Stanford University

Spring 2010-2011

Signal Processing and Linear Systems I

Lecture 8: The Continuous Time Fourier Transform

January 30, 2011

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Introduction to Fourier Transforms

• Fourier transform as a limit of Fourier series

• Inverse Fourier transform: The Fourier integral theorem

• Examples: the rect function, one-sided exponential

• Symmetry properties

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 Fourier Series

We can expand f (t) as a Fourier series in interval [−T /2, T /2):

∞
�
f (t) = Dnejnω0t
n=−∞

where �
1 T /2
Dn = f (t)e−jnω0 dt,
T −T /2
and
2π
ω0 = .
T
What happens if we let T increase?

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For example, assume y(t) = rect(t), and that we are computing the Fourier
series over an interval T ,

f (t) = rect(t)

−1/2 1/2 t
T

The fundamental period for the Fourier series in T , and the fundamental
frequency is ω0 = 2π/T .

From Lecture 7, the Fourier series is

1 �
∞ �n�
f (t) = sinc ejnω0t
T n=−∞ T

sin(πt)
where sinc(t) = πt is plotted below

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 1

-4 -2 0 2 4 t

If we plot the Fourier series coefficients as a function of ω = nω0 for T =

1, 2, and 4,

1/2
T =2
T =1 1 ω0 = π
ω0 = 2π
-8! -4! 0 4! 8! " = n "0

T =4 1/4
-8! -4! 0 4! 8! ω0 = π/2
" = n "0
-8! -4! 0 4! 8! " = n "0

More densely sampled, same sinc() envelope, decreased amplitude.

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Fourier Transforms

Given a continuous time signal f (t), define its Fourier transform as the
function of a real ω:
� ∞
F (jω) = f (t)e−jωt dt
−∞

if the integral makes sense.

This is similar to the expression for the Fourier series coefficients.
We can interpret this as the result of expanding f (t) as a Fourier series in
an interval [−T /2, T /2), and then letting T → ∞.
The Fourier series for f (t) in interval [−T /2, T /2):
∞
�
f (t) = Dnejnω0t
n=−∞

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where �
1 T /2
Dn = f (t)e−jnω0t dt.
T −T /2
Define the truncated Fourier transform:
� T
2
FT (jω) = f (t)e−jωt dt
− T2

so that
1
Dn = FT (jnω0).
T
The Fourier series is then

�∞
1
fT (t) = FT (jnω0)ejnω0t
n=−∞
T

T = ω0 → 0. Suppose that n increases with T so that

As T → ∞, then 2π
nω0 → ω where ω is fixed (this is only approximate except as limit).

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The limit of the truncated Fourier transform is

� T
2
� ∞
F (jω) = lim FT (jω) = lim f (t)e
−jωt
dt = f (t)e−jωt dt
T →∞ T →∞ − T2 −∞

Taking limit in Fourier series for f (t) using Riemann approximations to

integral with ∆ω = 2π/T and ω = 2πn/T = n∆ω

f (t) = lim fT (t)

T →∞
�∞
1
= lim FT (jn∆ω)ejn∆ωt
∆ω→0
n=−∞
T
∞
� ∆ω
= lim FT (jn∆ω)ejn∆ωt
∆ω→0
n=−∞
2π
� ∞
1
= F (jω)ejωt dω
2π −∞

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Which yields the inversion formula for the Fourier transform, the Fourier
integral theorem:

� ∞
F (jω) = f (t)e−jωt dt
−∞
� ∞
1
f (t) = F (jω)ejωt dω
2π −∞

Comments:

• There are usually technical conditions which must be satisfied for the
integrals to make sense in the mean square or ordinary improper Riemann
integral sense. Form of smoothness or Dirichlet conditions.

• As with Fourier series, inverse transform formula generally gives f (t)

accurately at points where f (t) is continuous, but yields the midpoint
when f (t) has jumps.

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• The intuition is that Fourier transforms can be viewed as a limit of

Fourier series as the period grows to infinity and the sum becomes an
integral.
�∞
• 1
2π F (jω)ejωt dω is called the inverse Fourier transform of F (jω).
−∞
Note it is identical in form to the Fourier transform except for the sign
in the complex exponential, and the factor of 1/2π.

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 Uniqueness of a Signal Recovered From its Fourier
Transform

• The signal recovered from its Fourier transform is not necessarily unique.
For example, consider the Fourier transform of (one definition of) the
rect signal
�
f (t) = rect(t/T ) = 1 |t| ≤ 2
T

0 otherwise
where
�
rect(t) = 1 |t| ≤ 12
0 otherwise

This is the same signal as considered in Fourier series, but now it is

defined for all real t and not just in a finite interval.

