Kusse B.R., Westwig E.A. Mathematical Physics: Applied Mathematics for Scientists and Engineers
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286
FOURZER
TRANSFORMS
Figure
8.41
Closure
of
the Fourier Inversion Integration
where the contour
C
is closed in the upper half @-plane for
t
>
0,
and in the lower
half @-plane for
t
<
0,
as shown
in
Figure
8.41.
For each contour, the integral can
be
determined by adding
up
the residues of
the enclosed poles. The
t
<
0
contour has no enclosed singularities, and the
t
>
0
contour
has
just one, with
a
residue of
e-%'.
Thus
our
final result
is
(8.132)
which is our
original
function.
Now, consider what happens as
a,
approaches zero.
This
makes the
f(t)
look
more and more like a step function,
as
shown in Figure
8.42.
For any
a,
bigger than
zero, we can obtain a valid Fourier transform, but if
q,
=
0,
we know we cannot take
the Fourier transform, because the integral
1"
F(w)
=
-l
dte-id
2Tr
(8.133)
does not exist.
1
Figure
8.42
The Exponential Function in the
Limit
of
a,
---t
0
FOURIER TRANSFORMS
-
EXAMPLES
L
-
3'
287
imag
@plane
"
a0
11
real
-
-
LJ
By
looking at what happens to the inversion contour for
this
function in the
a,
+
0
limit, we can gain a valuable alternative viewpoint.
As
long as
a,
is greater than zero,
no matter how small, the pole of the integrand of Equation
8.125
is in the upper half
-
o-plane, and the
F
contour can remain on the real @-axis. But when
a,
=
0,
the pole
sits right on top of the F-contour and the integral cannot be performed.
Notice, however, that if the T-contour is deformed to
T'
by dodging the pole, as
shown in Figure 8.43, the inversion integral
reproduces the unit step function
0
t<O
1
t>O
f(t>
=
{
(8.134)
(8.135)
This is no coincidence, and it will lead
us
to a more general integral transform called
the Laplace transform.
8.6.9 Damped
Sinusoid
As
a last example, consider the damped sinusoid
(8.136)
where
a,
and
o,
are both positive numbers. Figure 8.44 shows this function. The
Fourier transform is easy to calculate:
(8.137)
288
FOURIER TRANSFORMS
Figure
8.44
The
Damped
Sinusoid
where
g,
=
w,
+
ia,
and
g2
=
-0,
+
ia,.
Notice
f(w)
is a complex function
because
f(t)
has neither
odd
nor even symmetry.
The inversion
transfom
of Equation
8.137
is
As
with the exponential decay function of the previous section, the
best
way to
evaluate
this
integral is to extend it into the complex plane,
and
close the complex
contour. The complex version
of
the inversion integral is
(8.139)
where
7
is the Fourier contour along the real @-axis,
as
shown
in
Figure
8.45.
Also,
like the exponential decay example, we are forced to close the contour
in
the upper
half-plane for
t
>
0
and the lower half-plane for
t
<
0.
This
is due to the influence
of the
eig*
term in the numerator.
As
you will prove
in
one of the exercises
at
the end
of the chapter, Equation
8.139
returns the original function
f(t),
as long as
a,
>
0.
(8.140)
Figure
8.45
Inversion
of
the
Damped
Sinusoid
Transform
in
the g-Plane
FOURIER
TRANSFORMS
-
EXAMPLES
F‘
-
00
-A
289
*plane
imag
“0
hag
-
-
-
I
Figure
8.46
The
Undamped Sinusoid
The Fourier transform and inversion for the damped sinusoid work correctly for
all values of
a,
>
0,
no matter how close
to
zero
a,
gets. Exactly at
a,
=
0,
however,
we know we cannot have a valid Fourier transform, because now
f(t)
becomes
(8.141)
as shown in Figure 8.46. The sinusoid is no longer damped, and the Fourier integral
does not exist. This problem
is
also
evident from looking at the inversion contour.
