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GG 313 Lecture 26
11/29/05
Sampling Theorem
Transfer Functions
Homework review:
Data are 128 points, 20 samples/sec.
What is the time vector that matches the data?
Starting at t=0, we want a new point every 1/20=.05 s
time=[0:deltat:6.35]; % The last point is at 127*.05 sec.
How about the frequency vector?
Our sampling frequency is 20 Hz, so we expect the
frequency scale to be symmetric about the middle
and for it to go from 0 to the sampling frequency in
128 steps:
freq=[0:1/(N*deltat):1/deltat*(1-1/N)];
Now generate and plot the spectrum:
Y=fft(y);
p=Y.*conj(Y);
plot(freq,p)
Note the symmetry about the Nyquist (10 Hz) and peaks
at 1 and 4 Hz. We actually only need to plot the 1st half.
Now we apply a simple filter to the data by convolution:
h=[.5 -1 .5];
out=conv(h,y);
Now calculate the power spectrum of “out” and compare it to
the input to the filter “y”.
OUT=fft(out(2:129));
OUTP=OUT.*conj(OUT);
plot(freq,OUTP,'r')
But, let’s only plot from zero to the Nyquist:
plot(freq(1:65), OUTP(1:65),'r')
The red (filtered) spectrum has lost energy at the low
frequencies (attenuated) and gained (amplified) at the high
frequencies relative to the input signal. The peak at 1 Hz has
been effectively removed.
Now try the other filter:
hlp=[.25 .5 .25];
outlp=conv(hlp,y);
OUTLP=fft(outlp(2:129));
OUTLPP=OUTLP.*conj(OUTLP);
plot(freq(1:65),OUTLPP,'k')
In this case, both peaks
are present, but the
high frequencies have
been attenuated
relative to the input
signal.
Let’s plot the same data in dB (10*log10(p)):
blue: input
red: [.25 .5 .25]
black: [.25 -.5 .25]
Note that the peaks that look like they had been completely
wiped out are actually there - just much smaller. For
example the peal at 1 Hz is reduce by ~20 dB by the “black”
filter - that’s about a factor of 10 in amplitude.
The sampling theorem
The sampling theorem says that if all the energy in a
continuous signal is below the Nyquist frequency, then
all the information in that signal is captured in the
sampling process.
We can show this by using our last transform pair:
This is the comb function made up of an infinite string of
delta functions separated by T. It’s Fourier Transform is
also an infinite string of delta function separated by
frequency 1/T:

t  jT   (T )
j
(5.117)
Consider what happens to a continuous analog signal in the
time domain if we MULTIPLY it (NOT convolve) with a comb
function.
What we get is a SAMPLE of that function every T. Where T is
the sample period.
Now we need to look at the Fourier Transform of each of
these functions:
The greens are positive frequencies and oranges negative.
As long as the signal is BAND LIMITED to frequencies below
the Nyquist (1/2T), ALL of the information is preserved after
sampling.
No let’s see what happens if the signal is NOT band limited:
In this case, the signal contains high frequency energy above
the Nyquist. This energy shows up in the 0-Nyquist frequency
range as a peak of negative frequency energy. The arrow
shows where the peak originated.
Transfer Functions
Analysis of time series data often involves determination of
how the equipment and processing steps affect the signal as
it passes through various steps. For example, a particular
sound source will have a particular spectrum, and it will be
detected by hydrophones that change the spectrum of the
signal in a particular way. These functions must be
determined and removed if the characteristics of the “earth
filter” are to be determined.
There are many ways to determine these TRANSFER
FUNCTIONS. For example, we can make a simple filter from
a resistor and a capacitor:

This circuit has a well-known transfer function:
Vout
1

2
Vin
1 RC 
Notice that this function is already in the frequency domain,
and we can plot it using Matlab:
The slope is -6 dB/octave, and the corner frequency
(3dB down point) is 1/(2πRC), and the phase response
is =tan-1(1/(RC)
One way to define the transfer function of a system that is
not well known is to use an impulse for an input signal.
Recall that an impulse convolved with any function yields
the function itself. So the output of a system when an
impulse is the input is called the IMPULSE RESPONSE.
The Fourier transform of the impulse response yields the
transfer function of the system.
Output/Input=Transfer function=FFT(Impulse response)
The impulse response of a Linear Time-Invariant (LTI)
system yields all the information about transfer properties
that the system performs.
Consider the following function: out(i)=out(i-1)+in(i), where
out(i) is the output at time (i) and in(i) is the input.
USE MATLAB to:
1) get the impulse response of this function
2) find its response to a sine wave
in(i)=sin((ii-1)*8*pi*(f/(n*dt)));
f=8;
Notice several things: The impulse response of the function
is a step (=0 if ii<0, =1 if ii>=0)
What kind of function is this?
How does the output change as f changes for the sin wave
input?

Recall the integral of a sine wave:
 sin( t)dt    cos(t)
1
This says that the output of an integrator should
decrease in amplitude by a factor of 2 with each doubling
in frequency. That is, an integrator should modify the
spectrum by -6dB/octave.
Thus, to integrate a signal, all you need to do it to move it
to the frequency domain, subtract 6 dB/octave (with 0 dB
at =1) and return to the time domain.
Similarly, to differentiate a signal, ADD 6dB/octave in the
frequency domain.
Consider a “perfect” filter in the frequency domain, such a
filter passes all energy below a particular frequency and
none above. This is a “boxcar”. This is one of our
transform pairs, and we know the function in the time
domain will be a sinc function: y=sin(x)/x, y=1|x=0.