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GG 450
March 18, 2008
Convolution and Fourier
Transforms in Digital Signal
Analysis
What’s a “filter”?
A filter is a device or process that modifies
the input process such that the output has
different characteristics. The air filter in a car
changes dirty air into clean air by filtering out
particles. Soil filters out impurities in ground
water.
An electronic filter changes the characteristics of an
electronic signal.
A digital filter changes the characteristics of a digital
signal.
The characteristics of time-invarient filter don’t
ever change.
A linear filter has the same effects on a signal
regardless of the signal characteristics. For example,
a filter might always reduce the amplitude of
frequencies below 10 Hz by 6 dB/octave. This would
be a linear time-invarient filter.
When we pass a signal through a filter, or
through the earth, or through a seismometer
or amplifier, that device or operation
changes our signal into a new signal.
The purpose of signal analysis is to
apply operations (filters) to our
signals that make the information in
them easier to see and understand,
while reducing the effects of
undesirable components such as noise
and echoes.
For example, we will send a signal into the ground
with a hammer. The earth will filter that signal by
delaying it, reflecting part of the signal, and
absorbing other parts.
We will detect the signals with seismometers that
will further modify the signal.
What we WANT in this case are parts of the signal
that represent reflections or refractions through the
earth. In analysis we will attempt to remove the
effects of those filters that tend to distort or hide
these features.
Digital signal analysis is a very powerful
tool for providing insight into geological
problems.
There are several important concepts
used in digital analysis that we will look
at briefly The concept of the frequency domain
and
The process of convolution
We usually look at geophysical data in terms of
amplitude changes with time (or distance). Another way
to look at it is in terms of amplitude and frequency:
time
0
frequency
The heavy arrow implies that it is possible to go from the time
domain to the frequency domain and back without losing
any information. This transformation is accomplished with
the FOURIER TRANSFORM.
The change from time domain to frequency
domain is routine in nearly all signal analysis
of geophysical and other data.
Why is it so important?
Often, more insight is acquired by displaying
the desired information in one domain or the
other, and
many important mathematical operations are
far easier to accomplish in one domain than in
the other.
Transforming linear numbers into logarithms
changes multiplication and division into addition
and subtraction (which is how a slide rule works)
and exponents into multipliers:
A /B  C
log( A)  log( B)  log( C)
D E
n
n  log( D)  log( E)
The Fourier transform can be used to change the process of
CONVOLUTION into multiplication and division. It also
changes DIFFERENTIATION and INTEGRATION into
simple multiplication and division in many situations. More
on that later.....
So.... What's CONVOLUTION?? The convolution operation is
very important in seismology and geophysics in general because it
mathematically describes how the earth filters the signals we put
into it.
For example when we hit the earth with a hammer we generate an
IMPULSE, or Delta function. As that impulse passes through the
earth, what happens to that signal can be described by the operation
of CONVOLUTION:
Consider the seismic case:
The input to the ground is approximately an impulse:
g
ro
u
n
d
m
o
ti
o
n
hammer blow
t im e
If all the ground did was pass this impulse on at a particular velocity
with refractions and reflections arriving at various times, our
seismograms might look like the following:
a
This is called a response
function.
g
ro
u
n
d
m
o
ti
o
n
t im e
hammer blow
b
.
In fact, the above figure shows exactly what we want: all the
reflections and refractions, their amplitudes, signs, and times
of arrival.
But, the earth isn't so nice, and it changes the impulse by
absorbing some frequencies and delaying the signal in a process
called FILTERING.
A filter can change both the amplitude and phase of a signal in
different ways at different frequencies. The “earth filter” delayed
the hammer blow peak “a” above changing the phase and amplitude
of all frequencies the same way. The peak "b" did the same, except
amplitude was "convolved" with a negative number (note that it is
inverted with respect to peak “a”.
As usual, reality is more complicated, in that each refraction path
and each reflector, and each seismometer and source for that
matter, change our nice clean impulse into a train of waves, and
our hammer blow might actually look like:
called the
“source function”
every time this train of waves hits another interface it gets changes
by whatever "filter" that interface applies. If each interface above
applied only an "impulse" filter as shown, then our seismogram
might look like:
This looks more like a seismogram, and the process that has been
done is CONVOLUTION of the SOURCE function (hammer) with
the RESPONSE function of the ground. The IMPULSE
RESPONSE is defined as the output of a process if the input is an
impulse. If a system (described by a combination of filters) has a
LINEAR response then the impulse response of a system completely
defines the operation of that system.
The signal shown above is formed by replacing each of the impulses
in the response function by the source function multiplied by the
appropriate constant, then adding the results together.

