Please join me in welcoming Mr. Ben Barnes. Good morning, everyone. First of all, I want to thank you for that very kind introduction. I appreciate very much. General, it's an honor to be on the program with you and thank you for your service to our country. To all the employees of the National Park and the State Park here, you're part of a very important team that does a yeoman job and I welcome this opportunity to thank you. To Lucy and Ian and Pat, representatives of the Johnson family, I not only say thank you for being here today, but I think one of the greatest legacies that Lyndon Johnson had is the fact that he raised a family that was not going to come to this cemetery and mourn his death, as Dr. Graham delivered the eulogy here, and go back and live their lives and raise their children and their grandchildren and have a good life, a life as a result of the labors and the love of President Johnson. They didn't go back home and do that. They didn't get a rocking chair. They didn't get a bed that goes up and down and has all the comforts of home. They went back and went to work with the same spirit, the same drive and the same enthusiasm that President Johnson had. And Lady Bird, who was blessed to be here for two decades after President Johnson left this world, what asset she has been, not only to Lyndon Johnson, not only to Lucy and Linda and Ian and Pat and the entire Johnson family, but she was America's workmate and helpmate. And she was a person that set an example for us in every way that she lived her life. If someone asked me who the woman that I admired most in the United States, my answer would be my mother, who worked so hard to help me get out of those sand fields of Comanche County and off that peanut farm and in the gold places like the Johnson Ranch and the University of Texas and Texas State University, Mr. President, thank you for being here. And to get to work with the two marks who have been introduced to you or who are such talented young leaders and do such a great job for the foundation, for the women that were serving the food and the ice water and the coffee this morning, everyone turns out to do their part. And I believe that, and in my heart, that we're all here on this warm August Sunday because we want to do more because President Johnson wanted us to do more. He taught us that we should do more. He taught us that we had a responsibility.  You're looking a little older like me, but you've got a real responsibility. You represent the third generation and there's already the fourth generation and the fifth generation of the family. And the family has to continue to provide the leadership that we all need to remind us of Johnson's many contributions. I'm honored to be here for so many reasons. The anniversary of, first of all, it's something that's dear to my heart and the people that work for the State Park here. It's the 53rd anniversary of LBJ State Park. And I was in office when that was created and I was so happy that the legislature voted for it and appropriated the money. We were in session. The legislature meets in odd number of years in 1969 and we were in session in the closing hours of the legislative session. And my secretary sent someone out on the Senate floor and said, the President of the United States is on the phone for you, Ben. Of course, that pleased me a great deal. Anytime the President called me. But it really pleased me this time because I said, well, you know why what a thoughtful person he is. He's calling me to tell me thank you for getting to State Park. Well, I left the podium and went into the Lieutenant Governor's office and there was the President. And I waited for the word, Ben, thank you so much for what you're doing. And obviously, I was so hungry for that praise and that blessing that was gonna come through that AT&T line. Well, that wasn't a blessing. It wasn't a praise. I said, Ben, I've got a question for you. I said, yes, sir, Mr. President. He said, you know how hot it gets in August? July in Texas? And I said, well, yes, sir, I do. I grew up on the peanut farm here, remember?  He said, they're not gonna think a lot about you but they're gonna think more about me. But he said, they're gonna come down there to see our State Park. It's gonna be dry, it's gonna be hot. And you've only got three water wells in that appropriation bill. You understand how little water we're gonna get out of those three wells? Those children are gonna get off those school buses after riding. Some of them may come from the Panhandler where Joe Batson lives. They may come from deep east Texas. But they're gonna get off that school bus and the first thing, what are they gonna want, Ben? They're gonna want a cool drink of water. And I tell you, if it's in the summer, there's a lot of not to be any water there.  