Stability Analysis 1

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Stability Analysis (1)

• Digital Control Theory
• Lecture 8

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Outline

• Derivation of the characteristic equation and
  examples
• Bilinear transformation
• Routh-Hurwitz criterion and example
• Jury stability test and example

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Consider the closed-loop sampled-data system in
the figure.
Using partial-fraction expansion, we can write
C(z) as
CR(z) contains the terms which originate in the
poles of R(z).
These terms are the natural response terms. If
the inverse z-transform of these terms
tends to zero as time increases, the system is
stable, and these terms are called the transient
response.
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The transient response of the system is
determined by
If pilt1, this term approaches zero as k?8.
The factor (z-pi) originates in
or equivalently, originates in the system
characteristic equation
Thus, the system is stable provided that all the
roots of the characteristic equation lie inside
the unit circle in the z-plane.
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For the case that a root of the characteristic
equation is unity in magnitude, i.e., pi1, then
is constant in magnitude. Hence this natural
response term neither dies out nor becomes
unbounded as k?8. If the natural response
approaches a bounded nonzero steady state, the
system is said to be marginally stable. Thus for
a marginally stable system, the characteristic
equation has at least one root on the unit circle
but with no root outside the unit circle.
For some sampled-data systems, the TF cannot be
derived. How do we determine the characteristic
equation?
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Derivation of the characteristic equation
Since the stability of a linear system is
independent of the system input, we set input to
zero. In addition, we open the system in front
of a sampler and derive a TF at this open - this
TF is called the open-loop TF. The reason for
doing so is because we can always write a TF if
an input is sampled before being applied to a
continuous-time part of the system. We present
a procedure below for deriving the open-loop TF,
which easily leads to the characteristic equation.

• Set input R(s)0.
• Open the system in front of a sampler and draw
  the signal-flow graph. At this open, denote the
  sampler output as Ei(s) and sampler input as
  Eo(s).
• Express the sampler input in terms of each
  sampler output. Sampler output is treated as
  input in the flow graph.
• Find the TF Eo(z)/ Ei(z) at the open.

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Examples - derivation of the characteristic
equation
Example 1. Find the open-loop TF and
characteristic equation of the system in the
figure.

• Set input R(s)0.
• Open the system in front of a sampler and draw
  the signal-flow graph.

• Express the sampler input in terms of sampler
  output.

• Find the TF at the open.

Taking the z-transform of the above equation
yields Denote the open-loop TF as
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Examples - derivation of the characteristic
equation (contd)
For the closed-loop system, Ei(z)Eo(z), so the
equation
yields
This is the general expression of characteristic
equation.
which is the system characteristic equation, i.e.,
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Examples - derivation of the characteristic
equation (contd)
Example 2. Find the open-loop TF and
characteristic equation of the system in the
figure.

• Set input R(s)0.
• Open the system in front of the first sampler and
  draw the signal-flow graph. The effect of the
  second sampler is included by denoting its input
  as E1(s) and its output as E1(s).

• Express the sampler input in terms of each
  sampler output.

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Examples - derivation of the characteristic
equation (contd)
Taking the starred transform of
yields
Substituting E1, we obtain

• Find the TF at the open. The open-loop TF is

For the closed-loop system, Ei(z)Eo(z), thus
The characteristic equation is then
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Bilinear transformation
We have known that in the s-plane the stability
boundary is the imaginary axis. Based on this
property many analysis and design techniques,
such as Routh-Hurwitz criterion and Bode diagram,
have been used for LTI continuous-time systems.
But these techniques cannot be applied to LTI
discrete-time systems in the z-plane, since the
stability boundary is the unit circle. However,
through the use of the transformation
the unit circle of the z-plane transforms into
the imaginary axis of the w-plane, because on the
unit circle in the z-plane, zej?T,
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Bilinear transformation (contd)
The mappings of the primary strip of the s-plane
into both the z-plane (zeTs) and the w-plane are
shown in the figure. Let ?w be the imaginary part
of w. The expression
The stable region of the w-plane is the left half
plane.
gives the relationship between frequencies in the
s-plane and w-plane. For small values of ? such
that ?T is small,
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Routh-Hurwitz criterion
The Routh-Hurwitz criterion may be used to
determine if any roots of the characteristic
equation of an LTI continuous-time system are in
the right half of the s-plane. For LTI
discrete-time systems, if the characteristic
equation is expressed as a function of the
bilinear transform variable w, the stability of
the system may be determined by directly applying
the Routh-Hurwitz criterion. The stability
judgment is made based on the Routh array. The
following cases regarding the first column of the
Routh array may occur

