Impulse Functions
In this section : Forcing functions that model impulsive actions − external forces of very short duration (and usually of very large 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 ∫−∞ .
* It was introduced by, and is named after the British physicist Paul A. M. Dirac (1902 – 1984), co winner of the Nobel Prize in Physics in 1933. A pioneer of quantum mechanics, Dirac is perhaps best known (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,
∞ ∞ L δ (t − c) e−st dt = δ (t − c) e−st dt = e−cs {δ(t − c)} = ∫0 ∫−∞ .
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,
∞ −st −cs L δ (t − c) f (t) e dt = f (c) e {δ(t − c) f (t)} = ∫0 .
© 2008, 2012 Zachary S Tseng C 3 2 Example : y″ + 6 y′ + 5 y = δ(t) + δ(t − 2), y(0) = 1, y′(0) = 0.
Transform both sides and simplify:
2 (s L{y} − s y(0) − y′(0)) + 6( sL{y} − y(0)) + 5 L{y} = L{δ(t) + δ(t − 2)}
2 −2s (s L{y} − s) + 6( sL{y} − 1) + 5 L{y} = 1 + e
2 −2s (s + 6 s + 5) L{y} − s − 6 = 1 + e
2 −2s −2s (s + 6 s + 5) L{y} = 1 + e + s + 6 = e + s + 7
e−2s s + 7 + L{y} = (s +1)(s + )5 (s +1)(s + )5 .
Further simplifying the two parts using partial fractions:
Part 1, without the e−2s term, becomes
1 1 1 1 1 = − (s +1)(s + )5 4 (s + )1 4 (s + )5 ,
1 −t 1 −5t It has an inverse transform e − e . 4 4
Now apply the effects of the e−2s term, resulting in
1 −t+2 −5t+10 u2 (t)(e − e ). 4
© 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 − e . 2 2
Therefore, the solution is
3 −t 1 −5t 1 −t+2 −5t+10 y = e − e + u2 (t)(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″ + 2 y′ + 10 y = −δ(t − 4π), y(0) = 0, y′(0) = 1.
Transform both sides and simplify:
2 (s L{y} − s y(0) − y′(0)) + 2(sL{y} − y(0)) + 10 L{y} = L{−δ(t − 4π)}
2 −4πs (s L{y} − 1) + 2(sL{y} − 0) + 10 L{y} = − e
2 −4πs (s + 2 s + 10) L{y} − 1 = − e
2 −4πs (s + 2 s + 10) L{y} = 1 − e
1 −4πs 1 L − e {y} = s2 + 2s +10 s2 + 2s +10 .
−4πs Notice that, other than the extra factor − e , 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 = s2 + 2s +10 3 (s + )1 2 + 32 .
1 −t Its inverse transform is e sin(3t) , which corresponds exactly to 3 the first part of the expression.
© 2008, 2012 Zachary S Tseng C 3 5 −4πs The second part, with the extra − e term, gains its effects (plus a negative sign) of a unit step function and a translation by c = 4π. It becomes
1 −(t−4π ) − u4π (t) (e sin (3 t − 4π )). 3
Therefore,
1 1 y = e−t sin(3t) − u (t)()e−(t−4π ) sin (3 t − 4π ) 3 3 4π 1 1 = e−t sin(3t) − u (t) e−t+4π sin(3t) 3 3 4π
© 2008, 2012 Zachary S Tseng C 3 6 Resonance induced by a discontinuous forcing function
Example : u″ + u = δ(t) + δ(t − 2 π) + δ(t − 4 π) + … + δ(t − 2000 π),
u(0) = 0, u′(0) = 0.
Transform and simplify the equation:
2 −2πs −4πs −2000 πs (s L{u} − 0 s − 0) + L{u} = 1 + e + e + … + e ,
2 −2πs −4πs −2000 πs (s + 1) L{u} = 1 + e + e + … + e ,
1 1000 1+ e−2nπs L{u} = 2 ∑ . 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, become u2nπ (t ) sin( t − 2 nπ ) = u2nπ (t ) sin( t).
Therefore, sin(t), 0 ≤ t < 2π 2sin(t), 2π ≤ t < 4π 1000 u = sin(t) + u2nπ (t)sin(t) = 3sin(t), 4π ≤ t < 6π ∑ n=1 : : 1001 sin(t), 2000π ≤ t
Notice the increase in amplitude, not continuously but rather in discrete steps, that occurs linearly with elapsed time. This is a discrete version of the resonance behavior.
© 2008, 2012 Zachary S Tseng C 3 7
The graph of the previous solution:
Transfer Function and Impulse Response
When a system or device can be described by a linear differential equation (a mass spring system, or a RLC electrical circuit, for examples) its transfer function , H(s), is the ratio of the Laplace transform of the system’s output (its response to an input, expressed by the equation’s solution), Y(s), divided by the Laplace transform of the system’s input (the equation’s forcing function), G(s), given that all initial conditions are zero. It is a description of how the system would react to a unit measure of external stimulus.
Take, as an example, a system described by a second order linear equation
a y″ + b y′ + c y = g(t).
Transform both sides, with all initial conditions zero, it becomes
2 a (s L{y} − 0s − 0) + b (sL{y} − 0) + c L{y} = L{g(t)}.
© 2008, 2012 Zachary S Tseng C 3 8 Simplify to obtain
2 (as + bs + c) L{y} = L{g(t)} = G(s),