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An Engineer’s Guide to Mathematica®
190
is found that gn = fn /Δt. These relationships are valid when f(t) = 0 for t < 0. Furthermore,
it can be shown that the Fn are symmetrical about M/2. Therefore, when Gn is displayed as
the amplitude in a frequency band Δf centered at nΔf, it is displayed as Gn  = 2ΔtΔfFn  =
2Fn /M, where n = 1, 2, … , M/2.
In this notation, it can be shown that the total energy in the signal using the time domain
samples is
E = Δt
M
∑
fn2
(5.3)
n=1
This value is equal to the total power density obtained from the transformed quantities, which
is given by
E=
M∕2
2Δt ∑  2
F
M n=1  n 
(5.4)
Consequently, the power density Pn in the frequency band centered at nΔf is
Pn =
2Δt  2
F
M  n
n = 1, 2, ..., M∕2
The numerical implementations of Eqs. (5.1) and (5.2), respectively, are
Fn=Fourier[fn,FourierParameters>{a,b}]
and
fn=InverseFourier[Fn,FourierParameters>{a,b}]
In signal processing, auto and crosscorrelation on lists of signal amplitudes can be determined with
r12=Listcorrelate[list1,list2,{kL,kR}]
where list1 and list2 are lists of the same length and for signal processing kL = 1
and kR = 1 is adequate. When list1 = list2, we have autocorrelation; that is, r11.
Frequently, r12 is normalized as r12/(r11[[1]] r22[[1]])1/2 and r11 is normalized
as r11/r11[[1]], where the index 1 indicates the autocorrelation function at t = 0.
We now illustrate the use of these commands. Additional applications are given in Examples
10.3 and 10.4.
Numerical Evaluations of Equations
191
Example 5.27
Spectral Analysis of a Sine Wave
We shall examine the sampling of m periods of a sine wave of frequency fo and a maximum
magnitude of 2.5; that is,
fn = f (nΔt) = 2.5 sin(2𝜋fo nΔt)
1 ≤ n ≤ 𝛼m
where 𝛼 is the number of samples per period. Hence, is it seen that Δt = 1/(𝛼fo ) (since fh =
fo ) and M = m𝛼. It is assumed that fo = 10 Hz, 𝛼 = 50 and m = 4. We shall display the values
of the sampled signal, plot the magnitude of the components of the frequency spectrum given
by Gn , and take the inverse Fourier transform to recover the original signal. In addition, we
shall compute the total energy in the waveform using Eqs. (5.3) and (5.4).
In the program that follows, datp, datf, and ifdt are the coordinate pairs that are
placed in a format required by ListPlot and ListLinePlot. In addition, dat are the
values of the sampled waveform, ft are the complex values of the Fourier transform of the
sampled values, and ift are the values of the inverse of ft. In plotting Gn , only the first 20
data values are used. The option Filling is used to connect the data values with a straight
line that ends on the xaxis; this option is illustrated in Table 6.15. The results are shown in
Figure 5.20.
