The Bold Blueprint Podcast
The Bold Blueprint Avideh Zakhor Embrace failure as an opportunity to grow
Hey Amazon Prime members, why pay more for groceries when you can save big on thousands of items at Amazon Fresh? Shop Prime exclusive deals and save up to 50% on weekly grocery favorites. Plus save 10% on Amazon brands like our new brand Amazon Saver, 365 by Whole Foods Market, Aplenty and more. Come back for new deals rotating every week. Don't miss out on savings. Shop Prime exclusive deals at Amazon Fresh. Select varieties. We wear our work day by day, stitch by stitch. At Dickies we believe work is what we're made of. So whether you're gearing up for a new project or looking to add some tried and true work wear to your collection, remember that Dickies has been standing the test of time for a reason. Their work wear isn't just about looking good. It's about performing under pressure and lasting through the toughest jobs. Head over to Dickies.com and use the promo code WorkWear20 at Checkout to save 20% on your purchase. It's the perfect time to experience the quality and reliability that has made Dickies a trusted name for over a century. Let me start with a few announcements. There's a work to do on Friday. It's a written assignment. There'll be a new work to sign on Friday just so that it's a head up for that. There's a slight change to send these office hours scheduled just for this week. She has to attend a SPI conference to present her paper tomorrow so she won't be able to hold office hours and unfortunately I'll be also out of the country so I can hold office hours on her behalf. Instead, she's holding extra office hours today, Wednesday from 35. Are there any questions and comments? I realized that I had told you I will hand out the project descriptions, possible project suggestions today. Unfortunately, it didn't happen today and given that I'm leaving the country tonight, coming back tomorrow night, it doesn't seem like it's going to happen for Friday either. So I will come out with the proposed topics for your projects on Wednesday of next week which is about March 1. I don't think that it should be too much of difficulty for you. In that document, I also try to put some sort of a scope in it. I don't want you to go all out and spend umpteenth million hours on this project. It would have to be proportional to the number of weeks that you don't have a regular homework to do. That's kind of how it works out. So if I'm assigning the homework on Friday, that's due next Friday and then I'll count how many more homework beyond that we have. I think beyond the one that I assign on Friday, there's two more lab assignments and mostly that's about it. So then I'll put some guidelines in the handout. That's due roughly how many hours I should spend on term paper or project. Meanwhile, you should just start thinking in the back of your minds about the kinds of things that you might be interested in doing and doing something already for your thesis that you think is related to my passing. I'm more than happy to entertain the idea of using that for the term party. Some of you have already discussed the possibility with me. Are there any questions or comments? Has anybody finished the homework yesterday on Friday? Oh excellent. Now how many hours did it take? A lot. I thought that was an easy one. It was only four problems. Yeah. But that was deceptive. I take a long time to learn. To do it. Okay and you, Chris, as well? Okay, so it's a variable. Okay, I mean also different homework takes different kinds for different people. Some people are faster at the labs and slower at the Britain, somewhere about the other way around, etc. Okay, so where we left things last time was today I'm going to wrap up the discussion on filter design and this then wraps up or wraps up our 2D signal processing discussions for this course. Really, as of next Wednesday, it'll be a change of gear. We'll skip the special estimations after in Jay Lim's book and we'll talk about the image processing and I might follow some of the stuff in Lim's book and some on Castleman's book when we get to the image processing part. But for today we want to wrap up the 2DFR filter design. So we talked about windows design and also frequency sampling design and we talked about the transformation technique. So the rest of today's lecture is wrapping up the 2D filter design using the transformation technique. And the basic idea was, well, if we design a little filter, a little 2D filter, let's say 3x3 or 5x5 at most, and then use that as a transformation applied upon a one-dimensional filter to log one-dimensional filter, FIR filter, to build a large two-dimensional filter, then there might be a viable kind of an approach. So we went over this discussion and so the basic transformation that I proposed if you can bring the camera down here is okay and do a rotation. So the basic transformation that I proposed was something of this nature. We built the large two-dimensional filter that we're interested in by applying a transformation that's shown here to a one-dimensional filter. And the kind of the idea is that we're very good at designing 1D filters. We know it inside out, we have excellent optimal algorithms that can help us design one-dimensional filter in optimal way. And optimal, I mean, given a certain amount of ripple and cutoff frequencies, there's techniques that guarantees you to give you the FIR filters with the lowest number of taps. So we're all good as far as designing edge of omega is concerned. So then if we apply this transformation, this t of omega 1 and omega 2 is the frequency response of a smaller filter, t of n1 and n2, which is given by this expression. And so that's it. We make t of n1 and n2 just be a 3 by 3 filter, and if we apply this transformation, we get a big filter, h. And we also went through the fact that if if little t is zero phase, and if h of omega is zero phase, then h of omega 1 and omega 2 will be zero phase. And we wrapped up last time's lecture by kind of writing down the steps necessary for designing 2D FIR filters. So you start, choose or design, and we'll do both examples that do either one of those today. Choose or design little t, given that little t, and given the specs for h of omega 1 and omega 2, the specs stands for specifications, come up with the specifications for h of omega. Then, Hallelujah, we know how to design h of omega, even if the specs are tough, we can go to our remains, exchange algorithm in Matlab and design h of omega, given those specs. And then once we have h of n, then we combine it with little t to get the big h of omega 1 and omega 2 using this transformation technique. So what I'm going to do in today's lecture is first give you an example of these steps for a case where we choose t of n and 1 and omega 2 to be the McClellan transformation. So McClellan not only came up with this design methodology, step 1, step 2, step 3, step 4, but also he came up with a particular transformation filter that's shown here that it obviously bears his name. And this transformation is interesting, it's zero phase, it has real coefficients, it's symmetric, it also has fourfold symmetry. And generally speaking, if this filter has fourfold, if this little t filter has fourfold symmetry in the omega 1, omega 2 plane, the chances are that because of the way we do the transformation, the big filter will also have fourfold symmetry. So a lot of the cues as to how the big filter should perform, a lot of those requirements can be translated down to how the little filter looks like. In essence, we've transformed the problem of solving a big problem into a little problem which is designing this t guy. So let me now give you an example where this thing is used. Okay, if you can just