In analog form (integral calculus) the convolution of two
functions of time f and g is:
f g 


f
(

)g(t


)d





g(

)
f
(t


)d


Digital convolution is similar:
g[x] 
x w
 f [k]h[x  k]
k xw
Notice that convolution is NOT simple multiplication. Each
g[x] is the SUM of products of g.h with time offsets.
Convolution involves
1) multiplying each value of the input signal by the
values of the filter IMPULSE RESPONSE
2) adding the results together (this is the output for one

3) stepping the signal one time step and repeating the
operation.
A simple “convolution box” is shown below set up to
convolve x[n]= 0,0,1,1,1,1,1 with h[n]=1,-1 (We can
ignore zero values past the ends). The mathematical
notation for this operation is x[n]h[n]=v[n]
x[ n]
0
1
2
3
4
5
6
7
0
0
0
1
1
1
1
1
X
-1
X
1
+
y[ n]
What is y[n]?
h[ n] (f lipped)
Go to:
http://www.jhu.edu/signals/convolve/
and try their interactive convolution box.
1) Use your mouse to generate x(t) and h(t).
2) click on “x(t-v) and h(v)”. This places the functions in
the convolver.
3) Use your mouse to move x(t-v) over h(v).
4) The values of x(t-v)h(v) are shown as well as the
resulting integral.
5) Try functions like those suggested and:
Another convenient way to think about digital convolution is
using what is called the z transform.
Consider two discrete functions: x= 1, 2, 4, 0
and
y= -1, -1, 0, 1
where the four values are the measurement or value of four
consecutive samples.
The product of x and y is: z=-1,-2,0,0 (multiplying pointby-point)
but the convolution of these two series is more difficult.
An easy way to do it is to consider the values of x and y to
be coefficients of a polynomial:
x(z)= 1 + 2 z +4 z2+0z3
y(z)= -1 -1 z + 0z2 + 1z3
Where z is a unit time delay, and the power of z is the
sample number.
To convolve y and z, we simply find the produce of the two
polynomials:
c(z) = x(z) y(z) = 1 - 1 z - 7 z 2 -4 z3 + z4 + z5
and take the inverse z transform to obtain the time series:
c = 1, -1, -7, -4, 1, 1
Note that the convolution of two functions creates a function
that is longer than either of the original functions, thus
convolution process can spread signals out.
When one or both of the time series we want to convolve get
long, we have to do a huge number of multiplication's to
accomplish convolution, but.. there is a better way, and that is
to change to the frequency domain using the Fourier transform.
If you’d like to know how the Fourier transform is defined and
calculated mathematically and by Fast Fourier Transform,
check Google and MatLab functions for these terms. For our
purposes, we only need to know what happens when we go
from the time domain to the frequency domain and back
(USING the Fourier transform), not how it's done.
When we go from the time domain to the frequency domain the
signal properties do not change. All of the information present
in one domain is still there in the other. All that changes is how
the signal is presented. In the time domain, time is the
independent variable, and in the frequency domain, frequency
is the independent variable. In both cases, amplitude is the
dependent variable. In spatial data, wavenumber is often used
as the independent variable.
What happens to various signals on transformation?
i
The functions on the previous page are some famous
transform pairs. Each is worth understanding. Be aware
that there are many ways to display data in the frequency
domain, real and imaginary components, phase and
amplitude, power spectrum, etc. Those above show only
part of the complete spectrum for simplicity. If the signal in
the time domain is symmetric about zero, then the
imaginary part of the Fourier transform is zero. That is the
case in most examples above.
The comb function transform pair was included here
because it is critical in understanding of sampling of
continuous signals. Discussed below.
• Both domains have both positive and negative values.
Negative time is fairly easy to understand, but what does a
negative frequency imply?? One clue is the sin wave
transform. It's the only one of the above that isn't symmetric
in the time domain. What do you think the transform of a
cosine wave might be?
• Note the relationship between PERIOD in the time domain
with FREQUENCY in the frequency domain.
• There is complete symmetry between time and frequency
domain - that is, we could have called the time domain
"frequency domain" and visa versa.
