On the Zeros of the Fourier Transforms of Some Functions of Finite Duration

Abstract

This short note examines the zeros of the Fourier transforms of functions of finite duration. Two classes of functions are identified for which the zeros of the Fourier transform can be calculated without knowledge of the Fourier transform.

Full text

On the Zeros of the Fourier Transforms of Some
Functions of Finite Duration
Prof. Dr.-Ing. Johannes Wandinger
March 26, 2022
1 Introduction
A function of finite duration is a function of the form
x(t) = 


0t < 0
φ(t/T ) 0 ≤t≤T
0t>T
(1)
where Tis the duration of the function and φis an arbitrary function. Functions
of this type occur e.g. in structural dynamics where they describe impact loads.
In a frequency or transient response analysis, it is necessary to determine a cut-off
frequency, i.e. a frequency above which the magnitude of the Fourier transform is
negligible. In many cases, the magnitude of the Fourier transform has zeros that
are uniformly distributed, and the maximum of the magnitude between the zeros
rapidly decreases with increasing frequency. Therefore, the cut-off frequency can
be easily determined if the zeros are known.
This article shows that in some cases the Fourier transform can be expanded into
an infinite series depending on the values of the derivatives of function φ(t/T ) at
t= 0 and t=T. This is possible if the derivatives exist and the series converges.
Two special cases are identified where the zeros can be easily determined from the
series and are found to be uniformly distributed.
2 Series Expansion of the Fourier Transform
The Fourier transform of a function of finite duration Treads
X(f) = Z∞
−∞
x(t)e−2πif tdt =ZT
0
φ(t/T )e−2πif tdt. (2)
By partial integration,
ZT
0
dnφ
dtne−2πif tdt =−1
2πif dnφ
dtne−2πif tt=T
t=0
−ZT
0
dn+1φ
dtn+1 e−2πif tdt!.(3)
1

Repeated application of Eq. 3 to Eq. 2 leads to
X(f) = −1
2πif
∞
X
n=0 1
2πif ndnφ
dtne−2πif tt=T
t=0
(4)
or
X(f T ) = iT
2πf T
∞
X
n=0 −i
2πf T ndnφ
d(t/T )n(1)e−2πif T −dnφ
d(t/T )n(0).(5)
Eq. 5 shows that, if the series converges, the Fourier transform depends on fT
and on the values of φ(t/T ) and all its derivatives at t= 0 and t=T, provided
these derivatives exist. Under the same conditions, the Fourier transform X(fT )
can be seen to be proportional to the duration T.
3 Zeros of the Fourier Transform
In the following two cases, the zeros of the Fourier transform can be easily deter-
mined from Eq. 5.
Case 1
dnφ
d(t/T )n(0) = dnφ
d(t/T )n(1) ∀n∈N0(6)
With Eq. 6 Eq. 5 reads
X(f T ) = iT
2πf T e−2πif T −1
∞
X
n=0 −i
2πf T ndnφ
d(t/T )n(1).(7)
Eq. 7 shows that X(f T ) = 0 for
fnT=n, n ∈N,(8)
provided the infinite series converges for these values. In many practical cases, the
series converges for n > 1.
Case 2
dnφ
d(t/T )n(0) = −dnφ
d(t/T )n(1) ∀n∈N0(9)
With Eq. 9 Eq. 5 reads
X(f T ) = iT
2πf T e−2πif T + 1
∞
X
n=0 −i
2πf T ndnφ
d(t/T )n(1).(10)
Now, X(f T ) = 0 for
fnT=2n+ 1
2, n ∈N0,(11)
provided the infinite series converges.
2

