Let's consider a continuous time complex
exponential function. This is defined so x of t
is equal to e to the j two Pi f_0 of t. F_ is the frequency of
the complex exponential. Let's see the differences
with respect to the discrete time
complex exponential. The first is that
in continuous time, a complex exponential
is always periodic. The period is 1 over f_0, and since the argument
now t is a real number, there will always be
a value of t for which t times f_0 is going
to be an integer, and therefore the phase of the complex exponential
in multiple of 2 Pi. Secondly, all angular
speeds are allowed, there is no maximal
speed anymore. So f_0 can be as
large as we want. The Fourier transform
of a continuous time complex exponential
is a Dirac-Delta, a single Dirac-Delta not a train of Deltas centered at f_0. So by definition, a complex exponential is
a band-limited function. In this case, a 2f_0 band
limited function. Just like in discrete time, we can imagine
the complex exponential as point rotating around the unit circle at a frequency of f_0 revolutions per second. But contrary to discrete time, we have to imagine the movement has been smooth now rather than as a succession of hubs
around the unit circle. If we now take raw samples of the continuous time
complex exponential, we obtain a discrete
time sequence which is a discrete time
complex exponential. The samples are snapshots at regular intervals of
the rotating point. So the resulting
digital frequency of the sequence of positions
around the unit circle is given by Omega zero
which is equal to 2 Pi f_0 time T_s where T_s
is our sampling period. Or we can also write this
as 2 Pi that multiplies the ratio between
the original frequency of the complex exponential, and the sampling frequency
which is 1 over T_s. Now, we have three cases. The first case is the easy case. Suppose that the relationship between the frequency
of the continuous time complex exponential and
the sampling frequency of a raw sample, is such that f_0, the frequency of
the complex exponential, is less than half the
sampling frequency. In this case, the resulting
digital frequency of the samples will
be less than Pi, and we will have snapshots that go around
the unit circle and clearly indicate the movement of the complex exponential
and allows us to reconstruct
the original frequency. A tricky case is when the relationship between
the two frequencies is such that f_0 is larger than
half the sampling frequency, but still less than
the sampling frequency itself. The digital frequency will
be between Pi and 2 Pi. We already know what
happens in this case. At each step the movement of the discrete time
complex exponential will be larger than Pi, and as we accumulate
points in the sequence, they will appear to
be going backwards at a slow frequency rather than
forward at a fast frequency. So we already have
some ambiguity on the result in
discrete-time sequence with respect to
the original signal. Of course, the worst case is when the relationship between
the two frequencies is such that the frequency of the complex exponential is larger than the
sampling frequency. At this point, the
digital frequency will be larger than 2 Pi, but we know that we
cannot have that everything is modulo 2
Pi in discrete time. So we will be moving around the unit circle more
than 2 Pi at each step, and this sequence will be indistinguishable
from the samples of a continuous time
complex exponential at a much lower frequency
than the original one. So we have lost
the information on the frequency of
the original sequence. This is in essence
the concept of aliasing. Let's look at it from
a system point of view, we have an input signal, we have a raw sample operating
at a sampling period T_s. We obtain a discrete
time sequence. Now, we interpolate with
the sync again with interpolation period of T_s to obtain another
continuous time signal, and we want to know
the difference between the input signal and
the reconstructed signal. In the case of continuous time
complex exponential, we have seen the following cases. So let's pick a sampling period, and equivalently
a sampling frequency which is 1 over T_s. We take an input which is a complex exponential
at frequency f_0. The digital frequency of the intermediate
discrete time signal here will be Omega zero
equal to 2 Pi f_0 over F_s, and the cases that we have
seen are the good case. So when f_0 is less
than F_s over 2, then the digital frequency
will be between zero and Pi, and the reconstructed
signal will be exactly identical to
the input signal. This goes up to the case where f_0 is equal to F_s over 2. As f_0 increases, or F_s decreases according to how we want to look at the problem, and we are in
a situation in which f_0 is between F_s
over 2 and F_s, the resulting
digital frequency will be between Pi and 2 Pi. So the reconstructed
signal will be of the form e to the j 2
Pi f_1 t. So we have a different frequency
here f_1 rather than f_0 and f_1 is equal
to f_0 minus F_s. It's a negative frequency
which indicates a clockwise movement rather than a counterclockwise movement which is where we started from. The full aliasing
case, the third case, is when the relationship
between the two frequencies is such that f_0 is larger than F_s. In this case the intermediate
digital frequency is Omega zero larger than 2
Pi which results in a reconstructed continuous
time signal at a frequency F_2 where F_2 is equal to
f_0 module of F_s. Let's illustrate
the three cases graphically. We have three panels here, in the top panel we
will show the spectrum of the original continuous time
complex exponential, it will be a Dirac Delta
at a given frequency f_0. In the second panel,
will show the DTFT of the intermediate raw sampled
discrete time sequence. In the bottom panel, we will show the spectrum of the reconstructed
continuous time signal. So remember when we sample
at sampling frequency F_s, we're mapping the interval
from minus F_s over 2 to F_s over 2 to the minus Pi, Pi interval in the discrete time. Then vice versa when we
interpolate to continuous time, we're mapping the minus Pi, Pi interval to the minus F_s over 2 to F_s over
