APPPHYS 217 Tuesday 6 April 2010
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1 APPPHYS 7 Tuesday 6 April Stability and input-output performance: second-order systems Here we present a detailed example to draw connections between today s topics and our prior review of linear algebra and ODE s Of course this "example" of second-order systems is actually an important topic in and of itself as one often gains intuition about more complex systems by noting similarities to and differences from this canonical class of models The equation of motion we want to consider is q q q u where u is an input signal and are parameters of the model This equation of motion could represent for example a mass-spring-damper system for which F ma takes the form mẍ kx bẋ F ext ẍ m b ẋ m k x m F ext k m b 4km u m F ext where m is the mass k is the spring constant and b is the damping coefficient The same form of equation of motion arises for a series LCR circuit (inductor-capacitor oscillator with resistance) V C Q C V I L di LQ V dt R IR Q R V ext LQ Q R C Q Q R L Q LC Q L V ext LC R C 4L u L V ext where L is the inductance C the capacitance and R the resistance Here Q represents the charge on the capacitor General solution of the initial value problem for second-order systems We have already seen how to put this equation of motion into state-space form: q q q u x q x q d dt x x x x For the moment let s set u to zero and make sure we understand the behavior of the undriven system In this case we are looking at ẋ Ax with A the square matrix in the equation above and we know that we can solve all initial value problems by matrix exponentiation This in turn tells us that we should look at the eigenvalues of A and check whether it is diagonalizable Once we find the matrix exponential we can use it to find analytic expressions for the impulse step and frequency response We ll do the u
2 step response here in class today; the impulse and frequency responses are left as exercises Computing the eigenvalues of A for arbitrary and wefind deta I 4 4 Assuming and we can then identify three distinct "cases" for : : C : : R Traditionally one refers to the case as being underdamped as critically damped and as overdamped Let s try to figure out why The easiest case to analyze is (the undamped case) Then A i i i i so we have P i i P i i expat P e i t e i t P e it eit i e it i e it i e it i eit e it eit cos t sin t sin t cos t We see from this that for any non-zero initial condition the integrated trajectory will consist of undamped oscillations If we can still find general expressions for the eigenvectors In unnormalized form These two eigenvectors are independent as long as so let s assume that for
3 now Then expat P exp t exp t P E E E E E E where E exp t exp t E exp t exp t Note that if then i i so E e t e it e it ie t sint E e t cost expat e t sint cost sint sint sint cost This shows us that for arbitrary initial conditions the integrated trajectories look like exponentially damped oscillations Likewise if then so E e t e t e t e t sinht E e t cosht expat e t sinht cosht sinht sinht sinht cosht Taking into account the fact that we see that the integrated trajectories are exponentially damped without any oscillating factors Finally we consider the case Now A Looking at the eigenvalue equation A Ix x x we see that the only solutions are of the form x x and are thus all linearly 3
4 dependent (correspond to a single eigenvector) Hence we are in the case where A is not diagonalizable As mentioned in previous lecture notes however we can find a decomposition into Jordan form: A TJT where J S N and clearly SN since S is proportional to the identity Then expat TexpJtT TexpStexpNtT e t e t t e t t e t e t e t t e t te t e t t e t Hence we see that the integrated trajectories will contain exponentially damped terms as well as terms that intially t / grow linearly with time but then are dominanted by the e t factorsaswell Step response of second-order systems Our solutions to the initial value problem can directly be used to solve for the step response once we introduce a small trick Remember that the step response is defined as the system output corresponding to an input signal of the form ut t ut u t assuming the state variables are at equilibrium prior to t Looking at the second-order equation of motion in its original form q q q u we see that qq is the desired equilibrium for t and thus that the step response of the state space variables simply corresponds to an initial value problem q q q u q q If we can solve for qt and q t we can use them to compute any desired output signal 4
5 qt yt C q t At this point we note that the initial value problem we want to consider can be transformed by a simple change of variables q q q u q q q q into q q q u q q q u q q q with initial condition q u q Hence we can simply use the solutions we derived above for u withaninitial condition corresponding to the desired step size and then transform the integrated solution according to qt q t u q t q t In the undamped case we have so q t q t cos t sin t sin t cos t u qt u cos t q t sin t To help visualize this we can make a plot for u : u u 3 cos t sin t 5
6 u(t) in red q(t) in black t/ Note the serious overshoot and lack of any settling whatsoever! In the underdamped case we have q t q t u e t sint cost sint so qt q t u e t sin t cos t e t sin t For an example to plot we can choose u and take /3 : 6
7 u(t) in red q(t) in black t/ Here the overshoot is greatly reduced and the system settles within a time 4/ Next the overdamped case q t q t u e t sinht cosht sinht so qt q t u e t sinh t cosh t e t sinh t Again taking u and now 4/3 we can plot: 7
8 u(t) in red q(t) in black t/ Here there is no overshoot at all and settling takes 3/ Finally we consider the critically damped case qt u q t which for u looks like e t t e t e t e t t e t u(t) in red q(t) in black t/ Here again there is no overshoot and settling occurs within / Putting our cases together into one plot: 8
9 5 u(t) red =/3 green = blue =4/3 black t/ Exercise : How does the rise time vary with? Consider all values of Exercise : How does the settling time vary with? Consider all values of Exercise 3: Consider the following second-order control system: d dt q q q q q y q where ut is a scalar input signal and yt is a scalar output signal For a feedback law of the form ut Kyt explain the dependence of the closed-loop step response on the gain K for K u Transient versus steady-state response Looking at the general solution of a driven linear system that we discussed last week xt expatx t ds expat sb s we can now garner some insight as to transient versus steady-state response For our second-order systems with (ie any case but the completely undamped case) we know that expa is exponentially damping on long timescales Hence for t the initial conditions will be "forgotten" (will no longer make a significant contribution to xt) The driving input signal b s at early times such that t s will also be forgotten Hence if b t is a repetitive signal we can expect that xt settles into a steady-state regime in which x is forgotten and it doesn t matter exactly how b was 9
10 "switched on"
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