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MATH 2270 (28)
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Department
Mathematics
Course
MATH 2270
Professor
Matthew Demers
Semester
Fall

Description
1 4.5 Creating other Functions from the Heaviside Function The Heaviside function is really interesting, because it can be used to create many other functions. To do this, we add or subtract two or more di▯erent Heaviside functions to or from one another. Here are some examples: Rectangular Impulse Function If a < b and t ▯ 0, then a rectangular impulse r(t) can be created by r(t) = H(t ▯ a) ▯ H(t ▯ b) (4.97) 8 > < 0; 0 ▯ t < a = > 1; a ▯ t < b (4.98) > : 0; t ▯ b Figure 4.1: On the left, a graph of the rectangular impulse function r(t), with a = 1 and b = 2. This sort of function is sometimes depicted as appears on the right. The Laplace Transform is then given by e▯as▯ ebs Lfra;b)g = ; s > 0 (4.99) s 2 Staircase Function Let a > 0 and t ▯ 0. Then, let a staircase function be given by stair(t) =8H(t ▯ a) + H(t ▯ 2a) + H(t ▯ 3a) (4.100) > > 0; 0 ▯ t < a > < 1; a ▯ t < 2a = : (4.101) > 2; 2a ▯ t < 3a > : 3; t ▯ 3a Figure 4.2: On the left, a graph of stair(t) when a = 1. On the right, a di▯erent depiction. The Laplace Transform is then given by e▯a+ e▯2a+ e▯3as Lfstair(t)g = ; s > 0: (4.102) s This staircase function has three steps. We could easily create a staircase function with n steps by adding together n Heaviside functions. \Simple Switch O▯" Function Let a > 0 be the point of time at which we want a switch from 1 to 0, for t ▯ 0. Then, let a switching-o▯ function be given by 3 off(t) = 1 8 H(t ▯ a) (4.103) < 1; 0 ▯ t < a = : : (4.104) 0; t ▯ a Figure 4.3: Two depictions of the \Simple Switch O▯" Function, with a = 2. The Laplace Transform is ▯as Lfoff(t)g = 1 ▯ e ; s > 0: (4.105) s \Switch F on" Function Let a > 0 be the point of time at which we want a behaviour given by a function F(t) to switch on, for t ▯ 0. Then, we de▯ne the important \Switch F on" function by SwitchFOn(t) = H(t ▯ a)F(t ▯ a) (4.106) 8 < 0; 0 ▯ t < a = : (4.107) : F(t ▯ a); t ▯ a 4 Figure 4.4: The \Switch F on" Function, where a = 3 and F(t) = sin(t). The Laplace Transform for this special function is ▯as LfH(t ▯ a)F(t ▯ a)g = e f(s); (4.108) where f(s) = LfF(t)g. (We could derive this from ▯rst principles!) Going backwards, we also have ▯1 ▯as L fe f(s)g = H(t ▯ a)F(t ▯ a); (4.109) ▯1 where F(t) = L ff(s)g. 5 Example: 1 ▯2s Evaluate LfH(t ▯ 2)g = se . This is done using the basic Laplace transform of the Heaviside function. Example 2: Evaluate Lf(t ▯ 2)H(t ▯ 2)g = e ▯2s1 . s2 This is done using the identity (4.109). Example 3: Evaluate LftH(t ▯ 2)g. This is not quite as straightforward. We need to get this into a form we can use! It would help if the \t" multiplying in front was instead a \t ▯ 2". But, observe that LftH(t ▯ 2)g = Lf(t ▯ 2)H(t ▯ 2) + 2H(t ▯ 2)g (4.110) ▯2s 1 1 ▯2s = e 2 + 2 e : (4.111) s s Example 4: e▯6s Evaluate L ▯1f g. s + 9 We can rewrite this as ▯1 ▯6s 1 L fe s + 9 g (4.112) = 1 L▯1fe▯6s 3 g (4.113) 3 s + 9 1 = H(t ▯ 6)sin(3(t ▯ 6)) (4.114) 3 A rule for writing piecewise functions in terms of Heaviside functions: Given a piecewise function: ▯ Write down the ▯rst branch 6 ▯ For each value \a" that separates the branches, add H(t ▯ a)[The right branch’s function ▯ The left branch’s function] (4.115) A rule for writing combinations of Heaviside functions as a piecewise func- tion: Given a combination of Heaviside functions: ▯ Arrange the terms by the values a in ihe Heaviside functions H(t ▯ a ) that i appear, in order of increasing a . Put any terms not multiplied by a Heaviside i function before all of the others. ▯ Locate the ▯rst terms, those that are NOT multiplied