[BLANK_AUDIO] Welcome back. I hope you enjoyed the first two weeks of class. As I mentioned at the beginning of last week, in these two weeks, weeks two and three, we'll cover the fundamentals of two-dimensional signal systems. Last week, we get this analysis in the spatial domain. While this week we move to the frequency domain. We use two-dimensional sequencing systems in our presentation. However, all the information we provide is also applicable to one-dimensional signals. But those at the higher dimensional signals that is multidimensional signals in general. Thanks to Joseph Fourier, who lived in the late 18th, early 19th century, we can describe any signal of any dimensionality, and the operations of any linear and spatially invariant system in the frequency domain. The frequency domain is mostly familiar to us when we refer to one-dimensional signals. For example, we can distinguish between a baritone and the soprano or between low and high frequencies in a music piece. Images have similarly low and high frequencies, which you cannot hear, of course, but we can see. They're spacial frequencies. We'll get familiar with the concept first and then we will take the material and the operations of [UNKNOWN] systems we learned last week to the frequency domain. We'll also understand what is involved when we convert the continuous signal to discrete signal. What we called in week one sampling. And also when we changed the size of an image When for example we want to display a large image on a small screen, of our cell phone. Or a thumbnail image on a large screen. And this is called sample grade conversion. Equipped with this material you can understand what will follow in this class. But also when you are interested in learning new material outside this class. As for example, what's the process of acquiring an MRI image? Or an x ray image of a slice of a patient's body using computerized tomography. It's in fact to a CT or CT scan, a term you might have heard before. In this first segment we'll define the two dimensional continuous Fourier transform of a two dimensional discrete signal. And we will discuss some of its important properties. We will then work out a number of examples which will give us a good understanding of what constitutes a low pass and what constitutes a high pass filter. We'll finally talk about the important convolution theorem which states that the convolution operation in the special domain becomes a multiplication operation in the frequency domain. So, let us start covering this important material. Let us go see the linear especially in variance system with impulses response h, n1 and 2. We put in signal x as it's input and signal's a complex exponential with specific frequencies omega 1 prime, omega 2 prime. This is a two dimensional Q tone, since it only contains again this particular frequency, omega 1 prime, omega 2 prime. We want to find out what the output of the LSI system for this particular input is equal to. I did mention at some earlier point in the course that these complex exponentials are Eigenfunctions of LSI systems. Which simply means that such complex exponentials go through the system without their frequencies being altered. Of course by now we know how to find the output of an LSI system. Its y n 1 and 2 is equal to the convolution of the input to the system with the system's input response. [SOUND] If I apply directly the definition of the convolution this is equal to this double sum. [BLANK_AUDIO] Again. I just applied the diffusion of the revolution. Now we see that these terms. [BLANK_AUDIO] Are independent of K1 K2. Therefore they can come outside the summation, and all I'm left with these are these terms here. [SOUND] So we see that the complex exponentials appear at the output of the system. They just went through the system. And this term here tells me how the system responded to this particular complex exponential of this come, at this specific frequencies. This here is a complex quantity. It has a [UNKNOWN] phase. And we see that the LSI system is able to discriminate among sinusoidal systems on the basis of their frequencies. So if the magnitude of this term here is equal to one, then the sign [INAUDIBLE], goes through the system unchanged, while on the other hand, if the money is equal to zero, the complex [INAUDIBLE] is rejected by the system, so therefore this is the, [INAUDIBLE] of the LSI system. With [INAUDIBLE] h, n 1 and 2 at these frequencies omega 1 prime, omega 2 prime are denoted by capital H, omega 1 prime, omega 2 prime and this is the frequency that responds to the system. Now since this resolve is not going to change, if I consider additional frequencies, let's say omega one double prime, omega 2 double prime, we have that this is again the response of the system frequency and this capital H, omega 1 and omega 2 is the fully transform of the impulse response of the system H and 1 and 2 were going to say cause there will be more things about the fourier transfer in the next slides. We show here the two dimension of fourier transform pair. This is here the expression for the forward transform that will take an image from the special domain and map it to the frequency domain where omega 1, omega 2 continues of variables. So x omega 1, omega 2 is the two dimensional fourier transform or the spectrum of the image x n1 and 2. This is the expression of the inverse fourier transform that will take us from the frequency domain back to the spatial domain. We see with the expression on top that even when the image is real as it's usually the case. The fourier transform is complex due to the presence of this complex exponentials. Therefore, it [UNKNOWN] I can show the magnitude and the phase of the fourier transform. And this represents the polar