear difference equations become simple algebraic expressions...

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Z-transform

2 formsBilateral:

Requires knowledge of h for all values of k, including negative valuesCan be used for non-causal systems

Unilateral:Only requires knowledge of h for positive values of kCan only be used for causal systems without loss of information

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Z-transform: properties

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Transfer function (DT)

The transfer function of a discrete system is the Z-transform of the impulse response.H(z) = Z{h[k]}Recall: y[k] = h[k] * u[k]Let U(z) = Z{u[k]}

Y(z) = Z{y[k]}Thus, applying the convolution theorem:

Y(z) = H(z) . U(z)BUT: Only applies when system starts from a null state

(Reason: impulse response itself starts from a null state)

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Getting rid of convolutions (DT)

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List of common Z-transform pairs

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List of common Z-transform pairs

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For the Z-transform to converge the following must hold:

We will look at convergence separately for positive and negative k, splitting the convergence criterion in 2:

Using z = r ejϴ

with R+ as small as possible and R- as large as possible we get:

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Region of convergence





The sums are finite if andRegion of convergence:

R+< R-: RingR-< R+: No ROC

Causal system, for negative k:

cannot contain any poles of the systemROC of a stable system always contains

the unit circle

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R+

R-

Source: http://www.expertsmind.com/learning/z-transform-and-realization-of-digital-filters-assignment-help-7342873888.aspx





System 1: Causal: h[k] = { …, 0, 0, 0, 1, 1/3, (1/3)2

, (1/3)3

, … }System 2: Anti-causal: h[k] = { …, 3

3

, 32

, 3, 1, 0, 0, 0, … }Analytical representation: h[k] = (1/3)

k

After Z-transform: H(z) = z / (z – 1/3)2 systems with very different behaviors, but the same transfer function?Answer: different ROC:

System 1: |z| > 1/3System 2: |z| < 1/3

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Inverse Z-transform

Split the transfer function up in partial fractionsThis is done by first factorizing the denominatorIf all poles have multiplicity 1 then the following can be used:

The coefficients can be calculated by:

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Inverse Z-transform

If there are poles with multiplicity higher than 1 then the following approach is needed:

Where the highest coefficient for each pole can be calculated by:

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Inverse Z-transform

Any remaining coefficients can be found by evaluating the equation:

for a number of values of z.

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Inverse Z-transform

Because of the linearity of the inverse Z-transform, each partial fraction can be transformed individually and the results can be added together afterwards.The individual inverse Z-transforms can be found with the following:

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Example

Transfer function:Partial fraction decomposition:

Using the given formula’s:

This gives:

By evaluating the transfer function for z=1 we get:

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Example

The resulting transfer function is:

We can now find the inverse Z-transform for each individual fraction:

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Inverse Z-transform

Another technique for calculating the inverse Z-transform is direct divisionThe numerator of the transfer function is divided by the denominator via long division.Example:

⇒Z-1

{F(z)} = 1, 3, 12, 25, ...

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Solving difference equations with the Z-transform

A system is described by a difference equation of the following form:

After the Z-transform:

Rearranged:

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lWe’ll apply the following transformation of the double summations:

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The final simplified result is:

With this result it is easy to find the resulting output from a given input or vice-versa given a difference equation.Right-hand fraction = output resulting from starting conditions: will vanish with time = transient behaviorLeft-hand fraction = output resulting from input: will remain = steady state response

is the “transfer function” of the system

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Steady state behavior via Z-transform

Starting from the previous result:

We wish to find the resulting output from the input:

To simplify derivation, we use:

u[k] = ej(kα + θ)

With Z-transform:

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Filling in U(z) and splitting into partial fractions:

Calculating the coefficient c:

After the inverse Z-transform:

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Because of linearity we can ignore the imaginary component, leading to the result:

In most applications (= stable system) we can ignore transient behavior as it will quickly die outUsing the transfer function steady state behavior can easily be determined by converting sinusoidal signals to phasorsThe input: cos(kα + θ)Will produce steady state: |H(ejα )| cos(kα + θ + ∠H(ejα))

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Example

We’ll have a look at the steady state response to the input for the system:

Evaluating the transfer function for the exponential with pulsation 3 gives:

The resulting output has been reduced to a third in amplitude and has undergone a small phase shift.

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Interpreting the Z-transform

Z-domain = frequency domain, similar to LaplaceThe results used for calculating steady state behavior can be used to give a more concrete interpretation to the Z-domainA z-value is the phasor representation of a sinusoidal signal in the k-domainEvery signal in the k-domain can be decomposed into a sum of sinusoidal signals.Every signal in the k-domain can be analyzed in the Z-domain as a sum of sinusoidal signals.

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State space to transfer function

The transfer function can be derived directly from the state space model of a system:

The Z-transform gives:

Rearranged to have X(z) in explicit form:

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State space to transfer function

The result for Y(z):

Left hand term = effect on output from starting conditions = transient behaviorRight hand term = effect on output from input = steady state behaviorIf all previous effects have died out we can equate the starting conditions to 0:Y(z) = H(z) U(z)H(z) = C (zI – A )-1 B + D

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Link between Eigenvalues and poles

Are Eigenvalues of A poles of H(z)?

As z approaches an Eigenvalue of A, is no longer defined. may still be defined depending on the values of C and B.An Eigenvalue of A will sometimes, but not always, be a pole of H(z).If every Eigenvalue of A is also a pole of H(z) then a minimal number of internal states has been achieved

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Link between Eigenvalues and poles

Are poles of H(z) Eigenvalues of A?

B, C and D are matrixes with properly defined valuesIf H(z) is undefined then must be the causez must be an Eigenvalue of APoles of H(z) are always Eigenvalues of A

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Stability (DT)

BIBO-Stability (Bounded-Input Bounded-Output)Every bounded input results in a bounded output

Internal StabilityStricter than BIBO-StabilityAll possible internal states return to zero after a finite time in the absence of an input.All Eigenvalues of the matrix A are contained within the a circle of radius 1 around zero in the complex plane.

BIBO-Stability follows from Internal Stability, but the inverse is not necessarily true.

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Stability (DT)

A discrete system is BIBO-Stable if all poles of H(z) are within a circle of radius 1 around the origin.

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Overview

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Discrete Time Systems: Impulse responses and …maapc/static/files...Systems and Control Theory...

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