[BLANK_AUDIO] Welcome back. I hope you enjoyed the first week of class, during which you covered the introductory material. In this and next week, we will cover the fundamentals of two-dimensional signals and systems. We're interested in describing some important and useful signals. One of them, as you'll see, the complex exponential, is the building block of any signal, and we're also interested in the input-output relations of systems. This week, we will carry this discussion in the spatial domain, while next week we will move the discussion to the frequency domain. We use two-dimensional signals and systems in our presentation. However, all the information it provides is also applicable to one-dimensional signals, but also to higher-dimensional signals, that is, multi-dimensional signals in general. We will become now more specific and talk about important to the signals, the ones you encounter often and systems that process those signals. We're primarily interested in the algorithmic or the mathematical, or the software implementation of such systems in this course. We will then move on as specific class of two dimensional systems that have 2 important properties that these are linear and time invariant or special invariant. Such systems are very useful and popular for processing images and videos and they're friendly in the sense that they allow us to describe them in a succinct way in the special domain through their impulse response, but also allow us to look at their behavior in the frequency domain. This week's material forms the backbone of any course on multidimensional signal and systems. Equipped with the knowledge you'll acquire this week, you'll be able to understand any paper or work that deals with the processing of signals in any dimension. If you already have had a course on one dimensional signal processing, then the material I'll be presenting here these two weeks will reinforce your knowledge of the topic. We will simply add one independent variable to our signal representation, and we'll take everything from the one-dimensional line, to the two-dimensional plane. If this is the first time you are exposed to this material, don't be discouraged if everything is not clear right away. It might take some time and experience for everything to sink in. We will be utilizing some of these ideas throughout the course. So by the end of the course, everything should become much clearer. I will clearly be presenting only basic material which will be required for the remaining of the course. For those of you interested in this specific topic, there is a plethora of books and resources that will allow you to look deeper into the topic. In this first segment, we'll introduce some conventions representing 2D and 3D spaces and then, we'll introduce 2 basic two dimensional signals, the unit impulse and the unit step. We will see that both of these signals are separable signals. So let's proceed with the material of this second week. Let us first establish some notations and conventions when we deal with 2-D and 3-D discrete signals. Let x-n1 and -n2 be a two-dimensional signal. n1 and n2 are then dependent variables. They're integers, they take value 0, plus-minus-1, plus-minus-2. And mathematically speaking, a signal, an image can have infinite support. That is the values that n1 and n2 take are between minus infinity and plus infinity. When it comes to the orientation of the axis, I can use n1 as the horizontal axis and to the vertical one with this orientation. That's how we're used to presenting functions in calculus. So, here is 0. Here is the point n minus 1. Here's the point m minus 1. So this is an n by m image. So if this is n1 prime, this is m2 prime, then here is the pixel with coordinates n1 prime, n2 prime. The value of the pixel is in general a real number. Now drawing from the way matrices are represented on the computer, I can pick the n1 axis this way and the n2 axis gone, to have gone like this. So in this case if I pick the same n1 prime, it would be here, approximately here. The same n2 prime would be approximately here and therefore, this is now the pixel with coordinates n1 prime, n2 prime. Clearly as long as the orientation of the axis is known, the result of processing an image should be identical. It shouldn't matter which way I have chosen to represent the axis. When I deal with color images, then there are 3 channels involved, the blue, the green, the red channel. If these channels are added up, I end up with a color image. So if this is the n1 axis, and this is the n2 axis, again, the location, the n1 prime, n2 prime location of the pixel of the same, the pixel of the same location xn1 prime, n2 prime, but the difference now from the previous case is, is that x n1 prime, n2 prime is now a vector, a 3 by 1 vector where I have the value of the red channel at this location followed by the value of the green channel at the same location followed with value of the blue channel at same location. Okay, so the algebra representation. So x, therefore in the general presentation can be a scalar or it can be a vector. When I deal with three-dimensional discrete signals, then xn1 and 2 and 3 is the representation n1, n2, n3 are integers. And again, mathematically speaking, they can range from minus infinity to plus infinity. So, here is a collection of frames representing a video. So I can choose n1. Going