CS代写 Grandfather of control and telecommunications

Grandfather of control and telecommunications
Two Flavours of Stability Routh-

Motivation

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 Control systems must be designed to meet performance criteria – e.g. step response specs.
 If the control system will be operated over a long period of time, stability is usually also necessary to be able to meet these criteria
 We present two definitions of stability.
 For LTI systems, both conditions turn out to be very closely related

Stability w.r.t. Initial Conditions
𝐺𝐺𝑠𝑠 =𝐵𝐵(𝑠𝑠)
Arbitrary initial conditions
u(t)=0 Free response y(t)0?
If A(s) has order n, then initial conditions are 𝑑𝑑𝑗𝑗𝑦𝑦 (0 ), 𝑗𝑗 = 0,…,𝑛𝑛 − 1.
Say A(s), B(s) are given polynomials, e.g. from a diff. eq. model

Definition (input=zero)
 Stable if output converges to zero for any initial condition.
 Unstable if output diverges for at least some initial conditions
 Marginally stable if output stays bounded from all initial conditions and may converge to zero for some initial conditions.

Visualisation
Equilibrium
Stable Marginally Stable Unstable

Free Response
 Free response
∑𝑛𝑛 𝑎𝑎𝑑𝑑𝑖𝑖𝑦𝑦 =∑𝑚𝑚 𝑏𝑏𝑑𝑑𝑖𝑖𝑢𝑢=0,t≥0 𝑖𝑖=0 𝑖𝑖 𝑑𝑑𝑡𝑡𝑖𝑖 𝑖𝑖=0 𝑖𝑖 𝑑𝑑𝑡𝑡𝑖𝑖
⟺∑𝑛𝑛 𝑎𝑎 𝑠𝑠𝑖𝑖𝑌𝑌𝑠𝑠−∑𝑖𝑖−1𝑠𝑠𝑗𝑗𝑑𝑑𝑗𝑗𝑦𝑦0− =0 𝑖𝑖=0 𝑖𝑖 𝑑𝑑𝑗𝑗𝑦𝑦 𝑗𝑗=0 𝑑𝑑𝑡𝑡𝑗𝑗
∑𝑛𝑛−1𝑠𝑠𝑗𝑗 𝑑𝑑𝑡𝑡 0− ∑𝑛𝑛 𝑎𝑎 �=poly(𝑛𝑛−1) ⟺𝑌𝑌𝑠𝑠= 𝑗𝑗=0 𝑗𝑗 𝑖𝑖=𝑗𝑗+1 𝑖𝑖
∑𝑛𝑛 𝑎𝑎𝑖𝑖𝑠𝑠𝑖𝑖� = 𝐴𝐴(𝑠𝑠) 𝑖𝑖=0

LTI Stability w.r.t. Initial Conditions
All poles of A(s) strictly in the left-half
plane (LHP).
 Marginally stable:
Non-repeated pole(s) on imaginary
axis, all other poles strictly in the LHP
 Unstable:
Some pole(s) strictly in the right-half plane
OR repeated poles on the imaginary axis
You can understand this if you remember how terms in the partial fraction expansio affect the impulse response.

Graphical Interpretation of Stability
Complex domain (pole locations)
Brown colour represents a “forbidden (unstable) region”.

Bounded-Input Bounded-Output (BIBO) Stability
System Zero initial conditions
Bounded input Bounded output?

Signals as functions of time
delta function or impulse δ(t)
We measure the size of signal u using:

BIBO Stability Definition (zero initial conditions)
 Formally: For any 𝑢𝑢� > 0, there exists a finite 𝑦𝑦� > 0 such that for any input u with u(t) ≤ 𝑢𝑢�,
𝑦𝑦𝑡𝑡 ≤𝑦𝑦�, forall𝑡𝑡≥0.
 Informally: Bounded inputs yield bounded outputs.

BIBO stability for LTI systems
 BIBO stability holds if and only if (iff) the impulse response g is absolutely integrable,
 Proof of sufficiency ():
�∞𝑔𝑔𝑡𝑡 𝑑𝑑𝑡𝑡<∞ 0 BIBO Stability  Proof of necessity (): By contradiction. Suppose∫0∞𝑔𝑔𝑡𝑡 𝑑𝑑𝑡𝑡=lim∫0𝑇𝑇𝑔𝑔𝑡𝑡 𝑑𝑑𝑡𝑡=∞. For an arbitrary T>0, apply an input
𝑢𝑢 𝑡𝑡 =�𝑢𝑢.sign 𝑔𝑔(𝑇𝑇−𝑡𝑡) , 0≤𝑡𝑡≤𝑇𝑇,
0, else. ⇒ 𝑦𝑦 𝑇𝑇 = �0 𝑇𝑇 𝑔𝑔 𝑡𝑡 𝑢𝑢 𝑇𝑇 − 𝑡𝑡 𝑑𝑑 𝑡𝑡
= ∫0𝑇𝑇 𝑔𝑔(𝑡𝑡)𝑢𝑢. sign 𝑔𝑔(𝑡𝑡) 𝑑𝑑𝑡𝑡
= 𝑢𝑢 . ∫0𝑇𝑇 | 𝑔𝑔 𝑡𝑡 | 𝑑𝑑 𝑡𝑡 ⟶ ∞ . Q E D

