18.03 Class 29, April 16, 2010 Laplace Transform IV: The pole diagram 1. Another example 2. t-shift rule 3. Poles 4. What the pole diagram of F(s) says about f(t) [1] We saw that if p(D)w = delta(t) , i.e. a_n w^(n) + ... + a_1 w' + a_0 w = delta(t) with rest initial conditions, then p(s) W(s) = 1 , or W(s) = 1/p(s) , or w(t) = L^{-1}(1/p(s)) w(t) = weight function W(s) = transfer function This is another RULE. Other input signals? x" + 4x = cos(2t) rest initial conditions (No delta functions in this signal, so we know that x(0+) = 0 and x'(0+) = 0, but that will come out automatically.) X = W(s) LT[cos(2t)] = 2s/(s^2+4)^2 From our latest addition to the table, we find that x = u(t) 1/2 t sin(t) (Resonance!) GENERAL FACT: p(D) x = f(t) with rest initial conditions has Laplace transformed equation p(s) X(s) = F(s) with solution X(s) = W(s) F(s) In the s-domain, the system response is obtained by multiplying by the transfer function! [2] Another rule: Let a > 0 and define f_a(t) = 0 if t < a= f(t-a) if t > a (If f_s(t) contains c delta(t) , I also want f_a(t) to contain c delta(t-a) . So for example delta_a(t) = delta(t-a) .) If you know F(s) , what is the LT of f_a(t) ? integral_{0-}^infty f_a(t) e^{-st} dt = integral_{a-}^\infty f(t-a) e^{-st} dt Substitution: u = t-a , du = dt , ... = integral_{0-}^\infty f(u) e^{-s(u+a)} du = e^{-as} integral_{0-}^infty f(u) e^{-su} du = e^{-as} F(s) So we have the t-shift rule: f_a(t) -----> e^{-as} F(s) Eg u_a(t) ----> e^{-as}/s and delta_a(t) ----> e^{as} , as we know. [3] F(s) "essentially" determines f(t) , but as we have seen the path from F(s) to f(t) can be difficult. But there are certain features of f(t) which can be EASILY read off from F(s) . They have to do with the long term behavior of f(t) . Typical LTs are F(s) = 1/s or F(s) = 1/(s^2+2s+5) . F(s) is complicated because it assigns to each complex number s another complex number F(s) . Let's content ourselves with understanding the absolute value of F(s) : |F(s)| . This assigns a real number to each complex number. In 18.02 you learned how to think of functions on the plane. It has a graph which is a surface in 3-space. For example |1/s| = 1/|s| = 1/(distance from the origin). The graph of this function is radially symmetric: it is a surface of revolution of one branch of the hyperbola z = 1/x . It rises towards infinity as s goes to 0, and falls off to zero as s grows large. The point s = 0 is called a "pole" of F(s) = 1/s . Its presence, and location, is a simple gross feature of F(s) . In fact, it's the only thing you see from a distance! How about F(s) = 1/(s^2+2s+5) ? The poles occur at the roots of the denominator: s^2+2s+5 = (s+1)^2 + 4 has roots -1 +- 2i . This is a two-ring circus! And the sum has three poles. Have you figured out what functions have those two F's as LT's? L[1] = 1/s L[(1/2)e^{-t} sin(2t)] = 1/((s+1)^2 + 4) . Region of convergence again: The location of the poles explains the region of convergence.F(s) = integral_{0-}^infty e^{-st} f(t) dt Since |e^{-st}| depends only on Re(s) , the region of convergence will always be to the right of some line Re(s) = a , or empty, or the whole plane. The value of F(s) at a pole is infinite; this reflects the divergence of the improper integral. The region of convergence is the half plane to the right of the rightmost pole. [4] What the poles of F(s) tell us about f(t) Knowing where the poles are is just a small part of knowing the whole of F(s), and it doesn't tell you everything about f(t) . Two examples: (1) If f(t) = 0 for t > A , then the improper integral isn't so improper - F(s) = integral_{0-}^A e^{-st} f(t) dt - so it converges for ALL s : no poles at all. (2) e^{-as} is never zero and has no poles: so the poles of e^{-as}F(s) are the same as the poles of F(s) : time translation of f(t) doesn't affect the pole diagram of F(s) . So the pole diagram can't see phase. But the pole diagram does say a lot about the long term behavior of f(t). Example 1. f(t) = sin(2t). F(s) = 4/(s^2_4) has poles at 2i and -2i. Any f(t) such that F(s) has this pole diagram exhibits (for large t ) (a) oscillation with circular frequency 2 , and (b) neither exponential growth nor exponential decay. [The Laplace transform of f(t) = t sin(2t) is 4s/((s^2+4)^2) , which has the same pole diagram. This function does grow with t , but less than exponentially.] Example 2: f(t) = e^{-t} sin(2t) . F(s) = 2s/((s+1)^2+4) has poles at -1+2i and -1-2i . Any f(t) such that F(s) has this pole diagram exhibits (for large t ) (a) oscillation with circular frequency 2 , and (b) exponential decay on the order of e^{-t} . Example 3. f(t) = 1 + e^{-t} sin(2t) . F(t) = 1 + 2s/((s+1)^2+4) , which has poles at 0 , -1+2i, and -1-2i . Any f(t) such that F(s) has this pole diagram exhibits (for large t ) (a) sub-exponential growth/decay, i.e. for any a > 0 , e^(-at) < |f(t)| < e^(at) . (b) oscillation with circular frequency 2 but with amplitude decaying like e^{-t} .The rightmost poles give the dominant information. A major lesson: if all the poles of F(s) have negative real part, then the function f(t) decays to zero exponentially. The position of the poles of F(s) gives detailed information about the long-term rate of growth and oscillation of f(t) , information which it can be hard to glean from the graph of f(t) ... all exponentials look rather similar ... So when you solve an ODE using LT, it may be good enough to stop with X(s). Find its poles, and you know what the solution behaves like in the long run.MIT OpenCourseWare http://ocw.mit.edu 18.03 Differential Equations�� Spring 2010 For information about citing these materials or our Terms of Use, visit:
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