bring in the, thank you. So example of to the FIR filter design using the transformation scheme. Okay, let me see if I can get a slightly better hand here. So suppose that, no, this is not a better one. Suppose that we're asked to design a low-pass filter, LPF stands for low-pass filter, which is this, which is supposed to be circularly symmetric. In other words, it's, and it's going to have contour requirements as a function of omega 1 and omega 2. It's going to have specs that are as follows, which is delta p ripple in the passband is 0.05, delta s, which is ripple in stop band, is 0.025. Hey Amazon Prime members, why pay more for groceries when you can save big on thousands of items at Amazon Fresh? Shop prime exclusive deals and save up to 50% on weekly grocery favorites. Plus save 10% on Amazon brands, like our new brand Amazon Saver, 365 by Whole Foods Market, Aplenty, and more. Come back for new deals rotating every week. Don't miss out on savings. Shop prime exclusive deals at Amazon Fresh. Select varieties. We wear our work day by day, stitch by stitch. At Dickies, we believe work is what we're made of. So whether you're gearing up for a new project, or looking to add some tried and true work wear to your collection, remember that Dickies has been standing the test of time for a reason. Their work wear isn't just about looking good. It's about performing under pressure and lasting through the toughest jobs. Head over to Dickies.com and use the promo code Workwear20 at checkout to save 20% on your purchase. It's the perfect time to experience the quality and reliability that has made Dickies a trusted name for over a century. And C sub P, which is the passband contour, is just a circle. It's a circle of radius 0.45. And C sub S is a stopband contour. It's also a circle of radius 0.55. So this is C sub P here. This is C sub S. And this is a circle of radius 0.4 pi. And this is a circle of radius 0.55. And so pictorially, the passband looks like this. And stopband is this region. So we want this filter to pass signals in this frequency band and stop signals in this frequency band. And in this example, we're just going to follow the steps that I just talked about. But we're not going to design T of N1 and N2, the transformation filter. We're just going to assume that the Michelan transformation that I just showed you is good enough. Okay, so what we're going to do is assume or use T of N1 and N2 to be the Michelan transformation. If you can roll up, please, that would be good. Okay, so our next step is given this set of specifications in 2D, we have to transform the specs into 1D. And somehow figure out what our specifications in our 1D filter, H of omega, is going to be. So the goal here is to transform the specs from 2D to 1D. Okay, and so how do we go about doing that? Well, remember what our transformation is, this H of omega 1 omega 2 is equal to H of omega, let me write this down. So recall, instead of just talking about it, that H of omega 1 and omega 2 is just H of omega with cosine omega replaced by T of omega 1 and omega 2. Okay, so what does it mean? Well, it means that I can plot given T to be this Michelan filter that I just 3 by 3 Michelan transformation that I talked about, I can plot the isokon tours of T of omega 1 or omega 2. Okay, so here's and I'm just kind of exaggerating in terms of the amount of wiggles that I'm introducing. So for example, this is the isokon tour corresponding to 0.1 pi. This is the isokon tour corresponding to 0.2 pi. This is the isokon tour corresponding to 0.3 pi. This is 0.4 pi, 0.4 pi. And this would be the isokon tour corresponding to 0.5 pi. Okay. Yeah, yeah. Because remember, we've nailed T of N1 and N2 to be this. To be this filter, it has these tabs. And so if I plug it into T of omega 1 and omega 2, I get an expression which is minus 1/2 plus half cosine omega 1 plus half cosine omega 2 plus 1/2 cosine omega 1 cosine omega 2 equals cosine omega. So because I've nailed my little T to be these, I have an expression for T of omega 1 and omega 2. And if I pick omega to be 0.1 pi, that defines a contour of all the omega 1 and omega 2 for which, if I plug it into the left hand side of this equals cosine of 0.1 pi. Similarly, if I put let omega equals 0.2 pi, that defines another contour, right? And just by hand, almost as a caricature or as a illustration, I'm just by hand pretending to draw those isokon tours. It doesn't really matter if they exactly look like this or something else. So the question that after we've drawn these isokon tours, the question that comes up is how do I pick the passband and the stopband frequencies of my H of omega? Remember, as a function of omega, H of omega is going to let some frequencies pass through a certain amount of ripple all the way all the way through omega p and it's going to stop certain frequencies all the way through omega from omega s to pi. And the amount of ripple here is delta s and the amount of ripple here is 1, 1 minus delta p. So there's this much ripple here, this much ripple here. So the question I'm asking myself is how should I pick omega p and omega s? What values of omega p and omega s should I have? Well clearly, those need to be influenced by the requirement of my 2D filter. I need to design a passband filter where a circle of radius 4. pi is my passband. What does that mean? That means I need to ensure the ripple inside a circle of radius 0.4 pi does not exceed 0.05. So let me now then for simplicity, let me then superimpose this circle of radius 0.4 pi, which I also called C sub p, also circle of radius 0.5 pi, which I called C sub s superimpose that on top of this diagram. Let me do that in a right way. Okay, so let's say that I'm going to just draw this randomly. This is C sub p and this is C sub s. So these are just circles. Okay, so how should I pick omega p and omega s? Well clearly I want to pick omega p to be in such a way that the contour that, omega p, I mean for my h of omega, in such a way that the contour that corresponds to cosine omega p is entirely outside of C sub p. So let me write that down and then I'll justify it. So I have to choose omega p in such a way that the contour or the isocontour corresponding to omega p is entirely outside C sub p. And why is it that I want to do that? Well because if I pick my omega p that way and if I make sure that for my one dimensional filter between omega between 0 and omega p, my one dimensional filter is going to remain, take on the values between 1 and 1 minus delta p, then hopefully by picking my corresponding omega p to enclose C sub p, then I can make sure that all the points inside C sub p. Hey Amazon Prime members, why pay more for groceries when you can save big on thousands of items at Amazon Fresh? Shop Prime exclusive deals and save up to 50% on weekly grocery favorites. Plus save 10% on Amazon brands, like our new brand Amazon Saver, 365 by Whole Foods Market, Aplenty and more. Come back for new deals rotating every week. Don't miss out on savings. Shop Prime exclusive deals at Amazon Fresh. Select varieties. We wear our work day by day, stitch by stitch. At Dickies, we believe work is what we're made of. So whether you're gearing up for a new project or looking to add some tried and true work wear to your collection, remember that Dickies has been standing the test of time for a reason. Their work wear isn't just about looking good. It's about performing under pressure and lasting through the toughest jobs. Head over to Dickies.com and use the promo code Workwear20 at checkout to save 20% on your purchase. It's the perfect time to experience the quality and reliability that has made Dickies a trusted name for over a century. The P will also have ripple between 1 and 1 minus delta P. Remember, the significance