They'll probably forget about you. That really made me feel good. He said, but they're gonna blame me. And he said, I guess we'll put a sign out there by the State Park that you can't have a drink of water because of Ben Barnes, some town called De Leon and the President of the United States. We are not thinking about them. And I said, oh, Mr. President, I didn't even realize how many water wells were there. I didn't know anything about this. He said, oh, no, nothing happens in the Senate that you don't know about. He said, I know full well, but he said, what are we gonna do? And I said, well, Mr. President, but we gotta get some water. I didn't know what we're gonna do either. The appropriation bill had already passed and it was at the governor's office, Senator Johnson. And this is Senator Nathan Johnson sitting on the front row here. And his name is Johnson, spelled the same way the President spells his. And he's got the same fiery spirit for public service that Lyndon Johnson had. And I've watched politics now for 55 years from afar and I've not met a state senator or a state representative or a state congressman that's got the raw talent and the ability to provide leadership as Nathan Johnson and Senator. We're glad you're here today. And I want y'all to remember, you won't have trouble remembering his last name, but you'll remember his first name is Nathan Johnson. But anyway, Senator Johnson knows that when a bill goes to the governor, well, that's it, the session's ready to be over. So to put those water wells in there, we had to call the bill back from the governor by concurrent resolution. You had to pass the house and send it both. Well, it was gonna be pretty difficult to stand up and explain that I'd screwed up or the legislature screwed up and we'd not drilled enough water wells for President Johnson's Park. And I didn't think the president wanted that published and I didn't either, but so what we did, we just made a motion to recall the bill. We'd made some clerical errors.  But we said we'd made clerical errors. We called the bill back and put the money in there to drill those wells. And today, if you go to the state park and you get off a school bus, you can get a cold drink of water at that state park. And if those wells ever get low, I'll try to get some money someplace else to drill some more wells. Because I don't want Johnson calling me from heaven and telling me, Ben, you screwed it up again. So anyway, it was a great day. And I came and made a speech in August to dedicate that part. And I had an opportunity. Some of the papers read, wrote stories about it. And so this being the anniversary that they were published and they come up on my computer and I looked at them and I finally got a tape of the ceremony to watch it, just to bring back the pleasant memories. And what I saw in that tape was so interesting because it kind of hurt my feelings at first because I would have thought the tape would be all about my speech. Now I'm just sure they couldn't keep the cameras off me. Well, they did keep them off me. It was a great day of the time because I put them on President Johnson. But what the great thing was, watching he and Lady Bird sitting down the front row, what was the great thing was Lyndon Johnson was smiling during that 20 or 30 minute ceremony. Smiling like I had not seen him smile since he had come back to Texas from the presidency. Not seen him smile since he had to go through the angry and agony of Vietnam. But it was so wonderful to see him smile and it was a great day. And there were between six and 8,000 people there that day and they were there in appreciation and respect of what Lyndon Johnson had done for this state and this country and it was certainly a happy day and it pleased me so much. And I want to say this, I've bragged on everybody that's in leadership roles, but I want to say something to you, the people from Johnson City and people from Stonewall and the people from this area out here on this Sunday morning. You're all part of the Johnson family too. You're all part of the people that love Johnson and respected him and you have carried the torch. You've done your share of the pushing and pulling and lifting. And it does my heart so good. I've seen a couple of people I went to law school with, I've seen some people I rode in the parade with in Stonewall, I've seen, saw one guy that saw me get stung because they put me in a red ant bed and I didn't realize that I stand in a red ant bed and I made a little speech in Stonewall and I got bitten by red ants and I'll always remember that. But you are part of the Johnson family and you're part of Texans who appreciate the leadership that we received and I'm so proud. President Johnson was, he was the education president. He was the civil rights president. He was the consumer president. He was the president of the poor and the oppressed. He was the president of conservation. I could go stand here all morning long and talk about the different areas that Lyndon Johnson passed very important legislation in and it's too hot and I appreciate your time being here. I'm not gonna do that. But I'll tell you that the real story of the Johnson administration is not told with statistics. I believe that Lyndon Johnson's achievements can be measured in the terms of human lives, in the terms of hopes fulfilled, in the terms of bodies healed, of spirits lifted and of eager young minds educated and all of our lives enriched. President Johnson left us with two critical messages that we've really failed to live up to and in this day of praise and appreciation I wanna mention those two requests that Johnson made of each and every one of us. One, it relates to the temperature. 