Section 6.5 Impulse Functions

Transcription

1 Section 6.5 Impulse Functions Key terms/ideas: Unit impulse function (technically a generalized function or distribution ) Dirac delta function Laplace transform of the Dirac delta function IVPs with forcing functions that are impulses

2 For years physicists and engineers have found it useful to use the notion of an applied force at a particular instant in time or to view a mass concentrated in a single point. For example:. Concentrated load at a single point. 2. Electrical potential applied instantaneously in a circuit. 3. A collision or sharp blow in a mechanical system; a) A ball hitting a bat. b) Collision of billiard balls. Such applied forces are called IMPULSE FUNCTIONS. Technically they are not functions, but are "generalized" functions or "distributions". An impulse function, often called the (Dirac delta function), is defined not by its values but by its behavior in a limit and by its behavior under integration. The Dirac delta function provides a model for a force that concentrates a large amount of energy over a short time interval. Ref: Polking et. al.

3 In order to better understand the definition of the Dirac delta function, we first discuss the impulse of a force. (Keep in mind that the objective is to determine a way to represent a force that produces an impulse.) Definition: Suppose F(t) represents a force applied to an object at time t. Then the impulse of force F over time interval a t b is defined as impulse b a F(t) dt Geometrically, the impulse of F(t) is the area under the curve y = F(t) over interval [a, b]. From the point of view of mechanics, the impulse is the change in momentum of a mass as the force is applied to it over the time interval from a to b. We can see this as follows. Recall that momentum is the product of the mass with velocity. So by Newton s second law, the force F(t) = ma = m dv/dt. Therefore b b b dv impulse F(t) dt ma dt m b m v(t) mv(b) mv(a) a a a dt dt a which is the change of momentum from t = a to t = b.

4 Now let s consider a force of unit impulse acting over a short time interval. A force of unit impulse will have impulse b a F(t) dt = We consider a sequence of forces modeled by functions expressed in terms of unit step functions. For illustration purposes we will use such functions centered around t = 2 to visually display things which can then be reformulated in a sequence and a limit expression.

5 t = 2, t =, base = 2, height = /2, [. ) F(t) / 2, [, 3), [ 3, ) F(t) / 2u (t) / 2u (t) 3

6 t = 2, t = /2, base =, height =, [.. 5) F(t), [. 5, 2. 5), [ 2. 5, ) F(t) u (t) u (t)

7 t = 2, t = /3, base = 2/3, height = 3/ The expression for the newest F(t) is constructed in a similar fashion.

8 t = 2, t = /4, base = /2, height = The expression for the newest F(t) is constructed in a similar fashion.

9 t = 2, t = /5, base = 2/5, height = 5/ The expression for the newest F(t) is constructed in a similar fashion.

10 t = 2, t = /6, base = 2/6, height = The expression for the newest F(t) is constructed in a similar fashion.

11 t = 2, t = /7, base = 2/7, height = 7/ The expression for the newest F(t) is constructed in a similar fashion.

12 t = 2, t = /8, base = 2/8, height = The expression for the newest F(t) is constructed in a similar fashion. As the length of the base shrinks to zero, what is the intuitive value of the area of the rectangle? (In the limit the base is a single point!)

13 We use a force of unit impulse acting over a short time interval. A force of unit impulse will have impulse b a F(t) dt = Area of the rectangle is. Let ε (epsilon) be a small positive value and consider an interval from t = p to t = p + ε, then a force of unit impulse can be modeled by a rectangle with base from p to p + ε and height / ε. Force F e (t) can expressed in terms of step functions: / e, [p,p e) F e(t) u p(t) u pe(t), for t p and t p e e As epsilon varies so does the force function. But regardless of how small epsilon becomes the rectangle will have area, because as the base decreases the height increases. So we always have a force that produces a unit impulse.

