Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 1

Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU

SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

FOURIER TRANSFORM (FOURIER INTEGRAL)

The Fourier Integral does not converge for all functions.

The Dirichlet conditions,

 The function is absolutely integrable, that is

| | ∞

 has a finite number of maxima and minima and a finite number of

discontinuities in any finite interval,

provide a set of sufficient conditions for the existence of the Fourier Transform
.

If is absolutely integrable, then: lim → lim → 0

Intuitively this result derives from the fact that, for large , the exponential oscillates
faster than any length scale present in . Thus, for large enough, is
essentially constant over each interval 2 2 1 and the integral vanishes.

Functions that do not meet the Dirichlet conditions may still have a Fourier
Transform. These include periodic functions, whose transforms consist of impulses,
and functions whose Fourier Integral only converges as a limit.

__________

REFERENCES:

PAPOULIS, A. (1962). The Fourier Integral and its Applications,

McGraw-Hill New York

BRACEWELL, R.N. (1965). The Fourier Transform and its Applications,

McGraw-Hill New York

KRANIAUSKAS, P. (1992). Transforms in Signals and Systems, Addison

- Wesley
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 2
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

We use the notation:

↔

to indicate that the functions & form a Fourier Transform pair.

where: = Fourier Spectrum of

= Energy Spectrum of

= phase angle

Real time functions:

If f is real, then the real and imaginary parts of are given by:

cos sin

From the above expressions we conclude that is even and is odd, i.e.

Therefore,
∗
∗

∗
Conversely, if then = real.
∗
Thus, is a necessary and sufficient condition for to be real, i.e.

∗
⟺
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 3
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

EXAMPLE: Let , where: = Heaviside (unit step) function.

In the FIGURE below, we show the various ways of plotting :

1 1 ⁄
√
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 4
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

EXAMPLE: Find the Fourier Transform of the signum function defined as:

1 0
sgn
1 0

which is not absolutely integrable and its Fourier integral does not converge.

We form the auxiliary function: 0

0

which yields sgn as the limit: lim →

The Fourier transform of is:

and the transform of is obtained as the limit:

2 2
sgn lim
→

This yields the Fourier transform pair: sgn ↔

which is real odd in time, hence imaginary odd in frequency.

NOTE: Some authors denote the Fourier transform of by [instead of ]. This

is consistent with the fact that the Fourier Transform is a special case of Laplace
Transform.
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 5
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

SIMPLE THEOREMS:

The following is a list of simple theorems that can be easily derived from
the Fourier integral and its inverse; it is assumed that all functions under
consideration have Fourier integrals.

Most of these theorems are valid, in slightly modified forms, for the
Laplace Transform, the discreet classes of the Fourier
Transform and the z-transform [the z-transform is to discrete-
time signals what the Laplace Transform is to their continuous-time
counterparts].
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 6
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

LINEARITY:

Let: ↔ , ↔ , ⋯ , ↔

Then: ⋯ ↔ ⋯
where: , ,⋯, are arbitrary constants.

Proof: The theorem is self-evident and is based on the linearity of the Fourier Integral.
EXAMPLE: Find the transform of the unit step from the transforms of its even and odd
components.

NOTE: In general, any real function may be expressed as:

Therefore, in our example:

1 1 1
sgn ↔
2 2
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 7
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

TIME SCALING:

Let: ↔ then ↔ | |

Thus, expansion of the time scale (or time duration) leads to compression of the
frequency scale (or bandwidth) and vice versa. This is accompanied by an
inverse scaling of the amplitude.

Proof:
The change of variable , which implies ⁄ & ⁄ , yields:

. This is valid when 0. For 0 the integration limits

are inverted: . These are combined

into the single pair: ↔ | |
.

EXAMPLE:
1
1 | | sin
2 ↔ 2
1
0 | | 2
2
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 8
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

FREQUENCY SCALING:

Scaling the frequency variable by a factor has a similar effect to the scaling. Making the
substitution 1⁄ in expression ↔ | |
yields the expression:

↔
| |
which is completely symmetrical to the expression for ‘time scaling’ and, therefore, is an
expression of the duality of the Fourier Transform.

EXAMPLE:
sin
↔
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 9
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

SYMMETRY (DUALITY) OF TRANSFORM:

If is the Fourier Integral of , then:

↔

Proof:
The above follows from:
1
2

if we write it as follows:

and interchange and .

EXAMPLE: From ↔ 1 and the symmetry deduce that 1 ↔ 2 .