53 years ago in August, about the same time it was here today, the high that day was 90 degrees. The high today is predicted to be 105 or six degrees. And it's been in Austin, those of you that live there near here and I'm sure it's been that same way in Johnson City and Stonewall for 30 to 40 days it's been over 100 degrees. Matter of fact, maybe over 105 and 160 degrees. And it was President Johnson that said to us over 60 years ago that global warming is an incredible threat and that if we don't do something about it that we could, because of global warming lose our world as we have it today. Unfortunately, we're just now beginning to realize that. We didn't realize that 60 years ago when President Johnson said that to us. And we're gonna have to take the political action and build a coalition that Johnson asked us to build 60 years ago to do something about global warming is a challenge for each and every one of us. And the second thing is bipartisanship. Johnson used to say, we can't have division, division, division.  He wanted people to work together. And he set an example of Democrats and Republicans and independence working together. And it was Lyndon Johnson that took the minority leader of the United States Senate and forged a relationship with him and with the Republicans in the Senate and wrote the first civil rights and voting rights act. Unbelievable. You think about going to the United States Senate and the United States House today and passing that legislation. It would be absolutely impossible. But President Johnson did something for me that I remember all my life. And I wanna share it very briefly with you that I was on a committee. We were meeting in the afternoon in the old executive office building. That's the building right next to the White House. And one of the members of that committee was Everett Dirksen. And I don't know whether any of you have met Senator Dirksen or not, but you at least heard his voice on television. And I always thought if I would close my eyes and hear Senator Dirksen speak, that it was maybe the voice of God because he spoke in a rather crisp, hoarse voice. Ben came back in the room, called me young Ben. He said, young Ben, the president of the United States of America. This is from a lot of elected officials. And I was embarrassed, but about kind of secretly, I was enjoying the heck out of Berkson, saying this to all the people. He said, young Ben, the president of the United States wants you and me to come to the Oval Office and have a drink with him. And I don't know about you, but I plan to go. And he said, unfortunately for all of you, other leaders here, the president said to say hello, but he wants to speak to me and young Ben. Well, I don't know why I even got invited. In a few minutes, maybe I was gonna learn, but I jumped up and hit my knee on the table because I was so excited. But I jumped up and Senator Dirksen and I walked over to the Oval Office. There was President Johnson, the young man that was the steward of the office there. And he said, what do y'all want to drink? Well, I wasn't really experienced with the beverages at that time. I would drink some, but I more had a beer budget than I had a Scotch budget. But Dirksen said, I'll have Scotch. And the president said he'd have Scotch. And of course, I wasn't gonna be left out. I said, I'll have Scotch. Well, the steward brought Dirksen, Scotch, Johnson, Scotch, and me a Scotch. And I'll always believe, and I'm not being accusatory, but the statutes run, so the president couldn't be indicted for this or getting a minor and an older man. We beat him too much Scotch. But anyway, I always felt like he gave Senator Dirksen a double Scotch and me a double Scotch. And he had about a third of a Scotch, but I don't know that and I can't prove it. But anyway, we drank our Scotch. And Dirksen in the middle of that first Scotch said, now, Mr. President, I want to tell you something. I've got an idea.  Well, Johnson's, well, that's not true. Yeah, I want to talk to you about those three votes, but I invited you over here because you're my partner. You're the second man to Abraham Lincoln, whose name's gonna be inscribed in history because what Johnson did in the Emancipation Act for the slaves and what you did but joining me in passing the Civil Rights Act, there'll be two great heroes of America, Abraham Lincoln and Everett Dirksen. Well, that made the Center pretty good. But anyway, but he said, well, we're talking about that. I need those three votes every now. I've got to pass this tax bill. We've got to pay for this war. We've got to get this war over with. We've got to pay for it. And Dirksen said, well, I'm sorry, Mr. President. We had a Republican caucus this morning and we voted unanimously to not, any of us vote for the war. Well, Johnson changed the subject. He said, let's have another one. Well, we had another drink. And near the conclusion of completing that drink, Johnson said, let's have an honest have and then y'all can go. And I leaned over to Senator Dirksen, said, Senator, can I ride with you? That was two reasons. I'd gotten there in a cab. I didn't know where to go call a cab. I wasn't familiar with the White House. And the second reason I didn't know was that I was gonna be capable of walking and go up and get me a car. So anyway, Dirksen said, yes, I'll take you. Well, we were finished in that drink and Johnson said to Senator Dirksen, Senator, I'm so glad I thought of this. He said, we better have another half drink. Let's have a half drink and then I'll tell you and then y'all can go. He said, Senator, the commander of the Corb engineers was here in my office this morning in general. And he said that, unfortunately, there's not enough money for that dam in Southern Illinois from the