With
a,
=
0,
the Fourier transform in Equation 8.137 has become
F(w)
=
-
wJJ2.rr
(8.142)
(w
-
WoXW
+
00)’
and the two poles in Figure 8.44 now lie on the Fourier contour.
sinusoid
Notice, however, Equation 8.142 still can
be
used to generate the undamped
f(t)
=
-
dw
eiot
-00/J2.rr
(8.143)
if the Fourier contour is modified to
y’,
as shown in Figure 8.47. In fact, even
an
exponentially growing sinusoid,
AL.
-
(w-w0)(4!+
wo)’
290
3'
-
00
-I
-
v
-%
FOURIER TRANSFORMS
imag
*plane
00
real
-
-
-
-
Figure
8.48
Exponentially Growing Sinusoid
(8.144)
such
as
the one shown
in
Figure
8.48,
can be recovered from
this
inversion, if we let
the Y'contour dip below
the
poles of
Em,
as
shown in Figure 8.49.
These deformed contours seem to extend the Fourier inversion into realms where
you would not expect it to work. In the next chapter we formally introduce these
extensions
as
the basis for the Laplace transform.
8.7
THE
SAMPLING
THEOREM
One of the more important results of Fourier transform theory is the
Sampling
The-
orem.
This
theorem determines at what rate a signal must be sampled
so
enough
information is available to properly reproduce it.
This
is important, because sampling
too infrequently will cause the reproduced signal quality to degrade, while requiring
a sampling rate which is too high
is
inefficient.
Consider sampling an audio signal for recording onto a compact disc. Let the
original microphone signal
u(t)
look as shown in Figure
8.50.
Furthermore, let this
analog signal have a Fourier transform
&w),
as
shown in Figure 8.5
1.
This
spectrum
vanishes above and below some cutoff frequency
om,
which is perhaps imposed by
THE
SAMPLING
THEOREM
291
Figure
8.50
The
Original Analog
Audio
Signal
the mechanics of the microphone, For the purposes of this discussion, the transform
will be assumed to be pure real.
This
assumption is not necessary for the validity of
the Sampling Theorem, but is convenient for
our
drawings.
A
useful way to model the sampling process, and the signal recorded on the
CD,
is
to let the sampled signal, which we will
call
a,(t),
be
equal to
the
original signal
times a periodic set
of
&functions
m
as(t)
=
a(t)
c
s(t
-
nTo).
The time between the &functions is the sampling interval. Figure
8.52
shows
this
representation.
Because
a,(t)
is
the product of two functions, the spectrum of
a,(t),
which we will
call
A,(o),
is the convolution of
A(w)
with the transform of the train
of
$-functions:
(8.145)
"=
-m
J2?r
5
6(o
-
2nn/T0).
(8.146)
1
A,(W)
=
-A(w)
@
-
6
To
-m
When the sampling rate is very high, that is when
To
G
2?r/wm,,
A,(o)
has the form
shown in Figure 8.53. The spectrum of
as(t)
is not the same as the spectrum of
a(t),
because it has many more high frequency components. If it were reproduced through
-%ax
%lax
Figure
8.51
Fourier
Transform
of
the
Original Signal
292
FOURIER
TRANSFORMS
Figure
8.52
The
Sampled
Signal
an
amplifier and speaker system, it would not sound like
a(t).
However, sitting in the
spectrum of
a&),
from
-w,-
to
+w,-,
is the exact spectrum
of
a@).
Thus, if we
send
this
signal
through a low-pass filter before amplification, the result should be a
sound
that reproduces
a(t).
A
simple low-pass filter can be made from the capacitor and resistor circuit shown
in Figure 8.54. The input voltage
vdt)
and output voltage
vout(t)
are related by the
differential equation
(8.147)
Applying a Fourier transform to both sides
of
this
equation gives an algebraic equation
relating the Fourier transforms of the input and output waveforms:
K.Jw)
=
(iwRC
+
l)V-,ut(m)
(8.148)
01
(8.149)
0
-
-2f10
2flO
4f10
Figure
8.53
Spectrum
of
Sampled
Signal
2anU
-4flO
THE
SAMPLING
THEOREM
293
R
+-+
Figure
8.54
Simple
Low-Pass
Filter
The quantity
V-,,(w)/~.,(o)
is called the transfer function for the filter. The
magnitude
of
this function
is
shown
in
Figure
8.55.
