bibo stability

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BIBO STABILITY

• EFFECT OF POLES

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Problem 1

• Let us take the first problem. This is the
  system, whose governing equation is given
  by
• y
  ??(??)5y?(??)6??(??)??(??)
• This is a second order system. If we take the
  Laplace transform on both sides, we get
  
• ?? 2 ??(??)-????(0)-y?(0)5
  ????(??)-??(0)6??(??)??(??)

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• We collect terms and we will get
• ?? 2 5??6??(??)????(0)5??(
  0) y ?(0)??(??)
• So, now when we want to get the transfer
  function, we take all the initial conditions as
  0. we take ??(0)0 and y?(0)0.

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• Here we see that the order of the denominator
  polynomial of the transfer function ??2 and
• the order of the numerator polynomial of the
  transfer function ??0.
• And we solve the denominator polynomial equal 0
  to get the poles
• 
  ?? 2 5??60
• 
  ?? 2 3?? 2?? 60
• 
  ?? ??3 2 ??3 0
• The poles are -2 and -3 and there are no zeroes
  for this problem

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• Now let us calculate the unit step response of
  the system.
• For a unit step response ??(??)1 and
  ??(??)1/??.
• We can write
• A, B, C are called the residues.

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If we take the inverse Laplace transform This
is the unit step response of the system.
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• what we observe from this unit step response.
• As,???8,??(??)?1/6, this is called as the steady
  state value.
• 2) We locate the poles -2 and -3 on the negative
  real axis.
• 3)What about BIBO stability here?
• Step is a bounded input, and the corresponding
  ??(??) that we have calculated is bounded.
• The observation we can make here is that the
  system is bounded input bounded output stable.

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Problem 2

• ??
  (??)??(??)??(??)
• take the Laplace transform on both sides
• Now in order to find the transfer function take
  all initial conditions to be 0
  P(S)
  
• 
  
• n is 2, second order system, m is 0. What about
  poles

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• solve for ?? 2 10
• s j
• we need to get the unit step response
• Find A ,B,C?
• https//www.chilimath.com/lessons/advanced-algebra
  /partial-fraction-decomposition/
• A1 ,B-1 and C 0

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• Apply inverse Laplace transform
• y(t)1-cos(t)
• So, what happens as t?8?
• the magnitude of y(t) remains bounded,
• we are getting a bounded output
• so from this would we conclude that the system is
  BIBO stable.

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Problem 3

• 
  ??(??)???(??)??(??)
• take the Laplace transform on both sides,
• Initial condition 0
• n 2, m 0. What are the poles? The poles are
  at 0 and 1
• Apply unit step input ??(??)1/?? to find unit
  step response

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• A,B,C ? Using partial fraction decomposition we
  get
• A-1, B1,C 1
• Taking inverse Laplace transform
• So the magnitude of y(t) as t tends to infinity,
  what happens to its magnitude
• It tends to infinity, because we have a t term.
• we have found a step input for which the output
  is unbounded

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Problem 4

• ??(??)???(??)-2??(??
  )??(??)
• take the Laplace transform on both sides and
  putting initial condition equal to zero we get
• n 2, m 0, the poles are at -2 and 1.

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unit step response ??(??)??(??)??(??)
Taking inverse Laplace transform
t?8, the magnitude tends to infinity because we
have ?? ?? , the magnitude of ?? ?? tends to
infinity
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Problem 5

• 
  ??(??)??(??)??(??)
• take the Laplace transform on both sides
• Now in order to find the transfer function take
  all initial conditions to be 0
  P(S)
  
• 
  
• n is 2, second order system, m is 0.
• ?? 2 10
• s j

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• And we saw that in problem 2 when it was
  subjected to a unit step input, the output was
  bounded in magnitude
• Now subject it to a cosine input of a very
  specific angular frequency i.e ??1 rad/s. If
  ??(??)cos??, then ??(??) ?? ?? 2 1 .
• Now, we use a property of Laplace transform, the
  complex differentiation theorem.

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• Taking inverse Laplace transform
• We observe that, as ???8,??(??)?8.

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Problem 6

• same as problem 3 but with a different input
  
  ??(??)???(??)??(??) , u(t) cos(t),
• poles at 0,-1
• ??1 rad/s. If ??(??)cos??, then ??(??) ??
  ?? 2 1

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• A,B,C?
• Solving the partial fractions, we get ??1/2,
  ??-1/2 and ??1/2.
• Putting the values we get
• Taking inverse Laplace, we have
• Earlier we saw that, for the same system when we
  gave the unit step response, ??(??)?8.
• Now we gave another bounded input cos??, we are
  getting a bounded output.

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BIBO Stability

• A system is said to be BIBO stable, if the output
  remains bounded in magnitude for all time given
  any bounded input
• We will look into the location of poles and its
  influence on the stability of the system

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Conclusion
Problem System Poles Input output
Problem No 2 ??(??)??(??)??(??) ?? Unit Bounded as t?8
Problem No5 ??(??)??(??)??(??) ?? cost Unbounded as t?8
Problem No 3 ??(??)???(??)??(??) O , -1 unit Unbounded as
Problem No 6 ??(??)???(??)??(??) O , -1 cost Bounded as
t?8
t?8
if we use the definition of BIBO stability which
states that a system is BIBO stable if its output
is bounded for all possible bounded inputs, we
should brand these two systems as unstable
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conclusion

• For BIBO stability, all poles of the plant
  transfer function must lie in the LHP or in
  other words having negative real parts.
• 2. If there exists even one pole of the plant
  transfer function in the right half
• plane, that is have has a positive real
  part, then the plant or the system is
• not BIBO stable.
• 3. If there are repeating poles of the plant
  transfer function on the imaginary
• axis (j omega axis) with all remaining
  poles in the LHP, then system is not
• BIBO stable.
• 4. If there are non repeating poles of the plant
  transfer function on the imaginary
• axis with all remaining poles in the LHP
  then the system is unstable or critically
• stable.

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Effect of zeros

• We identified the fact that the location of the
  poles influences the stability of the system.
• And for BIBO stability all poles of the system
  transfer function should lie in left half complex
  plane.
• In other words, they should have negative real
  parts.
• Now to find out the impact of zeros on the
  system response, Let us consider two systems

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Problem 1
The first plant transfer function is The order
of these systems is 2 because the order of the
denominator polynomial is always going to be the
order of the system for the class of systems that
we study. For the first system, n2 and m0 . We
want to figure out the unit step response.
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• Evaluating the partial fractions, we get
• take the inverse Laplace transform, we get
• We observe that the two poles are in the left of
  complex plane. So the system is BIBO stable
• As t?8, y (t)? 1/10

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Problem 2

• n2 and m1
• The poles are at -1 and -10 there is a zero at
  -2.
• The difference between system 1 and system 2 is,
  we have introduced a zero at -2.
• We want to figure out the unit step response. U
  (s)1/s
• Evaluating the partial fractions, we get

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Evaluating the partial fractions, we get If
we take the inverse Laplace transform, we
get we can see that as t?8, y (t)?1/5 So, it
is bounded, the system is BIBO stable because the
poles are in the left half plane.
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• We observe that the zero is going to affect the
  dynamic response of the system but it is not
  going to have a direct impact on the stability.
  Because stability is dependent on the location of
  the poles.
• Zeros will affect the coefficients of the
  response function

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• Problem 4