representation of a complex number. Or equivalently I can show the real. And the marginal parts. I see here in the expression for [INAUDIBLE] fourier transform between the grade for minus pi to pi in the omega 1 and minus pi to pi in the omega 2 directions. And this is due to the first property of the fourier transform that it is periodic With periods 2 pi, 2 pi, in the omega 1, omega 2 directions. This is a straightforward expression to show, in essence, we showed it when we started the complex exponentials, we saw that they're periodic with period 2 pi, 2 pi, and this is what gives rise to this property. Another useful property is To consider what happens, when I shift the image by amounts m 1, m 2 in the special domain. And what happens in the frequency domain is I get the spectrum of the signal, but it's multiplied by this complex exponential. So since the magnitude of the complex exponential is one, the magnitude of the shifted image has not changed But I have this linear face component that, is due to the spatial shift. The reciprocal of this property is the modulation property that, tells me what happens if I shift the spectrum by 31, 32. So similarly, I see that I have this complex exponential appearing in the special domain. Another use for property is this which is typically the [UNKNOWN] property which tells us that when the signal is real, then the magnitude of the Fourier Transfer has even symmetry. While the phase of the Fourier Transform has odd symmetry. And finally this is the Parseval's relational theorem which tells me that the energy of the signal, which is found according to this expression in the spacial domain Can also be found in the frequency domain by integrating over one period to pi, to pi this energy density spectrum. There are clearly additional properties, quite a few of them, that one can find in a standard image processing textbook, but these are some of the more useful ones. Let us consider a simple example. We are given a system with an impulse response depicted here. And we want to find the frequency response of this system. So, we apply the definition of the Fourier transform and the frequency response is equal to this double summation [BLANK_AUDIO] [SOUND] h n1, n2, e to the minus j omega1 n1, e to the minus j omega2 n2. Now for the given impulse response, you see that we have only five samples. So we have the. H minus 1 0 sample. h 0 0, h 1 0, h 0 minus 1 and h 0 1. So out of this infinite sum, I only have these five terms. So h 0 0, multiplied by e to the Minus J zero which is one plus H minus one zero multiplied by this term [INAUDIBLE] [SOUND] And now I substitute the, I substitute the given values of h. So this is one-third and all the rest are one-sixth. [SOUND] So equal to 1 3rd. Now we will observe here that due to Euler's formula, e to the plus j omega one, plus e to the minus j omega 1 will give me two times cosine omega, and similarly, the other two terms will give me two times cosine omega 2, and they can further simplify this as. 1/3, 1 plus cosine omega 1, plus cosine omega 2. So we see that, for this particular example, the frequency response of the system is 3l. This should come as no surprise, since one of the propositions of the Fourier transform is that if the signal has even symmetry Then the fullier transfer is real. This means that the magnitude is the absolute value of the stem, and therefore the phase is zero at both frequencies except when the stem here becomes negative, in which case I have a plus or minus pi phase jumps, hence minus one has absolute value one And phase plus minus five. So lets see how this frequency response looks like. So I brought there the magnitude on the vertical axis and they show it over period the mega 1 for minus pi to pi and the mega to for minus pi to pi. This actual magnitude has even symmetry, which is also, should come as no surprise, because we saw in the previous slide that one of the properties of the Fourier transform is that if the signal is real, then the magnitude has even symmetry, and the phase odd symmetry. So if I look at this magnitude, I see that H Zero, zero is equal to one, right. Cosign zero is one so you have three over three. And also you observe that at certain frequencies the magnitude becomes exactly zero. So for example if I look at the frequency, mine was pie. Pi over 2. I have 1 plus cosine minus pi, which is minus 1, so they cancel out, 0. Plus cosine of pi over 2, which is 0. And therefore this is 0. This is actually the frequency over here, right? So by and large, this is a low pass filter. Is passes the 0, 0 frequency unchanged. Attenuate slightly the low frequencies, but as the frequencies increase, the continuation becomes, greater. We show here, and example similar to the previous one, where given the impulse response of a system, and well distributed to find its frequency response. The Form that h n 1 and 2 is the depicted slightly different from the previous case and shown here as a 3 by 3 matrix. It should be understood that this is the h 0 0 sample. This one, the h 1 1, while let's say, this one is Is the h minus 1, 1, and so forth. This actually approximates a Gaussian shaped function in this three by three window. After a point, when you look at this impulse response which is again, is symmetric, has even symmetry. One should be able to identify terms in the frequency response, so for example, this and this term, should give rise in the frequency response of a 2 times cosine omega 1 times the coefficient while the diagonal terms This term and this term should give rise to a term like this in the frequency response. 