Math 344
May 24, 2012
Complex Fourier Series
Part II: Application to Periodic Solutions to an ODE
Consider an object that is attached to a wall by a spring and forced to oscillate to and fro by a periodic driving
function f (t) = F0 cos(ωt). Let x(t) be the object’s displacement from equilibrium at time t. Under the assumption of
small displacements, the system is governed by the following second order, constant-coefficient, differential equation.
mx00 + bx0 + kx = F0 cos(ωt)
(1)
The letter m denotes the mass of the object, k is the spring constant, and the letter b denotes the damping constant.
All three of these constants, and the constants F0 and ω, are
p assumed to be positive; ω is called the forcing frequency.
Recall that the natural frequency of the system is ω0 = k/m . This is the frequency of the oscillations when the
system is undamped and unforced: mx00 + kx = 0 .
Steady-State Solutions
The general solution to Equation (1) is x(t) = xh (t) + xp (t) where xh (t) is the general solution to the associated
homogeneous equation and xp (t) is a particular solution to the forced equation. When the system is damped, xh (t)
contains decaying exponentials and it approaches zero as t → ∞. After sufficient time has passed, the motion will
be governed by xp (t) and the system is said to have reached steady-state. The steady-state solution, also denoted
xss (t), is periodic with the same frequency as the driver. This is the motion that interests us here.
The steady-state solution for the simple driver F0 cos(ωt) in Equation (1) can be found by substituting x =
A cos(ωt) + B sin(ωt) into (1) and solving for A and B. When the driver is a general periodic function of frequency
ω the steady-state solution can be found by a similar method involving the complex Fourier series for the driver.
Steady-State Solution for Any Periodic Driver
We wish to obtain the steady-state solution to
mx00 + bx0 + kx = f (t)
(2)
where f is periodic of period P and (circluar) frequency ω = 2π/P . Here is how.
1. Obtain the complex Fourier series representation for the driver.1
∞
X
f (t) =
cn einωt
n=−∞
2. Substitute x(t) = an einωt into the left side of (2) and solve for an , assuming that the driver is cn einωt . That
is, let x = an einωt in
mx00 + bx0 + kx = cn einωt
(3)
and solve for an . This turns out to be easy because all of the exponential terms cancel. For the n = 0 case
substitute x = a0 and the constant a0 will simply be c0 /k. However, for n ≥ 1, an will be a complex number.
3. Once an is found, let a−n = an , and the steady-state solution to (2) is
xss (t) =
∞
∞
X
c0
c0 X
+
an einωt =
+
2 <(an einωt ) .
k
k
n=−∞
n=1
(4)
n6=0
Here is an example.
Example 1. Obtain the steady-state solution to the mass-spring system x00 + x0 + 4x = f (t) where f is periodic of
period P = π defined on the interval 0 < t < π as follows.
1 , 0 < t < π/2
f (t) =
−1 , π/2 < t < π
1 Note
the switch from k to n for the summation index in the Fourier series. This is to avoid confusion with the spring constant.
1
Plot the steady-state solution and the driver, suitably scaled to the dimensions of position vs time.
Solution. The driver is odd. See its graph
P∞ below. Therefore, c0 = 0. Since P = π, ω = 2π/P = 2, so the Fourier
series for the driver has the form f (t) = n=−∞ cn e2int where
!
Z
Z π/2
Z π
1 π
1
−2int
−2int
−2int
f (t)e
e
cn =
dt =
dt +
(−1)e
dt
π 0
π
0
π/2
t=π/2
t=π !
1 e−2int e−2int =
+
π −2in t=0
2in t=π/2
1
=
(1 − e−πin ) + (e−2πin − e−πin )
2πin
1
=
(2 − 2(−1)n ))
2πin
(−1)n − 1
=
i.
πn
The coefficient cn is pure imaginary. This is always the case when the waveform is odd. Observe also that when n
is even, cn = 0. The following plot confirms that these are the correct coefficients.
1
0.5
Kp
K
0.5
p
K
K
0.5
0.5
p
p
1.5
p
2
p
t
1
The driver f and its Fourier approximation S9 (t).
Because there is no constant in the Fourier series for f , there is no constant in the steady-state solution. Therefore,
xss (t) =
∞
X
2 <(an e2int )
n=1
where an is found by substituting x = an e2int into the differential equation
x00 + x0 + 4x = cn e2int .
Doing so yields the following equation
−4n2 an e2int + 2inan e2int + 4an e2int = cn e2int ,
which simplifies to
(4 − 4n2 + 2in)an = cn .
Consequently, the nth complex Fourier series coefficient for the steady-state solution is an = cn /(4 − 4n2 + 2in).
Observe that these coefficients approach 0 very quickly, implying that the oscillations in the steady-state solution will
be completely determined by the first few non-zero harmonics. The amplitude spectrum for the driver is sketched
below. The amplitude spectrum for the steady-state solution follows.
0.6
0.4
0.2
K K K K K K K K K K
20
18
16
14
12
10
8
6
4
2
0
2
4
6
8
10
12
14
16
The amplitude spectrum for the periodic driver f .
