Fourier Transforms
“The whole is the sum of the parts”
Euclid
Fourier Series
As nmr spectroscopists working with modern digital equipment we
routinely acquire data in the form of the free induction decay. This
is the digitized form of the analog time domain data passing through
the receiver. We wish, of course to look at these data in the
frequency domain and also routinely transform the time domain data to
frequency domain data for display and further analysis:
t
w
Figure 7-1:Fourier transform
from time domain to frequency
domain
This transform of the data is accomplished using the ideas of Jean
Baptiste Joseph Fourier (and others). FOURIER, LAPLACE The basic idea
is that the application of infinite sums of sine and cosine functions
multiplied by suitable constants can be used to represent any
periodic function. To understand what is happening in our
spectrometer software to accomplish this remarkable feat we must
delve into the mystery's of the Fourier transform ...
Our analysis of vectors showed us that we could use unit vectors
multiplied by suitable factors to build up an overall vector that
spans the vector space. This is particularly easy to visualize in two
or three dimensional space. Mathematically, we represent this as:
⃗
A =∑ ai ⃗
ui
[7-1]
i
Thus, the vector A is sum of unit vectors which forms the basis set
from which we can build any vector.
Moreover, as is obvious in three-D (and two-D) space, these unit
u i , are orthogonal. That is, when we do a scalar product of
vectors, ⃗
any two unit vectors the result will be zero:
u⃗i⋅⃗
u j=∣ui∣∣u j∣cos( π )=cos ( π )=0
2
2
[7-2]
This is generally true of multi-dimensional inner product spaces. The
basis vectors are or can be made to be orthogonal if not already so.
Of course, since these are unit vectors with length one, the inner
product of a unit vector with itself is one:
ui⋅⃗
⃗
ui =∣ui∣∣u j∣cos (0)=cos(0)=1
[7-3]
Why not build up a function, F ,
way that is analogous to vectors?
vectors (section 3.1) you will see
rules that define vectors is
Now, we make a bit of a leap.
from a set of basis functions in a
If you look back at the chapter on
that, as long as each of the basic
followed, this is ok.
F=∑ ai f
[7-4]
i
This is what we do when using Taylor series to represent various
functions:
∞
x2 x 3 x 4
xn
+ + ⋯=∑
2 ! 3! 4 !
n=0 n !
2
3
4
n
∞
x x x
x
log(1+x)=x− + − ⋯=∑ (−1)n−1
2 3 4
n
n=1
e x =1+x+
[7-5]
Taylor series are, however, not applicable to functions that contain
discontinuities with finite magnitude. In other words, we could not
model a pulse or “on-off” function using a Taylor series. We could
however do so using a Fourier series. This type of series is
excellent for modeling periodic functions that may or may not have
discontinuities in them.
The Fourier series is applicable to a problem if it satisfies
the Dirichlet conditions:
1.the function must be periodic.
2.the function must be single valued and differentiable within one
cycle.
3.within a cycle the function must have a finite number of maxima and
minima.
As with vectors, we wish to be able to use a series of functions
that are orthogonal to one another. This is the foundation upon which
all of Fourier analysis is based. We can define such a set of
functions using complex exponentials:
−i3ω f t
A=⋯, e
−i2 ω f t
,e
−i ω f t
,e
,1, e
i ωf t
,e
i2ω f t
i3 ω f t
,e
⋯
[7-6]
where we define ωf to be the fundamental radial frequency of the
periodic function. To demonstrate that these functions are in fact
2π
orthogonal we do an integration over one period, τ= ω , of the
f
function:
τ
1
inω t* imω t
e
dt
τ ∫e
f
f
[7-7]
0
where we are using the complex conjugate of one of the functions.
Recall that this is what is analogous to what is done for vectors in
1
a formal sense. It will become apparent why we multiply by τ
momentarily.
2π
Where does τ= ω come from?
f
-t
t
0
Figure 7-2: Square wave function
We see from this figure that, starting at time 0, one complete cycle
or period of the function is finished at t. We can specify the
frequency of this periodic function in two ways by using hertz (or
cycles per second in the older nomenclature) or by using radians per
second. The two are, of course, related. One cycle per second would
be 2 π radians per second. So, to convert from hertz to radians per
second:
ωf =2 π ν
Now that we know what the frequency is in radians per second we ask
ourselves, “how long is one complete period?”. It is very simply the
inverse of the frequency:
1 2π
τ= ν = ω
f
Our integral,[7-7], becomes:
τ
τ
1
1
−i n ω t i m ω t
e
dt= τ ∫ e i(m−n)ω t dt
τ ∫e
f
f
f
0
[7-8]
0
We have two possible situations now. If m=n then the integral
evaluates to 1:
τ
[ ]
τ
1
t
τ 0
τ ∫ dt = τ 0= τ − τ =1
[7-9]
0
Thus the functions are normalized.
