Lecture 19: Impulse Function and its Laplace Transform ( 6.5) April 24, 2012 (Tue)
Impulse Phenomena of an impulsive nature, such as the action of very large forces (or voltages) over very short intervals of time, are of great practical interest, since they arise in various applications. This situation occurs, for instance, when a tennis ball is hit, a system is given a blow by a hammer, an airplane makes a hard landing, a ship is hit by a high single wave, and so on. Our present goal is to show how to solve problems involving short impulses by Laplace transforms.
Impulse Phenomena of an impulsive nature, such as the action of very large forces (or voltages) over very short intervals of time, are of great practical interest, since they arise in various applications. This situation occurs, for instance, when a tennis ball is hit, a system is given a blow by a hammer, an airplane makes a hard landing, a ship is hit by a high single wave, and so on. Our present goal is to show how to solve problems involving short impulses by Laplace transforms. (Impulse of a force) Suppose F (t) represents a force applied to an object m at time t. Then the impulse of F over the time interval a t b is defined as Impulse = b a F (t) dt. The analog for an electric circuit is the integral of electromotive force applied to the circuit.
Impulse Geometrically, the impulse of F (t) is the area under the curve y = F (t), for a t b. 1 Momentum represents the power residing in a moving object captured by mathematical definition P = mv, the product of the mass and velocity of an object.
Impulse Geometrically, the impulse of F (t) is the area under the curve y = F (t), for a t b. From the point of view of mechanics, the impulse is the change in momentum 1 of a mass as the force is applied to it over the time interval a t b. 1 Momentum represents the power residing in a moving object captured by mathematical definition P = mv, the product of the mass and velocity of an object.
Unit Impulse Function Of particular practical interest is the case of impulse over a very short time ɛ (and its limit ɛ 0); that is, the impulse of a force acting only for an instant.
Unit Impulse Function Of particular practical interest is the case of impulse over a very short time ɛ (and its limit ɛ 0); that is, the impulse of a force acting only for an instant. Let s consider a force, d ɛ acting over a short time interval (t 0 ɛ t 0 + ɛ): { 1 d ɛ (t t 0 ) = 2ɛ if t 0 ɛ t t 0 + ɛ 0 otherwise = 1 2ɛ [u(t (t 0 ɛ)) u(t (t 0 + ɛ))] = 1 2ɛ [u t 0 ɛ(t) u t0 +ɛ(t)]
Unit Impulse Function Of particular practical interest is the case of impulse over a very short time ɛ (and its limit ɛ 0); that is, the impulse of a force acting only for an instant. Let s consider a force, d ɛ acting over a short time interval (t 0 ɛ t 0 + ɛ): { 1 d ɛ (t t 0 ) = 2ɛ if t 0 ɛ t t 0 + ɛ 0 otherwise = 1 2ɛ [u(t (t 0 ɛ)) u(t (t 0 + ɛ))] = 1 2ɛ [u t 0 ɛ(t) u t0 +ɛ(t)] The (Dirac) Delta Function centered at t = t 0 is the limit δ t0 (t) = lim ɛ 0 d ɛ (t t 0 ). We will set δ t0 (t) = δ(t t 0 ) (so, when t 0 = 0, δ(t) = δ 0 (t)).
Laplace Transform of the Delta Function Properties:
Laplace Transform of the Delta Function Properties: (a) δ t0 (t) = { if t = t0 0 otherwise
Laplace Transform of the Delta Function Properties: (a) δ t0 (t) = { if t = t0 0 otherwise (b) δ t 0 (t) dt = 1
Laplace Transform of the Delta Function Properties: (a) δ t0 (t) = { if t = t0 0 otherwise (b) δ t 0 (t) dt = 1 (c) Suppose t 0 0 is any fixed point and let φ be any function that is continuos at t = t 0. Then δ t0 (t)φ(t) dt = φ(t 0 ).
Laplace Transform of the Delta Function Properties: (a) δ t0 (t) = { if t = t0 0 otherwise (b) δ t 0 (t) dt = 1 (c) Suppose t 0 0 is any fixed point and let φ be any function that is continuos at t = t 0. Then δ t0 (t)φ(t) dt = φ(t 0 ). The Laplace transform of the Delta Function: L{δ t0 (t)}(s) = e t 0s.
Example The Laplace transform of the Delta Function: L{δ t0 (t)}(s) = e t0s. Example: Find the solution of the initial value problems 2y + y + 2y = δ(t 5), with y(0) = 0 and y (0) = 0. Practice: Find the solution of the initial value problems y + 2y + 2y = δ(t π), with y(0) = 1 and y (0) = 0.