If this is true, though, how come the delta function transform
looks so much like the transform of the sin wave? HINT:
think about how the sine wave signal transform looks as the
sine wave frequency approaches 0.
Referring to the Fourier transform pairs in the figure,
1) What happens to the Sinc function as the width of the
boxcar (d) approaches zero?
2) Which of the other transform pairs does this limiting
case approach?
3) What happens to the cosine wave transform as the
frequency of the wave approaches zero? What other
transform pair does this case approach?
Convolution using the Fourier transform:
A very simple and time-effective way of performing the
convolution process is using the Fourier transform and the
convolution theorem.
If two functions are convolved in one domain (time or frequency),
then if we multiply the Fourier transforms of those functions
together, and take the inverse transform, the result is the same as
the convolution. As it is often easier to take a Fourier transform
than do a convolution, this method is important:
x(t)  X( f )
y(t)  Y ( f )
c(t)  C( f )
if x(t) y(t)  c(t) (convolution)
then X( f )Y ( f )  C( f ) (multiplication)
Let's use the above principles to derive the sampling
theorem, which says that a signal is completely determined if
it is sampled AT LEAST twice per cycle at its highest
frequency. A corollary of this theorem is that a signal will be
ALIASED if the signal is sampled at a lower rate.
Say we have the continuous analog signal below shown in
both the frequency and time domain that we want to digitize:
digitizing is the same operation as MULTIPLYING the signal in
the time domain by a COMB function with samples every T
seconds: (see the transform pairs)
The MULTIPLICATION of these signals in the time domain
results in a CONVOLUTION of the equivalent signals in the
frequency domain:
This analysis is worthy of lots of study and understanding. Note that
if the frequency of the original signal is less than the nyquist
frequency (as it is above), then, when we look at frequencies
between 0 and the Nyquist frequency in the sampled data, the signal
is identical to the analog signal. Notice that the frequency domain
signal consists of many repeated pairs of the basic signal symmetric
around zero. But we don’t care, since ALL of the information is
contained between zero frequency and the Nyquist frequency.
BUT, what happens when we look at a signal that has energy at
frequencies ABOVE the Nyquist?
f requency domain
t ime domain
•
samplin g f req uency
*
comb
•
•
•
• •
•
•
•
•
•
• •
•
•
•
•
•
Ny quist fr equ ency
•
•
• •
•
•
•
•
•
•
•
•
•
In this case, our signal contains energy ABOVE the Nyquist frequency.
Notice that the convolution in the frequency domain results in the high
frequency noise peak (light colored) appearing between 0 and the Nyquist
frequency ! The digitizing process has resulted in this energy being
ALIASED. It is IMPOSSIBLE to tell whether a sampled signal has been
aliased or not, thus it is VERY IMPORTANT to be sure that your sample rate
is high enough to prevent aliasing.
Or, you can FILTER with an ANTI-ALIASING filter. Such a
filter decreases the power at frequencies near and above the
Nyquist BEFORE sampling to insure that no significant energy
remains to be aliased.
t ime domain
*
f requency domain
•
Ny quist fr equ ency
The above is an example of a LOW PASS FILTER. Note that the
CONVOLUTION is done in the TIME DOMAIN, and the
MULTIPLICATION in the FREQUENCY domain. After
application of such a filter it is usually safe to digitize without
worries about aliasing.
Notice what happens when a signal is aliased – the energy in the
signal is STILL PRESENT, but the frequency has been
transformed to between zero frequency and the Nyquist frequency.
The new frequency is folded back around the Nyquist (N) and
zero frequency, one fold for every factor of a Nyquist. For
example a signal with a frequency of 1.2 N will end up at a
frequency of 0.8 N. A 2 N signal will be observed at zero
frequency, and a signal at 2.3 N will end up at 0.3 N. Thus, the
Nyquist frequency is also called the “folding” frequency.
The above statement on folding can be used to analyse the
wheel rotation sampling in slide 18 of Lecture 28 (Last
lecture).
If the rotation rate of the wheel is 32 cycles per second:
1) What is the sampling rate for each row?
2) What is the Nyquist frequency for each row?
3) What is the APPARENT rotation rate for each row.