If m≠n then:
τ
1
e i Δ ω t τ ei Δ ω t τ
i(m−n )ω t
e
dt=
|=
|
τ∫
τ i Δ ω f 0 2 πi Δ 0
0
f
f
f
[7-10]
where Δ=m−n . Finishing the evaluation of the definite integral:
i Δ ωf t
iΔ2π
τ
e
e
−1 1−1
|=
=
=0
2πiΔ 0
2π iΔ 2π i Δ
[7-11]
That e i Δ 2 π is equal to 1 is evident when you consider that using
integral multiples of 2 π in e ix must result in the cosine portion
being equal to 1 and the sine portion being equal to zero:
e i n2 π =cos (n 2 π)+i sin(n 2 π)
where n is any integer.
Ok ... we have a set
with them? We use them to
fashion to that of Taylor
periodic under one of the
of orthonormal functions. What do we do
model other functions, in a similar
series. Since Fourier series must be
Dirichlet conditions, we write:
f (t)=
∞
∑
[7-13]
C( n) e i n ω t
f
n=−∞
where f (t) is the function to be modeled and C(n) is a constant. This is
the synthesis equation since it is used to synthesize or build up the
function, f (t) .
The central problem is to assign values to all of the C(n) .This
is accomplished by multiplication of both sides of the series
equation by e−i m ωt and integrating:
τ
τ
1
1
−im ω t
dt = τ ∫
τ ∫ f (t)e
f
0
0
∞
C( n)e i n ω t e−i m ω t dt
∑
n=−∞
f
f
[7-13]
We now interchange the integral and summation on the right side to
get:
τ
τ
∞
1
1
−im ω t
f
(t)e
dt
=
∑ C(n) τ ∫ e i n ω t e−i m ω t dt
τ∫
f
0
f
n=−∞
f
[7-14]
0
There was, by the way, some considerable confusion over the validity
of this step resulting in many mathematicians pondering it for many
years. In fact, Fourier's original manuscript was submitted in 1806
but not published until 1822 because of this!
Recalling our recent discussion of the orthonormality of these
functions, the right side of the equation vanishes for m≠n leaving
only the term with m=n . Of course, when m=n , the integral
evaluates to 1, leaving one term:
T
[7-15]
1
−im ω t
dt=C (m)
τ ∫ f (t) e
f
0
This is the required evaluation of the terms in the Fourier series
and is called the analysis equation. Note that it is necessary only
to integrate over one complete period .. the actual starting and
ending points do not matter as long as it is over one complete
period. We could just as well write:
τ
2
1
f (t)e−i m ω t dt=C (m)
τ∫
−τ
[7-15a]
f
2
An alternate way of writing the synthesis equation is:
∞
∞
n=1
n=1
f (t)=a0+∑ an cos (n ω f t)+∑ bn sin(n ω f t )
[7-16]
Equation [7-16] can be easily shown to equivalent to [7-13]:
∞
∞
f (t )=a 0+∑ a n cos (n ω f t )+∑ b n sin (n ωf t)
n=1
i n ωf t
n=1
+e
ei n ω t−e−i n ω t
+∑ bn
2
2i
n=1
n=1
∞
∞
(a −ib n) i n ω t
( a +ib n ) −i nω t
=a 0+∑ n
e +∑ n
e
2
2
n=1
n=1
∞
(a n−ib n) i n ω t −1 ( an+ib n ) i n ω t
=a0 +∑
e +∑
e
2
2
n=1
n=−∞
∞
=a 0+∑ a n
(
−i n ω f t
e
∞
=
∑
)
(
∞
f
f
f
f
f
f
)
C (n)e i n ω t
f
n=−∞
where
(an −bn )
n>0
2
(a +b )
C(−n)= n n n<0
2
C (0)=a0 n=0
C(n)=
Sometimes we speak of harmonics, especially with respect to the
analysis of periodic functions involving sound waves. From the
Fourier series point of view the harmonics are:
h(0)=C(0)(fundamental or average value )
h(1)=C( 1)e iω t +C(−1)e−i ω t (1st harmonic)
h(2)=C (2)ei 2 ω t +C (−2)e−i 2 ω t (2nd harmonic )
⋮
inω t
h (n)=C (n)e +C (−n)e−i n ω t (nthharmonic )
f
f
f
f
f
f
Example 13.1
Given the periodic function,
⋮
f [t −τ ]=f [t ] −τ⩽t ⩽0
f [t ]=t 0⩽t ⩽τ
f [t +τ ]=f [t ] τ⩽t ⩽2 τ
⋮
we wish to find the Fourier series representation for f (t) . This is
equivalent to finding the Fourier coefficients, C(n ) , using equation
[7-15]. Note that since we are dealing with a periodic function, its
definition need only be specified over one period. Our current
function represents a “sawtooth” function with a dc offset such that
the amplitude of the function never dips below zero.