• No element in the first column is zero.
• There is a zero in the first column, but some
  other elements of the row containing the zero in
  the first column are nonzero.
• There is a zero in the first column, and the
  other elements of the row containing the zero are
  also zero.

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Routh-Hurwitz criterion (contd)
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Routh-Hurwitz criterion (contd)
When constructing the Routh array, the following
cases may occur

• No element in the first column is zero.
• Solution Apply the normal procedure to the
  Routh array.
• There is a zero in the first column, but some
  other elements of the row containing the zero in
  the first column are nonzero.
• Solution If only one element in the array is
  zero, it may be replaced by a small positive
  number, e, which is allowed to approach zero
  after completing the array.
• There is a zero in the first column, and the
  other elements of the row containing the zero are
  also zero.
• Solution This is treated by utilising the
  auxiliary polynomial, which is formed from the
  row preceding the row of zeros. The order of the
  auxiliary polynomial is always even. The
  auxiliary polynomial is a factor of the
  characteristic polynomial. In this case, the
  system has symmetrical roots.

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Example - Routh-Hurwitz criterion
Example. Consider the system in the figure.
Determine the values of K such that the system is
stable.
In a previous lecture, we have obtained
The characteristic equation is given by
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Example - Routh-Hurwitz criterion (contd)
Thus the characteristic equation is
The Routh array is then
Hence the system is stable for 0ltKlt2.39. When
K2.39, the w1 row of the Routh array has all
zero elements. To find the symmetrical roots, we
resort to the auxiliary polynomial, formed from
the w2 row of the array
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Example - Routh-Hurwitz criterion (contd)
Solving the equation
for w, we get wj1.549, which is the cross-over
point into the right half w-plane. That is, for
K2.39 the system is marginally stable (the
system output exhibits sustained oscillation for
a bounded input). ?w1.549 is the oscillation
frequency in the w-plane and the corresponding
frequency ? in the s-plane is obtained by
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Jury stability test
The Routh-Hurwitz criterion cannot be directly
applied to discrete-time systems if the system
characteristic equation is expressed as a
function of z. Jurys stability test is a
stability criterion for discrete-time systems. It
is similar to the Routh-Hurwitz criterion and can
be applied to the characteristic equation
expressed in z. Let the characteristic equation
of an LTI discrete-time system be
Form the array as shown in the table. The 1st
row is constructed from the coefficients of the
characteristic equation. The 2nd row is 1st row
reversed.
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Jury stability test (contd)
Elements of each of the even-numbered rows are
the elements of the preceding row in reverse
order. The elements of the odd-numbered rows are
defined as
The necessary and sufficient condition for the
characteristic equation Q(z) with angt0, to have
no roots outside or on the unit circle are
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Jury stability test (contd)

• For a second-order system, the array has only one
  row.
• For each additional order, two additional rows
  are added to the array.
• For an nth-order system, there are a total of
  (n1) constraints.

Jurys test may be applied in the following
manner

• Check the first three conditions Q(1)gt0,
  (-1)nQ(-1)gt0, a0ltan. Stop if any of them is
  not satisfied.
• Construct the array and check the condition as
  each row is formed. Stop if any condition is not
  satisfied.

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Example - Jury stability test
Suppose the characteristic equation for a
closed-loop discrete-time system is given by
The first three conditions of Jurys test are
Since the first three conditions of Jurys test
are met, we proceed to form the array. As the
system is of 3rd-order, there are 4 constraints
to be tested against, i.e., one more plus the
above three. The Jury array is constructed as
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Example - Jury stability test (contd)
According to the calculation rule
The elements b0, b1, and b2 in the array
are calculated as
Hence the last condition is b00.96 gt
b20.69. As all the conditions are satisfied,
the system is stable.