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# Notes-LT3.pdf

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Department
Mathematics
Course
MATH 250
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Unknown
Semester
Spring

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Impulse Functions In this section: Forcing functions that model impulsive actions − external forces of very short duration (and usually of verylarge amplitude). The idealized impulsive forcing function is the Dirac delta function * (or the unit impulse function), denotes δ(t). It is defined by the two properties δ(t) = 0, if t ≠ 0, and ∞ δ(t)dt =1 ∫ −∞ . That is, it is a force of zero duration that is only non▯zero at the exact moment t = 0, and has strength (total impulse) of 1 unit. Translation of δ (t) The impulse can be located at arbitrary time, rather than just at t = 0. For an impulse at t = c, we just have: δ(t − c) = 0, if t ≠ c, and ∞ ∫ −∞ δ(t − c)dt =1 . * of the Nobel Prize in Physics in 1933. A pioneer of quantum mechanics, Dirac is perhaps best known▯winner (besides for the delta function) for formulating the Dirac equation, which predicted the existence of antimatter. © 2008, 2012 Zachary S Tseng C▯3 ▯ 1 Laplace transforms of Dirac delta functions L{δ(t)} = 1, −cs L{δ(t − c)} = e , c ≥ 0. Here is an important and interesting property of the Dirac delta function: If f(t) is any continuous function, then ∞ ∫−∞ δ(t − c) f (t)dt = f (c) Therefore, for c ≥ 0, ∞ −st ∞ −st −cs L{δ(t − c)} = ∫0 δ(t −c)e dt = ∫−∞ δ(t −c)e dt = e . Note: Since the integrand in the above integrals is zero everywhere except at the single point t = c ≥ 0, the contribution from the interval (−∞, 0) to the definite integral is, therefore, zero. Hence the two definite integrals above will have the same value despite their different lower limits of integration. And, ∞ δ(t −c) f (t) e dt = f (c)e −cs L{δ(t − c) f(t)} = ∫0 . © 2008, 2012 Zachary S Tseng C▯3 ▯ 2 Example: y″ + 6y′ + 5y = δ(t) + δ(t − 2), y(0) = 1, y′(0) = 0. Transform both sides and simplify: (s L{y} − sy(0) − y′(0)) + 6(sL{y} − y(0)) + 5L{y} = L{δ(t) + δ(t − 2)} (s L{y} − s) + 6(sL{y} − 1) + 5L{y} = 1 + e −2s (s + 6s + 5)L{y} − s − 6 = 1 + e −2s 2 −2s −2s (s + 6s + 5)L{y} = 1 + e + s + 6 = e + s + 7 −2s e + s +7 L{y} = (s +1)(s +5) (s +1)(s +5) . Further simplifying the two parts using partial fractions: Part 1, without ehesterm, becomes 1 1 1 1 1 = − , (s +1)(s +5) 4 (s +1) 4 (s +5) 1 1 It has an inverse transform − e −5. 4 4 Now apply the effects of ehesterm, resulting in 1 u (t) e−t+2−e −5t+)0 4 2 . © 2008, 2012 Zachary S Tseng C▯3 ▯ 3 Part 2 becomes s +7 3 1 1 1 = − . (s +1)(s +5) 2 (s +1) 2 (s +5) 3 −t 1 −5t It’s inverse transform is− e . 2 2 Therefore, the solution is 3 −t 1 −5t 1 −t+2 −5t+10 y = e − e + u 2t) e −e ). 2 2 4 Note that the first term in the forcing function, δ(t), has the same effect to the solution of this initial value problem as it would if the second initial condition was y′(0) = 1. (Check this: remove the first term of the forcing function and change the second initial condition to y′(0) = 1 and see that this new initial value problem will have the same solution as the above problem.) © 2008, 2012 Zachary S Tseng C▯3 ▯ 4 Example: y″ + 2y′ + 10y = −δ(t − 4π), y(0) = 0, y′(0) = 1. Transform both sides and simplify: 2 (s L{y} − sy(0) − y′(0)) + 2(sL{y} − y(0)) + 10L{y} = L{−δ(t − 4π)} (s L{y} − 1) + 2(sL{y} − 0) + 10L{y} = −e −4πs 2 −4πs (s + 2s + 10)L{y} − 1 = −e (s + 2s + 10)L{y} = 1 −e −4πs 1 −4πs 1 L{y} = 2 −e 2 . s + 2s +10 s + 2s +10 Notice that, other than the extra−eact, the second part contains the same expression as the first part. This fact simplifies our task somewhat. Complete the squares in the denominator and rewrite the expression as: 1 1 3 2 = 2 2 s + s + 10 3 (s 1) + 3 . 1 e sin(3t) Its inverse transfor3 is , which corresponds exactly to the first part of the expression. © 2008, 2012 Zachary S Tseng C▯3 ▯ 5 The second part, with the ex−ea−4πsterm, gains its effects (plus a negative sign) of a unit step function and a translation by c = 4π. It becomes − u (t) e −(t−4sin3(t − 4π) ) 3 4 . Therefore, 1 −t 1 −(t−π ) y = e sin(3t) − u (4π (e sin3(t − 4π) ) 3 3 1 −t 1 −t+4π = e sin(3t) − u (4πe sin(3t) 3 3 © 2008, 2012 Zachary S Tseng C▯3 ▯ 6 Resonance induced by a discontinuous forcing function Example: u″ + u = δ(t) + δ(t − 2π) + δ(t − 4π) + … + δ(t − 2000π), u(0) = 0, u′(0) = 0. Transform and simplify the equation: (s L{u} − 0s − 0) + L{u} = 1 + e −2πs+ e−4π+ … + e −2000,s 2 −2πs −4πs −2000πs (s + 1)L{u} = 1 + e + e + … + e , 1000 1  −2nπs 2 1 + ∑ e  L{u} = s + 1  n=1 . The inverse transform of each term contains sin(t). The terms inside the summation, after applying the required step functions and translations, becoue2nπt) sin(t − 2nπ= u2nπ(t) si
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