2 times cosine omega 1 plus omega 2. So if you work out the example you should find out that this is the frequency response of this system as shown here. [BLANK_AUDIO] It's again, a real signal which should come as no surprise because hn1 and 2 shares even symmetry. If we plot the magnitude of the stem It looks like this. Here a plot of the vertical axis that 10 times log base 10 of the magnitude, so one can see more clearly small values of the magnitude of the frequency response. Again, if I look at H(0,0) this is the sum of the values of the impulse response and they sum up to 1. And therefore, the logarithm of this value is equal to zero. So this term here, that's the zero, zero, it's equal to zero. We'll see again that this is a low pass filter. It, Attenuates slightly the low frequencies and considerably the higher frequencies. And of course the zero zero frequency goes through the system unchanged of, of the input signal. Here's a final example. We're given a system with impulses [UNKNOWN] depicted here And, we're going to find its frequency responds. The comments I made earlier apply here as well. So this is the eight zero zero sample. We see actually that this simple response differs in some sense from the previous ones in the sense that I take differences now between the center Sample here nine in the neighboring sample. So we should expect that the shape of the frequency response should be different. So following exactly the same steps that we followed in the previous two examples one can find that the frequency response of this filter of this to the system is shown here. Its real again because h(n1,n2) has even symmetry If I plot this. So I plot here the magnitude in the vertical dimension and I plot again the 10 log 10 magnitude so that I am able to see smaller difference. Anylsis. We see that now the shape is different in the sense that the low frequencies are attenuated while the high frequencies are amplified. We can easily that the h 0 0 is equal to 1, and therefore the log of this Is equal to zero. While if I look at, some frequencies. For example, the h 0 pi frequency, which is the frequency over here. This is equal to 13, while the H pi pi is equal to nine. So, considerably higher values than 1 at high frequencies and small values At zero. Again low frequencies are attenuated, high frequencies are amplified high special frequencies and [UNKNOWN] are represented by the edges. Therefore the edges are allowed to go through the system while the flat regions are rejected or greatly attenuated. And therefore, a filter like this performs edge detection on an image. One of the important properties of the Fourier transform is the convolution property, or, as also referred to, the convolution theorem. It describes the input output relationships of an LSI system in the frequency domain. So given a system with impulse points h, and x in, the input, y, the output, we know that, the input output relationships in the spatial domain are through the to the convolutions. So, the output equals the convolution of the input with impulse response of the system. Clearly, I can take each of these quantities through the frequency domain. So I can find the spectrum of the signal. I can find the fully transform of the impulse response, which is the frequency response of the system. And I can find the fully transform of the output signal. Since the signals are clearly related in the spatial domain one expects that they should be related in the frequency domain. And what the Convolution Theorem tells us is that indeed the spectrum of the output signal equals the spectrum of the input signal times multiplied by the frequency response of the system. So convolution in the spatial domain becomes multiplication. The frequency domain, and actually the reverse is also true. Multiplication, the time domain, becomes convolution in the [UNKNOWN] domain. Of course, here, we have to talk about periodic convolution since the signals are periodic. So, let's see that indeed, this is the case. So, if I, derive the output of the system, as the system T, let me call the system T operating on the input signal x n1, n2, and then I'll replace the signal by its inverse Fourier transform. So this is equal to this double integral of X omega 1 omega 2 e to the J omega 1 n 1 and to the the J omega 2 n 2 d omega 1 d omega 2. Actually, the inverse Fourier transform tells me that You can think of any signal x, n1, n2 as the weighted sum of this complex exponentials where the weights are indeed the values of the Fourier transform. Now since the system is, is linear, and assuming all the signals here are well-defined, I can Interchange the order of integration and, the operation of the system, and the system operates on signals that are functions of N1 and 2, therefore they're fully transferred the weight as I called them. constant, as constants as far as the system is concerned. So, I am left with this expression. Now, we have started The, expression, this expression we know that this complex exponentials are Eigen functions of lsi systems. So they go through the systems unchanged and they're multiplied by the frequency response of the system, So therefore, this expression. Becomes equal to this one, so if the frequency response of the system times the complex exponentials. [SOUND] So now this is the expression for the inverse for the transfer of y, and therefore this signal here is the fully transfer of y, so this is y omega 1, omega 2. So, indeed, we showed that y omega 1, omega 2 equals x, omega 1, omega 2 times the frequencies. And so this is. [SOUND] We are rooted to this operation when we talk about the example. So therefore all this tells me that looking at the frequency response of the system, I have to just process each frequency one at a time, right? So if H here is 0 at certain frequencies then the output Y is going to be 0. Therefore those frequencies of the input signal would not go through the system.