down this way, n2 is the second axis and 3 is the third axis, it represents discrete values of the time domain. And now, nx, n1 prime and 2 prime and 3 prime would be a pixel, let's say, n3 prime is here, it will be a pixel in this frame with the other 2 coordinates n1 prime, n2 prime. A little bit hard to exactly represent it in this case, but you do get the idea. We'll just now talk about some specific signals that we would be making use of in this class. The first one is the unit impulse. It will be noted by delta n1, n2, and by definition is equal to 1 when the arguments are equal to 0. So if here on the 2d plain, here's n1, here's n2, then I'll denote delta like this, and the height of the delta equals 1. I can shift the delta around, so I can write delta n1 minus n1 prime, n2 minus n2 prime. And then according to the definition, it's equal to 1 as long as the arguments are equal to 0 and 0 otherwise. So from here, we see that it's 0 when n1 equals n1 prime and n2 equals n2 prime. So again if I'm to sketch it, depending on the values of n1 and 2, I have not restricted them. So let's say n1 prime is here and n2 prime is here, therefore in this case I have a signal like this, so it's a delta, its value's 1, and it's 0 everywhere else, everywhere else on the plane. I can talk about the three-dimensional delta by introducing an n3 there, so the definition will extend to be 1 when n1 equals n2 equals n3 equals 0. An interesting notion is the notion of separable signals. A signal g is separable if it can be written as f1 which is a function only of the one variable, n1 times a function of 2 which is a function of the second variable, omni. The question then is whether the 2D delta is separable, separable signal, and the answer is yes. We can if we let's, let's look at it pictorially first, right? If I have a signal but it looks like this. So it's a line of deltas. So it's 1 on the vertical axis and 0 everywhere else and all these values are equal to 1, right? And then if I have a signal that, here's a line of impulses on the horizontal axis, so, with solid dots here, [UNKNOWN] value of 1, and the signal is 0 everywhere else, right? So now if I multiply these two signals, then it should be clear by just these pictures I'm drawing here that I multiply it like this, one is multiplied by 0 here, this one is multiplied by 0 over here. This one is multiplied by 0 at this location and it's only this one here at the origin that is multiplied by 1 here and therefore the result is just this signal which of course is just, we have to find if this is delta n1 and 2. Now what is the first signal equal to, this is a delta n1, right? Because this is by definition equal to 1 when the argument is equal to 0 and 0 otherwise. But I'm on the two-dimensional plane, right, so delta n1 is 1 for n1 equals 0 and any value of n2, no restriction on n2. So therefore n1 equals 0 on the vertical axis, n1 equals 0 here, n1 equals 0 here, here, here, here. So this line of impulses, the first one is represented by this automatically by this function, delta, delta n1 and similarly, this line of impulses is a delta n2 signal. Therefore delta n1 and 2 can be written as delta n1 times delta n2 and it is therefore separable according to the definition, right? So delta n1 takes the place of f1 n1 in the definition, right? So this, I'll be writing the definition here times f2 n2. Another useful function is the discrete unit step. It's denoted by u. And by definition, it's equal to 1 as long as the its arguments are not negative. So n1 is greater equal than 0, n2 greater equal than 0. So from the schedule here, is again 1 as long as n1 is greater equal than 0, but also n2 greater equal than 0. [SOUND] So the values of the solid circles is 1, and then the signal extends from 0 to infinity in the n1 direction, and 0 to infinity the n2 direction. We often label the quadrant; this is the first quadrant, this is the second one, this is the third one, this is the fourth one. And in this case we say that the, the discrete unit step here support in the first quadrant. We can shift the unit step function around, so if I have n1 minus n1 prime, n2 minus n2 prime, then by definition, this is equal to 1, as long as its arguments are not negative and 0 otherwise. So from here, we see that n1 is greater or equal to n1 prime. n2 is greater or equal to n2 prime. So, if I'm to sketch this shifted step depending on where n1 prime and n2 prime are located, so let's say, n1 prime is here and n2 prime is here, right? So then. [BLANK_AUDIO] Shift it. The step function looks like this, and extends between this way, and continues this way. So this is, the coordinates of this point is n1 prime, n2 prime. The next question is whether the unit step is separable as was the delta and the answer is yes. So, un1, n2 can be written as un1 times un2 and pictorially, un1 is 1 as long as n1 is greater or equal than 0. So, it looks like this. [SOUND] Again, these solid circles have a value 1. So it extends to infinity this way. This way and this way, right? So this is un1, un2 is the signal that is 1 as long as n2 is non-negative, so it will look like this. Right? So we fix them this way, this way, this way, infinity. So this is un2 and clearly if I multiply these two signals, then what I'll end up with is the unit step, right? So, it's going to be non zero only in the first quadrant. So, this is un1, n2.