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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 tt=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 ndnφ

dtne−2πif tt=T

t=0

(4)

or

X(f T ) = iT

2πf T

∞

X

n=0 −i

2πf T ndnφ

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

φt

T= sin2πt

T.(12)

The derivatives are

dφ

d(t/T )= 2πsin πt

Tcos 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 )2n

.

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 )2n

=1

1−1

(f T )2

−1 = 1

(f T )2−1.

Thus,

X(f T ) = iT

4πe−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

φt

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 )2n

.

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 )2n

=1

1−1

(2f T )2

=(2f T )2

(2f T )2−1

entailing

X(f T ) = T

πe−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