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Mathematics

MATH 250

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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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