Corb engineers. And I know, Senator, how important that dam is to you. And I know how it's important to those farmers in Southern Illinois that their crops were washing away. And I had an argument with him. I said, that dummy is Senator Dirksen's most able man in America and he deserves that and his constituents deserve it. And he said there still wasn't any money in there. Well, they handed us our third drink. I would see Dirksen, he was just seething. He'd be drinking it slowly. And he said now, if you want me to get him on the phone, Senator, I'll get him on the phone and you and I can talk to him together. Dirksen just waved his hand like that, took another sip of that scotch. And he said, Mr. President, you know how much I respect you. You know how much I love you even. But, and I never had heard anybody say this to the President of the United States, but I was later to learn it was out of love and affection and you ladies, you pardon me when I say this, but Dirksen said to the President of the United States, my hero, you know, Mr. President, you are a SOB, but I love you. Well, we finished the drink. We got up, Dirksen walked around, walked around the desk in the Oval Office and kissed Johnson on the cheek and said, Mr. President, you know I love you and respect you so much. And I think about now somebody in another party even going over and shaking hands and doing a responsible thing. But we completed our meeting. Johnson had his three votes from the Republicans for the tax bill. Dirksen had his dam for Southern Illinois and I'd had way too much scotch. But it was a wonderful experience that I saw the Johnson treatment in person and saw him with the majority leader, the minority leader of the United States Senate. But that was Lyndon Johnson. Lyndon Johnson put his country ahead of his party. He put his values always to the individual and to the person that needed help and that needed his government. That person was in President Johnson's mind and not the person that had the advantages of what our society was giving each of us and that Johnson was largely responsible for. Last night, Sam and I, I've been out of town for some time and if I'm addicted to anything, it's Mexican food, it's Tex-Mexi. Anybody, I've talked to my doctors about how I get rid of this addiction but I love Mexican food. So I said, Sam, let's go have Mexican food early because there's a huge crowd bills at the entrance to El Rancho on South Lamar and if I have to show my hands, I bet nearly everybody has eaten there. General UN, I'd have, but it's really worth the drive from Santa Tony. But anyway, we go out and have my tamale and tequila and taco and it was a great experience and we were leaving the restaurant. Even though we went at 5.30, we were leaving at about 6.30 and it was packed. It was just jammed and we were fighting to get our way out and I got out of the restaurant and was getting in the car and I could hear someone saying, Ben Barnes. And I turned around and the guy walked up and stuck his hand into my hand and said, I just wanted to shake your hand. This man was about probably about 55 years old and he shook my hand and when he put his palm in my palm, he said, Ben Barnes and LBJ, back when politics was done right. And I couldn't think about how fate played a role and that man working his way through that crowd and not that he said anything about me and my name didn't even be mentioned in the same breath with Johnson. But he said something that really sums up President Johnson as we stand here in this family cemetery and stand here laying this reef on President Johnson's grave, that Lyndon Johnson was there when politics was done right. What he did in my story about Dirksen, what he did about the story about the state park, what he did with all the stories that you as your friends and neighbors could tell where you individually had the experience. You know what Lyndon Johnson was and what made him tick. You know why he hurried through the presidency. You know why he hurried through life is that he wanted to do more and he went to his grave disappointed that he didn't have more time. He left Washington disappointed that he didn't have more time. But I think that history is gonna become kinder and kinder to President Johnson and I think he's gonna rise to the top or near the top of the presidency. But I do know this, that when the history books are written of the past 300 years, there's not gonna be any public leader, any president of the United States who was in a bigger hurry and who used every single second of his time to do good for other people and do good for this country. Lucy and Patrick and Ian and the extended members of the Johnson family. We have an example every day and you can walk into the library that Mark presides over and see that those reams of statute holders of the thousands of bills he passed. He did so much in such a short period of time and we've been blessed by giving more time by our Lord and I want to remind you that we have that responsibility. Dr. Graham stood in this cemetery and gave the eulogy and I remember this quote that is very appropriate today on this anniversary. Billy Graham said and I quote, LBJ was a powerful, gigantic personality whose charisma dominated a room the many he entered it. The focus of the attention and even the balance of power automatically shifted to him. He could be coarse and charming at the same time and certainly profane. Almost