# STABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse

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1 SIGNALS AND SYSTEMS: PAPER 3C1 HANDOUT 4. Dr David Corrigan 1. Electronic and Electrical Engineering Dept. STABILITY Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse responses Use of the Laplace Transform may be summarized through Transfer functions. The contents of the blocks in the block diagrams are Transfer functions. Looked at alternative system analysis in using convolution. Found that it may sometimes be more complicated than Laplace analysis, however does yield useful insights to system behaviour. Now we need to assess the stability of systems. In general, only stable systems are useful to us. System stability can be assessed in both s-plane and in the time domain (using the system impulse response). This handout will 1. Define asymptotic and Bounded-input, Bounded-output (BIBO) stability. Then examine equivalence for Linear Systems. 2. Relate system stability to poles of transfer function. 1 This handout is based on the set of notes produced by Prof. Anil Kokaram 3C1 Signals and Systems 1

2 1 THE NEED FOR STABILITY 3. Relate transient response to poles of transfer function. 4. Show how the Final Value Theorem can be used to determine the steady state response of the system. 1 The need for stability Some questions may occur to you: Who cares if a system is unstable? Is being unstable bad? Why is being stable good? What s the big deal? What does it mean for a system to be stable anyway? Consider a Hovering Harrier. The pilot switches on hover mode. He expects the control system to adjust the lift surfaces and the directional thrust jets so that the aircraft remains in the same place. Why is this so difficult? Its because without the engine thrust, the vehicle will fall. Because of wind gusts you have to compensate for the applied force to keep the aircraft in the same place. If the system is not designed correctly, the counter thrust could overcompensate and the harrier may never return to the equilibrium position. Such a system is unstable, and when designing systems we have to ensure that the system is stable. In this case stability is the ability to return to the equilibrium position regardless of external noises. It ll be pretty sad if testing a system for stability implies you have to test it with all possible inputs to see if it performs as you expect or not. Thankfully, the unifying idea in the LTI systems above is that when a stable system is in some steady state and you kick it (within some reasonable bounds), it returns to a steady state after a little while. The response of a system to a kick (an impulse) is its impulse response. Stabilty can therefore be determined by a system impulse response. 3C1 Signals and Systems 2

3 Asymptotic Stability 1 THE NEED FOR STABILITY Definition: A linear system is asymptotically stable if its impulse response h(t) satisfies the condition where B is some positive number. Examples: h(t) dt = B < (1) 1. LCR circuit: h(t) = e 2t sin(3t + 4) 2. Delay line with lossy reflections: 1 h(t) = 2kδ(t kt ) k= In both these cases h(t) is bounded all the time and always decays to zero. 3C1 Signals and Systems 3

4 2 BIBO 2 Bounded-Input Bounded-Output (BIBO) Stability Definition: A linear system is BIBO stable if there is a positive number B such that, for any bounded input signal x(t), x(t) < X, the resulting output signal y(t) is bounded by: y(t) < XB. Theorem: Asymptotic and BIBO stability are equivalent for linear systems. Note: They are not equivalent for non-linear systems. 3C1 Signals and Systems 4

5 3 MARGINAL STABILITY 3 Marginal Stability Definition: A linear system is marginally stable if it is not asymptotically stable and one can find A, B < such that T h(t) dt < A + BT for all T 3.1 Examples: 1. Integrator T h(t) = u(t) h(t) dt = 2. Delay line with lossless reflections h(t) = δ(t k) T h(t) dt = k= T T 1dt = T δ(t k)dt = floor(t ) + 1 k= < T + 2 (2) 3. Undamped second order system h(t) = cos(3t) 3C1 Signals and Systems 5

6 4 INSTABILITY 4 Instability Definition: A system is unstable if it is neither asymptotically stable nor marginally stable. 4.1 Examples: 1. Inverted Pendulum: h(t) = e 4t + e 4t 2. Two integrators in series: h(t) = t 3. Unstable oscillator: h(t) = e.1t sin(.3t) Warning: Different people use different definitions of stability. In particular, systems which we have defined to be marginally stable would be regarded as stable by some and unstable by others. For this reason we avoid the term stable without qualification. 3C1 Signals and Systems 6

7 4.1 Examples: 4 INSTABILITY Poles and Zeros The zeros of a transfer function G(s) are those values of s at which G(s) becomes zero, and its poles are those values of s at which G(s) becomes infinite. If a transfer function is rational, i.e. it can be written as the ratio of two polynomials G(s) = n(s) d(s) then the zeros of G(s) are the roots of the numerator polynomial n(s), and its poles are the roots of the denominator polynomial d(s). d(s) is also known as the characteristic polynomial. For physically realisable systems: deg[n(s)] deg[d(s)] The degree of the denominator polynomial is known as the Order of the system. The poles of a transfer function are also called the characteristic roots or auxiliary roots. We can also speak of the poles and zeros of the Laplace transform of a signal. 3C1 Signals and Systems 7

