Topic 2
From Complex Fourier Series
to Fourier Transforms
2.1 Introduction
In the previous lecture you saw that complex Fourier Series and its coecients
were de ned by as
f (t ) =
1
X
n= 1
Cn ein!t
1
Cn = T
where
Z T=2
T=2
f (t )e
in!t dt
:
However, we noted that this did not extend Fourier analysis beyond periodic functions and discrete frequency spectra. What would x this?
2.1.1 What happens if we let T get very large? A thought ...
Suppose the period T tended <to in nity. The fundamental frequency ! would
become so small that it might properly be called an in nitesimal d!. To reach any
nite frequency !, m would have to tend to in nity. Then m d! ! !, and the Cm
might best be called C (!)d!. (This point is tricky. It is to preserve the integral
as the discrete spectrum morphs into a continuous one!)
So in this regime,
1
C (!)d! = Tlim
!1 T
But as T
Z T=2
T=2
f (t )e
i!t dt
:
! 1, T = 2=d!
d!
C (!)d! = 2
Z 1
1
f (t )e
i!t dt
)
;
1
1
C ( ! ) = 2
Z 1
1
f (t )e
i!t dt
:
2/2
The expression for f (t ) has to be changed too. First change Cn ! C (!)d!, then
change from a sum to an integral:
f (t ) =
Z 1
1
C (!)d!ei!t
=
Z 1
1
C (!)ei!t d! :
2.1.2 We (NEARLY) get to the Fourier Transform!
Our last thought was a good one. By allowing T to tend to in nity, we seem
to have a method of going from the non-periodic time domain signal f (t ) to a
frequency domain spectrum C (!). Everything is correct, EXCEPT the actual
Fourier transform is, by modern convention, 2C (!).
So forget the thinking, let's use the de nition.
2.2 The de nition of the Fourier Transform
The Fourier Transform of a temporal signal f (t ) is the frequency
spectrum
F (! ) =
Z 1
1
f (t ) e
i!t
dt :
Given a frequency spectrum, the equivalent temporal signal is given by the
Inverse Fourier Transform
Z
1 1
f ( t ) = 2
F (!) ei!t d! :
1
Because Fourier Series have a continuous time signal but discrete frequency spectrum, it is rather natural to think about the signal, f (t ), as the \proper thing",
and the coecents An and Bn (or Cn ) as slightly \derived".
With Fourier Transforms, however, there is a real equality between the representations. Indeed the time signal and frequency spectrum are described as a Fourier
transform pair.
2/3
A Fourier Transform Pair is denoted by
f (t ) , F (! ) :
This means
F (!) = FT [f (t )] and f (t ) = FT
2.3
1 [F (! )] :
| Example
[Q] Determine the Fourier transform F (!) of the function shown in Figure 2.1(a),
f (t ) = u (t )e bt , where u (t ) is the Heaviside step function, and plot the amplitude
spectrum jF (!)j.
[A] The Heaviside step function is zero for t < 0 and unity thereafter, so
FT u (t )e
bt =
Z 1
0
e
bt
1
=
b + i! e
e
i!t dt
(b+i!)t
1
0
=
1
b + i! :
The frequency spectrum F (!) is complex,
so it has amplitude and phase. The
p 2 and
2
amplitude spectrum is jF (!)j = 1= b + ! , and is plotted in Figure 2.1(b).
F(ω)
f(t)
1/b
t
ω
0
(a)
Figure 2.1: (a)
f (t ) = u (t )e
(b)
bt (b) Amplitude spectrum
F (!)j = j1=(b + i!)j.
j
Notice that negative values of ! are permitted. There is nothing deep happening:
recall that cos( !t ) = cos(!t ) and sin( !t ) = sin(!t ).
2.4
| Example
[Q] Determine the Fourier transform of the unit \top hat" function of width a:
jt j < a=2
f(t ) = 10 otherwise
2/4
F( ω)
Π (t)
a
1
t
ω
−2π/ a
a /2
−a/2
2π/ a
Figure 2.2:
[A] The Fourier transform is
F (! ) =
Z a=2
2
a=2
e
i!t dt
=
1
i!
e
i!a=2
ei!a=2
! sin (!a=2)
sin (!a=2)
sin (!a=2)
= a
=a
!a=2
!a=2
= a sincN(!a=2) = a sincH(!a=2)
=
The Fourier Transform is real, and the frequency spectrum, rather than the amplitude spectrum, is shown in Figure 2.2.
Why do we write the answer in two ways? Unfortunately the sinc() function has
two de nitions in common use.