every time he swore in my presence, he'd quickly look over his shoulder and say, excuse me, preacher, that was LBJ and Dr. Graham summed him up as the general has summed him up as the person that not only has inspired us that we can look to, but his record in his life, I think will be an inspiration for people throughout America for the next century and the next century. And I hope there's someone else that there's a young politician or a former politician that's eating at L Rancho in another 50 years and somebody walks up and says, Tom Smith and LBJ, y'all did politics right. Johnson did do it right. Thank you all for being him here in God bless America and God bless President and Mrs. Johnson for what they've done for all of us. Thank you very much. That may be a sound from God that I wasn't invited to take this podium. But invited, excuse me, invited or not, I just had to let everybody who is sitting in this hot sun know that there was a reason why Ben Barnes was asked to speak here today. And that was because Shirley James who was referred to earlier as my mother's executive secretary and had been my own as well, lovingly look back through her memory and said, you know, there's somebody that your mother always said she wanted to speak at this service because as a young man, he had embraced the values that mattered so much to Lyndon Johnson. And he never, ever, ever forgot.  And we are so grateful that you have not forgotten us but far more importantly than that, we're glad you never forgot those values that daddy and you shared of wanting to do as you so ably alluded to, politics right in wanting to put the people first, their hopes, their dreams, regardless of whether they came from the homes of mighty mansions or the humble hill country that he had. We were all fellow Americans and we needed to look at each other with respect and dignity and try to give everybody the best chance they could have to be the best that they could be. Ben, thank you for continuing daddy's work. Thank you for coming here today. Thank you for never forgetting and thank you most especially for never forgetting that all of the people who've come here today have remembered what Lyndon Johnson used to say.  Thank you for coming home to the hill country, to the values that mattered to Lyndon Johnson on his 115th birthday just as he did on the day that he discovered America. The work is not done. The needs are still there. We had yet another mass shooting yesterday in Jacksonville, Florida. The need for healing continues. Let us all hope to be a part of that healing and do the best we can as God gives us the energy and the insight to serve our fellow man. Lyndon Johnson only had 64 years to do so but he did the best he could and we thank you for not forgetting that that best as Lyndon Johnson would say was not good enough. The work continues. The answer is in each of us. Thank you, Ben.


Welcome everyone to this lecture. Today we will start with a new topic called describing functions. This technique has widely been used since perhaps 1930s, 1940s or a little later and it is a method for finding out periodic orbits that is where we will see. It is an approximation method. So, the periodic orbit values, the amplitude and the frequency, those values are expected to be approximate. So, more precisely we can say to obtain likelihood, likelihood meaning it is not a method that guarantees easily. There are sufficient conditions for guaranteeing periodic orbits and also sufficient conditions for ruling out existence of periodic orbits, but otherwise what we will study in this course is more about the likelihood of periodic orbits and amplitude frequency. This is something in continuation of what I told in the very start of this course that we need non-linear systems for sustained oscillations for robust sustained oscillations. Keeping that in mind, we will see an example of how non-linear systems can achieve this and the technique used for finding the amplitude and frequency goes through this particular method called describing function method. This describing function method is not to be confused with another word. There is something called descriptor which is also called a singular system. This one is a linear system. So, we have all seen state space LTI system. This is LTI system in which E is singular, singular matrix. Determinant of E is equal to 0. This is descriptor system. This descriptor is not the same as describing function. Descriptor system is not the same as describing function. When I started studying describing function many years ago, many places on the net had interchanged these words. So, please read very critically. Whatever you read on the web, one should go for established authors, textbooks or research papers by established authors rather than relying on the net. If one wants to use the net, of course one should because so many things there are all free. That time one should read everything very critically. So, please note that descriptor system is different from describing function. This is a linear time invariant system while this is an approximation method in non-linear systems. Approximation method for finding out periodic orbits, their frequency and amplitude. So, let us take a specific example. Let us begin with a specific example. So, consider a system G, a transfer function, a linear system which has in the feedback path the saturation non-linearity. So, this