2 in continuous time. So if we start with a complex exponential at frequency zero,
everything works fine. It gets transformed
to a discrete time complex exponential
of frequency zero, and reconstructed to
a continuous time complex exponential
of frequency zero. As we increase the frequency
of the complex exponential, everything goes well,
this gets mapped to this, and then reconstructed to this, so the frequency
remains the same. So everything is
fine until we reach a complex exponential
whose frequency is equal to half the
sampling frequency. Here, we are in the limit case, the discrete time
complex exponential will have a frequency of Pi, and the reconstruction will be the last good reconstruction. As we increase the original
frequency even further, so here, you see we
move past F_s over 2. The modular operation in the digital frequency
makes the Delta in discrete time pop back on the other side on the
negative part of the spectrum. This will be reflected in the reconstruction which will
give us a continuous time complex exponential at a
negative frequency of f_0 minus F. So on and so forth as we increased the
frequency more and more, see here, we're always moving in the space of positive
frequencies. But here, we are in the negative frequency
space which will be reflected in
the reconstruction. As we pass F_s, we're back again to the positive frequencies
in the discrete time, but the reconstructed
signal will have a frequency that will
always be module F_s. So we will keep going around in this space of frequencies for
the reconstructed signal. Another way to look
at the problem is to keep the frequency of the continuous time
complex exponential fixed, and change the
sampling frequency. Instead, we obtain
exactly the same results. So here again, the
same three panels, the original continuous time
Fourier transform, the DTFT of the
intermediate signal, and the Fourier transform of the reconstructed
continuous time signal. So the frequency of our
complex exponential is fixed, and is equal to f_0,
and we will change F_s. So if we start with the sampling frequency
that is large enough, so the frequency of
the complex exponential is comfortably within the interval minus F_s over 2 to F_s over 2, that everything goes well. We get an intermediate
discrete-time signal that has digital frequency
between zero and Pi, and the reconstructed signal will be identical to the input. Now, we reduce
the sampling frequency, and as long as the sampling
frequency is such that f_0 is less than F_s over 2, everything will be fine. When F_s over 2 is equal
to f_0 we have once again the limit case which will be the last time we get
the reconstruction. As we reduce F_s even more, we start to have a frequency
that should be positive, and gets mapped to
negative frequency here. We reduce it even more, and even more now, we're in a situation where
f_0 is larger than F_s. So here we have
a positive frequency that gets mapped to f_0 modulo F_s. So we reduce
the sampling frequency, we get a narrower interval
for reconstruction, and the frequency will
always fall within this interval independently
of its original value. So a summary of the same problem, the look from
the point of view of choosing the right sampling
frequency can be the following: If
the sampling frequency is larger than
twice the frequency of the complex exponential where within the conditions of
the sampling theorem, so everything will be okay. The sequence of sample fully captures the information
in the sinusoid, and we will be able to
reconstruct it exactly with an interpolator at the same
interpolation frequency. If the sampling frequency is less than two times the frequency of the complex exponential, but still larger than the frequency of
the complex exponential, we will be mapping
the positive frequency to a negative frequency
by this formula. The reconstructed
frequency will be the original frequency
minus the sample frequency. Finally, in the case
of full aliasing when the sampling frequency
is less than the frequency of
the complex exponential, we will have that
the reconstructed frequency will be the original
frequency module F_s. To drive the point
home even further, let's look at aliasing in the time domain rather
than the frequency domain. So here, we have a simple sinusoid x(t) is equal
to cosine of 6 Pi. So we have a sinusoid at a
frequency of three hertz. We now sampled sinusoid at a sampling frequency
of a 100 hertz. So we have a 100 samples
over one second. So here, we are showing
one second of a sinusoid, three periods as you can see, and a 100 samples. It's clear from this image that these samples will allow certainly a perfect
reconstruction. Incidentally, note that although it looks like the density of the samples varies along
the curve of the sinusoid, the samples are all
equally spaced, it is just the shape of
the sinusoid that makes it appear as if there were more points around
the peaks here. Anyway, we can lower
the sampling frequency. With 50 samples per second, we still have plenty
of data to go by, and the reconstruction
seems very easy to perform. We can go down to 10 hertz, and that's still enough data
to reconstruct the sinusoid. As a matter of fact, six hertz
is still enough data, because it's twice the
sampling frequency. As you can see,
the result is that every sample will be a
plus one or minus one, which is coherent with the fact that the digital frequency of the sample sinusoid
will be Omega equal to Pi. But now, let's go to
the full aliasing situations. So we take a sampling
frequency that is less than the frequency
of the sinusoid. So we pick F_s
equal to 2.9 hertz. So we will take 2.9
samples per second. Now from this image, you wouldn't be able really to understand
what's going on, but let's zoom out and show more of the sinusoid and
more of the samples. So here, we show in
two seconds of the sinusoid, here we show in four seconds. Now if you show 10 seconds
of the sinusoids, so we have a 30 cycles of the continuous-time sinusoid,
and 29 samples. You can see that
these samples describe a sinusoid which completes
one cycle in 10 seconds. Therefore, has a
frequency of 0.1 hertz, which by the way is
exactly equal to three hertz modulo 2.9 hertz.