f(t)
Gn
2.5
2
2.0
1
1.5
0.1
0.2
0.3
0.4
t
1.0
–1
0.5
–2
10
20
30
(a)
40
f (Hz)
(b)
f(t)
2
1
0.1
0.2
0.3
0.4
t
–1
–2
(c)
Figure 5.20 (a) Sampled waveform, (b) spectral plot of the sampled waveform, and (c) inverse Fourier
transform
192
An Engineer’s Guide to Mathematica®
m=4.0; α=50.0; nn=m α; fo =10.0; dt=1/(α fo );
dat=Table[2.5 Sin[2 π fo n dt],{n,nn}];
datp=Table[{k dt,dat[[k]]},{k,1,nn}];
ListPlot[datp,AxesLabel>{"t","f(t)"}]
ft=Fourier[dat,FourierParameters>{1,1}];
datf=Table[{(k1)/(nn dt),2. Abs[ft[[k]]]/(m α)},
{k,1,nn/21}];
ListPlot[datf[[1;;20]],PlotRange>All,Filling>Axis,
AxesLabel>{"f (Hz)","Gn "}]
ift=InverseFourier[ft,FourierParameters>{1,1}];
ifdt=Table[{k dt,ift[[k]]},{k,1,nn}];
ListLinePlot[ifdt,AxesLabel>{"t","f(t)"}]
daf2=Table[Abs[ft[[k]]]ˆ2,{k,1,nn/21}];
Print["E from time data = ",dt Total[datˆ2]]
Print["E from transformed data = ",2 dt/nn Total[daf2]]
The following values of the total energy are displayed
E from time data = 1.25
E from transformed data = 1.25
Example 5.28
Spectral Analysis of a Sine Wave of Finite Duration
We again consider the sign wave of Example 5.27, except it is now sampled over a time T =
Km, where K ≥ 1 is an integer and m is the number of periods of the sine wave. We assume
that fo = 10 Hz, 𝛼 = 50, m = 4, and K = 8. It is mentioned that as K increases, the frequency
resolution increases; that is, Δf becomes smaller. We shall obtain a plot of the amplitudes Gn 
comprising the components of its amplitude spectrum and display them over the range 0 ≤
f ≤ 20 Hz. The maximum number of amplitudes to be plotted is chosen to be the integer value
of 20Km/fo . Thus,
kk=8; m=4.0; α=50.0; nn=kk m α; fo =10.0; dt=1/(α fo );
dat=Table[Piecewise[{{Sin[2 π fo n dt],1<=n<=α m},
{0,α m<=n<=nn}}],{n,nn}];
ft=Fourier[dat,FourierParameters>{1,1}];
datf=Table[{(k1)/(nn dt),2. Abs[ft[[k]]]/nn},{k,1,nn/2}];
last=Round[20 kk m/fo ];
ListPlot[datf[[1;;last]],Filling>Axis,
AxesLabel>{"f (Hz)","Gn "}]
produces the result shown in Figure 5.21.
Numerical Evaluations of Equations
193
Gn
0.12
0.10
0.08
0.06
0.04
0.02
5
10
15
20
f (Hz)
Figure 5.21 Amplitude spectral plot of a 10 Hz sine wave with a duration of four periods that is
sampled over a time interval of thirtytwo periods
Example 5.29
CrossCorrelation of a Signal with Noise
We shall obtain the normalized crosscorrelation of a sine wave with a sine wave with noise.
The amplitude of the noise will not exceed 20% of the sine wave amplitude. We use a
procedure employed in the previous two examples to generate the sampled waveform and
we use RandomReal to create the noise. The signal with noise is displayed along with the
crosscorrelation of the signals. The program that produces Figure 5.22 is as follows.
m=4.0; α=50.0; nn=m α; fo =10.0; dt=1/(α fo );
dat=Table[Sin[2 π fo n dt],{n,nn}];
datn=dat+0.2 RandomReal[{1,1},nn];
datp=Table[{k dt,datn[[k]]},{k,1,nn}];
ListLinePlot[datp]
datauto=ListCorrelate[dat,dat,{1,1}];
datnauto=ListCorrelate[datn,datn,{1,1}];
datcross=ListCorrelate[dat,datn,{1,1}];
datplot=Table[{k dt,datcross[[k]]/
Sqrt[datauto[[1]] datnauto[[1]]]},{k,1,nn}];
ListLinePlot[datplot]
An Engineer’s Guide to Mathematica®
194
1.0
0.5
0.1
0.2
0.3
0.4
0.3
0.4
–0.5
–1.0
(a)
1.0
0.5
0.1
0.2
–0.5
–1.0
(b)
Figure 5.22 (a) Sine wave with noise and (b) cross correlation of a sine wave with a sine wave with
noise
5.9
Functions Introduced in Chapter 5
The functions introduced in this chapter are listed in Table 5.1.