particular device, suppose this output is input to this saturation non-linearity is y and output is u, then the graph of this is up to some value. It is linear, outside some value it saturates. We will call this saturation non-linearity. For making this standard, we can make this equal to 1, this equal to 1, this is minus 1, this is minus 1. So, it is linear in the range minus 1 to 1, that time it has slope 1, outside that range it has saturated to that value plus 1 here, minus 1 here. So, this is an example of a memory less non-linearity. What is memory less about it? The output to this device, the output u depends only on the value of y now. It does not depend on the rate of change of y, does not depend on the second derivative of y. In other words, whether y is increasing or decreasing, it takes this value. Further, this dependence of u on y does not depend on time explicitly. It is memory less and time invariant. Many of the commonly encountered non-linealities are like this. Memory less and time invariant non-linearity. Another example is the dead zone. This is a dead zone. Because we have this non-linearity in the feedback path, we prefer calling input to this non-linearity as y and output is u. So, that output of the original system continues to be y. So, this is another example of a memory less time invariant non-linearity. This is called dead zone. It can also be understood as play in the system. This is more correctly called as dead zone non-linearity. This is saturation non-linearity. These are extremely widely encountered in practice. So, let us, if one wants to increase the slope, if one wants to change the slope of this, all one can do is just multiply this output u by a constant k, so that the slope for this range is equal to k. So, that can easily be accommodated. Let us see how it can be accommodated. So, let us take this example of a g of s in the feedback path. It has the standard saturation non-linearity. Standard is not a very standard term. I am calling it standard to say that it is exactly slope 1 in the range plus and minus 1 and then it is saturating plus or minus 1. Saturation, what comes as output goes through a linear block k. So, one can easily consider an amplification by k and then by such an amplification, one can get any other slope also. If you want a slope of k in this range, if you want a slope of, if you want slope of value k, one can easily make this change. By that, you can get this by just making this output go through a constant gain called k. So, let us look into whether there can exist periodic solutions. Let us take an example, g of s equal to 1 over s plus 1, s plus 2, s plus 3. Let us take this particular transfer function and see for what value of k one might expect periodic orbits. For doing this, we will use some linear systems theory. We will use a development of Nyquist criteria, Nyquist plot and Nyquist criteria for stability, which is all applicable only for linear systems from that we will find a value of k, which we will, beyond that value of k, we expect that there will be periodic orbits, there will not be instability. That is what we will calculate in detail now. So, as I said, we will first calculate for a linear time invariant system in which g of s equal to 1 over s plus 1, s plus 2, s plus 3. Find value of k for stability. This is the problem. So, what one can do is, one can plot the Nyquist plot of g. Nyquist plot of g with this negative unity feedback will give, and depending on the number of encirclements of the point minus 1, we can get lot of information about whether the closed loop is stable for unity gain and even if it is stable for unity gain, for what value of gain in the feedback path, it will become unstable. See, it is only the loop gain that matters. This negative sign is important. This k could have been either in the feedback path or in the feed forward path. So, let us consider k in the feedback path now. So, this Nyquist plot, if it goes like this, what is this value? It is 1 by 6. Why 1 by 6? Because this is where g of 0, where omega equal to 0 starts. This is omega equal to 0, then it becomes like this and this is where for omega equal to infinity and then from omega equal to minus infinity onwards, it comes like this. So, we are interested in calculating this point precisely where it intersects the negative real axis because for larger values of k, we expect that it will encircle this point minus 1. So, first let us find out whether it is whether the closed loop is stable. Of course, it is easy to infer that the open loop is stable because the poles are all nothing but minus 1, minus 2, minus 3. All the 3 poles are in the left half complex plane. That does not mean that the closed loop will also be stable. Closed loop is stable by that we mean the transfer function from r to y. For that purpose, we are interested in finding the number of encirclements of the point minus 1 by the Nyquist plot and the number of open loop unstable poles. So, we will quickly review the Nyquist criteria for stability also. So, we have the Nyquist criteria of stability as count number of anti-clockwise encirclements of point minus 1. So, we are speaking y minus 1 because we have negative unity feedback. Let us take h because that is more standard in various textbooks. So, consider this feedback. We are going to plot the Nyquist plot of G h actually. So, first we will review the Nyquist criteria for stability for this case and then we will replace h by k. So, plot the Nyquist plot of G h on the complex plane and then count the number of anti-clockwise encirclements of the point minus 1 and call this integer as n. If the encirclement is clockwise, then you count the number and just put minus sign to it. So, this integer n can be positive or negative. If it is positive, then it has indeed the Nyquist plot has encircled upon minus 1 anti-clockwise. If this n is negative, that would be because we have clockwise encirclements of the point minus 1 and y is minus 1 important because we have negative sign here. That is why minus 1 is important. Then we also have two more integers. P is number of unstable open loop poles, open loop poles of which transfer function of 1 plus G h actually. 