Fourier Transforms. Figure 7-1:Fourier transform from time domain to frequency domain

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:

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

Figure 7-1:Fourier transform from time domain to frequency

domain

t

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=

_∑

i

a_i⃗u_i [7-1]

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

vectors,⃗u_i, are orthogonal. That is, when we do a scalar product of

any two unit vectors the result will be zero:

⃗

u_i⋅⃗u_j=

∣

u_i

∣∣

u_j

∣

cos( π

2)=cos ( π2)=0 [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

∣∣

uj

∣

cos (0)=cos(0)=1 [7-3]

Now, we make a bit of a leap. Why not build up a function, F,

from a set of basis functions in a way that is analogous to vectors? If you look back at the chapter on vectors (section 3.1) you will see that, as long as each of the basic rules that define vectors is

followed, this is ok.

F=

∑

i

a_if [7-4]

This is what we do when using Taylor series to represent various functions: ex_=1+x+x2 2 !+ x3 3!+ x4 4 !⋯=

∑

n=0 ∞ _xn n ! log(1+x)=x−x 2 2 + x3 3− x4 4⋯=

∑

n=1 ∞ (−1)n−1x n n [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:

A=⋯, e−i3ωft, e−i2 ωft, e−i ωft,1, ei ωft, ei2ωft, ei3 ωft

⋯ [7-6]

where we define ω_f to be the fundamental radial frequency of the

periodic function. To demonstrate that these functions are in fact

orthogonal we do an integration over one period, τ=2 π_ω

f , of the function: 1 τ

∫

0 τ ei n ωft*ei m ωftdt [7-7]

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

a formal sense. It will become apparent why we multiply by 1_τ

momentarily.

Where does τ=2 π_ω

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:

ω_f=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:

τ=1_{ν =}2 π_ω f

Our integral,[7-7], becomes:

1 τ

∫

0 τ e−i n ωftei m ωftdt=1 τ

∫

0 τ ei(m−n)ωftdt [7-8]

We have two possible situations now. If m=n then the integral

evaluates to 1: 1 τ

∫

0 τ dt=

[

t_τ

]

0 τ = ττ−0τ =1 [7-9]

Thus the functions are normalized.

Ifm≠nthen: 1 τ

∫

0 τ ei(m−n )ωftdt= e i Δ ωft τi Δω_f|0 τ = e i Δ ωft 2 πi Δ|0 τ [7-10]

whereΔ=m−n. Finishing the evaluation of the definite integral:

Figure 7-2: Square wave function

0

t

ei Δ ωft 2 π i Δ|0 τ =e i Δ 2 π −1 2 π iΔ = 1−1 2 π i Δ=0 [7-11]

Thatei Δ 2 πis equal to 1 is evident when you consider that using

integral multiples of 2π in eix must result in the cosine portion

being equal to 1 and the sine portion being equal to zero:

ei n2 π=cos (n 2 π)+i sin(n 2 π)

where n is any integer.

Ok ... we have a set of orthonormal functions. What do we do with them? We use them to model other functions, in a similar fashion to that of Taylor series. Since Fourier series must be periodic under one of the Dirichlet conditions, we write:

f (t)=

∑

n=−∞

∞

C(n)ei n ωft [7-13]

wheref (t)is the function to be modeled andC(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:

1 τ

∫

0 τ f (t)e−im ωftdt=1 τ

∫

0 τ

∑

n=−∞ ∞ C(n)ei n ωfte−i m ωftdt [7-13]

We now interchange the integral and summation on the right side to get: 1 τ

∫

0 τ f (t)e−im ωftdt=

∑

n=−∞ ∞ C(n)1_τ

_∫

0 τ ei n ωfte−i m ωftdt [7-14]

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

1 τ

∫

0 T f (t)e−im ωftdt=C (m) [7-15]

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:

1 τ

∫

_{− τ} 2 τ 2 f (t)e−i m ωftdt=C (m) _[7-15a]