-t
0
We begin with the analysis equation:
t
τ
1
−in ω t
C(n)= τ ∫ f (t)e
dt
f
(n≠0)
0
2π
ωf
=
ωf
−i n ω t
te
dt
∫
2π 0
f
2π
ωf
ωf
=
∫ t [cos(n ω t )+isin (n ωf t)]dt
2π 0
2π
ωf
2π
ωf
ωf
ωf
=
t cos(n ωf t)dt +i
∫
∫ t sin(n ωf t)dt
2π 0
2π 0
The integrals are solved by integration by parts to give:
[
] [
]
2π
ω t sin(n ω f t ) cos (n ω f t )
ω t cos (n ω f t) sin (n ωf t) ω
= f
−
−i f
−
|
2π
n ωf
n ωf
2π
nω f
n ωf
0
f
The first term evaluates to zero after insertion of the integration
limits. The second evaluates to
−i
n ωf
for n≠0
For n=0 the analysis equation is:
τ
1
C (n)= τ ∫ f (t )e−i n ω t dt
f
( n=0)
0
2π
ωf
=
ωf
t e−i n ωt dt
∫
2π 0
2π
ωf
ω
= f ∫ t [cos(n ω t)+i sin(n ω f t)]dt
2π 0
2π
ωf
ωf
∫ t dt
2π 0
2π
ωf t 2 ω
=
|
2π 2 0
=π
2
=
f
Thus, our Fourier series that models this periodic function is:
∞
−i −in ω t
A= π + ∑
e
( n≠0)
2 n=−∞ n ωf
∞
−i
= π+ ∑
[ cos (n ω f t )−isin (n ωf t )]
2 n=−∞ n ω f
f
In terms of harmonics this is:
∞
A=h (0)+ ∑ h(n)
n=−∞
∞
−2i
= π+ ∑
cos (n ω f t )
2 n=−∞ n ω f
Properties of the Time Domain Functions
The sine and cosine functions have the characteristics of being
odd and even, respectively. What is meant by this is that for sine:
f (t )=−f (−t )
or
sin( ϕ)=−sin (−ϕ)
[7-17]
and for cosine:
f (t )=f (−t )
or
cos( ϕ)=cos(−ϕ)
[7-18]
Thus, sine is anti-symmetric about the y-axis and cosine is
symmetric. This is very easily seen in the following
sin(wt) figure:
t
cos(wt)
t
Figure 7-3: Sine and cosine functions
Each of these types of functions has a useful property:
α
α
f (x)dx=2∫ f (x ) dx
∫
−α
α
∫ f ( x)dx=0
−α
(even functions)
0
[7-19]
(odd functions)
Referring to figure 13-1, we can see that
sin( ωt )+sin (−ω t )=0
cos(ω t )+cos (−ω t )=2⋅cos(ω t )
This also follows from the definitions of even and odd functions, [716, [7-17]:
f (t )odd +f (−t )odd =0
f (t)odd −f (−t)odd =2f (t)odd
f (t)even+f (−t)even =2f (t ) even
f ( t)even −f (−t )even=0
[7-20]
Now, suppose we have a periodic function that is neither even nor
odd. Is it reasonable to presume that it can be broken down into the
sum of even and odd parts? Let's try a very simple linear
combination:
g(t)=Af (t )even+Bf (t )odd
We can get even simpler by assuming the constants A and B are
incorporated directly into the functions:
g(t)=f ( t)even+f (t)odd
Given this we write:
[7-21]
g(t)+g (−t )=[ f (t)even+f (−t)even ]+[ f (t)odd +f (−t)odd ]
= [ f (t )even+f (−t)even ]+0
=2 f (t )even
1
f (t)even= [ g(t )+g(−t) ]
2
[7-22]
g(t)−g (−t)=[ f (t)even −f (−t)even ]+[ f (t)odd −f (−t)odd ]
=0=[ f (t)odd −f (−t)odd ]
=2 f (t)odd
1
f (t )odd = [ g (t)−g (−t) ]
2
[7-23]
also:
We can also say some things about the Fourier coefficients, C(n) .