of C sub P is that we need to make sure all the points inside it are going to have ripple between 1 and 1 minus delta P. So if I pick omega P here in such a way that contour corresponding to omega P is sometimes inside C sub P, sometimes outside it, there might be some points inside C sub P that will not satisfy this requirement. So in order to make sure that all the points inside my isocontour corresponding to omega P are between 1 and 1 minus delta P, I have to make sure that the isocontour corresponding to omega P completely encircles C sub P. Is that clear? Because that's the only way I can be sure that all my points inside C sub P satisfy this constraint question. Yeah, like I totally follow what's going on here but I just, I feel like there wasn't enough motivation leading me to believe that this is like really going to work. I mean, we just sort of talked about this Maclinton filter and then it's like the ideal case and I don't know, I just... My client filter is just one example and in five minutes I'll give you another example but we don't even start with that. We designed that filter from scratch or so, the little filter, the transformation filter. Okay. Let me go through this example and then are you trying to say that you filled in my clean filter that was pulled out of some hat? Or is that the problem you have? I'm not saying pulled out of a hat necessarily, like I just, I don't... Yeah, okay, you could say that like maybe it was kind of a hand wave or something but... Right, right. Well, in a minute we'll just going to throw that away and just systematically design our own three by three filters which is not, doesn't even come from my phone. It's just that Maclinton used this in his classic paper as an example so, so first we'd have to start walking before we have to, we'd have to start running and so for now I'm doing a simpler example where I'm assuming I know what T is and I'm just showing you how to pick omega P. And then now, similarly, how do we pick omega S? You want to pick omega S such that the isocontal corresponding to cosine omega S is barely inside C sub S. In other words, C sub S encircles the cosine omega S. Okay, so I choose, similarly choose, omega S so that contour, isocontal rather, corresponding to omega S is barely inside C sub S. So the truth is that I can pick omega S to be really small and so what I want to do is pick the largest possible omega S. In other words, if I pick omega S to be really small I can easily be inside C sub S but then by doing that I have made my omega S to be very close to omega P and I made the transition band of my one-dimensional filter really small and that makes the specifications for H of omega very tough and that results in a long filter. So for that reason I want to choose kind of the largest possible omega S so that it's isocontour corresponding to it is barely inside C sub S. So C sub S, I want to make omega S as large as possible without that contour getting out, so any portion of it getting outside C sub S. And why is that necessary? Well because we want to make sure that for all the omega is larger than omega S or one-dimensional filter is going to have a ripple of smaller than delta S. By doing that we will make sure that all the points outside of contour corresponding to cosine omega S will have ripple smaller than omega S and because we have chosen C sub S to be outside of the isocontour of omega S as therefore all the points outside of C sub S will also have ripple smaller than omega S, delta S. And again you want to pick the largest omega S to do that because if you pick the rather small one you can satisfy the fact that the contour corresponding to omega S is inside C sub S but what you don't want omega S to come too close to omega P because that's detrimental. Is that clear? This is kind of the key point of almost all of today's lecture, if you understand this the rest of the lecture is self-explanatory. Yeah, is that clear for most people? So I want to pick omega P let me now maybe use a different color to do that. So I want to pick omega P such that for example this happens. It could touch, so this is my cosine omega P contour and I'm going to design, so it has to encircle C sub P because inside C sub P everything between 1 minus delta P has to be between 1 and 1 minus delta P by making the contour cosine omega P in circling C sub P, everything inside the blue contour will also be between 1 minus delta P and therefore if everything inside the blue contour is between 1 and 1 minus delta P and everything inside C sub P is also between 1 and 1 minus delta P and as far as the second requirement is concerned we want to pick again I'm exaggerating here but just bear with me. I want to pick an omega sub S that is just barely, this is contour corresponding cosine omega S such that it's omega S is large enough to increase the distance between these guys but because outside of C sub S I'm required to have ripple of delta sub S by making my cosine omega S this way and ensuring that outside of omega S everything is smaller than delta S automatically I've ensured that everything outside C sub S is also, has also ripple of delta S question. So once we pick the contours that are like either inside or outside the past band stop back on how do you, how do you make them have the small ripple that they need? Oh well we're getting there, we're getting there. Oh I just sort of, I don't know, it just sounded like you just said it. You mean for the two dimensional filter? Well now okay so so the next steps is is we pick omega P, we pick omega S this way and now we have the necessary specs for our edge of omega, we know what omega P is when we want omega S is and delta P and delta S are going to be same delta P and delta S from our two dimensional design. So if delta P here was 0.05, delta S was 0.025 then we're going to pick delta P to be 0.05, the same as our 2D filter, delta S is 0.025, the same thing as from our one dimensional filter. Now we have the complete set of specifications for H of omega, our one dimensional filter and we can just plug it into the MATLAB one-dimensional optimal filter design routine and you input delta P delta S omega P omega S into MATLAB one-dimensional filter and it spits out the FIR filter coefficients for H of omega that satisfies the specs for the one-dimensional case. So for example let me come back here, roll up please. So so for this example the one-dimensional specs are as follows, H of omega, there's an omega P, there's an omega S and for omega P we need to have this kind of specs for omega S all the way to pi we need to have this kind of specs. So delta P is again we choose the delta P for our one-dimensional filter to be the same delta P as we wanted for our two-dimensional filter is 0.05, it's from our 2D spec and delta S is 0.025, this is also from our 2D spec. And omega P given that C sub P is a circle of radius 4.4 pi and given that we have chosen a particular transformation filter, i.e. my clinical transformation filter, the omega P that does what we wanted to do is so is 0.4 pi. So we find numerically using the computer for example or by just keeping drawing if you didn't have a computer, the ISO contour corresponding to omega equals 0.4 pi, i.e. cosine 0.4 pi to... Hey Amazon Prime members, why pay more for groceries when you can save big on thousands of items at Amazon Fresh. Shop Prime exclusive deals and save up to 50% on weekly grocery favorites. Plus save 10% on Amazon brands, like our new brand Amazon Saver, 365 by Whole Foods Market, Aplenty and more. Come back for new deals rotating every week. Don't miss out on savings. Shop Prime exclusive deals at Amazon Fresh. Select varieties. We wear our work, day by day, stitch by stitch. At Dickies, we believe work is what