Numerical Evaluations of Equations
Table 5.1
195
Commands introduced in Chapter 5
Command
Purpose
Break
FindFit
Exit a Do or While command
Finds the numerical values of the parameters appearing in an
expression that will provide the best fit to a set of data
Searches for a numerical approximation to a local maximum of
an expression in a specified region
Searches for a numerical approximation to a local minimum of
an expression in a specified region
Searches for a numerical approximation to a root of an
expression in a specified region
Determines the discrete Fourier transform of a list of numerical
values
Constructs a function that approximates a given set of data
Determines the discrete inverse Fourier transform on a list of
numerical values
Determines the crosscorrelation of two lists of numerical
values
Finds the numerical solution to one or a set of ordinary
differential equations or to a set of partial differential
equations with two independent variables
Obtains a numerical approximation to a definite integral or
definite multiple integral
Attempts to obtain a numerical solution to an equation or a
system of equations
Finds the numerical solution to one or a set of ordinary
differential equations or to a partial differential equation with
two independent variables as a function of one or more
parameters
Suppresses the display of any messages generated by the
implementation of a Mathematica function
Generates a pseudorandom real number
FindMaximum
FindMinimum
FindRoot
Fourier
Interpolation
InverseFourier
ListCorreleate
NDSolveValue
NIntegrate
NSolve
ParametricNDSolveValue
Quiet
RandomReal
References
[1] S. Chatterjee, A. K. Mallik, and A. Ghosh, “On impact dampers for nonlinear vibrating systems”, Journal of
Sound and Vibration, 1995, 187(3), pp. 403–420.
[2] F. C. Moon, “Experiments on chaotic motions of a forced nonlinear oscillator: strange attractors,” ASME Journal
of Applied Mechanics, 1980, 47(3), pp. 638–644.
[3] Y. Zhu, H. N. O˘guz, and A. Prosperetti, “On the mechanism of air entrainment by liquid jets at a free surface,”
Journal of Fluid Mechanics, 2000, 404, pp. 151–177.
[4] S. Timoshenko and S. WoinowskyKrieger, Theory of Plates and Shells, 2nd edn, McGraw Hill, New York,
1959.
[5] F. P. Incropera and D. P. Dewitt, Introduction to Heat Transfer, 4th edn, John Wiley & Sons, New York, 2002.
[6] T. Dahlberg, “Procedure to calculate deflections of curved beams,” Journal of Engineering Education, 2004,
20(3), pp. 503–513.
[7] S. Kosti´c, I. Franovi´c, K. Todorovi´c, and N. Vasovi´c, “Friction memory effect in complex dynamics of earthquake
model”, Nonlinear Dynamics, 2013, 73(3), pp. 1933–1943.
An Engineer’s Guide to Mathematica®
196
[8] B. S. Shvartsman, “Analysis of large deflections of a curved cantilever subjected to a tipconcentrated follower
force,” International Journal Nonlinear Mechanics, 2013, 50, pp. 75–80.
[9] A. Belendez, A. Hernandez, T. Belendez, M. L. Alvarez, S. Gallego, M. Ortuno, and C. Neipp, “Application
of the harmonic balance method to a nonlinear oscillator typified by a mass attached to a stretched string,”,
J. Sound Vibration, 2007, 302, pp. 1018–1029.
[10] A. S. dePaula, M. A. Savi, and F. H. I. PereiraPinto, “Chaos and transient chaos in an experimental nonlinear
pendulum,” Journal of Sound and Vibration, 2006, 295, pp. 585–595.
[11] Y. Starosvetsky and O.V. Gendelman, “Interactions of nonlinear energy sink with a two degrees of freedom
system: Internal resonance,” Journal of Sound and Vibration, 2010, 329, pp. 1836–1852.
[12] J. J. Thomsen, Vibrations and Stability, 2nd edn, Springer, Berlin, 2003.
[13] A. A. AlQaisia and M. N. Hamdan, “Subharmonic resonance and transition to chaos of nonlinear oscillators with a combined softening and hardening nonlinearities,” Journal of Sound and Vibration, 2006, 305,
pp. 772–782.
[14] S. S. Rao, Engineering Optimization, John Wiley & Sons, New York, 1996, p.706.
[15] I. G. Currie, Fundamentals of Fluid Mechanics, McGrawHill, New York, 1993, p. 181.
[16] B. R. Munson, D. F. Young, and T. H. Okiishi, Fundamentals of Fluid Mechanics, 4th edn, John Wiley & Sons,
New York, 2002, p. 641.
[17] J. H. Duncan, Chapter 11, Fluid Mechanics, in E. B. Magrab, et al., An Engineer’s Guide to MATLAB, 3rd edn,
Prentice Hall, Upper Saddle River, New Jersey, 2011, p. 618.
[18] E. E. Lundquist and E. Z. Stowell, “Critical compressive stress for flat rectangular plates supported
along all edges and elastically restrained against rotation along the unloaded edges,” NACA Report 733,
1942.