1 plus G h is the important thing and you see if you add a number 1, the poles of this is same as of G h which is a loop gain. The poles do not change by just adding a fixed number. If you add a fixed number, of course, zeros might change that is indeed important, but the number of unstable open loop poles of 1 plus G h and G h are the same. That is because it is of G h, this open loop should be replaced by loop gain. If it is constant gain K in the feedback, then this is nothing but G in that case. So, it is number of unstable poles of 1 plus G h is how the Nyquist criteria, Nyquist plot, Nyquist criteria reads, but it is exactly loop gain because unstable open loop, the number of unstable poles do not change by adding a constant and when the feedback path is a constant either K or 1, then it is nothing but number of unstable poles of G and hence it is also called the number of unstable open loop poles. Open loop poles that is before you close the loop. If you close the loop with a dynamic transfer function called h, then of course you should be counting the number of unstable poles of G h which is same as 1 plus G h and why is 1 plus G h important? Because 1 plus G h comes in the denominator in the transfer function from R to Y. Consider this. So, Z is the eventual unknown. We are using Nyquist criteria for stability to obtain this number Z. This is number of unstable zeros of 1 plus G h. This is the most important index integer in the three things. It is to find Z that we are trying to use Nyquist criteria for stability. Z is used because of it is zeros, but zeros of which that is where many, many books. In fact, I have seen textbooks also that give this particular criteria wrongly. It is number of unstable zeros, but not of open loop transfer function G, not of loop gain G h, but of 1 plus G h. This it is number of unstable zeros of this which is actually unknown. It is unknown to whom now? It is unknown in this Nyquist criteria for stability. It is unknown that is what I will first clarify. If these are three integers, then Nyquist criteria for stability says n is equal to P minus Z. We are supposed to use the Nyquist criteria for stability this particular equation. This equation, there are three variables, only one equation. So, in three variables one equation, we expect two to be known values so that we will use the equation to find out the third variable which can be unknown. So, n comma P known and Z unknown. Unknown means what? Use equation to obtain. So, please note that this is extremely important. This is often told wrongly in various books. So, Nyquist criteria for stability correctly is like this. It is n is equal to P minus Z. Three integers get related by the Nyquist plot. What are these three integers? n is the number of anticlockwise encirclements of the point minus 1. Anticlockwise refers to that these encirclements should be counted positive. If it is anticlockwise, it should be counted negative if it is clockwise. P is the number of unstable poles of 1 plus G h and Z is the number of unstable zeros of 1 plus G h. So, why is Z important? Z is number of unstable zeros of 1 plus G h, that is what I have explained just now. But why should we be concerned about the number of unstable zeros of 1 plus G h? That is because transfer function from R to Y is nothing but G over 1 plus G h. So, the zeros of the denominator are precisely the poles of the transfer function from R to Y. So, please come back to this particular figure. In this figure, the transfer function from R to Y, the transfer function is nothing but G over 1 plus G h. One can obtain this by very straightforward block diagram reduction procedures. In this transfer function from R to Y, 1 plus G h comes in the denominator and if you do not want unstable poles in the transfer function from R to Y, then this denominator 1 plus G h should not have unstable zeros and it is to count unstable zeros of 1 plus G h that we are using the Nyquist criteria for stability. So, we will conclude that the closed loop is stable if Z is equal to 0. Please do not confuse Z as the number of zeros, number of unstable zeros of G nor of G h. Z is the number of unstable zeros of 1 plus G h, that is why Z, but it is the number of unstable closed loop poles. There, when you are speaking of the closed loop, that time Z indeed denotes the number of unstable closed loop poles. So, it is to obtain Z that we are doing this exercise. N and P are known. We will use G, knowledge of G and H to find out P. We will plot the Nyquist plot and obtain N. N refers to the number of anticlockwise encirclements of the point minus 1. Minus 1 point significance comes because of this