we're made of. So whether you're gearing up for a new project or looking to add some tried and true work wear to your collection, remember that Dickies has been standing the test of time for a reason. Their work wear isn't just about looking good. It's about performing under pressure and lasting through the toughest jobs. Head over to Dickies.com and use the promo code WorkWear20 at checkout to save 20% on your purchase. It's the perfect time to experience the quality and reliability that has made Dickies a trusted name for over a century. I'll completely encircle CSOTP, i.e. Circular Radius 4.5. Radius 0.4 pi. This is a bit of a coincidence that this is 0.4 pi and this is 0.4 pi. In real life, it never happens so nicely like that. Don't get confused by that. And then also find numerically isocontal corresponding to omega equals 0.486 pi i.e. cosine 0.486 pi is entirely inside CSOTP, i.e. a circle of radius 0.55. So this and this are omega p and omega s for our one-dimensional filter. So this is 0.4 pi and this is 0.486 pi. And delta p we've already decided 0.05 and delta s is 0.025. Now, given all of these things, we cannot design h of omega using any of your favorite techniques. For example, we can use Rhemeze Exchange Algorithm from MATLAB. I'm not sure MATLAB causes optimal for a particle to design a Rhemeze Exchange Algorithm. They're all the same thing. So when I said design h of omega, I mean design h of n. I figured out the filter type coefficients for our one-dimensional filter h of n. And once we have h of n, then we can combine it with the t of n1 and n2. So here it comes. You can roll up please. Let me show it like this. Okay. So once we design h of n and we know t of n1 and n2, we can combine these two things together to get h of n1 and n2. And the way you do it is you say by just the basic relationship, h of omega 1 omega 2 is cosine omega. I mean h of omega with cosine omega replaced by t of omega 1 and omega 2. So we know this, we know this, plug it into this and you get h of omega 1 omega 2. And from that you compute the coefficients for that. And just to show you that this technique works for this example. Yeah. So, so. So for this example, the particular filter that we get, so first of all, the isocontours of t of omega 1 and omega 2 look like this. So instead of my stupid hand drawn thing, this is actually accurate, computer generated. And the two dimensional filter that you end up designing using this scheme is going to have, the one dimensional filter is going to be 31 taps and is going to have a frequency response that looks like this. Okay. And then finally, after you combine that 1D filter with the transformation, McLaren transformation filter, you get a 2D filter that's going to have isocontours, that's going to have a frequency response like this. And isocontours like this, which not too amazingly enough, not too surprising, are circular, just like we wanted them to be. Okay. So the final filter ends up, ends up being 31 by 31. Remember, we went over this analysis last time. If the 1D filter is 2N plus 1, in this case, N is 15, because the 1D filter, so once you, once you design, once you have the specification for your H of omega, and you plug it into the Rameza exchange algorithm, it comes back and tells you how many taps you need to precisely meet those specs, not over satisfy, not under satisfy. That's the beauty of one dimensional F.R. filter design. It's a unique filter with minimal number of taps that exactly satisfies the specs, no more, no less. And there's no ambiguity, nobody else can come up with something better, fewer taps, or anything like that. So once you design your H of omega, and in this case, the Rameza comes back and tells us, okay, it's a 31 point filter. So, and we already went through this discussion, if that, if H of omega is 2N plus 1, has 2N plus 1 taps, and if the T of N1 and N2 has 2N plus 1 times 2N plus 1 taps, in this case, for us, M is equal to 1, because T is 3 by 3, then the final filter is 2MN plus 1 times 2MN plus 1. So it becomes, the final filter becomes 31 by 31 tap, and this is what this filter here looks like, and that's the ISO contours for that. Okay, so before I go, so the next thing I'm going to do is I'm going to not assume that I know T of N1 and N2, and even start design, not use my clone transformation, not use the T that my clone gave me, and just from scratch, design that. It's kind of like, you know, making chocolate chip cookies from scratch versus just getting the powder and adding the eggs on. Here, in this example, we just added the eggs on top of my clone filter, right? We started with the powder, with the mix, and we just added the eggs on it. Now, we're not even going to start with that powder, we're going to design T from scratch ourselves. So, any questions? Yeah. Yeah, just, so once you have H of omega getting the two dimensional filter, that transform cosine omega equals, that's not so straightforward to me, like I'm just curious about that. We just did that last time, let me flash it on so that you remember it. So, if you can zoom in here, so this is our H, let me go back a few steps. This was our H, in terms of the little H's, this is, then we took advantage of the symmetry in H, because we wanted H to be zero phase to write it down as this, then we wrote it down as A of n times cosine omega n, then we wrote it down as B of n times cosine omega to the power of n, and you can see how M, B are related, because this is a temperature polynomials, and you can write down cosine of omega times n in terms of a polynomial in, in powers of cosine omega, right? And that, you can look up in any textbook or CRC handbook, and we went, we went over temperature polynomials. So, combining all of those things together, you can, you can write down combining, I guess, this thing, this expression with, let me see if I can fold this, with this expression, you can replace the cosine omega with T, right? So now, H of omega 1 on omega 2 is related to this thing. So, if I know my H, I know my A's, right? If I know my H of n here, that lets me know what my A's are. If I know my A's, I know what my B's are, because this is temperature polynomials, and those are known, right? And if I know my B, and I plug it into, and if I know my T, because T of omega 1 omega 2 is related to T, so if I know my B's and my T's, I can compute what H of n1 and n2 is. But, essentially, by equating the powers of e to the minus j omega 1 on the right-hand side of the e to the minus j omega 1 on the left, and vice versa. Is that clear? Okay. Any other questions? Okay. So, let's now move on to the slightly more complicated case. So, example two, I'm going to use this. So, example two for F5 for third design, and in this example, we want to make a note of the following observation, in that the shape of the contour for T of n1 and n2, i.e. the shape of the contour for T of omega 1 on omega 2, that transformation filter, is very similar, ultimately, to the shape of the ultimate filter, H of omega 1 on omega 2, we're going to have. Okay. Let me see if I can motivate that by showing a picture. Yeah. Hey, Amazon Prime members. Why pay more for groceries when you can save big on thousands of items at Amazon Fresh? Shop Prime exclusive deals and save up to 50% on weekly grocery favorites. Plus, save 10% on Amazon brands, like our new brand, Amazon Saver, 365 by Whole Foods Market, Aplenty and more. Come back for new deals rotating every week. Don't miss out on savings. Shop Prime exclusive deals at Amazon Fresh. Select varieties. We wear our work day by day, stitch by stitch. At Dickies, we believe work is what we're made of. So, whether you're gearing up for a new project or looking to add some tried and true work wear to your collection, remember that