4 Examples
Example 1
A function frequently occurring in structural dynamics or aeroelasticity is
φt
T= sin2πt
T.(12)
The derivatives are
dφ
d(t/T )= 2πsin πt
Tcos πt
T=πsin 2πt
T
d2φ
d(t/T )2= 2π2cos 2πt
T
d3φ
d(t/T )3=−4π3sin 2πt
T
d4φ
d(t/T )4=−8π4cos 2πt
T
d5φ
d(t/T )5= 16π5sin 2πt
T
. . .
The values at t= 0 and t=Tare
φ(0) = 0, φ(1) = 0
dφ
d(t/T )(0) = 0,dφ
d(t/T )(1) = 0
d2φ
d(t/T )2(0) = 2π2=1
2(2π)2,d2φ
d(t/T )2(1) = 2π2=1
2(2π)2
d3φ
d(t/T )3(0) = 0,d3φ
d(t/T )3(1) = 0
d4φ
d(t/T )4(0) = −8π4=−1
2(2π)4,d4φ
d(t/T )4(1) = −8π4=−1
2(2π)4
. . . . . .
or d2nφ
d(t/T )2n(0) = d2nφ
d(t/T )2n(T) = −(−1)n
2(2π)2n, n ∈N.(13)
The values of the derivatives at the two ends of the interval are identical, so this
is case 1. Therefore, the zeros are
fnT=n, n ∈N
if the series converges. With Eq. 13, the series reads
∞
X
n=0 −i
2πf T ndnφ
d(t/T )n(1) = −1
2
∞
X
n=1 −i
2πf T 2n
(−1)n(2π)2n
=−1
2
∞
X
n=1 1
(f T )2n
.
3

This geometric series converges if
fT > 1.
Thus the zeros of the Fourier transform are
fnT= 2,3,4, . . .
For f T > 1 the series converges to
∞
X
n=1 1
(f T )2n
=1
1−1
(f T )2
−1 = 1
(f T )2−1.
Thus,
X(f T ) = iT
4πe−2πif T −1
fT (1 −(f T )2).(14)
In fact, Eq. 14 is also valid when fT < 1 and the value at fT = 1 exists as a limit.
It is identical to the Fourier transform of x(t) obtained by elementary calculations.
Example 2
A function often used in structural dynamics to describe shock loads is
φt
T= sin πt
T.(15)
The deriatives are
dφ
d(t/T )=πcos πt
T
d2φ
d(t/T )2=−π2sin πt
T
d3φ
d(t/T )3=−π3cos πt
T
d4φ
d(t/T )4=π4sin πt
T
The values at t= 0 and t=Tare
φ(0) = 0, φ(1) = 0
dφ
d(t/T )(0) = π, dφ
d(t/T )(1) = −π
d2φ
d(t/T )2(0) = 0,d2φ
d(t/T )2(1) = 0
d3φ
d(t/T )3(0) = −π3,d3φ
d(t/T )3(1) = π3
d4φ
d(t/T )4(0) = 0,d4φ
d(t/T )4(1) = 0
. . . . . .
4

or d2n+1
d(t/T )2n+1 (1) = −d2n+1
d(t/T )2n+1 (0) = −(−1)nπ2n+1, n ∈N0.(16)
Here, the values of the derivatives at the two ends of the interval have opposite
sign, so this is case 2. Therefore, the zeros are
fnT=2n+ 1
2, n ∈N0(17)
if the series converges. With Eq. 16, the series reads
∞
X
n=0 −i
2πf T ndnφ
d(t/T )n(1) = −
∞
X
n=0 −i
2πf T 2n+1
(−1)nπ2n+1
=i
2fT
∞
X
n=0 1
(2f T )2n
.
This geometric series converges if
2fT > 1.
Thus the zeros of the Fourier transform are
fnT=3
2,5
2,7
2. . .
For 2f T > 1 the limit of the series is
∞
X
n=0 1
(2f T )2n
=1
1−1
(2f T )2
=(2f T )2
(2f T )2−1
entailing
X(f T ) = T
πe−2πif T + 1
1−(2f T )2.(18)
Again, Eq. 14 is also valid when 2fT < 1 and the value at 2fT = 1 exists as a
limit. Elementary calculation of the Fourier transform also yields Eq. 18.
5