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The Fourier transform is

� ∞
F (jω) = f (t)e−jωtdt
−∞
� T
2
= e−jωt dt
− T2
�T
e−jωt �� 2
=
−jω �− T
2

1 � �
= e−jωT /2 − ejωT /2
−jω
1
= (−2j sin(ωT /2))
−jω
2 sin(ωT /2)
=
ω
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 sin(π(ωT /2π))
= T
π(ωT /2π)
= T sinc(ωT /2π)
sin(πt)
where sinc(t) = πt . In particular

rect(t/T ) ⇔ T sinc(ωT /2π)

and
rect(t) ⇔ sinc(ω/2π).
If we use any of the alternative definitions for rect which define the value
at the jump points ±T /2 to be 1 or 0 or 1/2, we get the same transform.
The inverse transform of F (jω) will be 1/2 at t = ±T /2, so the signal
is not uniquely recoverable (unless its values were 1/2 at the jumps).
It is common to define rect(1/2) = 1/2 so that the inverse FT agrees
with the original signal.

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Another Perspective on the Signal Recovered from its

Fourier Transform

Assume f (t) has a Fourier transform

� ∞
F (jω) = f (t)e−jωt dt
−∞

for all real ω.

Define truncated inverse transform
� a
1
fa(t) = F (jω)ejωt dω
2π −a

Will argue that

lim fa(t) = f (t)
a→∞

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at points of continuity of f ; that is, that the integral
� ∞
1
F (jω)ejωt dω
2π −∞

exists and has the desired value.

Substituting the expression for the Fourier transform into the expression for
the truncated inverse Fourier transform
� �� ∞ �
1 a
fa(t) = f (s)e −jωs
ds ejωt dω.
2π −a −∞

Assume we can change the order of integration (important detail)

� ∞ � � �
1 a
fa(t) = f (s) ejω(t−s)
dω ds
−∞ 2π −a

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Evaluating the integral in parentheses:

�a
ejω(t−s) �� sin(a(t − s)) � a � sin(π πa (t − s))
= =
j2π(t − s) �−a π(t − s) π π πa (t − s)
a �a �
∆
= sinc (t − s) = Fa(t − s),
π π

where �a �
a
Fa(t) =
sinc t
π π
is called the Fourier integral kernel

Fa(t) is symmetric about the origin and has unit integral. Thus
� ∞
fa(t) = f (s)Fa(t − s) ds,
−∞

which is a convolution integral!

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Example: If f (t) = rect t, then fa(t) is:

f (t)

∗
Fa (t)

=
fa (t) = (f ∗ Fa )(t)

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As a increases, Fa(t) gets narrower and taller,

4

3.5

2.5

1.5

0.5

-0.5

-1
-10 -8 -6 -4 -2 0 2 4 6 8 10
t

The Fourier kernel Fa(t):

a = π (solid line), 2π (dashed line),4π (dash-dot line)

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As a → ∞ the kernel behaves as an impulse function:

• kernel is zero except near origin

• always has area 1

Thus
� ∞
lim fa(t) = lim f (s)Fa(t − s) ds.
a→∞ a→∞ −∞
= f (t)

so fa(t) → f (t) as a → ∞.

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This argument explains the behavior at discontinuities. If f (t) = rect(t),

Fa (t − τ ) f (τ ) t1

t2

t3

t3
t1
t2 fa (t) = (f ∗ Fa )(t)

At t2, half of the sinc() overlaps the rect(), so the value of the convolution
is 1/2.

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Suppose that f (t) has a simple jump from f (t−) on the left to f (t+) on
the right. If f (t) is continuous near t except at the jump, for very large a
and very small �
� ∞
fa(t) = f (s)Fa(t − s) ds
−∞
� t � t+�
≈ f (s)Fa(t − s) ds + f (s)Fa(t − s) ds
t−� t
� t � t+�
≈ f (t ) −
Fa(t − s) ds + f (t ) +
Fa(t − s) ds
t−� t

f (t−) f (t+)
≈ +
2 2

Which explains why the value of inverse Fourier transforms (and for similar
reasons, Fourier series) yield midpoints at jumps.

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Fourier Transform Example: Exponential Signal

1.2
�
1
f (t) = e−at t ≥ 0
0.8
0.6
0 otherwise
0.4 = e u(t); all t
−at
0.2
0

where a > 0.
!0.2
!4 !3 !2 !1 0 1 2 3 4

1.2
1
0.8 e−at u(t)
f (t) 0.6
f(t)

0.4
0.2
0
!0.2
!2 !1 0 1 2 3 4
tt

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By the definition of the Fourier transform
� ∞
F (jω) = f (t)e−jωtdt
−∞
� ∞
= e−ate−jωt dt
0
� ∞
= e−(a+jω)t dt
0
�∞
e−(a+jω)t ��
=
−(a + jω) �0
1
=
a + jω

Thus
1
e−atu(t) ⇔
a + jω

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For a = 1:
1.2
|F( j!)|
11
0.8
0.6
0.4
0.2

00
!0.2

−4!
!4 !3
−2!
!2 !1 0
0 1 2
2! 3
4!4
!
0.6
∠F( j!)
!/20.4
0.2

00
!0.2
!0.4
−!/2
!0.6

!4 !3 !2 !1 0 1 2 3 4
−4! −2! 0 2! 4!
!

