Introduction to Sequences and Difference Equations

Transcription

1 6.30.0x, January 016 Notes Set 1 1 MASSACHVSETTS INSTITVTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science x Introduction to Feedback System Design January 016 Notes Set 1 Introduction to Sequences and Difference Equations Questions are interspersed with these notes, they are intended to help you test your understanding. We experience a world that evolves in continuous time, full of physical objects whose trajectories are accurately modeled with differential equations. We drop a ball, and gravity accelerates it. The ball s acceleration is the time derivative of the its velocity, and its velocity is the time derivative of its position. But from the perspective of a computer, the continuous-time world must be reconstructed from a series of samples collected at discrete time-points. A falling ball s velocity must be approximated from differences in successive position samples, its acceleration approximated from the change in position differences from one pair of samples to the next. One might wonder if the computer s discrete-time perspective is more accurate. Perhaps continuous time is just a convenient model for what is actually discrete, just as Newtonian mechanics is a convenient model for what is actually quantized. But such musings are best left to scientists and philosophers. As engineers, we have work to do. Computers, in their many forms (digital signal processors, microprocessors, micro-controllers, etc), are powerful tools that are pervasive in engineering design. We can work with them more effectively if we understand their perspective. On the computer, it is more natural to think in terms of sequences. Derivatives in time become differences between samples, integrals become sums, and physical system trajectories governed by differential equations become sample sequences governed by difference equations. Adopting this perspective from the beginning can help us design better systems. And as an added benefit, you can study this subject even if you have not yet studied calculus or differential equations. In this first set of notes, we will use a number of examples to introduce: 1. our notation for sequences and linear difference equations (LDEs),. how to model a physical problem using LDEs, 3. how to solve LDEs by computing natural frequencies, 4. some important properties of complex numbers. This class focuses on control problems that can be modeled with linear difference equations (LDE s), a surprisingly rich topic both theoretically and practically. And in this first set of notes, we study homogenous LDE s, LDE s with initial conditions but no inputs, so that we can emphasize three central concepts: LDE s have natural frequencies, If an LDE has a natural frequency with magnitude greater than one, the LDE is unstable. Sequences associated unstable LDEs can grow exponentially.

2 6.30.0x, January 016 Notes Set 1 Sequences For us, sequences can be considered a set of values indexed by an integer. Such sequences are useful for representing discrete-time data in variety of applications, including digital audio, stock prices, bank balances, and discrete-time control. For example, the Arduino-based controller for your copter-levitated arm has three sequences: two sequences of samples (the analog inputs from the propellor-motor and angle-sensor) and a sequence of outputs (the analog output to the motor driver transistor). We will use, and translate between, three representations for sequences: explicit enumeration, closed-form expression, and implicit definition using difference equations and initial conditions. 1 As an example, let y be the sequence of powers of 1. The sequence can be represented by explicit enumeration {y[0] = 1, y[1] = 1, y[] = 1,...}, (1) 4 or by closed-form expression, ( ) 1 n y[n] =, n 0, () or with a difference equation and an initial condition, y[n] = 1 y[n 1], y[0] = 0. (3) Note that we have chosen one of the standard notations for indicating sequence values, though our choice is far from universal. Throughout this course we will use the notation that indices for sequence values appear inside a pair of brackets. In some cases, it can be clearer to graph short explicitly-enumerated sequences using lollypop or stem plots. For example, below is a stem plot (created using MATLAB) for 00 samples of a sequence generated by roughly twenty milliseconds of a sound sampled eight thousand times a second. As must be obvious from the plot, stem plots can be cumbersome for long explicitlyenumerated sequences. They are not used, for example, in the software that plots real-time results from the copter-levitated arm. Linear Difference Equations As noted above, sequences can be represented implicitly, where an equation is given that relates one element of a sequence to other elements. For example, the elements of a power sequence satisfy y[n] = α 1 y[n 1] or y[n] α 1 y[n 1] = 0. (4) Implicit representations of the above form are referred to homogenous first-order linear difference equations (LDEs): difference equation because we are computing differences between sequence values, linear because we are using linearly scaled sequence values, first-order because only one difference is being computed, and homogenous because the difference equation is equivalent to insisting that a particular weighted combination of adjacent sequence values must sum to zero (as shown above). 1 Difference equations are a special case of the recurrances commonly studied in discrete mathematics.