The HLT de nition is sincH(x ) = sin(x )=(x ), but often elsewhere you'll see sinc
de ned as sincN(x ) = sin(x )=x , where the N stands for Not-HLT.
Because of the confusion we will try to avoid using it entirely, but if it is used we
should try to stick to the HLT de nition. Conversion is easy. If you read sinc()
in non-HLT-compliance-mode replace it with sinc(=).
2/5
2.5
| Example
[Q] Determine the Fourier transform of the unit \triangle" function of width 2a:
f(t ) =
at<0
0ta
1 + t=a
1 t=a
f(t)
F( ω)
1
a
t
−a
a
ω
−2π/ a
(a)
2π/ a
(b)
Figure 2.3:
[A] The Fourier transform is
F (! ) =
Z 0
= 2
a
(1 + t=a)e
Z a
0
Z a
0
i!t dt
(1 t=a)e
+
i!t dt
(1 t=a) cos !t dt
= grind using integration by parts
sin2(!a=2)
= a
(!a=2)2
2.6 Further examples
There are further examples of the Fourier transforms of important functions later
in the notes. But a more complete list is in HLT.
2/6
2.7 More realistic signals
In case you had started to think that Fourier Transforms were merely for signals
that could be described in as functions, Figure 2.4 shows a computational example (using methods described later) of the transform of signal which has equal
amplitude 50Hz and 120Hz harmonic components but is corrupted with noise.
Signal Corrupted with Zero−Mean Random Noise
Frequency content of y
5
80
70
60
50
0
40
30
20
10
−5
0
10
20
30
time (milliseconds)
40
50
0
0
100
200
300
frequency (Hz)
400
500
Figure 2.4:
2.8 \Routine" Properties of the FT
To start applying the Fourier transform to engineering problems will require a few
more mathematical tools to be in place.
However, we can immediately explore some of the basic properties of the Fourier
transform | those that can be derived from straightforward mathematical properties of the integral.
These are all in HLT. Deriving them is a bit tricky in places, and might seem a
slightly tedious, but it does provide practice at recognizing the Fourier integral.
2.8.1 Property #1: Linearity
Linearity Property: If FT [f (t )] = F (! ) and FT [g (t )] = G (! ), then
FT [ f (t ) + g (t )] = F (!) + G (!)
Proof:
Integration is a linear operation.
That is
R1
R1
i
!t
i!t
1 [ f (t ) + g (t )]e dt =
1 f (t )e dt +
R1
1 g (t )e
i!t dt .
2/7
2.8.2 Property #2: Duality
Dual Property If FT [f (t )] = F (! ), then
FT [F (t )] = 2f ( !)
Yes, weird! Watch out for the
!
Proof: We know that f (t ) is the FT 1 [F (!)] | that is
Z
1 1
f ( t ) = 2
F (!)e i!t d!
1
But t and ! are just symbols, so replace t with ( !) and ! with t .
NB! Replacing ! with t does NOT ip the limits of integration. So
Z
Z
1 1
1 1
i
t
(
!
)
F (t )e
F
(t )e i!t dt
dt =
f ( ! ) = 2
2 1
1
) 2f ( !) = FT [F (t )] \comme il faut", as Mrs Fourier would say
(If you did not like the swapping of symbols in one go, try t ! p and ! ! q , then
p ! ( !) and q ! t :
1
f ( t ) = 2
Z 1
1
Z 1
#
f ( p ) = 2
#
1
f ( ! ) = 2
then rearrange ...)
1
1
Z 1
1
F (!)e i!t d!
# # #
F (q )e iqp dq
# # #
F ( t ) e i t ( ! ) dt
2/8
2.8.3 Property #3: Similarity Property
Parameter Scaling or Similarity Property: If FT [f (t )] = F (! ), then
1 ! FT [f (at )] = F
jaj
a
Proof: The appearance of jaj hints that this proof needs to consider the
ranges a > 0 and a < 0 separately.
For a > 0:
FT [f (at )] =
Z 1
1
f (at )e
i!t dt
:
Write p = at , and note that the signs on the limits do NOT change
because a is positive. Then
Z
1 1
1 ! i(
!=a
)
p
FT [f (at )] =
f (p)e
dp = F
a
1
a
a
For a < 0:
Substitute p = jajt , and remember to change signs on the limits |
when t = 1, p = 1:
Z 1
Z 1
1
1
f (p)e i(!=a)p dp =
f (p)e i(!=a)p dp
FT [f (at )] =
=
jaj 1 !