An alternate way of writing the synthesis equation is:

f (t)=a0+

∑

n=1 ∞ ancos (n ωft)+

∑

n=1 ∞ bnsin(n ωft ) [7-16]

Equation [7-16] can be easily shown to equivalent to [7-13]:

f (t )=a₀+

∑

n=1 ∞ a_ncos (n ω_ft )+

_∑

n=1 ∞ b_nsin (n ω_ft) =a₀+

∑

n=1 ∞ a_n

(

e i n ωft₊e−i n ωft 2

)

+

∑

n=1 ∞ b_n

(

e i n ωft₋e−i n ωft 2i

)

=a₀+

∑

n=1 ∞ _(a n−ibn) 2 e i n ωft₊

∑

n=1 ∞ _(a n+ibn) 2 e −i nωft =a₀+

∑

n=1 ∞ _(a n−ibn) 2 e i n ω_ft +

∑

n=−∞ −1 _(a n+ibn) 2 e i n ω_ft =

∑

n=−∞ ∞ C (n)ei n ωft where C(n)=(an−bn) 2 n>0 C(−n)=(an+bn) 2 n<0 C (0)=a0 n=0

Sometimes we speak of harmonics, especially with respect to the analysis of periodic functions involving sound waves. From the

h(0)=C(0)(fundamental or average value)

h(1)=C(1)eiωft₊C(−1)e−i ωft(1st harmonic)

h(2)=C (2)ei 2 ωft₊C (−2)e−i 2 ωft(2nd harmonic )

⋮

h (n)=C (n)ei n ωft₊C (−n)e−i n ωft₍nthharmonic )

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.

We begin with the analysis equation:

0

t

C(n)=1_τ

_∫

0 τ f (t)e−in ωftdt (n≠0) = ωf 2 π

∫

₀ 2 π ω_f t e−i n ωftdt = ωf 2 π

∫

₀ 2 π ωf t [cos(n ω t )+isin (n ω_ft)]dt =ωf 2π

∫

₀ 2 π ωf t cos(n ω_ft)dt+i ωf 2 π

∫

₀ 2 π ωf t sin(n ω_ft)dt

The integrals are solved by integration by parts to give:

=ωf 2π

[

t sin(n ω_ft ) n ωf −cos (n ωft) n ωf

]

−i ωf 2 π

[

t cos (n ω_ft) nωf −sin (n ωft) n ωf

]

| 0 2 π ωf

The first term evaluates to zero after insertion of the integration limits. The second evaluates to

−i

n ω_f for n≠0

Forn=0the analysis equation is:

C (n)=1_τ

_∫

0 τ f (t )e−i n ωftdt (n=0) = ωf 2 π

∫

₀ 2 π ωf t e−i n ωt_dt =ωf 2π

∫

₀ 2 π ωf

t [cos(n ω t)+i sin(n ω_ft)]dt

=ωf 2 π

∫

₀ 2 π ωf t dt = ωf 2 π t2 2 |0 2 π_ω f =π 2

A= π 2+n=−∞

∑

∞ −i n ω_f e −in ωft ₍n≠0) =π 2+n=−∞

∑

∞ _−i n ωf

[

cos (n ω_ft )−isin (n ω_ft )

_]

In terms of harmonics this is:

A=h (0)+

∑

n=−∞ ∞ h(n) =π 2+_n=−∞

∑

∞ _−2i n ω_fcos (n ωft )

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 figure:

Figure 7-3: Sine and cosine functions

t

cos(wt)

t

sin(wt)

Each of these types of functions has a useful property:

∫

−α α f (x)dx=2

∫

0 α f (x )dx (even functions)

∫

−α α f (x)dx=0 (odd functions) [7-19]

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, [7-16, [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 [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 f (t)_even=1 2

[

g(t )+g(−t)

]

[7-22] also:

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

f (t )_odd=1

2

[

g (t)−g (−t)

]

[7-23]

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):

C(n)* =

[

1_τ

∫

0 τ f (t )e−i n ωt_dt

]

* =1_τ

∫

0 τ f (t)* ei n ωtdt =1_τ

∫

0 τ f (t )ei n ω tdt =

C (−n)

[7-24]

We can break the coefficient,C(n), into real and imaginary

coefficients:

C(n)=a(n)+ib(n) [7-25] 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)

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 iff (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.