Dickies has been standing the test of time for a reason. Their work wear isn't just about looking good. It's about performing under pressure and lasting through the toughest jobs. Head over to Dickies.com and use the promo code WorkWear20 at checkout to save 20% on your purchase. It's the perfect time to experience the quality and reliability that has made Dickies a trusted name for over a century. Well, I'm not sure if Tom would figure for 12 and Jay Limbs would might be a good motivational thing, but not quite. But basically, because of the way the transformation is, which is, this expression here, you can imagine the iso-contours of this thing have the more or less the same shape as the iso-contours of this final thing that we're going to get. Right? So, if the final filter, for example, needs to have iso-contours that are elliptical, not circular, it doesn't make sense for t of omega 1 and omega 2 to be circular, have circular iso-contours. It makes sense for t to also have elliptical iso-contours. So, generally speaking, there's a correlation between these iso-contours and the other ones. You could start with a circular one, but what you end up doing is having, end up with more filter tap coefficients in your final filter than you would have had otherwise. So, generally speaking, one observation to walk away is that shape of the iso-contours for t of n1 and n2, i.e., the shape of the contours of t of omega 1 and omega 2 are similar to shapes of iso-contours of the final filter we end up with, which is h of omega 1 and omega 2. And so, this motivates us to think is, if we don't want to use the given my colloidal filter, how do we design t of n1 and n2? So, how to, from scratch, design t of n1 and n2. And the answer is use the shape of c sub p and c sub s. That means the specification contours to design t of n1 and n2. That's the general idea. OK. So, how do we, how do we, how do we go about doing this? Well, ideally speaking, this is again under ideal conditions. What we want is, if you have omega 1, omega 2, and some sort of c sub p and some sort of c sub s with delta p specification here and delta s specification here, ideally, you want the contour corresponding to cosine omega p. And if this, and if the one dimensional filter that we designed, h of omega is going to have something called omega p and omega s like this. Ideally, what we want, we want that cosine omega p to be equal to, which is equal to the transformation of t of omega 1 and omega 2. We want this relationship to hold true for all omega 1 and omega 2 in c p in this contour. So, this is ideal here. Furthermore, ideally, we want the same thing to hold true at, at, at, at, stop and cosine omega 1 and omega 2. We want that to hold true for all omega 1 and omega 2 in c sub s. So, in other words, we don't want to play this game of, oh, cosine omega p has to be chosen with omega p such that the contour corresponding to, oh, cosine omega p is entirely outside c sub p. We want this blue curve to entirely coincide with the red curve, the c sub p curve. And we want this other blue curve here, cosine omega s contour, to entirely coincide with c sub s contour, the red circle, okay. Of course, that's, that's a bit of a wishful thinking, but, but that's what we're trying to strive towards, okay. And in addition, not only do we want cosine omega p to correspond to this and cosine omega s to correspond to that, we want all the points inside the, the passband contour correspond to values of omega between 0 and omega p. So, furthermore, let's roll up just a little bit. So, furthermore, it would be nice, I'm not saying we could have it, but it would be nice if we could have the following two relations to hold true. For cosine omega, for omega between omega p and 0, we want t of omega 1 and omega 2. We want all the omega 1s and omega 2s inside c sub p corresponding to that to result in, result in omega values between 0 and omega p. In other words, what I'm trying to say is that I want, as I traverse in my omega axis from 0 to omega p, I want all these frequency points in one dimensional to correspond to all the omega 1s and omega 2s inside c sub p. Not only, not only do I want cosine omega p to correspond on the contour to correspond to c sub p, not only do I want that i sub contour to be the same as c sub p and this i sub contour to be the same as c sub s. I want all the points inside c sub p to only correspond to 0 and omega p and all the points, all the points outside c sub s here to correspond to omega s to correspond to omega ranging between omega s and pi. So let me write that down also. So I want all the cosine and omega's for omega between pi and omega s to correspond to values of omega 1 and omega 2, where omega 1 and omega 2 are outside c sub s. These are kind of nice things to have and I put these asterisks because I'm going to refer to these equations in just a second. So the question is, I mean, I know what you're thinking about, say, look, she's daydreaming, this is a wishful thinking. How on earth are we going to make all of these things happen? It's close to impossible. But maybe not. So pictorially, maybe I can show this figure because that's a good one to show if you can zoom in as much as you possibly can. So I know, I realize this is a band pass example and what we're talking about is a low pass but nevertheless it drives the point. So this is h of omega as a function of omega. This is the omega 1, omega 2 plane. This is a two dimensional filter that we design, right? And basically what we're doing with the transformation design is, for example, the contour corresponding to omega naught prime in this figure, which is shown by g of omega 1, omega 2 equals omega naught prime, which is this contour here, they're all going to take on the same value as omega naught prime as h of omega 2 at omega equals omega naught prime. Furthermore, if I go to omega 1 prime, which is, which in 2D corresponds to this other contour shown here, all these points are going to take on the same height as h of omega did at omega 1 prime. And hopefully all the frequencies between omega naught prime and omega 1 prime, all the isoconters in between are going to correspond to the isoconters in here. And so the height variation in between these omega naught prime and omega 1 prime contour here is going to correspond to the height variation in h of omega here. Is that clear? This actually has nothing to do with the point I was talking about earlier. It's just a nice figure that I feel sorry not to have shown it. I should have probably shown it right at the beginning of the class or even, yeah, question. So you're saying if you take like cross sections like that, they'll all look the same, basically? Like cross section of this? Yeah, like that picture on the side will always look the same for every. Yeah, it's essentially that's what what's happening is that the transformation is cosine omega, isoconters corresponding to cosine omega equals this thing is going to have the same height variation. Exactly, the height variation here is going to be reflected by the height variation here. And how that is is through that isoconters given by cosine omega equals t of omega 1 on omega 2, right? So that's why the shape of the contours here is very much dominated by shape of the contours in t of n1 and n2. It's just as to which. Hey Amazon Prime members, why pay more for groceries when you can save big on thousands of items at Amazon Fresh? Shop Prime exclusive deals and save up to 50% on weekly grocery favorites. Plus save 10% on Amazon brands, like our new brand Amazon Saver, 365 by Whole Foods Market, Aplenty and more. Come back for new deals rotating every week. Don't miss out on savings. Shop Prime exclusive deals at Amazon Fresh. Select varieties. We wear