First, the complex conjugate of the the nth coefficient
is equal to the -nth coefficient (assuming f (t) real):
[
τ
1
C( n) = τ ∫ f (t )e−i n ωt dt
*
0
τ
*
]
1
= τ ∫ f (t)* e i n ωt dt
0
τ
[7-24]
1
= τ ∫ f (t )e i n ω t dt
0
= C (−n)
We can break the coefficient, C(n) , into real and imaginary
coefficients:
C( n)=a(n)+ib(n)
The complex conjugate of this is:
C( n)* =a(n)−ib( n)
Using [7-23] we can play with this:
C (n)*=C(−n)
a( n)−ib( n)=a(−n)+ib ( n)
The real parts are equivalent, as are the imaginary parts:
a(n)=a(−n)
b( n)=−b(−n)
[7-25]
Thus, a(n) is even and b (n) is odd (from [7-17] and [7-18]).
With respect to our functions and constants, f (t ) , a(n) and b (n) , a
little contemplation (and a bit of help from figure 13-2) will
convince you that the following are true:
odd×odd=even
even×odd=odd
even×even=even
[7-26]
Thus if f (t) and g(t) are odd and even respectively, their product is a
new function that is odd.
Example 13-2
For the function:
f (t )=1 0⩽t⩽1
f (t)=−11 1<t<2
f (t+2)=f (t)
which is defined over one period with τ=2, ω f =π , we wish to calculate
the corresponding Fourier series terms.
0
1
2
Visually, you can see that this is an odd function. The analysis
expression is:
1
2
1
1
C (n)= τ ∫ f (t )e−i n ω t dt+ τ ∫ f (t) e−i n ω t dt (n≠0)
f
f
0
1
2
1
1 −i n πt
1
e
dt+ ∫ −e−i n πt dt
∫
20
21
−i n π t 1
1e
1 e−in π t 2
=−
|+
|
2 i nπ 0 2 inπ 1
−i n π
0
−i n 2 π
−e−i n π ]
1 [ e −e ] 1 [ e
=−
+
2
in π
2
i nπ
1 [ cos (n π)−isin( n π)−1 ] 1 [ cos (n2 π)−isin(n2 π)−cos( n π)+isin(n π) ]
=−
+
2
inπ
2
inπ
1 cos(n π)−1 1 [ +1−cos (n π) ]
=−
+
2
inπ
2
inπ
1 cos (n π)−1 1 [ cos(n π)−1 ]
=−
−
2
inπ
2
inπ
1 2⋅cos (n π)−2
=−
2
inπ
1−cos( n π)
=
in π
=
Note that cos(n2 π) will equal +1 for all values of n and that sin (n2 π) and
sin (n π) will equal zero for all values of n. For sequential values of n
the cosine term will alternate between +1 and -1:
1−(−1)n
C( n)=
inπ
These are the Fourier constants for n≠0 . For n=0 :
1
2
1
1
C( 0)= τ ∫ f (t) e−i0 ω t dt+ τ ∫ f (t)e−i 0 ω t dt
f
0
1
=
f
1
2
1
1
1⋅dt+ ∫ −1⋅dt (n=0)
∫
20
21
1 1 1 2
= t |− t |
2 0 2 1
1
1
= −0−1+
2
2
=0
The Fourier series is now obtained from the synthesis equation:
∞
f (t)=
∑
C(n)e i n ω t
f
n =−∞
∞
=
∑
n=−∞
∞
=
∑
n=−∞
1−(−1)n i n π t
⋅e
i nπ
1−(−1)n
⋅[ cos(n π t)+isin(n πt ) ]
in π
For any value of n, the sine term must vanish:
n
1−(−1)
f (t)= ∑
⋅cos (n π t)
in π
n=−∞
∞
The C(n) term is an odd function and cosine is an even function. Thus
the overall function, f(t) is odd, as expected from [7-25].