8 4.1 Examples: 4 INSTABILITY s-plane surface for G(s) = (s+1.5)(s2 +s+1) (s+2)(s 2 +.1s+4) Re Im C1 Signals and Systems 8

9 5 POLES AND STABILITY 5 Poles and Stability Theorem: The transfer function of an asymptotically stable system cannot have any poles in the right half of the complex plane or on the imaginary axis. Proof A rational transfer function can always be expressed as: N r i H(s) = r + (s p i ) v i (Usually the poles (i.e. the values p i ) are distinct so that v i = 1). i=1 To work out the impulse response note from tables that L { t } v i 1 = (v i 1)! s v i { } t v i 1 L = 1 (v i 1)! s v i Using shift theorem { t v } i 1 1 L exp(p i t) = (v i 1)! (s p i ) v i (3) Hence the impulse response: N h(t) = i=1 r i t (v i 1) (v i 1)! ep it 3C1 Signals and Systems 9

10 5.1 Stable Systems 5 POLES AND STABILITY 5.1 Stable Systems By definition, for a system to be stable h(t) dt <. Since h(t) is the sum of a number of terms, a system will be stable if and only if each individual term of the sum corresponds to the impulse response of a stable system. Lets consider a single term of this sum: t (v i 1) h i (t) = r i (v i 1)! ep it Let p i = σ i + jω i. Then e p it = e σ it e jω it e p it = e σ it h i (t) = r i (v i 1)! tv i 1 e σ it If σ i < then h i (t) as t. (4) So h(t) returns to eventually if the pole has a negative real part, but remember this is not sufficient to prove that is stable. We need to show that h i (t) dt <. if σ i < : h i (t) dt = r i (v i 1)! tv i 1 e σ it dt Evaluating this integral is quite tricky because you have to do intergration by parts v i 1 times. It is possible to evaluate the integral using the following trick. 3C1 Signals and Systems 1

11 5.1 Stable Systems 5 POLES AND STABILITY h i (t) dt = h i (t) e st dt = L { h i (t) }. s= s= Hence { } ri L { h i (t) } = L (v i 1)! tvi 1 e σ it = r i (v i 1)! L { t vi 1 e } σ it from tables L { t } v i 1 = (v i 1)! s v i Shift Theorem L { t vi 1 e } σ it = (v i 1)! (s σ i ) v i r i L { h i (t) } = (s σ i ) v i and therefore h i (t) dt = h i (t) e st dt { = L h i (t) } s= r i = (s σ i ) v i = r i ( σ i ) v i < s= s= Hence h i (t) is stable if the real part of its pole is negative and hence a system will be stable if all its poles have negative real parts. 3C1 Signals and Systems 11

12 5.2 Marginal Stability and Instability 5 POLES AND STABILITY 5.2 Marginal Stability and Instability If σ i = and v i = 1, then from Equation (4): Hence: but: T h i (t) = r i (a constant) h i (t) dt = h i (t) dt r i T So the system exhibits marginal stability. Therefore it is marginally stable if there is a number distinct poles on the imaginary axis However, this does not apply if we have repeated poles on the imaginary axis. ie. If σ i = and v i > 1: T and the system is unstable. h i (t) dt = r i T v i v i! If σ i > then the system is also unstable. 3C1 Signals and Systems 12

13 5.2 Marginal Stability and Instability 5 POLES AND STABILITY Stability Theorem 1. A system is asymptotically stable if all its poles have negative real parts. 2. A system is unstable if any pole has a positive real part, or if there are repeated poles on the imaginary axis. 3. A system is marginally stable if all the poles on the imaginary axis are distinct, and all the remaining poles have negative real parts. 3C1 Signals and Systems 13

14 6 Poles and transient responses 6 POLES AND TRANSIENT RESPONSES Impulse response: h(t) = r δ(t) + N i=1 r i t (v i 1) (v i 1)! ep it Consider e p it. 6.1 Real Poles p i real: real exponential with time constant 1/p i. In other words the decay rate is propostional to the absolute value of the pole. The nearer it is on the real axis to the origin the longer the impulse response will take to decay. 6.2 Complex Poles p i complex: always has complex conjugate pole p i. These combine to give a damped or growing sinusoid. Assume v i = 1, then for a second order system (i = 2): { } Ae h(t) = L 1 jϕ + Ae jϕ s p i s p = Ae jϕ e pit + Ae jϕ e p i t i Note: For a real second order system the numerator coefficients are also complex conjugates. Recall that we defined p i = σ i + jω i { } h(t) = Ae σ it e j(ωit+ϕ) + e j(ω it+ϕ) h(t) = 2Ae σ it cos(ω i t + ϕ) 3C1 Signals and Systems 14