BIBO for LTI Systems with Rational Transfer Functions
 BIBO stability iff
 Transfer function is proper (why?), and
 All poles lie strictly in the LHP (why?).
 If no unstable pole-zero cancellations in B(s)/A(s), BIBO stability  Stability w.r.t. initial conditions
(but the reverse implication is not true)
 If no unstable pole-zero cancellations, and B(s)/A(s) is proper
BIBO stability  Stability w.r.t. initial conditions

Why pay heed to a pole if cancelled by a zero anyway?
 Pole-zero cancellation said to occur if there is a zero 𝛽𝛽 which equals a pole 𝛼𝛼 .
 In this case, the factor (𝑠𝑠 − 𝛽𝛽𝑘𝑘) in the
(𝑠𝑠 − 𝛼𝛼 ) in the denominator A(s), and both 𝑗𝑗
numerator B(s) formally cancels out the factor
factors disappear from the transfer function
G(s) = B(s)/A(s).
 So why do we care…?

Why Cancelled Poles Still Count
 A cancelled pole 𝛼𝛼𝑗𝑗 is no problem if initial conditions are zero. But in practice, initial
conditions may not be exactly zero.
 Recall y(t) = free response (due to nonzero initial conditions) + forced response (due to nonzero
 Cancelled pole 𝛼𝛼 appears in the free response!
If𝛼𝛼 unstableTroubleast∞ 𝑗𝑗
Could affect transient performance if too slow.

Example of How Pole-Zero Cancellations Arise
Consider a DE system 𝑦𝑦̈−𝑦𝑦=𝑢𝑢̇−𝑢𝑢,where𝑢𝑢𝑡𝑡 =0,𝑡𝑡<0. (𝑠𝑠2−1)𝑌𝑌(𝑠𝑠)−(𝑠𝑠𝑦𝑦̇(0−)+𝑦𝑦(0−))= 𝑠𝑠−1 𝑈𝑈 𝑠𝑠 𝑌𝑌 𝑠𝑠 = 1 𝑈𝑈 𝑠𝑠 + 𝑠𝑠𝑦𝑦̇(0−)+𝑦𝑦(0−) 𝑠𝑠+1 (𝑠𝑠+1)(𝑠𝑠−1) System is BIBO stable, but not stable w.r.t. initial conditions Example of System That’s Stable w.r.t. Initial Cond’s, but isn’t BIBO stable Determining stability for LTI systems  Consider a transfer function  Poles are the solutions of either equation: The second polynomial is “monic” and often used in tests. Radical Facts  A quadratic polynomial  Roots of cubic and quartic polynomials can be found in terms of 3rd and 4th roots (radicals). But the formulas are impossible to remember  Galois (1811-32): Polynomial equations of order 5 or higher cannot generally be expressed in terms of radicals! So how to manually check stability for systems of order>2 ?

A necessary condition for stability
 If a monic polynomial
has all roots with negative real parts, then all
its coefficients are positive; that is
 If a coefficient is negative or zero, then some roots have positive or zero real part.
 This is also sufficient condition for first and second order polynomials.

 Recall that an arbitrary monic polynomial in s with real coefficients can always be written as a product of monic linear and quadratic terms with real coefficients:
 Both linear and quadratic terms have all roots with strictly negative real parts iff all their coefficients are positive.

 If some coefficient in the characteristic polynomial is negative or zero, then there are some poles with positive/zero real part.
 If all coefficients in the characteristic polynomial are all positive, then use Routh- Hurwitz criterion to check whether there are any poles with positive/zero real part.

 Consider
This polynomial has all positive coefficients but two roots on the imaginary axis:

Routh-Hurwitz test

 For polynomial with positive coefficients: we form the following (Routh) array:

 Example of computing entries in the third row

 Example of computing entries in the third row

Routh-Hurwitz test
 The polynomial with positive coefficients
has all roots with strictly negative real parts if and only if all elements in the first column of the Routh array do not change their sign.
 No need to solve for the roots!

 For a first order polynomial Routh array is

 For a quadratic polynomial Routh array is

 For third order polynomials Routh array is

 For fourth order polynomials we have:

 Consider a feedback system with
 The characteristic polynomial is
 Find all values of K for which the system is stable.

 We use formulas for 3rd order polynomials:
 The system is stable if

 Consider the same system.
 For which values of K would all roots of the characteristic polynomial have real parts strictly smaller than -1?
 We modify the polynomial by introducing

 Routh array is
 Hence, we have the conditions: or

 Form the Routh array for the following polynomial:

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