our work day by day, stitch by stitch. At Dickies, we believe work is what we're made of. So whether you're gearing up for a new project or looking to add some tried and true work wear to your collection. Remember that Dickies has been standing the test of time for a reason. Their work wear isn't just about looking good. It's about performing under pressure and lasting through the toughest jobs. Head over to Dickies.com and use the promo code Workwear20 at checkout to save 20% on your purchase. It's the perfect time to experience the quality and reliability that has made Dickies a trusted name for over a century. Contour corresponds to what value, what height, that's what H of Omega dictates. Through that relationship. This is just a nice full picture. It doesn't directly correspond to what I was talking about. I think it's never too late to talk about it. That is the point I was making clear here. For example, Omega 2 Prime, which is shown, this corresponds to this isocontour here. The height of that isocontour for the two dimensional filter is the same as the height of the H of Omega at Omega 2 Prime, which is a small value. That's why this is low here. Coming back to this thing, what I'm arguing is that in this general picture here, it would be good if cosine Omega P isocontour, exactly match C sub P cosine Omega S contour, exactly match C sub S. All the points between zero and Omega P were corresponding inside of this C sub P. All the points between Omega S and pi are corresponding to the contours outside of C sub. All of these things are nice things to have, but let's start stop daydreaming. So the question is given all these niceties, how to design T of N1 and N2 to make all these nice things happen. This is a degree of freedom we have. The coefficients of T of N1 and N2 are the knobs we can turn to make some of these wishful thinking come through. And here's my four-point proposal for designing T of N1 and N2. So steps for designing T of N1 and N2. And this is assuming that we don't want to use a canned McClellan filter. We don't want to use brownie mix or a chocolate chip cookie mix. We want to design it from scratch, whatever that goes into it. And I'm afraid I don't really know what goes into any of those mixes, so I can't give you a good analogy. I think it's flour, must be flour. Okay, so here's my steps. Step 1.1, assume some sort of a shape and size and region of support for T of N1 and N2. And this could come from the specs. For example, we might want to say, okay, we're just going to make T to be three by three and no more than that. And we're going to make it, you know, symmetric or half-fourfold symmetry. If our final filter is going to have to have four-fold symmetry, we might as well make T have four-fold symmetry, okay. Then step 1.2, impose certain amount of constraints. And what that does is reduces the number of unknowns. And I'll go through a concrete example to show this in just a second. Step number three, choose omega p and omega s tentatively in our 1D design. Such that cosine omega p is approximately equal to T of omega 1 and omega 2 for all omega 1 and omega 2 in Cp. It can't be exactly but approximately. Okay. And such that cosine omega s is approximately equal to T of omega 1 and omega 2 for all omega 1 and omega 2 in C sub s. So these two are the same, two of our requirements in our rich list, right. And step 1.4, start with T over N1. Start with T of N1 and N2 in step 1.3. Okay. So this thing I didn't write properly. So choose omega p and omega points tentatively in our 1D filter. Design our 1D filter and then step within here, we want to say design T of N1 and N2 in such that this relationship holds true. So this comes out of, coming out of step 1.3, you already have a tentative design for T of N1 and N2. And then start with T of N1 and N2 instead of 1.3 and design a slightly modified version of it, which I call T prime of N1 and N2 to ensure that those star relationships in the previous page in my wish list are satisfied. And this looks awfully abstract. So before it scares the daylight out of you, let me just give you an example to make this a bit more concrete. Okay. And if you can roll up, please. Okay. So here's an example. Suppose that I want to design a 2D filter with elliptic pass band and stop bands. Okay. So my specs in omega 1, omega 2, I'm going to be like this. Okay. So it'll be 0.5 pi here, 0.6 pi here. 0.6 pi here and 0.7 pi here. And I can already understand why you would want to shoot me because my figure is entirely out of scale, but just bear with me. Pretend like these are ellipses with, this is 0.55, that's not 0.65, but just pretend it is. Okay. And delta P is 0.08 and delta S is 0.08. And this is our 2D specs. So essentially, mathematically, the C sub P is given by an ellipse with the major and the minor axis given by 0.55 and 0.6 pi. So it's omega 1 over 0.5 pi square plus omega 2 over 0.6 pi square equals 1. And C sub S is omega 1 over 0.6 pi square plus omega 2 over 0.7 pi square equals 1. So these are mathematical shapes of our contours. So that's it. This is a spec and we're not going to go through the steps 1.1 to 1.4 to design the T that does the job with this thing. And ultimately, to design the H, H of omega 1 and omega 2 that does the job with this. So if you can roll up, please. So what do we do next? Well, start with step 1.1. Assume some sort of a shape and size and region of support for our T of n1 and n2. And we're going to assume the region of support is 3 by 3. That means we have nine unknowns. And if I were to, as a function of n1 and n2, write down the coefficients. If you can roll down, please write down the coefficients for T. I'm going to have nine unknowns that I call A, B, C, D, E, E, F, G, H and I. And I won't. So you can imagine, I would have A, B, C, D, E, F, G, H, I, just nine coefficients for different numbers. So now step 1.2, according to my recipe here, I'm going to impose some constraints so that I don't have to deal with nine unknowns. So I'm going to impose the linear phase constraint. And so, or rather zero phase constraint because that's easy. And I want to assume we have a real impulse response. And as soon as you do that, then you have to have H of n1 and n2, or rather T of n1 and n2, to be equal to T of minus n1 comma minus n2. So that means that if I have nine coefficients here, then this point has to be equal to this point, this point has to be equal to this point, this equal to that, and this equal to that. And automatically, when I do that, because of this additional constraint, I end up having five unknowns. And I'm going to call them A prime, B prime, C prime, B prime, and E prime. And I'm going to show them here. A prime is the same as A prime here because I'm going to have this constraint imposed. B prime is the same as B prime here. C prime is the same as C prime here. D prime is the same as D prime here. And finally, E prime is the point in the middle. I'm going to impose more constraints if nothing else to reduce the number of unknowns I have. Hey Amazon Prime members, why pay more for groceries when you can save big on thousands of items at Amazon Fresh? Shop Prime exclusive deals and save up to 50% on weekly grocery favorites. Plus save 10% on Amazon brands, like our new brand Amazon Saver. 