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.