 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 10
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

SYMMETRY (DUALITY) OF TRANSFORM: (continued)

EXAMPLE: From knowledge of sin find the inverse transform of a

rectangular frequency domain pulse:
1
1 | |
2
0

EXAMPLE: Gaussian Function (represents the ultimate in symmetry and duality):

⁄
↔

For :
⁄
↔ √2
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 11
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

TIME SHIFTING:

Let ↔ then:
↔
i.e., if the function is shifted by a constant, , then its Fourier spectrum remains the
same, but a linear term is added to its phase angle.

Proof:

EXAMPLE:
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 12
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

FREQUENCY SHIFTING:
↔

Proof:

EXAMPLE:
1 1
↔ ⟹ ↔
2 2
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 13
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

TIME CONVOLUTION:

The Fourier Transform of the convolution of two functions & equals

the product of the Fourier Transform & of these two functions. Thus:

If ↔ & ↔

Then ∗ ≝ ↔

Proof:

Clearly,

Changing the order of integration, we obtain:

Therefore:

EXAMPLE:

COMMENT: In the above proof it was assumed that the order of integration can be
changed. This is true if the functions & are square-integrable in the sense
| | ∞ , , i.e., & have finite energy.
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 14
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

FREQUENCY CONVOLUTION:

From the time convolution theorem and the symmetry (duality) theorem it follows
that the Fourier Transform of the product of two functions equals the
convolution ∗ of their respective derivatives divided by :

One could also give a direct proof as in the time-convolution theorem.

 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 15
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

TIME DIFFERENTIATION:

Let ↔ then ↔

For 1 ⟹ ⟷ i.e., a time differentiation therefore causes a

frequency domain rotation of , and a linear scaling by .

Proof:

Taking the derivative of both sides of leads to:

1
⟹ ⟷
2

FREQUENCY DIFFERENTIATION:

Let ↔ then ⟷
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 16
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

INTEGRATION:

Let ↔ then ⟷

Proof:
We interpret the integral of as a convolution with the unit step function
, i.e.

Recall that:
1

Let:
↔

Then, invoking the convolution in time theorem, we get:

1
0

The term ⁄ represents the inverse of the differentiation property. If ,

then the properties (of differentiation and integration) are fully recoverable, in
the sense that the function can be recovered from through division by
.

In contrast, if the function contains a non-zero . . component, represented

by a non-zero value of , the transform of its integral contains an additional
impulse of strength at the origin.
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 17
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

EXAMPLE:
sin
↔

The . . value of the function is . Thus,:

sin
↔

The frequency representation consists of a hyperbola in the imaginary plane, truncated by

the pulse , and a real impulse at the origin of magnitude . This impulse
transforms back to the time domain as the constant ⁄ , which represents the
. . value of . Even symmetry of makes this value coincide with the value at the
origin, ⁄ , and leads to the asymptotic value ∞ .
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 18
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

EXAMPLE:

Let:
sin
↔ 2
2

The . . value of this function is . Thus:

sin
↔ 2 2

The integral of is a truncated ramp, whose value at the origin 0 corresponds to

the function’s . . value and related to the frequency domain impulse of
strength .
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 19
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

CONJUGATE FUNCTIONS:

∗ ∗
Let ↔ then ↔
∗
i.e., the Fourier Transform of the conjugate of a complex function
is given by ∗ .

Proof:
From

∗
⟹

∗
⟹

∗ ∗
⟹ ↔
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 20
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

PARSEVAL’s FORMULA:

If ↔ then:

| |

Proof:
∗ ∗
↔ ⟹ ↔

Therefore:
∗
1 ∗
| | ↔ ∗
2

i.e.
1 ∗ | |
2

1 ∗
⟹ | |
2

Therefore:
1
| |
2

The following is a more general form of Parseval’s formula:

If ↔ & ↔ then:

If & are real functions then:

NOTE: = energy spectrum of

∗
= cross-energy spectrum of &
 Lecture Notes: STRUCTURAL DYNAMICS / FALL 2011 / Page: 21
Lecturer: Prof. APOSTOLOS S. PAPAGEORGIOU
SEOUL NATIONAL UNIVERSITY
PART (07): FOURIER TRANSFORM (FOURIER INTEGRAL)

FOURIER TRANSFORMS INVOLVING IMPULSES:

We derive the Fourier Transforms of the time domain functions:

cos & sin

We can easily derive: ↔ & ↔

Therefore: cos ↔

and: sin ↔

1 1 1 1
↔ & ↔
2 2 2 2
1 1
↔ cos
2 2
1 1
↔ sin
2 2