365 by Whole Foods market, a plenty and more. Come back for new deals rotating every week. Don't miss out on savings. Shop Prime exclusive deals at Amazon Fresh. Select varieties. We wear our work day by day, stitch by stitch. At Dickies, we believe work is what we're made of. So, whether you're gearing up for a new project or looking to add some tried and true work wear to your collection, remember that Dickies has been standing the test of time for a reason. Their work wear isn't just about looking good. It's about performing under pressure and lasting through the toughest jobs. Head over to Dickies.com and use the promo code Workwear20 at checkout to save 20% on your purchase. It's the perfect time to experience the quality and reliability that has made Dickies a trusted name for over a century. This was the zero phase real impulse response was constrained A, I'm going to impose 4-fold symmetry. Well, if I have 4-fold symmetry and why do I do that? Well, because if I look stare at my contours, this 4-fold symmetry, this part is symmetric to this part, this is symmetric with that, so if I want the shape of my Tm and 1n and 2 contours to look like the final filter, I might as well impose the characteristics of the final filter onto the isocontours of my T. So, if I do that, then essentially that means A prime is equal to B prime. That means now I have 4 unknowns and those 4 unknowns are A prime, C prime, D prime and E prime. And that means that now if I write T of omega 1 and omega 2, it's by just plugging into the equation of T of n1 and then 2, e to the minus j omega 1, come up, e to the minus j omega 2, and I get something like E prime plus 2, C prime cosine omega 1 plus 2, D prime cosine omega 2 plus 2, A prime cosine omega 1 cosine omega 2. That's now the shape of my T, in terms of the 4 unknown I have A prime, C prime, D prime and E prime. And I'm going to add one more constraint and that is I'm going to map omega equals 0 to the center point here. Okay, so remember we're doing this, right? 0 omega p omega s pi, at the end of the day we're going to design a one-dimensional filter that does that and all I'm saying is I'm going to map this point into that point. And that's kind of a nice constraint to have because it's going to make sure that at least that omega equals 0, my h of omega takes on a high value the same way as I want that 0 and 0 to have. So part c, constraint number c, if you roll up please, condition number c is that I'm going to map omega equals 0 in my 1D space 2, omega 1 omega 2 equals 0, 0 in 2D space. So in other words I'm going to say cosine omega, add omega equals 0, which is really 1, is equal to T of omega 1 and omega 2 at omega 1 on omega 2 equals 0 comma 0, which means that if I now plug in omega 1 and omega 2 equals 0 to this one, what I get is E prime plus 2C prime plus 2D prime plus 2A prime equals to 1. That means I have one less unknown, one fewer unknown. In other words, that means that the four unknowns now have three unknowns. So the more constraints I impose the fewer unknowns I have. So now given this constraint here, I can get rid of E prime. So from here we can say that E prime is just 1 minus 2C prime minus 2D prime minus 2A prime. And now I plug this thing into T that I have up there and this becomes T of omega 1 and omega 2 is, well, 1 minus 2C prime minus 2D prime minus 2A prime, which was this guy, plus the rest of what it was before, plus 2C prime cosine omega 1 plus 2D prime cosine omega 2 plus 2A prime cosine omega 1 cosine omega 2. And now staring at this thing, now I have just three unknowns and those three unknowns are A prime, C prime and D prime. And now I'm going to choose, now I'm going to move on to the next step, step 1.3. And that is choose omega pair and omega s tentatively or arbitrarily in one dimensional space. We'll fix it in just a second. Don't worry about being arbitrary. We'll refine it, if you will, and here's the second. So coming down here, if you can roll up just a little bit, we're moving to step 1.3, which is choose omega p and omega s in 1D kind of arbitrarily. And staring at this diagram here, I don't know, what do you think we ought to pick omega p and omega s? How about if I just pick omega p to be 0.5 pi and omega s 0.6 pi? I could also go omega p 0.6 pi, omega s 0.7 pi. It doesn't make too much of a difference, but for the sake of the argument, just to say, we pick omega p to be 0.5 pi and omega s to be 0.6 pi. So we say omega p is 0.5 pi and omega s is 0.6 pi. And now step 1.3 here says after you pick these omega p and omega s's, design t such that these two conditions hold true. In other words, isocontor and omega p corresponds to the passband contour, c sub p, and isocontor and omega s, now that we picked omega p and omega s randomly, we want the isocontor corresponding to omega s to correspond to all the points in omega 1 and omega 2 that are a member of the stopband, c sub s. So now design t of n1 and n2, such that cosine omega p is approximately t of omega 1 and omega 2 for all omega 1 and omega 2 in c sub p. Furthermore, cosine omega s is approximately equal to t of omega 1 and omega 2 for all omega 1 and omega 2 in c sub s. This is my wishful thinking part, the daydreaming part. And how do I go about doing that? Well, it's not, it's kind of a standard technique in optimization. I'm going to come up with the error term, I call it error sub p, which is e sub p, which is, I want the difference between this and this squared immigrated over omega 1 and omega 2 over along this contour to be minimized. So I want to minimize an error where error is defined as the double integral of the square of the difference between t and cosine omega p over d omega 1 over omega 2, where this double integral is over a particular contour, omega 1 and omega 2 in c p. So for all the points on c p on the passband contour, you want to minimize this quantity. And why am I squaring it? Because if I don't square it, the pluses cancel the minuses and the two things won't be close to each other. So, you know, standard practice. And then I wanted that, that makes me satisfy or get close to satisfying this top condition. And the bottom condition I can do the similar thing, error sub s, e sub s is the double integral of omega 1 and omega 2 in c sub s of the same thing squared, the omega 1, the omega 2. And now it's t of omega 1 and omega 2 minus cosine omega s. And I'm going to convince you that this is not a very difficult optimization job. If I want to minimize, I want to minimize error sub p and error sub s with respect to what? Yeah, my unknowns, a prime, c prime, d prime. Remember, the t guy here is t of omega 1 and omega 2 is linear in terms of a prime, c prime and d prime. So, if I have a quantity, if you can roll up, please, if I have a quantity that I want to, so essentially I want to minimize e sub p and e sub s with respect to a prime, c prime and d prime. These are my unknowns, right? That's the only knobs I have to change. I have no other control over the universe other than a prime, c prime and d prime. And how does one minimize something with respect to something else? You take the derivative and set it equal to zero. So, what that means is that d e prime over d a prime is equal to zero, sorry, there's no prime here, d e p with respect to d c prime is zero, d e p with respect to d d prime is equal to zero, the partial's. And the same thing for e sub s, d e sub s, d a prime is equal to zero, d e sub s, d c prime is equal to zero. Hey Amazon prime members, why pay more for groceries when you can save big on thousands of items at Amazon Fresh? Shop prime exclusive deals and save up to 50% on weekly grocery favorites. Plus, save 10% on Amazon brands, like our new brand Amazon Saver, 365 by Whole Foods Market, Uplenty and more. Come back for new deals rotating every week. Don't miss out on savings. Shop prime exclusive deals at Amazon Fresh. Select varieties. We wear our work day by day, stitch by stitch. At Dickies, we believe work is what we're made of. So, whether you're gearing up for a new project or looking to add some tried and true work wear to your collection, remember that Dickies has been standing the test of time for a reason. Their work wear isn't just about looking good. It's about performing