15 6.2 Complex Poles 6 POLES AND TRANSIENT RESPONSES The system pole locations are related to various impulse responses below. This is an important figure. Note: Like the first order system, the real part of the pole, σ i determines stability and the time constant, 1/σ i. The imaginary part ω i determines the the oscillation frequency ω i (rad/sec) x x x x 15 You can interactively explore the effect of pole and zero positions on impulse responses at signals/explore/index.html 3C1 Signals and Systems 15

16 6.3 Standard Form of a 2 nd Order System 6 POLES AND TRANSIENT RESPONSES 6.3 Standard Form of a 2 nd Order System The standard form of a second order system (with no zeros) is 2 H(s) = ω2 n + 2ζω n s + ωn 2 s (ζ: damping factor, ω n : natural frequency (ie. oscillation frequency when there is no damping).) we see: 2Re{p i } = 2ζω n and p i 2 = ω 2 n = σ 2 i + ω 2 i Hence: ζ = Re{p i} p i and ω n = p i Im Re Figure 1: Second order system showing relationship of pole locations to ζ, ω n. 3C1 Signals and Systems 16

17 6.3 Standard Form of a 2 nd Order System 6 POLES AND TRANSIENT RESPONSES Observations As we have seen previously the frequency of oscillation is equal to the value of the imaginary part of the pole. This is not the same thing as the natural frequency which is equal to the magnitude of the pole. By definition, these two quantities are only equivalent when ζ =. In general, increasing the natural frequency (for constant ζ) increases the frequency of oscillation, but also increases the decay rate. Intuitively, if the damping ratio is increased the oscillations will decay faster. However, increasing the damping ratio (for fixed ω n ) also reduces the frequency of oscillation. A system is UNDER-DAMPED if ζ < 1. There are two complex poles in this case. A system is CRITICALLY DAMPED if ζ = 1. There are two real poles at the same point on the real axis. There is no oscillatory behaviour. A syetem is OVER-DAMPED if ζ > 1. The system has two distinct real poles and behaves like a cascade of first order systems. 3C1 Signals and Systems 17

18 7 FINAL VALUE 7 A last word on time domain behaviour: THE FINAL VALUE THEOREM Transient system behaviour is only part of the story as far as time domain system behviour is concerned. Of equal interest is finding out the final value of a signal when the system has settled down to steady state behaviour. The final value theorem is a simple mechanism for using the Laplace Transform of a signal to predict its final value as t. Given some signal f(t), the final value theorem relates the steady state behaviour f(t) to the behaviour of F(s) in the neighbourhood of s =. It states that lim f(t) = lim sf (s) (5) t s The conditions that need to be obeyed for this theorem to be successfully applied are as follows. 1. lim t f(t) exists. Which just means that f(t) does indeed converge to some definite value as t. 2. All poles of sf(s) are in the left half plane. Note that we are talking here about s F(s) not just F(s). 3. sf(s) has no poles on the imaginary axis. Note that we are talking here about s F(s) not just F(s). Proof Recall that (from tables for instance) { } d L dt f(t) = sf(s) f() 3C1 Signals and Systems 18

19 7.1 How to use it 7 FINAL VALUE We want to examine what happens when this first differential tends to zero, since then that would mean that the signal is no longer varying with time. Consider ( ) d lim sf(s) f() = lim s s dt f(t) e st dt ( ) d = dt f(t) lim s e st dt ( ) d = dt f(t) dt [ ] = f(t) = lim t f(t) f() We can cancel f() as it appears on both sides and we are left with lim sf(s) = lim f(t) s t 7.1 How is this theorem actually used? Well, its like this. The theorem relates the steady state value of any signal to the behaviour of sf(s) near s =. So if you are asked for instance, what is the steady state value of the step response of some system, you first have to calculate the Laplace transform of that step response. This would mean calculating the product of the system transfer function with 1/s, or integrating the time domain impulse response and take the Laplace Transform. Then you can apply the theorem to work out what the final value of the step response is. Alternatively, you can forget about the theorem and hope that you can always spot the final value of a time domain signal by manipu- 3C1 Signals and Systems 19

20 7.1 How to use it 7 FINAL VALUE lating it so that you can work out lim t f(t). For the step response example, that would mean working out the time domain signal output by taking the inverse laplace transform. Then trying to re-write the signal expression to handle t for all the terms in t. Sadly, it is not always easy to massage the final value out of a time domain expression. Hence you tend to have to know the final value theorem. Example Find the steady state response of a system with transfer function H(s) = 5s 2 + 7s s 3 + 4s 2 + 7s + 6 when the input signal x(t) is the ramp function r(t). X(s)= L {R(s)} = 1 s 2 Applying the Final Value Theorem we get lim y(t)= lim sy (s) = lim t s = lim s s. 1 s 2 sx(s)h(s) s 5s 2 + 7s s 3 + 4s 2 + 7s + 6 = lim s 5s + 7 s 3 + 4s 2 + 7s + 6 = 7 6 3C1 Signals and Systems 2

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