3 6.30.0x, January 016 Notes Set 1 3 Figure 1: Stem plot of 00 samples of a sound (roughly 0 milliseconds) Note that a first-order LDE does not completely specify a sequence, one must also specify the value of y[n] for one value of n. If one specifies y[0], as is most common, then y[0] is referred to as an initial condition for the first-order LDE. As an example, if y[0] = 5 and α 1 = 1 then the solution to the homogenous first-order LDE is ( ) 1 n y[n] = 5, (5) but if y[0] = 3, then the solution to the same LDE is ( ) 1 n y[n] = 3. (6) Another well-know sequence with a homogenous LDE representation is the Fibonacci sequence, in which the next element in the sequence is the sum of the previous two. That is, y[n] = y[n 1] + y[n ] or y[n] y[n 1] y[n ] = 0. (7) The LDE for the Fibonacci sequence has two differences, so we refer to its associated LDE as secondorder. And to completely specify the sequence generated by a second-order LDE, it is necessary to provide two initial values. To generate the classical Fibonacci numbers, set y[0] = 0 and y[1] = 1. Then, using the LDE, it is easy to explicitly enumerate the rest of the sequence values, {y[] = 1, y[3] =, y[4] = 3,...}. (8) In general, an LDE, together with initial conditions, defines an infinite sequence of values, y[0], y[1], y[],.... We say an LDE is kth order when y[n] depends only on the k previous values of the sequence y.

4 6.30.0x, January 016 Notes Set 1 4 That is, a kth order homogenous LDE has the general form i=k y[n] = α 1 y[n 1] + α y[n ] +..., +α k y[n k] = α i y[n i], (9) where α 1,..., α k are LDE-specific constants. The above LDE is referred to as homogenous because we can rewrite the equation as i=k y[n] α 1 y[n 1] α y[n ]..., α k y[n k] = y[n] α i y[n i] = 0. (10) This rewrite means the LDE can be interpreted as requiring that a particular weighted combination of adjacent sequence values sum to zero, and therefore the sequence of all zeros will always satisfy the LDE (but not necessarily the initial conditions). Or said another way, homogeneity means that if y[n 1], y[n ],... y[n k] are zero, then so it y[n]. When we add inputs, later, they will replace the zero on the right side of the rewritten homogenous LDE. As we will see below, it is easy to solve and reason about systems described by LDEs. So even though most systems are not precisely linear, we often model them with LDEs even if the model is not an exact fit. An easily analyzed, but imperfect, LDE model is likely to be more helpful in design than an accurate non-linear model that is nearly impossible to analyze. i=1 i=1 First-Order Homogenous LDE s The simplest LDE is the homogenous first order LDE, whose form we have already seen, y[n] = α 1 y[n 1]. (11) We already know that we must specify an initial condition, y[0], for the solution of the LDE to be unique. There are many ways to calculate the values of a difference equation, so if we wanted to know y[n], we could compute it relatively easily by starting with the initial conditions, and propagating forward. Such a method is referred to as plug and chug, and as the name suggests, is more perspiration than inspiration. In particular, plug-and-chug offers little insight into the sequence s general behavior. Instead, we can derive a closed form for y, one that provides insight in to the general behavior of the sequence. In the first-order case, we can derive a general form almost by inspection. To start, we expand the LDE as in y[n] = α 1 y[n 1] = α 1 (α 1 y[n ]) = α 1 (α 1 (α 1 y[n 3])). = α 1... (α 1 (α 1... y[0])) = y[0]α n 1.