The Fourier Integral
We will derive the Fourier integral using a simple example.
Consider the single pulse function:
f (t)=1 −1<t<1
f (t)=0 1<t <3
-1
1
2
3
We imagine this to be part of a periodic train of pulses. Using our,
by now, standard method of finding the Fourier coefficients:
1
1
C( n)= τ ∫ f (t) e−in ω t dt
f
−1
1
1
= τ ∫ e−i n ω t dt
f
−1
Note that we do not have to integrate over the region 1 to 3 since
f(t) is zero in this region and the integral collapses to zero. We
proceed with the integration:
1
1
C( n)= τ ∫ e−i n ω t dt
f
−1
1
1
C(0)= τ ∫ dt
−1
1 1
=τt|
−1
For n≠0 :
2
=τ
1
1
C (n)= τ ∫ e−i n ω t dt
f
−1
−i n ω f t 1
1e
=− τ
|
in ω −1
1
−i n ω
inω
=−
(e
−e )
τ i n ωf
1
=−
[−i2sin(n ωf )]
τ i n ωf
2 sin(n ω f )
=τ
n ωf
2
= τ Sa (n ωf )
ω
π)
= πf Sa( n ωf ) ( τ=2 ω
f
f
f
The last line in this part of the derivation includes the “sa”
function, pronounced “sah x”, whose definition is:
Sa (x)=
sin( x )
x
This is distinct from the sinc function:
sinc ( x)=
sin(π x)
πx
Obviously, since C(0)= 2τ , Sa (0)=1 . This result can also be obtained
using L'Hopital's rule.
Well, ok, so far so good. Nothing really new here (except sa).
We have derived the Fourier coefficient expression for f(t) in the
manner that we have discussed for a series of discreet frequencies,
nω f . It will help us at this point to have a picture of what we
τ
have. Specifically, plots of 2 C(n) vs nω f for increasing values of τ ,
10, 30, 60:
Figure 7-4a
Figure 7-4b
Figure 7-4c
You can see that as τ → ω the curve becomes smoother as the density of
nω f increases. So as τ → ω , nω f → ω and we write:
lim τ C (n)=Sa( ω) −∞<ω<∞
τ →∞
nω f changes to ω and the C(n) become a continuum of values related to ω
for our square wave function:
F( ω)=Sa (ω)
We can generalize this as follows beginnning with our discreet
Fourier series analysis equation:
τ
2
1
1 ∞
−i n ω t
−i n ω t
C( n)= τ ∫ f (t)e
dt= τ ∫ f (t) e
dt
f
−τ
2
f
−∞
or
∞
τ C (n)= ∫ f ( t)e−i n ω t dt
f
−∞
We can write this with infinity integration limits because our
−τ
τ
function's value from minus infinity to 2 and from 2 to infinity is
zero. As τ → ω , τC (n) becomes F (ω) and nω f becomes ω and we write:
∞
F( ω)=∫ f (t )e−i ωt dt
−∞
[7-27a]
This is the general Fourier transform integral which transforms a
time domain function, f(t) to a frequency domain function, F(w).
f(t) and F(w) are said to be a Fourier pair and is often written as:
f (t ) ← → F (ω)
The double arrows indicate that one can go either way; the right
arrow indicate the transformation from the time domain (t) to the
frequency domain (w) using the analysis equation just derived. The
left arrow indicates the opposite transformation from the frequency
domain to the time domain using the corresponding synthesis equation.
We do not derive this here since it is very rarely used in nmr
spectroscopy however standard texts such as those indicated in the
references will do this for you. We show it however for completeness:
f (t)=
1 ∞
F (ω) e iω t d ω
2π ∫
−∞
[7-27b]
Properties of Fourier Transforms
Linearity
The Fourier transform is a linear operation. So, for Fourier
transform pairs:
x (t ) ← → X (ω)
y (t ) ← → Y (ω)
we can write:
ax (t )+by (t ) ← → aX (ω)+bY (ω)
[7-28]
This is easy enough to show. We simply do the Fourier transform of
the left side of [7-27]:
∞
[ ax(t )+by (t )] e−i ω t dt
∫
−∞
∞
∞
= ∫ ax (t)e−iω t dt + ∫ by (t) e−iω t dt
−∞
−∞
∞
−i ω t
=a ∫ x (t )e
−∞
∞
−i ω t
dt +b ∫ y (t )e
−∞
= aX (ω)+bY (ω)
dt
Left or Right Shift in Time or Frequency
Again, starting with the Fourier pair:
x (t ) ← → X (ω)
we add (or subtract) a constant to t in x(t) and then do the Fourier
transform:
∞
∫ x (t+k )e−i ω t dt
−∞
Transforming variables:
̄t =t+k
t=̄t −k
∞
−i ω(̄t −k)
x (̄t )e
d ̄t
∫
−∞
∞
=
[∫ x (̄t )e
−i ω ̄t
−∞
= X (ω)e
]
i ωk
d ̄t e
iωk
Thus, shifting the time-domain function left or right corresponds to
a phase shift in the frequency spectrum and our Fourier pair is:
iωk
x (t +k ) ← → X (ω) e
[7-29]
What happens if we multiply x(t) by a complex exponential?