under pressure and lasting through the toughest jobs. Head over to Dickies.com and use the promo code Workwear20 at checkout to save 20% on your purchase. It's the perfect time to experience the quality and reliability that has made Dickies a trusted name for over a century. And the next question to ask yourself is that an easy optimization to do? And I'll argue that it's extremely easy. Why? Because T is linear in A prime, C prime, D prime. If you don't believe it, stare at this. T is linear in C prime, D prime, A prime. In our words, there's no C prime squared. There's no D prime squared. There's no A prime squared. There's not even A prime times D prime. There's no square root of A prime. It's linear in A prime. And C prime and D prime. Therefore, this quantity inside the parentheses here is quadratic with respect to A prime, C prime, D prime. And if I take the derivative of this term or this term with respect to A prime, because of the linearity, I can bring the derivative of E sub P inside this integral. So if this thing is quadratic in A prime, C prime, D prime, and I take the derivative with respect to A prime, C prime, D prime, that thing is going to be linear in A prime, C prime, D prime. If something is quadratic in some, quadratic means power of two. If this quantity here is quadratic in A prime, C prime, D prime, and I take the derivative with respect to them, the final answer is going to be linear in A prime, C prime, D prime. And therefore, if I'm setting, and therefore, if I'm setting this thing equal to zero, the derivative of E sub P equal to zero, it's a linear term equal to zero. It's another term equal to linear term equal to zero. So at the end, I have a linear system of equations that I'm setting equal to zero, and that's very easy to solve. So let me provide you with a little bit more details of that. Did everybody follow that intuitive discussion? Okay, so let's just write that down. So we know that T of omega 1, omega 2 is linear in A prime, C prime, and D prime. And therefore, T of omega 1, omega 2 minus cosine omega P squared is quadratic in A prime, C prime, D prime. What that means is that, if you can roll up, please, D E P, D A prime is equal to the double integral. Can you roll up, please, omega 1 and omega 2? Thank you. So remember, I'm taking a derivative of E sub P with respect to A prime. I can just bring in the A prime as T of omega 1, omega 2 minus cosine omega P squared, D omega 1, D omega 2. Okay, and I can apply now the chain rule, D E P, D A prime, is equal to, this is omega 1 and omega 2 in C sub P. So it's double integral of omega 1 and omega 2 in C sub P. So how do I take the derivative as that? Well, it's twice T of omega 1 comma omega 2 minus cosine omega P. And then times the derivative of this thing of this thing coming down here, D omega 1, D omega 2. So because T is linear with respect to, I'm just going to make all my argument respect to A prime, the same thing holds to it, but C prime and D prime, all of these things. And the same thing holds to it, but E sub S. Because T is linear with respect to A prime, that means that D D A prime of that thing is, is, is, is, is a constant, is a constant. And what is that constant? Well, it's equal to minus 2 plus 2 cosine omega 1 cosine omega 2. So that's what this thing is equal to. So plugging that into this, this equation from here, we get D E P, D A prime is equal to double integral of omega 1 and omega 2 in C P, right down this thing here, which is 2 cosine omega 1 cosine omega 2 minus 2 times 2 times this thing in the parentheses, which is T of omega 1 comma omega 2 minus cosine omega P. All of that, D omega 1, D omega 2, all of that is equal to 0. We're taking the derivative and setting it to 0. So now you can see that because, again, since T is linear in A prime, C prime, D prime, this thing is a, is a linear equation equals 0. So A prime, C prime, D prime in the right hand side of this thing appear on the right hand side in a linear fashion. So, so how many, so how many linear equations do I have? I have six of them. Why? Because I have D E P over D A prime, D C prime, D, D, D prime, that's three of them and then E sub S to six. So bottom line is that I have six linear systems of equations and how many unknowns and three unknowns and how do I solve that? Apply, pardon? Linear squared, right? Because it's an over-determined. So apply some sort of a simple thing like linear squared and that gives you A prime, C prime, D prime and then from there you can get E prime, from there you can get B prime, then from there you get T of N1 and N2. I'm not going to be able to finish this argument here but the reason I have to at least give you a one minute explanation why we're not done, why there is a need to go to step 1.4 which is, which is refining T and the best way to do that is for me to just show you the isocontours corresponding for this example anyway, isocontours corresponding to T of N1 and N2. And this is shown here, zoom in please, zoom in, okay great. So, so now that I computed my T of N1 and N2, my transformation is cosine omega, I mean H of omega 1, omega 2 is equal to H of omega where in H of omega I replaced cosine omega with T of omega 1, omega 2, right? And these are isocontours corresponding to different omegas in omega 1, omega 2 plane. And you might be wondering what happened here, why is there no isocontours here? Well the simple answer is that not all the T that we just designed nowhere in that design did we guarantee that T of omega 1 and omega 2 should be between minus 1 and plus 1. So for these values, there's no corresponding values of omega, real values of omega, that results in T being between plus 1 and minus 1. So we're messed up. So what I'm going to do on Friday's class because we run out of time in today's class is tell you, but that is not a big disaster, there's an easy fix for it. So what I learned in Friday's class is come up with a quick fix so that the resulting T of omega 1 and omega 2 is always between plus 1 and minus 1. Therefore, there's always cosine, there's always omega values in one-dimensional case that correspond to that, that correspond to these things. Why is that necessary? Because going back, so let me just show you, so this is how this diagram looks like, why is that necessary? Well because going back to this diagram, remember the values of h of omega determine the height variation of h of omega, determines the height variation of my final h of omega 1 and omega 2. And if there are some points in this frequency domain, in this two-dimensional frequency plane, omega 1 and omega 2, that don't correspond to any omega here, all hell could break loose there. The height for these guys can be anything. Hey Amazon Prime members, why pay more for groceries when you can save big on thousands of items at Amazon Fresh? Shop Prime exclusive deals and save up to 50% on weekly grocery favorites, plus save 10% on Amazon brands, like our new brand Amazon Saver, 365 by Whole Foods Market, Aplenty and more. Come back for new deals rotating every week, don't miss out on savings. Shop Prime exclusive deals at Amazon Fresh. Select varieties. We wear our work, day by day, stitch by stitch. At Dickies, we believe work is what we're made of. So whether you're gearing up for a new project, or looking to add some tried and true work wear to your collection, remember that Dickies has been standing the test of time for a reason. Their work wear isn't just about looking good, it's about performing under pressure and lasting through the toughest jobs. Head over to Dickies.com and use the promo code Workwear20 at checkout to save 20% on your purchase. It's the perfect time to experience the quality and reliability that has made Dickies a trusted name for over a century.