5 6.30.0x, January 016 Notes Set 1 5 Question 1. How does a computer calculate α n 1? Is it faster than n multiplications by α 1? The above derivation yields a direct formula for y[n] as a function of y[0] and α 1, a formula that can be evaluted far more efficiently than plug and chug, particularly when n is large. But the derivation also reveals a more general idea, that changing the initial condition only scales the associated sequence, but changing α 1 alters how the sequence evolves. For this reason, we refer to α 1 in the first-order LDE as its natural frequency. Example 1 Suppose you put $100 into a high-interest bank account 30 years ago and forgot about it. The account paid 1% interest, how much do you have now? If we assume that the interest was compounded annually (meaning the interest is added to your account balance once a year), we can represent your bank balance as a sequence, y, where y[n] is your bank balance n years after you opened the account. Then y[n] can be related to y[n 1] by adding the 1 % interest, leading to a first-order LDE, y[n] = y[n 1] + 0.1y[n 1] = 1.1y[n 1]. (1) The initial condition for the LDE is the initial bank deposit, so y[0] = 100. The bank balance LDE has a natural frequency of 1.1, so we know that y[30] = $3000. Not bad. But what if one of your parents had opened the account for you 60 years ago? Then you would have y[60] = $ 90,000! Such exponential growth is expected whenever an LDE has a natural frequency greater than one. Although in this case, the natural frequency is 1.1, and is not that much larger than one, so the exponential growth is not that fast. Question. Suppose the bank pays 1 % interest per year, but compounds the interest monthly. That is, suppose 1 % interest was added to your account every month. Then how much would you have after 30 years and after 60 years? Question 3. What if the bank pays the 1 % interest per year, but compounds it daily, so that % in interest was added to your account every day? How would your balance change after 30 and 60 years? Question 4. And what if the bank compound interest hourly? How would your balance change after 30 and 60 years? Example Suppose you put $100 into a checking account 30 years ago and forgot about it. If the bank charges checking accounts a 5% fee, and deducts it annually, how much is still in the account? The checking account can also be modeled as a first-order LDE, just like before, but with funds removed rather than added, y[n] = y[n 1] 0.05y[n 1] = 0.95y[n 1].. (13) The natural frequency of this LDE is 0.95, and we can use it to determine the amount left after 30 years, y[30] = Not much. From the two bank balance examples, we can see that the natural frequency of a first-order LDE, or equivalently α 1, governs the evolution of its associated sequences. In particular,

6 6.30.0x, January 016 Notes Set 1 6 If α 1 > 1, the sequence grows monotonically and without bound, and we say that the first-order LDE is unstable. If 0 < α 1 < 1, the sequence decreases monotonically to zero, and we say that the first-order LDE is stable. If α 1 = 1, then the sequence does not change value with index, and the first-order LDE is neither stable nor unstable. What if the natural frequency of a first-order LDE is negative? To answer that question mathematically, consider two first-order LDE s, y a [n] = 4 3 y a[n 1] y b [n] = 3 4 y b[n 1]. These two first-order LDE s have negative natural frequencies, 4 3 and 3 4, respectively. Question 5. Assuming an initial condition of one, how do the stem plots of the sequences y a and y b compare (hint: you can use plug-and-chug to determine the values for the plots)? What you should notice by plotting y a and y b is that when the natural frequency is negative, the associated sequence values oscillate between negative and positive. In addition, when the magnitude of the natural frequency is greater than one, the associated sequence values grow in magnitude, and when the magnitude of the natural frequency is less than one, the sequence values decay to zero. In particular, If α 1 < 1, the associated sequences oscillate, the magnitude of their values grow without bound, and the first-order LDE is unstable, If 1 < α 1 < 0, the associated sequences oscillate, their values decay to zero, and the first-order LDE is stable, If α 1 = 1, the associated sequences oscillate, the magnitude of their values neither decay nor grow, and the first-order LDE is neither stable or unstable. Example 3 A robot that measures obstacle distances ten times a second is placed five meters from a wall. The robot is programmed to move towards the wall with a velocity proportional to its distance from the wall, to slow it down as it gets close. But what really happens? To model how the robot moves towards the wall, we start by representing its distance measurements as a sequence, d, where d[n] is the n th measured distance and d[0] = 5. The difference between d[n] and d[n 1] can be related to the robot s velocity and the time between measurements, d[n] = d[n 1] 0.1v[n 1], (14) where 0.1 is time between distance measurements, and v[n 1] is the robot velocity at the time of sample n 1. Please take note of the negative sign in front of 0.1v[n 1]. It is there because we have chosen to define positive velocity as corresponding to the robot moving towards the wall, and decreasing its distance to the wall. Since the robot s velocity was programmed to be proportional to its distance to the wall, v[n 1] = γd[n 1], where γ is the proportionality factor. Combining equations leads to a first-order LDE