i ω0 t
x (t ) e
The Fourier transform of this is:
∞
x (t)e i ω t e−i ω t d t
∫
−∞
0
∞
= ∫ x (t )e
−i(ω−ω 0)t
−∞
dt
= X (ω−ω 0)
Thus, our Fourier pair is:
i ω0 t
x (t) e
← → X (ω−ω0 )
[7-30]
Which corresponds to a frequency shift in X(w) when x(t) is
multiplied by a complex exponential. We will use this property later.
Some Important Fourier Pairs
The Rectangular Pulse
We have already seen a bit of an example of this in the above
derivation of the Fourier integral. We define a rectangular function:
1 if |t |< τ
t
2
Rect ( τ )≡
0 if |t |> τ
2
(
)
Then, using the Rect function:
τ
2
∞
F( ω)=∫ f (t)e−i ωt dt= ∫ e−i ω t dt
−∞
−τ
2
τ
2
= ∫ [cos (ω t )−isin(ω t)]dt
τ
2
−τ
2
τ
2
= ∫ cos (ω t)dt−∫ isin(ω t )dt
−τ
2
−τ
2
Recalling [7-19] and observing that cos is an even function and sin
is an odd function, this becomes:
τ
2
τ
sin (ω t) 2
F( ω) = 2∫ cos (ω t) dt = 2 ω |
0
0
τ
τ
sin(ω )
sin(ω )
2
2
= 2
=
2
ω
2τ
ω×
2τ
τ
sin(ω )
2
= τ
= τ Sa(ω τ )
τ
2
ω
2
The Fourier pair is:
t
Rect ( τ ) ← → τ Sa( ω τ )
2
[7-31a]
or in longer form:
t
1 ∞
Rect( τ )=
τ Sa( ω τ ) ei ω t d ω
∫
2π −∞
2
and
∞
t
τ Sa( ω τ )=∫ Rect ( τ )e−i ωt dt
2
−∞
from [7-27a] and [7-27b].
What happens if we start with F(w) and want f(t)? Specifically,
if our frequency function is:
F( ω)=Rect ( ω
τ)
what is f(t)? To do this we invoke the inverse Fourier transform [727b]. The procedure is very similar to the preceding and the result
is the Fourier pair:
τ Sa( t τ ) ← → Rect ( ω )
τ
2π
2
[7-31b]
Or, given the linear property of transforms:
τ Sa(
tτ
)← → 2 π Rect( ω
τ)
2
Contrast this with [7-31a].
Single Side Exponential Decay
We begin with the definition of the function U(t) called the
Heaviside unitary step function:
(
0 for t<0
U (t)= 1 for t =0
2
1 for t>0
)
1
0
Now we define our single-sided exponential function using U(t):
−α t
f (t )=e
U (t) (α>0)
1
0
and finally, the Fourier transform:
∞
F( ω)=∫ e−α t U ( t)e−i ω t dt
−∞
Since the integral from
−∞
to 0 is obviously zero we write:
∞
F( ω)=∫ e−α t e−i ωt dt
∞
0
=∫ e−(α+i ω)t dt
0
e−(α +iω)t ∞
=
|
−( α+i ω) 0
1
=
α+i ω
and our Fourier pair is:
−αt
e
U (t ) ← →
1
α+i ω
[7-32]
From the plot you can see that f(t) is neither even nor odd. We can
extract the even part of the frequency function by restating it in
rectangular format (ala [7-25]):
1
α−iω
×
α+i ω α−iω
α−iω
= 2
α +ω2
α
= 2
+i 2 ω 2
2
α +ω
α +ω
F( ω)=
The real part is:
a(ω)=
α
2
α +ω
2
and is even. We shall be interested in this a bit later.