For frequencies much less than
w
=
1/RC,
the output voltage and input voltage are essentially identical.
For
fre-
quencies much above
o
=
I/RC,
the output is zero.
This
transition frequency is
called the cutoff frequency of the filter,
wco.
For a simple
RC
filter, such as this
one, the transition at the cutoff is not
very
sharp.
Better filters can be made with
a more complicated combination
of
components, or with more sophisticated digital
processing components. The ideal low-pass filter
has
the transfer function shown in
Figure
8.56.
If
we pass the sampled signal through an ideal low-pass filter with
o,,
=
om,
the output of the filter will have a Fourier transform
L&(w)
that
is
exactly equal to
-
A(o),
and consequently the signal coming out of the filter
af(t)
will sound exactly
like the original
a(t)
signal from
the
microphone. The filter has removed the spurious
high frequency components introduced by the sampling.
This
process is shown in
-1mc
j
1mc
Figure
8.55
Transfer Function
for
a Simple
RC
Low-Pass Filter
-00
I
0,
Figure
8.56
Ideal Low-Pass Filter
294
FOURIER TRANSFORMS
Low
Pass
-1-J
Figure
8.57
Frequency
and
Time Domain Pictures
of
the
Filtering
Process
Figure
8.57.
The top picture in Figure
8.57
shows the filtering process in the frequency
domain, while the bottom picture shows the same process
in
the time domain.
In
the previous arguments, we
made
the important assumption that the original
signal was sampled at a
high
rate. The spectrum for
AJo)
is composed
of
a
series
of
A(o)
spectra
replicated every
o
=
2n?r/T0.
As
long
as
r/To
>
w,-,
these
spectrum
do
not
overlap, and
our
previous analysis works fine. But when
To
grows
so
large that
r/T,
<
(l)mar,
the individual spectra overlap each other and
&(w)
looks as
shown
in
Figure
8.58.
Now, the output of the low-pass filter with
w,,
=
o,,
gives
the spectrum shown in Figure
8.59.
In
this
case the signal coming out of the filter
would no longer sound like the signal from the microphone.
Thus we
see,
in
order
to
preserve the information in a signal, it must be sampled
such that
r/T0
>
(l)mar.
This
means the sampling frequency
2r/T0
must be at least
twice the highest frequency present
in
the sampled signal. The minimum sampling
frequency is often called the Nyquist sampling rate.
Keep
in
mind that we have drastically simplified the process of
CD
sampling in
our
previous argument. Some practical issues that introduce complications should
be
pointed
out. First,
in
real
CD
sampling, the recorded audio signal is digitized
.....
.....
Figure
8.58
Spectrum
of
an
Undersampled Signal
EXERCISES
295
-%
%ux
Figure
8.59
Spectrum
of
the Filtered, Undersampled Signal
and can take on only discrete values.
This
requires the use of an analog-to-digital
converter
(ADC).
The recreation of the original signal then requires the inverse step
of a digital-to-analog conversion
(DAC).
Second, for an audio signal, the practical
limit for
wmx
is
about
20
kHz,
the frequency limit of the human ear. The music
industry has settled on a
44
Khz
sampling standard, just above twice the maximum
frequency we can hear. However, the microphone signal and its electronic handling
before sampling may very well have frequency components well above
22
kHz.
To
prevent the overlap in the convolved spectra for
&(o)
that would result by sampling
such a signal,
a
low-pass filter with a cutoff frequency near
22
lcHz
is inserted just
before the sampling electronics. Finally, there is no such thing as an ideal filter with
a transfer characteristic like that shown
in
Figure
8.56;
there will always be some
“rolloff.” The better the filter, the better the sound reproduction.
EXERCISES
FOR
CHAPTER
8
1.
What is the Fourier Transform of the function
where
a
is
real and positive?
2.
Find the Fourier transfonn of the function
Is
there a way
to
evaluate
this
without actually calculating the transform integral?
3.
Find the Fourier transform
of
the signal shown below,