by a Heaviside function. This is your ▯rst branch: Those terms, if 0 ▯ t < (the lowest value of a apiearing in any term): (4.116) ▯ Now, add a branch for each new term. Each branch should be of the form The function for this term PLUS all previous functions, if a previous▯ t < a next biggest (4.117) ▯ In this way, continue until the last branch, which should be the sum of ALL of the functions appearing in each term, for t ▯ a biggest 4.6 Using Laplace Transforms to Solve DEs with Discontin- uous Forcing Terms Some DEs have forcing terms f(t) with discontinuities/jumps at particular val- ues of the independent variable. Such DEs in general have solutions that are not di▯erentiable at those points, so we might be hesitant to talk about such di▯erential equations. We require that solutions for these DEs are: 7 ▯ di▯erentiable on each open interval where the forcing term is continuous; and ▯ continuous at each point at which there is a discontinuity in the forcing term. It is important that discontinuous forcing terms are written in terms of Heaviside functions if they are not already, so that we can take the Laplace transforms easily. Example: Find the solution to the IVP 00 y + 4y = H(t ▯ ▯) ▯ H(t ▯ 2▯) (4.118) y(0) = 1 (4.119) y (0) = 0: (4.120) We take the Laplace transforms of each part. ▯▯s ▯2▯s 2 0 e ▯ e s Lfyg ▯ sf(0) ▯ f (0) + 4Lfyg = s (4.121) ▯▯s ▯2▯s 2 e ▯ e (s + 4)Lfyg = s + s (4.122) ▯▯s ▯2▯s Therefore, Lfyg = e ▯ e + s : (4.123) s(s + 4) s(s + 4) s + 4 We can use partial fraction decomposition on the ▯rst two terms. This yields ▯ ▯ ▯ ▯ 1 e ▯▯s se▯▯s 1 e ▯2▯s se▯2▯s s Lfyg = ▯ ▯ ▯ + : (4.124) 4 s s + 4 4 s s + 4 s + 4 Taking inverse transforms, 1 1 1 y(t) = 4H(t ▯ ▯) ▯ 4(t ▯ ▯)cos(2(t ▯ ▯)) ▯ H4t ▯ 2▯) + (4.125) 1H(t ▯ 2▯)cos(2(t ▯ 2▯)) + cos(2t) (4.126) 4 Or, grouping by Heaviside function, ▯ ▯ ▯ ▯ 1 1 1 1 y(t) = cos(2t) + H(t ▯ ▯) ▯ cos(2t) + H(t ▯ 2▯) ▯ + cos(2t) ; (4.127) 4 4 4 4 8 noting that cos(2t) has a period of ▯ and so cos(2(t ▯ ▯)) = cos(2t). How can we make sense of this? Sort the di▯erent terms by the values of a in each of the Heaviside functions, and construct the piecewise function of t from left to right. Analyzing it this way, we see that: ▯ We have cos(2t) alone to begin with, until we get to t = ▯. ▯ At t = ▯, we add 1 ▯ 1cos(2t). 4 4 ▯ At t = 2▯, we add ▯ + 1 1cos(2t). 4 4 So, at t = 2▯, the changes from before are undone, and the function once again becomes cos(2t). A graph of this solution function can be seen in Figure 4.5. Figure 4.5: A graph of the solution y(t) of the IVP described in Equations (4.118 - 4.120). 4.7 The Dirac Delta Function Another function that is often seen in applications is an impulse function, repre- senting a sudden behaviour at a particular moment of time. To create this function, 9 we ▯rst de▯ne: 8 1 < 2▯; if ▯▯ < t < ▯ ▯▯(t) = : : 0; if jtj ▯ ▯ Note that the area under ▯ ▯t) is always equal to 1, no matter the value of ▯, since Z Z 1 ▯ 1 1 ▯▯(t) dt = dt = (2▯) = 1: (4.128) ▯1 ▯▯ 2▯ 2▯ As ▯ decreases, the height of the step increases, while the width decreases, but the area underneath maintains a constant area of 1. We may decrease ▯ arbitrarily, and even speak of a limit. Let ▯(t) = lim▯▯(t); (4.129) ▯!0 Z 1 with the property that ▯(t) dt = 1. Then, we call this function the Dirac Delta, ▯1 or unit impulse function at t = 0. The function ▯(t▯t )0is therefore a unit impulse at time t = t0instead. Let’s ▯nd the Laplace Transform of this function! Lf▯(t ▯ t0)g = limLf▯ (t ▯ t )g0 (4.130) ▯!0 Z 1 = lim e▯st▯▯(t ▯ 0 ) dt (4.131) ▯!0 0 Z 0 +▯ ▯st1 = li▯!0 e 2▯ dt (4.132) 0 ▯▯ ▯ 1 ▯ e▯st▯▯t0+▯ = lim ▯ (4.133) ▯!0 2▯ ▯s ▯ t0▯▯ 1 ▯ ▯s(t0+▯) ▯s(0 ▯▯) = lim ▯e + e (4.134) ▯!
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