7 6.30.0x, January 016 Notes Set 1 7 for d[n], d[n] = d[n 1] 0.1v[n 1] = d[n 1] 0.1γd[n 1] = (1 0.1γ)d[n 1], where (1 0.1γ) = α 1 is the LDE s natural frequency. If we do not want the robot to crash in to the wall, or equivalently, if we do not want d[n] to become negative, we now know that the first-order LDE s natural frequency must be non-negative. In addition, if we want d[n] to decay to zero (so that we end up next to the wall), The natural frequency must be less than one. Applying these natural frequency constraints leads to constraints on the the proportionality factor γ, γ < 1 0 < 0.1γ 1 0 < γ 10 Question 6. What happens if γ is very near but > 0? Question 7. What happens if γ is exactly 10? Question 8. What happens if γ is 11?. Second-Order Homogenous LDEs Sequences described by second-order homogenous LDEs, like the Fibonacci sequence, have the property that y[n] is given by a weighted combination of y[n 1] and y[n ]. Their general form is y[n] = α 1 y[n 1] + α y[n ], (15) and to uniquely specify the associated sequence, it is necessary to specify two initial conditions, y[0] and y[1]. Unlike the first-order case, a closed-form solution for a second-order homogenous LDE is not readily apparent from inspection. We know that the general solution to the first-order homogenous LDE is n y[n] = y[0]α 1 (16) and even though it is not the solution to the second-order LDE (unless α = 0, of course), it does suggest a form for a guess at the solution. We are going to guess that second-order homogenous LDE s have solutions of the form y[n] = cλ n, (17) and then we are going to plug this form back in to the homogenous LDE to determine values for λ and c. And just to spoil the surprise, it is going to turn out that we must allow both c and λ to be complex numbers. Recall that a complex number can be written as a + bi where a and b are real and i = 1. But since i is so often used to mean electric current, we prefer to use a + bj where j = 1

8 6.30.0x, January 016 Notes Set 1 8 Substituting y[n] = cλ n in to the general form for a second-order homogenous LDE yields cλ n = α 1 cλ n 1 + α cλ n. (18) Simplifying by dividing through by cλ n results in a quadratic equation for λ, λ = α 1 λ + α or λ α 1 λ α = 0, (19) which nearly always has two distinct solutions, often referred to as roots, (λ 1, λ ) = α 1 ± α 1 + 4α. (0) Assumption 1 In the following approach for solving second-order homogenous LDEs, we assume that the quadratic equation λ has two distinct roots, that the are two distinct natural frequencies. It can happen that λ 1 = λ, and in this generally-rare repeated-root case, the approach below must be replaced with a distractingly complicated alternative. In our case, we are the ones creating the models and designing the controllers, so it is particularly easy for us to avoid the rare repeated-root case. By construction, since both λ 1 and λ satisfy the above quadratic equation, the second-order homogenous LDE will be satisfied by y[n] = c 1 λ n 1 or y[n] = c λ n, regardless of the values chosen for c 1 and c. For this reason, we refer to λ 1 and λ as the two natural frequencies for the secondorder homogenous LDE. Like the first-order case, natural frequencies give us whole families of sequences that satisfy the homogenous LDE. But unlike the first-order case, we have two families that can be used in any combination. That is, y[n] = c 1 λ n 1 + c λ n satisfies the second-order homogenous LDE, for any c 1 and c, as we show by plugging in, y[n] α 1 y[n 1] α y[n ] c 1 λ n 1 + c λ n α 1 (c 1 λ n c λ n 1 ) α (c 1 λ n 1 + c λ n ). Question 9. Complete the proof that if y[n] = c 1 λ n 1 and y[n] = c λ n both satisfy the secondorder homogenous LDE, then y[n] = c 1 λ n 1 + c λ n satisfies the second-order homogenous LDE. For the last step, you need only separate the terms in to two groups, one group involving λ 1 and a second group involving λ, and show that the assumptions imply each group sums to zero on its own. Now that we have the two natural frequencies, there is the matter of initial conditions. How to incorporate the initial conditions is best seen by example, so we will examine solving the secondorder homogenous LDE for the Fibonacci sequence. Since the next value in the Fibonacci sequence is the sum of the previous two, or α 1 = α = 1 in our general for for a homogenous LDE. y[n] = y[n 1] + y[n ], (1) Using the quadratic formula above, we find that the Fibonacci LDE has two distinct natural frequencies, (λ 1, λ ) = 1 ± 5, ()