Exercises
Section 5.2
5.1
Show numerically that
√
6
√
du
= cot −1 2
√
∫
6 − u2
2
5.2
The total emissive power of a black body is determined from [5, p. 676]
∞
Eb =
c1 d𝜆
) = 𝜎T 4
(
∫ 𝜆5 ec2 ∕(𝜆T) − 1
0
5.3
where c1 = 3.742 × 108 W μm4 ⋅m−2 , c2 = 1.4388 × 104 μm⋅K, and 𝜎 = 5.670 ×
10−8 W⋅m−2 ⋅K−4 . Show that for T = 500 K the difference between the exact solution
(righthand side) and the approximate solution (numerical solution to the integral) is
0.006272%.
In determining the deflection of curved beams, the following relationship is obtained
[6]
𝛿(𝛽) = I1 (𝛽) + 1.3I2 (𝛽)
Numerical Evaluations of Equations
197
where 𝛿 is the nondimensional deflection, 𝛽 ≥ 0, and
1
I1 (𝛽) =
∫
0
[
]2
−𝛽 2 𝜉 − (1 − 𝜉)
d𝜉
√
1 + y2
[ √
]2
𝛽 1 − 𝜉 2 + (1 − 𝜉)y
I2 (𝛽) =
d𝜉
√
∫
1 + y2
1
0
−𝛽𝜉
y= √
1 − 𝜉2
5.4
Find the value of 𝛿(1.5).
Show numerically for z = 0.5 and y = 0.15 that
𝜋
(√
)
1
ey cos 𝜃 cos(z sin 𝜃)d𝜃 = J0
z2 − y2
𝜋∫
0
where J0 (x) is the Bessel function of the first kind of order zero.
Section 5.3
5.5
Determine the solution to the following system of nonlinear ordinary differential equations
dy1
= y2
dt
(
)
dy2
1
2 dL
=−
+ 𝛾L y2 − sin(y1 )
dt
L dt
L
over the range 0 ≤ t ≤ 30 where
L = 1 + 𝜀 sin7 (𝜔t + 9𝜋∕8)
dL
= 7𝜀𝜔 sin6 (𝜔t + 9𝜋∕8) cos(𝜔t + 9𝜋∕8)
dt
5.6
The initial conditions are y1 (0) = −1 and y2 (0) = 1 and the constants have the following
values: 𝜀 = 0.16, 𝛾 = 0.4, and 𝜔 = 0.97. Plot y1 (t) versus y2 (t) using ParametricPlot.
The nondimensional axial velocity Uz of blood in a large artery undergoing sinusoidal
oscillations can be can be determined from
𝜕Uz
= − sin 𝜏 + 𝛽
𝜕𝜏
(
1 𝜕
𝜂 𝜕𝜂
(
))
𝜕Uz
𝜂
𝜕𝜂
An Engineer’s Guide to Mathematica®
198
where 0 ≤ 𝜂 ≤ 1, 𝜏 is a nondimensional time, and 𝛽 is a constant. The initial condition
is
Uz (𝜂, 0) = 0
and the boundary conditions are
Uz (1, 𝜏) = 0
𝜕Uz 

=0
𝜕𝜂 𝜂=0
5.7
Solve this system and plot Uz as a function of 𝜂 and 𝜏, 0 ≤ 𝜏 ≤ 10, using Plot3D
when 𝛽 = 0.05. This equation has a discontinuity at 𝜂 = 0 so that the technique given
in Example 5.16 will have to be used.
The nondimensional equation for heat conduction in a slab is given by
𝜕𝜃 𝜕 2 𝜃
=
𝜕𝜏 𝜕𝜉 2
where 𝜃 = 𝜃(𝜉,𝜏) is proportional to the temperature in the slab. We are interested in the
region 0 ≤ 𝜏 ≤ 4.0 and 0 ≤ 𝜉 ≤ 1. The initial condition is 𝜃(𝜉,0) = 0 and the boundary
conditions are
𝜃(0, 𝜏) = 𝜏e−𝜏
𝜃(1, 𝜏) = 0
5.8
Solve this system and use Plot3D to display 𝜃.