Visually, you can see that this is an odd function. The analysis expression is:

C (n)=1_τ

∫

0 1 f (t )e−i n ωftdt+1 τ

∫

1 2 f (t)e−i n ωftdt (n≠0) =1 2

∫

₀ 1 e−i n πt_dt+1 2

∫

₁ 2 −e−i n πt_dt =−1 2 e−i n π t i n π |0 1 +1 2 e−in π t i n π |1 2 =−1 2

[

e−i n π −e0

_]

in π + 1 2

[

e−i n 2 π −e−i n π

_]

i n π =−1 2

[

cos (n π)−isin(n π)−1

]

i n π + 1 2

[

cos (n2 π)−isin(n2 π)−cos(n π)+isin(n π)

]

in π =−1 2 cos(n π)−1 i n π + 1 2

[

+1−cos (n π)

]

i n π =−1 2 cos (n π)−1 i n π − 1 2

[

cos(n π)−1

]

i n π =−1 2 2⋅cos (n π)−2 i n π =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 (nπ)will equal zero for all values of n. For sequential values of n the cosine term will alternate between +1 and -1:

C(n)=1−(−1)

n

i n π

These are the Fourier constants for n≠0 . For n=0 :

C(0)=1_τ

∫

0 1 f (t)e−i0 ωftdt+1 τ

∫

1 2 f (t)e−i 0 ωftdt =1 2

∫

₀ 1 1⋅dt+1 2

∫

₁ 2 −1⋅dt (n=0) =1 2t|0 1 −1 2t|1 2 =1 2−0−1+ 1 2 =

0

f (t)=

∑

n =−∞ ∞ C(n)ei n ωft =

∑

n=−∞ ∞ ₁₋₍₋₁₎n i n π ⋅e i n π t =

∑

n=−∞ ∞ ₁₋₍₋₁₎n in π ⋅

[

cos(n π t)+isin(n πt )

]

For any value of n, the sine term must vanish:

f (t)=

_∑

n=−∞

∞ ₁₋₍₋₁₎n

in π ⋅cos (n π t)

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

We imagine this to be part of a periodic train of pulses. Using our, by now, standard method of finding the Fourier coefficients:

C(n)=1_τ

∫

−1 1 f (t)e−in ωftdt =1_τ

∫

−1 1 e−i n ωftdt -1 1 2 3

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:

C(n)=1_τ

∫

−1 1 e−i n ωftdt C(0)=1_τ

∫

−1 1 dt =1_{τ t}| −1 1 =2_τ Forn≠0: C (n)=1_τ

∫

−1 1 e−i n ωftdt =−1_τ e −i n ωft in ω −1| 1 =− 1 τi n ω_f(e −i n ω_f −ei n ωf₎ =− 1 τi n ω_f[−i2sin(n ωf)] =2_τ sin(n ωf) n ω_f =2_{τ Sa (n ωf}) =ω_{π Sa( n ω}f _f) ( τ=2 π_ω 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

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,

nωf . It will help us at this point to have a picture of what we

have. Specifically, plots of ₂τC(n)vs nω_f for increasing values of τ ,

10, 30, 60:

Figure 7-4a

You can see that as τ → ω the curve becomes smoother as the density of

nω_f increases. So as τ → ω , nω_f→ ωand we write:

lim

τ →∞ τC (n)=Sa(ω) −∞<ω<∞

nω_f changes toωand the C(n) become a continuum of values related toω

for our square wave function:

F(ω)=Sa (ω)

We can generalize this as follows beginnning with our discreet Fourier series analysis equation:

C(n)=1_τ

∫

−τ 2 τ 2 f (t)e−i n ωftdt=1 τ_−∞

∫

∞ f (t)e−i n ωftdt or τC (n)=

∫

−∞ ∞ f (t)e−i n ωftdt

We can write this with infinity integration limits because our

function's value from minus infinity to −τ₂ and from ₂τ to infinity is

zero. As τ → ω , τC (n)becomesF (ω)and nωfbecomes ω and we write:

F(ω)=

∫

−∞ ∞ f (t)e−i ωt dt [7-27a] Figure 7-4c

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

2 π

∫

−∞ ∞

F (ω)eiω td ω [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 =a

∫

−∞ ∞ x (t )e−i ω t_dt+b

∫

−∞ ∞ y (t )e−i ω t_dt =aX (ω)+bY (ω)

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

∫

−∞ ∞

x(̄t)e

−i ω(̄t−k)

d ̄t

=

[

∫

−∞ ∞

x (̄t)e

−i ω ̄t

d ̄t

]

e

i ωk

=

X (ω)e

i ω k

Thus, shifting the time-domain function left or right corresponds to a phase shift in the frequency spectrum and our Fourier pair is:

x (t+k )← → X (ω)ei ω k [7-29] What happens if we multiply x(t) by a complex exponential?

x (t )ei ω0t

The Fourier transform of this is:

∫

−∞ ∞

x(t)e

i ω0t

e

−i ω t

dt

=

∫

−∞ ∞

x(t )e

−i(ω−ω0)t

d t

=

X (ω−ω

₀

)

Thus, our Fourier pair is:

x (t)ei ω0t_{← →}X (ω−ω

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:

Rect(t_{τ )≡}

(

1 if|t|< τ2 0 if|t|> τ

2

)

Then, using the Rect function:

F(ω)=

∫

−∞ ∞ f (t)e−i ωt dt=

∫

−τ 2 τ 2 e−i ω t dt =

∫

−τ 2 τ 2 [cos(ω t)−isin(ω t)]dt =

∫

−τ 2 τ 2 cos(ω t)dt−

∫

−τ 2 τ 2 isin(ω t )dt

Recalling [7-19] and observing that cos is an even function and sin is an odd function, this becomes:

F(ω) = 2

∫

0 τ 2 cos (ω t)dt = 2sin (ω t)_ω | 0 τ 2 = 2 sin(ω τ 2) ω = 2 sin(ω τ 2) ω×2 τ 2 τ = τ sin(ω τ 2) ω τ 2 = τSa(ω τ 2)

Rect(t_{τ ) ← → τ Sa(}ω τ 2 ) [7-31a] or in longer form: Rect(t_{τ )=} 1 2π−∞

∫

∞ τSa( ω τ 2 )e i ω t d ω and τSa( ω τ 2 )=

∫

−∞ ∞ Rect (t_{τ )e}−i ωt_dt

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( ωτ)

what is f(t)? To do this we invoke the inverse Fourier transform [7-27b]. The procedure is very similar to the preceding and the result is the Fourier pair:

τ 2 πSa(

t τ

2 ) ← →Rect( ωτ ) [7-31b]

Or, given the linear property of transforms:

τSa(t τ

2 )← →2 π Rect( ωτ)

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:

U (t)=

(

0 for t<0 1

2 for t=0 1 for t>0

)

1

0

Now we define our single-sided exponential function using U(t):

f (t)=e−αt

U (t) (α>0)

1

0

and finally, the Fourier transform:

F(ω)=

∫

−∞ ∞ e−αt U (t)e−i ω t dt

Since the integral from −∞ to 0 is obviously zero we write:

F(ω)=

∫

0 ∞ e−αt e−i ωt dt =

∫

0 ∞ e−(α+i ω)t dt = e −(α +iω)t −( α+i ω)|0 ∞ = 1 α+i ω

and our Fourier pair is:

e−αt

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]):

F(ω)= 1 α+i ω× α−iω α−iω = α−iω α2+ω2 = α α2+ω2+i ω α2+ω2

The real part is:

a(ω)= α α2+ω2

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:

I

_τ

(

t)=

1

_{τ Rect(}

t

_{τ )}

[7-33]

t/2 -t/2

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

τ →0

Iτ(t )

The area under It(t) is equal to one for all values of t:

∫

−∞ ∞

Iτ(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:

Iτ(ω)=

∫

−∞ ∞ Iτ(t )e −i ωt dt =(1_{τ )[τ Sa(}ω τ 2 )] =Sa( ω τ 2 )

and the corresponding Fourier pair is:

Iτ(t)← →

sin( ω τ 2 ) ω τ

2

Now we ask what happens as t approaches zero:

F(ω)=lim τ →0 sin( ω τ 2 ) ω τ 2 =1

by L'Hopital's rule (Appendix II). Thus we now have the Fourier pair:

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 τ →0 Iτ(t) = lim τ →0 f (t)Iτ(t ) = lim τ →0 f (t)1_{τ Rect (τ )}t

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:

lim

τ →0

f (t)1_{τ Rect (τ )=f (0) δ(t )}t

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.

Our impulse function would then be written:

Iτ(t−ζ)

t/2+z -t/2+z

1/t

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 dt =e−i ω ζ

∫

−∞ ∞ 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:

ei ω ζ=

∫

−∞ ∞ δ(t−ζ )e−i ω t dt and δ(t−ζ)= 1 2 π

∫

−∞ ∞ ei ω ζe−i ωt d ω

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 ωft

or:

f (t)=1⋅ei ωft₌g(t)ei ωft

and we know that the Fourier transform of 1 is 2pd(w) ([7-31b]) and

that multiplication of g(t) by ei ωft _{gives G(}w _– w

f) from [7-30].

Thus, we write the Fourier pair as:

ei ωft_{← →}2 π δ(ω−ω

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 (ωft ) F (ω)=

∫

−∞ ∞ f (t)e−i ω t dt =

∫

−∞ ∞ cos (ω_ft)e−i ω t dt =

∫

−∞ ∞

_[

_ei ωft₊e−i ωft

]

2 e −iω t_dt =1 2−∞

∫

∞ ei ωfte−i ω tdt+1 2−∞

∫

∞ e−i ωfte−i ωtdt =π δ (ω−ω_f)+π δ(ω+ω_f) [7-38a] and:

f (t)=sin(ωft) F (ω)=

∫

−∞ ∞ f (t)e−i ω t dt =

∫

−∞ ∞ sin (ω_ft )e−i ωt dt =

∫

−∞ ∞

_[

_ei ωft₋e−i ωft

]

2i e −i ω t_dt = 1 2i−∞

∫

∞ ei ωfte−i ω tdt−1 2i−∞

∫

∞ e−i ωfte−i ωtdt =π i δ (ω−ωf)− π_i δ(ω+ωf) [7-38b]

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:

The sine function plot is:

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.

wf

-wf

wf

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 ω0t

or

g(t)=U (t) exp(−α t) f (t )=g(t) ei ω0t

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(w

-w0). So:

F(ω)=G(ω−ω0)

= 1

α+i(ω−ω0)

As before, we can break this up into real and imaginary parts:

F(ω)= 1 α+i(ω−ω₀)× α−i(ω−ω0) α−i(ω−ω₀) =α−i(ω−ω0) α2+(ω−ω₀)2 F(ω)_real= α α2+(ω−ω₀)2 F(ω)_imaginary= −i(ω−ω0) α2+( ω−ω₀)2 [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.