9 6.30.0x, January 016 Notes Set 1 9 therefore any sequence of the form will satisfy the Fibonacci LDE. y[n] = c 1 ( ) n ( ) n c, (3) In order to produce the classical Fibonacci sequence, we need to find values for c 1 and c so that y[0] = 0 and y[1] = 1. To start, consider how y[0] = 0 constrains c 1 and c, y[0] = c 1 ( Including the constraint from y[1] = 1, y[1] = c 1 ( So, the final equation for the Fibonacci sequence is Evaluating the numeric constants (approximately), ) 0 ( ) c = 0 c 1 + c = 0 c 1 = c. ) 1 ( ) c = c 1 c 1 = 1 c 1 ((1 + 5) (1 5)) = c 1 5 = c 1 = 1 5 y[n] = 1 ( ) n ( ) n 1 5. (4) 5 5 y[n] = n ( 0.618) n. (5) Question 10. Are there values of y[0] and y[1] for which the sequence given by decays to 0 as n approaches? y[n] = c 1 ( ) n ( ) n c (6) Complex Natural Frequencies Consider modifying the Fibonacci LDE by a single sign change, as in y[n] = y[n 1] y[n ]. (7)

10 6.30.0x, January 016 Notes Set 1 10 Figure : Plots of the two components of the Fibonacci sequence, and their sum. The component in the leftmost graph is due to the negative natural frequency, and we can clearly see its oscillation, but note that its magnitude is very small. The lower sequence is the sum of the first two, and we can clearly see that the exponential growth associated with the large positive natural frequency dominates, and reveals the familiar large n approximation for the Fibonacci sequence.

11 6.30.0x, January 016 Notes Set 1 11 As we will see, things turn out quite a bit differently. For instance, the natural frequencies are complex numbers, (λ 1, λ ) = 1 ± 3j. (8) You might worry that complex natural frequencies will lead to sequences with complex values, disturbing because real systems should be described by sequences with real values. If so, your concerns are misplaced, things go surprisingly smoothly. Our initial values are real, and all subsequent values are sums of previous values, so they have to be real, too. A remaining concern might be that there not be values for c 1 and c that produce a a sequence matching the given real initial values. As an example of this not being an issue, suppose are our initial conditions are still y[0] = 0 and y[1] = 1 as before. A quick check will assure you that y[0] = 0 still results in the constraint that c 1 = c. And satisfying the y[1] = 1 constraint leads to y[1] = c 1 ( 1 + 3j The values of the sequence are then given by which you can easily verify is always real valued. 3 ) 1 ( ) 1 1 3j + c = 1 c 1 = j 3 y[n] = j ( ) n 1 + 3j + j ( ) n 1 3j, (9) 3 3 Why are the sequences always real, even when the natural frequencies are complex? Part of the reason is that α 1 and α are real, so λ 1 and λ are roots of a quadratic with real coefficients, and therefore must both be real, or must form a complex conjugate pair (a + bj and a bj). And if the roots, or natural frequencies, form a complex conjugate pair, then constraining c 1 and c to be a complex conjugate pair guarantees that the resulting sequence will be real. But even with the complex conjugate constraint on c 1 and c, there will still be two degrees of freedom (one real part and one imaginary part for the complex conjugate pair), just enough to match any two real values for y[0] and y[1]. Complex Natural Frequencies and Qualitative Behavior Representing complex natural frequencies in a + bj form is not as helpful as representing them in Me jθ form, where M is the magnitude of the complex number (sometimes written as a + bj and given by a + b ), and θ is its angle, (and given by arctan(b/a)). If one thinks of (a, b) as a point in the two-dimensional complex plane, then M, θ is the representation of that point in polar coordinates. The main advantage of the magnitude/angle form is the simpler formula for computing powers of a complex number. In the a + bj form, what is (a + bj) n? It might be possible to compute the 3 Of course, if you compute this numerically, you might end up with vanishingly small imaginary parts.