F
jaj a
1
So, for both cases, one can write
1 ! FT [f (at )] = F
jaj
a
jaj
1
2/9
2.8.4 | Property #3: Example
[Q] The Fourier transform of f (t ) is F (!). What the the FT of g (t )?
g(t)
f(t)
t
t
a
0
(a)
[A]
b
0
(b)
1
! FT [g (t )] = FT [f (St )] = jSj F S
So the real question is what is the scale S ?
VERY tempting to say that g (t ) looks wider so that S > 1. Now b > a so that
) S = b=a.WRONG
Something interesting happens to g (t ) when t = b.
This point matches to f (a).
So Sb = a, and
) S = a=b. CORRECT
2.8.5 Property #4: Parameter Shifting
Parameter Shifting Property: If FT [f (t )] = F (! ), then
FT [f (t a)] = exp( i!a)F (!)
Proof:
FT [f (t a)] =
Substitute p = t
FT [f (t a)]
Z 1
1
f (t a)e
i!t dt
:
a. Then
=
Z 1
= e
1
f (p)e
i!a F (! )
i ! ( p + a ) dp
=e
i!a
Z 1
1
f (p)e
i!p dp
2/10
2.8.6
| Properties #1,3,4: Example
[Q] Find the Fourier Transform of the function shown (a) below.
f(t)
2
f(t)
Π (t)
1
+1
1
t
t
t
−1
−a
a /2
−a/2
−2
0
(a)
(b)
(c)
[A] We know the FT of the tophat in (b) is !1 sin( !a2 ).
The version in (c) is time-shifted \to the left" and involves t + 2a , so its FT is
i!a 1 !a sin
exp
2 !
2
Subtract a copy time-shifted to the right ... and set a = 1 ...
i!a
exp
2
exp
i!a
2
1
!a ! sin 2
h
!a i 1
!a 2i
! 2
= 2i sin
sin
= sin
2 !
2
!
2
Now scale the amplitude by a factor of 2 ...
Ans =
4i
!
sin2
! 2
We've done the scaling in time by setting parameter a = 1. Suppose you applied
the scaling property. What would be the correct scaling | a or (1=a)?
2.8.7 Property #5: Frequency Shifting
FT [f (t )] = F (!), then
FT [f (t ) exp(i !S t )] = F (! !S )
Frequency Shifting Property: If
2/11
Proof:
FT [f (t ) exp(i !S t )]
=
Z 1
1
Z 1
f (t )ei !S t e
i!t dt
f (t )e i(!!S )t dt
1
= F ( ! !S )
=
2/12
2.8.8 Property #6: Amplitude modulation by a cosine
Amplitude modulation by a cosine If FT [f (t )] = F (! ) then
FT [f (t ) cos !0t ] = 21 [F (! !0) + F (! + !0)]
Proof: Write
1
cos !0t = ei!0t + e i!0t
2
then use the Frequency shifting property.
2.8.9 Property #7: Di erentiation wrt time
Di erentiation Property in time: If FT [f (t )] = F (! ), then
n
d
FT dt n f (t ) = (i!)n F (!)
Proof:
so tempting to start by writing
R 1 dn It is i!t
1 dt n f (t ) e dt . Resist. Instead, write
1
f ( t ) = 2
Z 1
1
FT
dn
dt n
f (t )
=
F (!)ei!t d!
Now di erentiate w.r.t. t . Because the integral is w.r.t. !, the di erentation can move through the integral sign:
Z
dn
1 dn 1
f
(t ) =
F
(!)ei!t d!
n
n
dt
2 dZt 1
1 1
dn i!t
=
F ( ! ) dt n e d!
2 Z 1
1 1
=
F
(!)(i!)n ei!t d!
2 1
This last expression says that
dn
f (t ) = FT
dt n
1 [F (! )(i! )n ]
n
d
) FT dt n f (t )
= F (!)(i!)n :
2/13
2.8.10 Property #8: Di erentiation wrt frequency
Di erentiation Property in frequency: If FT [f (t )] = F (! ), then
dn
n
FT [( it ) f (t )] = d!n F (!)
Proof: For you to colour in.
2.9 The Fourier Transform of Complex Fourier Series
This requires knowledge of the -function. See Lecture 3.
2.10 Summary
By allowing the period T to tend to in nity, and the fundamental frequency tend
to an in nitesimal d!, we have given a reasoned argument for the origin of the
Fourier Transform which allows non-periodic functions to be represented in the
continuous (and complex) frequency domain.
However, there is no need to reproduce the argument and you might just as well
take the de nition for granted.
We have worked out the Fourier Transform for a couple of signals.
We have gone through the proofs of several basic properties of the Fourier Transform.
We are nearly at the point of being able to do things, but not quite!
In the next lecture we must learn about the -function and the general technique
of convolution, and then applications begin to open up for us.