The Dirac Delta Function
This function is defined for our purpose as:
(∫
δ(t )≡
∞
−∞
0
t≠0
δ(t)dt=1
)
and is a strange way to define a function since there is no
assignment statement for t=0. This makes it hard to do the Fourier
transform directly. Nonetheless, it has some very interesting and
crucial properties for nmr spectroscopists.
Since the definite integral of the function is equal to one and
the function is equal to zero everywhere but at t=0, it is reasonable
to think that this represents a rectangular pulse (or impulse) of
infinite height and infinitely small width at zero. We can model this
using the Rect(t) function:
1
t
I τ (t)= τ Rect( τ )
1/t
-t/2
t/2
[7-33]
It(t) is our impulse function. As t is decreased, the width of the
rectangle decreases and its height increases. Therefore, in the limit
as t approaches zero It(t) approaches the Dirac Delta function:
δ(t )=lim I τ (t )
τ →0
The area under It(t) is equal to one for all values of t:
∞
∫ I τ (t )=1
−∞
which makes our impulse function definition consistent with the Dirac
Delta function's definition.
We now do the Fourier transform on the impulse function,
recalling the Fourier pair for the Rect function and the fact that
Fourier transforms are linear:
∞
I τ (ω)=∫ I τ (t )e−i ωt dt
−∞
1
=( τ )[τ Sa( ω τ )]
2
= Sa( ω τ )
2
and the corresponding Fourier pair is:
sin( ω τ )
2
I τ (t ) ← → ω τ
2
Now we ask what happens as t approaches zero:
sin( ω τ )
2
F( ω)=lim ω τ
τ →0
2
=1
by L'Hopital's rule (Appendix II).
Thus we now have the Fourier pair:
δ(t ) ← → 1
[7-34a]
By reasoning similar to that leading to [7-31b] we can also come up
with:
1 ← →2 π δ( ω)
[7-34b]
The Sample Property of the Dirac Delta Function
Let's multiply the delta function by another function of time:
f (t)δ(t)=f (t )lim I τ ( t)
τ →0
=lim f (t) I τ (t )
τ →0
1
t
=lim f (t ) τ Rect ( τ )
τ →0
This is just the limit as t approaches zero of f(t) and Rect(t/t)
multiplied together. As t gets smaller the upper and lower limits of
f(t) get smaller until t reaches zero and:
1
t
lim f (t) τ Rect ( τ )=f (0) δ(t )
τ →0
The delta function is said to be weighted by f(0).
We can generalize this property even more. First consider our
initial approach to the delta function using It(t) (equation [7-33]).
The rectangular 'boxes' defined by this function are centered about
t=0 and the delta function is centered about zero, but this need not
be so. Consider the 'boxes' being shifted to the right and centered
about z instead of zero.
1/t
-t/2+z
z
Our impulse function would then be written:
I τ (t−ζ)
t/2+z
Proceeding from here with the limiting process as t approaches zero,
as before, we end up now with a delta function located at t = z.
Evidently the delta function can be located at any point along the t
axis. Furthermore, we need not limit the delta function to the time
domain .. we can use it in any domain including the frequency domain.
So, we generalize the sampling property in the time and
frequency domains:
f (t )δ( t−ζ)=f ( ζ) δ(t −ζ )
F( ω) δ( ω−β)=F (β) δ( ω−β)
[7-35]
Now we use the sampling property (in step 2) as follows:
∞
∫ δ(t−ζ)e−i ω t dt
−∞
∞
= ∫ δ(t −ζ)e
−∞
−i ω ζ
=e
−i ω ζ
dt
∞
∫ δ(t −ζ) dt
−∞
=e
−iω ζ
which gives us another Fourier pair:
δ(t−ζ) ← → e−i ω ζ
[7-36]
This is really just a generalization of our first delta function
Fourier pair. If z is zero in [7-36] the delta function is centered
about t=0 then the exponential evaluates to one and we have:
δ(t ) ← → 1
as before.
Equation [7-36] is just a compact way of saying:
∞
e i ω ζ=∫ δ(t−ζ )e−i ω t dt
−∞
and
1 ∞ i ω ζ −i ωt
δ(t−ζ)=
∫ e e dω
2 π −∞
from the definitions of the analysis and synthesis equations, [7-27a]
and [7-27b].