12 6.30.0x, January 016 Notes Set 1 1 many product terms if n is two or three, but what if n = 100? Alternatively, consider squaring a complex number in the magnitude/angle representation, (Me jθ) = Me jθ Me jθ = M e jθ And for arbitrary n, we have that ( Me jθ) n = M n e jnθ, (30) which is not only easy to compute, but easy to reason about for large n. For example, what happens to the n th power of the magnitude/angle representation as n approaches? The magnitude, M n, will grow or shrink depending on whether M is larger or smaller than one. And e jnθ only changes the relative contribution and sign of the real and imaginary parts. So when we represent a natural frequency as Me jθ, we can easily answer questions about high powers of the natural frequency. If we are concerned with what a natural frequency tells us about the stability of the associated homogenous LDE, or want to know whether the associated sequence grows exponentially or decays to zero, then we need only compare the natural frequency s magnitude to one. In second-order homogenous LDEs, we have two natural frequencies, λ 1 and λ. And regardless of whether they are real or complex, positive or negative, their magnitudes are the main issue. If the magnitudes of both natural frequencies are greater than one, then any associated sequence will grow exponentially and the associated LDE is unstable. If both magnitudes are less than one, then any associated sequence will decay to zero, and the associated LDE is stable. And if the magnitude of one natural frequency is greater than one, and the other less than one, then the behavior of an associated sequence will depend on initial conditions, but the associated LDE is unstable. Higher-Order Homogenous LDEs The approach for solving second-order homogenous LDEs can be easily generalized to arbitrary order. The general form for a homogenous LDEs of order k is y[n] = α 1 y[n 1] α k y[n k].. (31) To find solutions to the k th -order homogenous LDE, one again assumes solutions of the form y[n] = cλ n, and substitutes that form in to the LDE to generate its characteristic equation, λ n α 1 λ n 1... α k λ n k = 0., (3) The roots of the characteristic equation, λ 1,..., λ k, are the k natural frequencies of the LDE. Given k initial conditions, and assuming k distinct natural frequencies, one can solve k linear equations in k unknowns, y[0] = c c k y[1] = c 1 λ c k λ k y[k 1] = c 1 λ k c k λ k 1 k

13 6.30.0x, January 016 Notes Set 1 13 Figure 3: Crane Seeking a Beacon to determine values for c 1,..., c k. Then the closed form representation for the values of the associated sequence is given by y[n] = c 1 λ n c k λ n k. (33) There are some complications in the higher-order case. The roots of higher-order polynomials are almost always computed numerically. And with more natural frequencies, there are more ways for repeated natural frequencies to make life complicated. For the case of control system design, difference equations of order higher than three are rarely used. Instead, one switches to state-space methods, a topic you will certainly learn if you continue studying control theory. Crane Mutiny A heavy cargo crane travels on track above a buried beacon, as shown in, and measures its signe d distance to the beacon 100 times a second (by signed distance, we mean that the distance is positive if the crane has not yet reached the beacon and negative if the crane has traveled past it). A human operator usually drives the crane and parks it just above the beacon. The crane then keeps itself in place by automatically, by always accelerating towards the beacon at a rate proportional to its distance from the beacon On one unfortunate day, the operator noticed that the crane seemed to