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Notes:

• The inverse Fourier transform of F (jω) will equal f (t) at all values
of t except possibly at the origin, where the inverse Fourier transform
evaluates to 1/2. Whether or not this equals f (t) for t = 0 depends on
the definition of u(t).
• The magnitude of the transform is symmetric about the origin,

|F (−jω)| = |F (jω)|; all ω

• The phase of the transform is antisymmetric about the origin,

� F (−jω) = −� F (jω); all ω

Will see Fourier transforms have many such symmetry properties

• As we progress, will compute and collect Fourier transforms of many
common signals.

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• So far have the FT for a rect signal and an exponential signal.

• In the real world, usually look transforms up in book or compute
numerically.

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 Symmetry Properties

Useful for checks on results, finding computational shortcuts, or for signal

encoding and reconstruction.

Recall: Even and Odd Functions

Any signal f (t) can be uniquely decomposed into an even part and an odd
part; that is,
f (t) = fe(t) + fo(t)
where fe(t) is even and fo(t) is odd.

Construction

1
fe(t) = (f (t) + f (−t))
2
1
fo(t) = (f (t) − f (−t))
2
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If e1(t) and e2(t) are even functions and o1(t) and o2(t) are odd functions,
then

• e1(t) ± e2(t) is even

• e1(t)e2(t) is even
• o1(t) ± o2(t) is odd
• o1(t)o2(t) is even
• and e1(t)o2(t) is odd

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 Fourier Transforms of Even and Odd Functions.

Assume f (t) is a possibly complex function, f (t) = fe(t) + fo(t), where

fe(t) and fo(t) are the even and odd components of f (t).
� ∞
F (jω) = f (t)e−jωtdt
−∞
� ∞
= (fe(t) + fo(t)) (cos(ωt) − j sin(ωt)) dt
−∞
� ∞ � ∞
= fe(t) cos(ωt)dt − j fe(t) sin(ωt) dt
−∞ −∞
� ∞ � ∞
+ fo(t) cos(ωt) dt − j fo(t) sin(ωt) dt
−∞ −∞
� ∞ � ∞
= fe(t) cos(ωt) dt − j fo(t) sin(ωt) dt
−∞ −∞

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= Fe(jω) + Fo(jω)

where Fe(jω) is the cosine transform of the even part of f (t) and Fo(jω)
is −j times the sine transform of the odd part.

Note:

• if f (t) is an even function of t, then F (jω) is an even function of ω.

• If f (t) is an odd function of t, then F (jω) is an odd function of ω.

Here f (t) could be real, imaginary, or complex.

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 Fourier Transform of Real Functions

If f (t) is further restricted to be real-valued, then fe(t) and fo(t) are also
both real. Then
� ∞ � ∞
F (jω) = fe(t) cos(ωt) dt − j fo(t) sin(ωt) dt
� −∞ �� � � −∞ �� �
real imaginary

and
� ∞
�(F (jω)) = fe(t) cos(ωt) dt
−∞
� ∞
�(F (jω)) = − fo(t) sin(ωt) dt.
−∞

The real part of F (jω) is even in ω, and the imaginary part of is odd in ω.

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• If f (t) is real and even in t, then F (jω) is real and even in ω.

• If f (t) is real and odd in t, then F (jω) is imaginary and odd in ω.

Combining these two facts,

F (−jω) = �(F (−jω)) + j�(F (−jω))

= �(F (jω)) − j�(F (jω))
= F ∗(jω),

The Fourier transform of a real-valued signal is Hermitian.

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 Fourier Transform of Imaginary Functions

If f (t) is purely imaginary then fe(t) and fo(t) are also imaginary, and

� ∞ � ∞
F (jω) = fe(t) cos(ωt) dt − j fo(t) sin(ωt) dt .
� −∞ �� � � −∞ �� �
imaginary real

Then � ∞
�(F (jω)) = −j fo(t) sin(ωt) dt
−∞

is odd in ω, and

� ∞
�(F (jω)) = −j fe(t) cos(ωt) dt
−∞

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is even in ω. Combining these

F (−jω) = �(F (−jω)) + j�(F (−jω))

= −�(F (jω)) + j�(F (jω))
= − (�(F (jω)) − j�(F (jω)))
= −F ∗(jω);

The Fourier transform of an imaginary-valued signal is anti-Hermitian.

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 Summary of Symmetry Properties

• For any real, imaginary, or complex signal,

– An even signal has an even transform, and
– An odd signal has an odd transform.
• A real signal has a Hermitian symmetric transform,

F (−jω) = F ∗(jω).

• An imaginary signal has an anti-Hermitian symmetric transform,

F (−jω) = −F ∗(jω).

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