2
18
20
0.3
0.2
0.1
K K K K K K K K K K
20
18
16
14
12
10
8
6
4
2
0
2
4
6
8
10
12
14
16
18
20
The amplitude spectrum for the steady-state solution xss .
The following picture shows the driver and the steady-state solution. Five harmonics were used to plot xss . The
picture obtained using just the first harmonic in the solution looks essentially the same.
1
0.5
0
K
K
0.5
0.5
p
p
1.5
p
2
p
2.5
p
3
p
t
1
The driver f and the steady-state solution xss .
A formula for the first harmonic in xss
First observe that cn is zero when n is even. For odd n, cn = −2i/nπ . Therefore,
an =
−2i
cn
=
, n odd .
2
4 − 4n + 2in
nπ(4 − 4n2 + 2in)
In particular, when n = 1, a1 = −1/π. Therefore, using just the first harmonic,
xss (t) ≈ 2 <(a1 e2it ) = −
2
cos(2t) .
π
2
The amplitude of the first harmonic is 0.637. The next non-zero harmonic is − 3π√9+16
cos(6t − δ3 ) where δ3 =
2
2
arctan(16 /9). Its amplitude is 0.013.
A General Formula
The method illustrated in Example 1 is completely general, applying to any piecewise smooth driver of period P . In
summary, it proceeds as follows. Begin by expressing the system in the form
mx00 + bx0 + kx =
∞
X
cn einωt , ω = 2π/P .
(5)
n=−∞
The steady-state solution will then have the form
xss (t) =
∞
X
an einωt
n=−∞
where x = an einωt is a particular solution to
mx00 + bx0 + kx = cn einωt .
To find an simply substitute x = an einωt into the last equation to obtain
−mn2 ω 2 an einωt + ibnωan einωt + kan einωt = cn einωt
which implies that
an =
Consequently,
xss (t) =
cn
.
k − mn2 ω 2 + ibnω
∞
X
cn
einωt .
2 ω 2 + ibnω
k
−
mn
n=−∞
Three comments:
3
(6)
1. The form of the nth complex Fourier coefficient in Equation (6) makes it clear that the higher-order harmonics
have almost no effect on the steady-state solution.
2. If information about a particular harmonic is needed, say the nth one, its amplitude is 2 |an | (why?). Therefore,
nth Amplitude = An = p
2|cn |
(k − mn2 ω 2 )2 + b2 n2 ω 2
.
3. An explicit formula for the nth steady-state harmonic is given by
nth Harmonic = An cos(nωt − (δn − γn ))
where δn is the argument of k − mn2 ω 2 + ibnω and γn is the argument of cn .
The following example illustrates these observations.
Example 2. Consider the mass-spring system
x00 + 0.1x0 + 24x = f (t)
where f is the period 2 driver defined in Example 1 of the Complex Fourier Series, Part I handout.
f (t) =
t
0
, 0<t<1
, 1<t<2
1.0
0.5
0
1
2
3
4
5
t
The driver f has period 2.
Since P = 2, ω = 2π/P = π, and f (t) =
c0 =
P∞
n=−∞ cn e
1
4
iπt
and cn =
1
2
where
(−1)n
(−1)n − 1
i+
nπ
n2 π 2
.
Therefore, the steady-state solution xss has the complex Fourier series
xss (t) =
∞
X
cn
1/4
+
einπt
2
2 + ibnπ
k
k
−
mn
π
n=−∞
n6=0
where m = 1, b = 0.1, and k = 24. That is,
∞
X
1
cn
xss (t) =
+
einπt .
96 n=−∞ 24 − n2 π 2 + 0.1inπ
n6=0
The following picture shows the amplitude spectrum of the driver.
0.2
0.1
K p K p K p K p K p K p K p Kp
8
7
6
5
4
3
2
0
p
2
p
3
p
4
p
The driver’s amplitude spectrum.
And here is the amplitude spectrum of the steady-state solution.
4
5
p
6
p
7
p
8
p
6
0.015
0.01
0.005
K p K p K p K p K p K p K p Kp
8
7
6
5
4
3
2
p
0
2
p
3
p
4
p
5
p
6
p
7
p
8
p
The amplitude spectrum of the steady-state solution.
The steady-state amplitude spectrum shows that the constant and the first three harmonics are enough to build an
almost perfect picture of the steady-state motion. See the plot below, showing a scaled version of the driver and
the steady-state approximation using 3 harmonics. The effects of the first two harmonics are especially evident. See
Exercise 11 in Exercise Set 5.
0.04
0.03
0.02
0.01
K
K
K
0
1
2
3
0.01
4
5
6
t
0.02
0.03
The scaled driver (dotted) and the steady-state solution
using the constant term and the first three harmonics.
Interesting Observation. Using the constant and the first three harmonics yields only a mediocre approximation
of the driver. See the plot in Example 1 of the Part I handout. However, the driver’s power spectrum on page 4 of
Part I clearly shows that its first three harmonics in the driver carry almost all of its power.
5