14 A reasonable way to model a sharp, instantaneous force at time t = p is to take the limit of force F e (t) as epsilon goes to zero. We next make this the definition of the Dirac delta function. Definition: The Dirac delta function δ(t p) centered at t = p is the limit of / e, [p,p e) F e(t) u p(t) u pe(t), for t p and t p e e as epsilon goes to zero. We have δ(t p) lim F (t) lim / e, [p,p e) e e e, for t p and t p e When p = we use the notation δ(t ). Sometimes the limit is expressed as by writing, for t p δ(t p), for t p which gives the feeling of an instantaneous blow by a hammer, but is not mathematically correct even though we use the term Dirac delta function. It is not a function in the usual sense of calculus. It is more precise to call it a generalized function or a distribution.

15 As we stated previously the Dirac delta function is defined not by its values but by its behavior in a limit and by its behavior under integration. With this in mind we can use the limit definition δ(t p) lim F (t) lim / e, [p,p e) e e e, for t p and t p e to develop properties and compute its Laplace transform. Property. (We state the following without proof which involves the definition and a manipulation involving an integral.) If p is any fixed value so that p and g(t) is any function continuous near t = p, then sp Property 2. L (t p) e δ(t p)g(t)dt g(p) Proof: By definition of the Laplace transform sp Now use Property. This gives L (t p) e s If p = we have L (t) e st L (t p) (t p)e dt

16 Property 3. The derivative of unit step function u p (t ) is δ(t p). Proof: Recall that δ(t p) lim F (t) lim / e, [p,p e) e e e, for t p and t p e d lim u p(t) u pe(t) u p(t) e dt e This is expression is just the form of the definition of a derivative from calculus: lim x f(x x) f(x) x

17 Example: A mass-spring system remains at rest until struck by a sharp blow at time t = 2. We describe this by the IVP 4y'' + 36y = δ(t - 2), y() =, y'() =. Find the equation of motion of the mass. Taking Laplace transforms of both sides we have: Solving for Y(s): So taking inverse transforms we have e2 s 3e2 s 3e2 s 4 s s s2 9 y(t) L L L, t2 u 2(t)sin( 3(t 2)) 2 sin( 3(t 2)),t 2 2 #3 in our table So until t = 2 we have no motion. Then the impulse δ(t - 2) causes a subsequent sinusoidal oscillation.

18 IVP 4y'' + 36y = δ(t - 2), y() =, y'() = , t 2 y(t) u 2(t) sin( 3(t 2)) 2 sin( 3(t 2)),t 2 2

19 Example: A mass-spring system remains at rest until struck by a sharp blow at time t = 2 and a second blow of twice the magnitude at t= 5. We now describe this by the IVP 4y'' + 36y = δ(t - 2)+ 2 δ(t 5), y() =, y'() =. Find the equation of motion of the mass. Taking Laplace transforms of both sides we have: Solving for Y(s): So taking inverse transforms we have e2s 2e5s 4 s29 s29 y(t) = L - = 2 u(t 2)sin( 3(t 2)) u(t 5)sin( 3(t 5)) 2 2 #3 in our table, [,2) = sin(3(t - 2)), [2,5) 2 2 sin(3(t - 2)) + sin(3(t - 5)), [5, ) 2 2

20 . What s happening here? IVP 4y'' + 36y = δ(t - 2)+ 2 δ(t 5), y() =, y'() = y(t) u(t 2)sin( 3(t 2)) 2 u(t 5)sin( 3(t 5)) 2 2, [,2) = sin(3(t - 2)), [2,5) 2 2 sin(3(t - 2)) + sin(3(t - 5)), [5, ) 2 2

21 2 y(t) u(t 2)sin( 3(t 2)) u(t 5)sin( 3(t 5)) 2 2 Lets look at the derivative of From MATLAB y'(t) is (heaviside(t - 2)*cos(3*t - 6))/4 + (heaviside(t - 5)*cos(3*t - 5))/2 + (dirac(t - 2)*sin(3*t - 6))/2 + (dirac(t - 5)*sin(3*t - 5))/6 Look at the discontinuities in the derivative.

22 Example: y'' 5y' 6y (t 3), y( ), y'( ) Y(s) 3s e (s 2)(s 3) Solution: 93t 62 y(t) u 3(t) e e t

23 Example: y'' 5y' 6y (t 3), y( ), y'( ) Y(s) e S 3s (s 2)(s 3) Solution: y(t) e e u (t) e e t 2t 9 3t 6 2t 3