14 6.30.0x, January 016 Notes Set 1 14 be rocking back and forth over the beacon, and tried to fix the problem by flipping the the sign of the distance sensor. The crane almost immediately sped off down the track, halted only by the quick-thinking operator, who shut down the crane. Why did this happen? Modeling the crane s motion is a little more complicated than the earlier robot example, but we will start in the same way. We will represent the crane s measured distance to the beacon as a sequence, d, where d[n] is the n th measured distance. Then d[n] can be related to d[n 1] by d[n] d[n 1] 0.01v[n 1], (34) where 0.01 is the number of seconds between distance measurements, v[n 1] is the crane velocity at the time of the (n 1) th distance measurement, and we note that the relationship is approximate because the velocity of an accelerating crane varies continuously. Since crane acceleration is proportional to measured distance, it does not vary between samples, so we have an exact relation between velocity and acceleration. That is, v[n] v[n 1] = 0.01a[n 1] = 0.01γd[n 1], (35) where a[n 1] is the crane acceleration at the time of the (n 1) th distance measurement, and γ is the proportionality factor relating acceleration to measured distance. There are now two sequences, v and d, and two first-order LDEs. We would like to determine natural frequencies, but of what? If you continue to study control theory, you will learn that transform techniques considerably simplify solving this problem. But for now, we will lean on some of the LDE properties we have learned. To begin, consider that an LDE holds for all n, so we can shift the index by one and write equivalent LDEs for d and v, d[n 1] d[n ] 0.01v[n ]. (36) and v[n 1] v[n ] = 0.01γd[n ]. (37) Subtracting the shifted from the unshifted LDE for d yields a second-order LDE for d and v, (d[n] d[n 1]) (d[n 1] d[n ]) 0.01 (v[n 1] v[n ]). (38) The shifted LDE for v can be used to eliminate velocity, and after that elimination, (d[n] d[n 1]) (d[n 1] d[n ]) (0.01) γd[n ]. (39) Gathering terms considerably simplifies the equation, and the result is a relatively simple secondorder homogenous LDE for d[n] alone, d[n] d[n 1] + (1 + (0.0001γ)) d[n ] 0. (40) Whew! Maybe we should learn transforms? Regardless, we are now ready to compute natural frequencies. Plugging the coefficients of the difference equation in to the quadratic formula for the natural frequencies (note that α 1 = and α = ( γ) when using the quadratic formula above), (λ 1, λ ) = 1 ± 0.01 γj, (41)

15 6.30.0x, January 016 Notes Set 1 15 where we have used j = 1. We could compute the crane s trajectory for a specific initial condition by determining c 1 and c in the closed-form expresion for d, d[n] = c 1 ( γj) n + c ( γj) n, (4) but we can learn more from just examining the natural frequencies. Question 11. For the crane problem, suppose γ = 10, d[0] = 0.5 meters, and the initial velocity is zero, v[0] = 0. Determine a value to use for d[1], and give the equations you would use to compute c 1 and c in the general formula for d[n]. When γ > 0, the two natural frequencies have equal magnitudes, given by M = γ. (43) Since M is clearly larger than one for any positive γ, our second-order homogenous LDE is certain to be unstable, and the associated sequence of distances should have grown exponentially. So the crane s automatic system should have always caused the crane to run away, but it did not. Why? The problem is that we are asking too much from our approximate model. The magnitude of the natural frequencies are greater than one, but just barely. For example, when γ = 0.1, M , and M <, so only a factor of two increase in distance to the beacon after one thousand seconds. Our second-order homogenous LDE model for the crane is unstable, and does produce sequences that correspond to slowly growing crane position oscillations. But that growth is so slow, any losses neglected in our model could easily overwhelm that growth, or render it unnoticeable. Perhaps the crane is usually parked very close to the beacon, and the oscillations went unnoticed, but on the day when rocking was observed, maybe the operator parked unusually far from the beacon? If our LDE model predicts such slow growth that it goes unnoticed, why did the crane run away so dramatically when the operator flipped the sign of the distance sensor? In our model, flipping the sign of the distance sensor is the same as flipping the sign of γ. So what happens if we repeat the example, but with γ = 0.1, flipped in sign. Our two natural frequencies are now real, and the magnitude of the largest one is given by M (44) This seems very similar to the positive γ case, M is larger that one, though like before, not much larger. It turns out that M is large enough, as M 3000 > That is, after 30 seconds, the model predicts that the initial distance to the beacon will increase by a factor of ten thousand. Good thing the operator acted fast! Our model may have been imperfect, but it was easily analyzed, and it helped us understand what happened to the crane. As you will see when you start working with the copter-levitated arm, you can learn a lot from approximate LDE models. And as you gain experience, you will be able to decide for yourself what models to use, and when to trust them.