The Complex Exponential Function
As nmr spectroscopists this is probably the single most
important transform to learn about. Our function is:
f (t )=ei ω t
f
or:
f (t)=1⋅ei ω t=g(t) ei ω t
f
f
and we know that the Fourier transform of 1 is 2pd(w) ([7-31b]) and
that multiplication of g(t) by ei ω t gives G(w – wf) from [7-30].
Thus, we write the Fourier pair as:
f
e i ω t ← → 2 π δ(ω−ωf )
f
[7-37]
The Sine And Cosine Functions
The Fourier transform of these functions follows immediately
from [7-37} and Euler's Formula:
f (t )=cos (ωf t )
∞
F (ω)= ∫ f (t)e−i ω t dt
−∞
∞
= ∫ cos (ω f t) e−i ω t dt
−∞
∞
=∫
[ ei ω t +e−i ω t ] e−iω t dt
f
f
2
−∞
1 ∞ i ω t −i ω t
1 ∞ −i ω t −i ωt
= ∫ e e dt+ ∫ e
e
dt
2 −∞
2 −∞
=π δ (ω−ωf )+π δ(ω+ωf )
f
and:
f
[7-38a]
f ( t)=sin(ωf t)
∞
F (ω)= ∫ f ( t) e−i ω t dt
−∞
∞
= ∫ sin (ωf t )e−i ωt dt
−∞
∞
=∫
[ e i ω t −e−i ω t ] e−i ω t dt
−∞
∞
=
f
f
[7-38b]
2i
1
1 ∞ −i ω t −i ωt
i ω t −i ω t
e
e
dt−
∫
∫ e e dt
2i −∞
2i −∞
= π δ (ω−ωf )− π δ(ω+ωf )
i
i
f
f
These results are of pivotal importance to us. Recalling that d(t-z)
represents a shift of t to the right by z units, d(w-wf) will
similarly represent a shift of the delta function to the right on the
frequency scale by wf units. Therefore, [7-38a] is the Fourier
transform of the cosine function and tells us that there is delta
function response at +wf and -wf:
-wf
wf
The sine function plot is:
-wf
wf
There is a simple way to extract either of the frequencies out
of these results. To get wf we simply add the cosine transform and
sine transform. Similarly, to get - wf we subtract the sine transform
from the cosine transform. This is the basis of the quadrature
detection method used in nmr spectroscopy.
The Free Induction Decay
This is our ultimate destination in this section. The modern FT
spectrometer collects data from the sample in the form of an
exponentially decaying signal called a free induction decay or FID.
The signal is a mixture of sine and cosine signals; in other words
neither even nor odd. We can, of course separate it out into even and
odd parts. In the parlance of nmr spectroscopy these would be the
real and imaginary signals.
The question is: how do we model this signal and what is its Fourier
transform? Modeling the signal is easy. Referring back to the
discussion of the one-sided exponential decay we use the Heaviside
unit step function to 'turn off' everything before zero time and then
the signal is just a cosine or sine function modulated (multiplied)
by an exponential function such that the signal decays with time.
Here we combine the sine and cosine functions into a complex
exponential for convenience:
f (t )=U ( t ) exp(−α t ) ei ω
or
g(t )=U (t ) exp(−α t )
iω t
f (t )=g(t ) e
0
0
t
We already know what G(w) is from [7-32] and from [7-30] we know
that multiplication of g(t) by a complex exponential results in G(ww0). So:
F(ω)=G(ω−ω0 )
1
=
α+i (ω−ω 0)
As before, we can break this up into real and imaginary parts:
α−i(ω−ω 0)
1
×
α+i(ω−ω0 ) α−i(ω−ω 0)
α−i(ω−ω0 )
= 2
2
α +(ω−ω0)
F( ω)real = 2 α
2
α +(ω−ω 0)
−i(ω−ω 0)
F(ω)imaginary = 2
α +( ω−ω 0)2
F( ω)=
A plot of each shows:
[7-39]
The real part is the Lorenzian lineshape equation.
The Discrete Fourier Transform
The modern high-resolution nmr spectrometer takes the analog FID
and digitizes it for storage in a computer. These digital data are
then transformed via software to frequency domain data for further
analysis by the user.
Impulse Sampling
Having looked at the sampling property of the Dirac delta
function ([7-35]) we are in a position to begin considering discrete
sampling of data and how to transform it. We begin by defining a
bandlimited spectrum
Problems
References
1. N. Morrison, Introduction to Fourier Analysis, John Wiley and
Sons Inc., 1994.