Math 2280-001
Fri Apr 17
Continue the extended discussion of Laplace transform techniques in Wednesday's and today's notes:
, Using the unit step function to turn forcing on and off - Exercise 5 in Wednesday's notes.
, Convolution table entry, and resulting formulas to solve any inhomogeneous constant coefficient
linear DE, Exercises 6, 7 in Wednesday's notes
, Studying resonance phenomena with "engineering function" forcing functions and convolutions ...
today's notes.
Laplace table entries for today:
f t % CeM t
f t with
Fs d
N
f t eKs t dt
for s O M
comments
0
u tKa
eKa s
s
unit step function
f tKa u tKa
t
eKa sF s
f t g t K t dt
Fs Gs
for turning components on and
off at t = a .
more complicated on/off
convolution integrals to invert
Laplace transform products
0
Convolutions and solutions to non-homogeneous physical oscillation problems (EP7.6 p. 490-492) See
Wednesday's notes. Then do this exercise:
Exercise 1. Let's play the resonance game and practice convolution integrals. We'll stick with our earlier
swing, but consider various periodic forcing functions f t .
x## t C x t = f t
x 0 =0
x# 0 = 0
a) Find the weight function w t (see Wednesday notes.)
b) Write down the solution formula for x t as a convolution integral of w with f.
c) Work out the special case of X s when f t = cos t by hand if we still want to, and verify that the
convolution formula reproduces the answer we would've gotten from the table entry
t
sin k t
2k
s
s2 C k2
2
t
>
0
t
sin t cos t K t dt;
cos t sin t K t dt; #convolution is commutative
0
>
d) Then play the resonance game on the following pages with new periodic forcing functions ...
We worked out that the solution to our DE IVP
x## t C x t = f t
x 0 =0
x# 0 = 0
with arbitrary forcing function f will be
t
x t =
sin t f t K t dt
0
Since the unforced system has a natural angular frequency w0 = 1 , we expect resonance when the forcing
2p
= 2 p. We will discover that there is the possibility
w0
for resonance if the period of f is a multiple of T0 . (Also, forcing at the natural period doesn't guarantee
resonance...it depends what function you force with.)
function has the corresponding period of T0 =
Example 1) A square wave forcing function with amplitude 1 and period 2 p. Let's talk about how I
came up with the formula below (which works until t = 11 p ) .
> with plots :
10
> f1 d t/K1 C 2$
>
K1 n $Heaviside t K n$Pi
:
n=0
plot1a d plot f1 t , t = 0 ..30, color = green :
display plot1a, title = `square wave forcing at natural period` ;
square wave forcing at natural period
1
0
10
K1
20
30
t
1) What's your vote? Is this square wave going to induce resonance, i.e. a response with linearly growing
amplitude?
t
> x1 d t/ sin t $f1 t K t dt :
0
plot1b d plot x1 t , t = 0 ..30, color = black :
display plot1a, plot1b , title = `resonance response ?` ;
Example 2) A triangle wave forcing function, same period
t
> f2 d t/ f1 s ds K 1.5 : # this antiderivative of square wave should be triangle wave
0
plot2a d plot f2 t , t = 0 ..30, color = green :
display plot2a, title = `triangle wave forcing at natural period` ;
triangle wave forcing at natural period
1
0
K1
10
20
30
t
2) Resonance?
t
> x2 d t/ sin t $f2 t K t dt :
0
plot2b d plot x2 t , t = 0 ..30, color = black :
display plot2a, plot2b , title = `resonance response ?` ;
Example 3) Forcing not at the natural period, e.g. with a square wave having period T = 2 .
20
> f3 d t/K1 C 2$
>
K1 n $Heaviside t K n :
n=0
plot3a d plot f3 t , t = 0 ..20, color = green :
display plot3a, title = `out of phase square wave forcing` ;
out of phase square wave forcing
1
0
K1
3) Resonance?
5
10
t
15
20
t
> x3 d t/ sin t $f3 t K t dt :
0
plot3b d plot x3 t , t = 0 ..20, color = black :
display plot3a, plot3b , title = `resonance response ?` ;
Example 4) Forcing not at the natural period, e.g. with a particular wave having period T = 6 p .
10
> f4 d t/
>
Heaviside t K 6$ n$p K Heaviside t K 6$n C 1 $p
:
n=0
plot4a d plot f4 t , t = 0 ..150, color = green :
display plot4a, title = sporadic square wave with period 6p ;
sporadic square wave with period 6p
1
0
0
50
t
100
150
4) Resonance?
t
> x4 d t/ sin t $f4 t K t dt :
0
plot4b d plot x4 t , t = 0 ..150, color = black :
display plot4a, plot4b , title = `resonance response ?` ;
>
Hey, what happened???? How do we need to modify our thinking if we force a system with something
which is not sinusoidal, in terms of worrying about resonance? In the case that this was modeling a swing
(pendulum), how is it getting pushed?
Precise Answer: It turns out that any periodic function with period P is a (possibly infinite) superposition
P P P
of a constant function with cosine and sine functions of periods P,
,
,
,... . Equivalently, these
2 3 4
functions in the superposition are
2$p
1, cos w t , sin w t , cos 2 w t , sin 2 w t , cos 3 w t , sin 3 w t ,... with ω =
. This is the
P
theory of Fourier series, which we will study next week. If the given periodic forcing function f t has
P
non-zero terms in this superposition for which n$w = w0 (the natural angular frequency) (equivalently
n
2$p
=
), there will be resonance; otherwise, no resonance. We could already have understood some of
w0
this in Chapter 3, for example:
Exercise 3) Discuss whether or not resonance will occur with the following forcing functions, keeping the
left side of the DE the same as before. Notice that the period of the forcing function is 6 p in a (not the
natural period), but does equal 2 p in b. And yet, it is in a that we will have resonance, not in b.
a)
x## t C x t = cos t C sin
t
3
.
b)
x## t C x t = cos 2 t K 3 sin 3 t C cos 6 t
Exercise 4) Visually compare the graph of the following function (a truncated Fourier sine series), with
the square wave forcing function in Example 1. Believe it or not, the corresponding infinite sum
converges precisesly to the square wave for each t (except at the jumps, where t is a multiple of p, where
the series evaluates to zero). It's the first term in the series for the forcing square wave, namely the
4
$sin t term, which is causing the resonance in Example 1. To be continued ...
p
20
4
1
> sq d t/ $
sin
p k = 0 2$k C 1
>
2$k C 1 $t ;
20
4
sq := t/
> sin
k=0
2 kC1 t
2 kC1
(1)
p
> plot sq t , t = 0 ..30, color = green ;
1
0.5
0
K0.5
K1
>
10
t
20
30