The equilibrium equations in terms of nondimensional quantities that govern the large
deflections of a solid circular plate of constant thickness are given by [4, p. 402]
( )2
d2 u 1 du
1 − 𝜈 dw
dw d2 w
u
+
−
− 2 =−
2
𝜂 d𝜂 𝜂
2𝜂
d𝜂
d𝜂 d𝜂 2
d𝜂
[
( )2 ]
𝜂
u 1 dw
d2 w 1 dw w
dw du
+ qo
+
− 2 = ha
+𝜈 +
2
𝜂
d𝜂
d𝜂
d𝜂
𝜂
2
d𝜂
2
d𝜂
𝜂
If the plate is clamped along the outer boundary, the boundary conditions are
u(1) = 0,
w(1) = 0,
w′ (1) = 0
At the center of the plate, symmetry is used to require that
u(0.001) = 0,
w′ (0.001) = 0
Numerical Evaluations of Equations
199
where, to avoid the discontinuity at 𝜂 = 0, the center of the plate is approximated by a
hole of radius 𝜂 = 0.001.
The displacement of a solid circular plate using linear theory that is clamped at its
outer edge and subjected to a constant load is given by
w(𝜂) =
5.9
)
qo (
1 − 2𝜂 2 + 𝜂 4
64
Use these equations to plot the linear and nonlinear transverse displacements of the
plate when 𝜈 = 0.3, ha = 0.001, and qo = 10000.
The following three coupled nonlinear equations in terms of nondimensional quantities
are given [7]
d𝜃
= −v [𝜃 + (1 + 𝜀) ln v]
dt
du
= v−1
dt
[
]
dv
= −𝛾 2 u + (𝜃 + ln v)∕𝜉
dt
5.10
Obtain the solution to these equations when u(0) = v(0) = 1 and 𝜃(0) = 0 and for
𝜀 = 0.5, 𝛾 = 0.8, and 𝜉 = 0.5. Display the results two ways: as a plot of v(t) versus
t using Plot and as phase portrait of u(t) versus v(t) using ParametricPlot. In
both displays, let 0 ≤ t ≤ 1000.
Consider a curved cantilever beam of initial radius of curvature R, length L, Young’s
modulus E, and moment of inertia I. A load P is applied to the free end of the beam at
an angle 𝛼, where 𝛼 = 0 indicates a load that is parallel to the axis of the beam and an
angle 𝛼 = 𝜋/2 indicates a load perpendicular to axis of the beam. If s is the arc length
of the beam and 𝜑(s) is the slope of the centroidal axis of the beam, then the equation
from which the deflection of the beam can be determined is obtained from [8]
𝜃 ′′ (𝜂) + po 𝜃(𝜂) = 0
where 𝜂 = s/L, the prime denotes the derivative with respect to 𝜂, and
𝜑(𝜂) = 𝜃(𝜂) − 𝜃(L)
po =
PL2
EI
The boundary conditions are given by
𝜃(0) = 𝛼
and
𝜃 ′ (0) = −
L
R
An Engineer’s Guide to Mathematica®
200
The nondimensional Cartesian coordinates of a point 𝜂 on the beam are given by
1
x(𝜂) =
cos(𝜑(𝜌))d𝜌
∫
𝜂
1
y(𝜂) =
∫
sin(𝜑(𝜌))d𝜌
𝜂
which describe the deformed shape of the beam.
Determine the shape of the beam for the case of 𝛼 = 0 and po = {0.02, 3, 9,
12} and for the case of 𝛼 = 𝜋/2 and po = {0.02, 5, 15, 25, 35}. The results should
look like those shown in Figure 5.23. Assume that the initial shape of the beam
is a semicircle so that L/R = 𝜋. The sign of the ycoordinate has been changed to
obtain the curves shown in this figure. The topmost curve belongs to the largest value
of po .
0.6
0.4
0.5
0.2
–0.2
0.2
0.4
0.6
0.0
0.2
0.4
0.6
0.8
1.0
–0.2
–0.5
–0.4
–0.6
(a)
(b)
Figure 5.23 Solution to Exercise 5.10: (a) 𝛼 = 0 and po = {0.02, 3, 9, 12} (b) 𝛼 = 𝜋/2 and po = {0.02,
5, 15, 25, 35}