negative sign in the unity feedback. Because of this negative sign, we have got 1 plus G h as the denominator. That is how we will find, using the Nyquist plot, we will find P and then we will use this equation, plug in N and P and obtain Z. If Z is equal to 0, we will conclude that the closed loop is stable. Closed loop is stable, no unstable closed loop poles, which is same as no unstable zeros of 1 plus G h, which is same as Z equal to 0. We want to conclude closed loop is stable or not. For that, that is same as saying that the closed loop has zero number of unstable poles, which is same as the number of unstable. This n o stands for number, n o stands for no unstable zeros of 1 plus G h. 1 plus G h comes in the denominator. Hence, we are interested in the zeros of 1 plus G h and this Z stands for zeros, but not of G h, but of 1 plus G h. So, closed loop is stable. In other words, if N is equal to P. So, this is the famous closed loop is stable if N is equal to P. If the open loop has unstable poles, open loop G is all of concern if h is a constant. Otherwise, you should be counting the number of unstable poles of the loop gain G h. So, if that P is not equal to 0, then you had better have that many anticlockwise encirclements of the point minus 1 in order to get the closed loop to be stable. That is how the Nyquist criteria for stability should be read as. If the open loop is already stable and if you have only constant gain feedback in the feedback path, then P is equal to 0, then indeed you want N equal to 0 if you want closed loop stability for that constant gain in the feedback path. So, coming back to our example, we had been working on the specific transfer function G of S equal to 1 over S plus 1, S plus 2, S plus 3. For this transfer function, I already drew, I already sketched that this is 1 by 6. It starts from here. Since we have only poles, the body plot is expected to only have its phase plot decreasing. So, which means that the Nyquist plot is going to go on coming closer to the origin because the magnitude is also falling and also the phase is decreasing from 0 eventually to minus 270 that is when it reaches 0 amplitude and for the other part is just reversal. Point minus 1 we expect is very far. Why do we expect it is very far? Because it has started at 1 by 6 here and the magnitude is only falling which means that it is coming closer and closer to the origin. So, it cannot have crossed minus 1, but nevertheless it is very important to know at which point it intersects the negative real axis that is what we will calculate quickly. So, we will say find omega value where G of j omega is real which means that imaginary part of G of j omega equal to 0. One might constantly ask why are we doing this for describing function method, but I should tell you that this same example we will use for describing function. We will use it for sector non-linearity calculations also for circle criteria. So, for all these purposes same example will serve a very key role and of course we have reviewed a very important topic in linear system theory namely the Nyquist criteria for stability. So, we expect that for gain larger than some value even for saturation non-linearity we expect periodic solutions. We will quickly see that it cannot be unstable hence it has to be periodic. So, we are going to now calculate those values of omega where the imaginary part is equal to 0 and for that value of omega when we substitute we will get the real part that will give us this real axis intersection. One of course we expected 1 by 6, the other values if any will be on the negative real axis this is what we will conclude very soon by equating imaginary part of G of j omega to be equal to 0.  So, which is nothing but equal to 1 over why do not we expand one these two brackets together, multiply it inside this gives us minus omega square j omega there is 2 j omega coming from here and this this gives 3 j omega plus 2 this remains as j omega plus 3 this is equal to 1 over please do this bracket opening very slowly one can easily make mistakes and that will affect all the conclusions later. So, mistakes here are kind of unpardonable because we have to keep in mind that the consequences because of small calculation mistakes are very bad later hence more care had better be given here no excuses for not giving care here.  So, that that gives minus 6 omega square then we have the j omega the j omega how many of them is what we have to calculate.  So, if we have calculated correctly then this is equal to this. So, imaginary part of j of j omega equal to 0 means you see the numerator is real denominator is all that is making this whole number a complex number. So, if the imaginary part of the whole j of j omega should be equal to 0, imaginary part of the denominator should be equal to 0. So, imaginary part of this equal to 0 is if and only if minus j omega cube plus 11 j omega equal to 0 in other words omega equal to 0 or omega square equal to 11. These are the two situations that cause the imaginary part of the denominator equal to 0. Hence, these are the only two situations which can cause the entire j of j omega to be a real number. So, at omega equal to 0, j of j omega is equal to 1 over 1 over what? In this part we expect that the imaginary part do not have to be evaluated because they will all cancel off that that should indeed be checked once again, but please do this part of the calculation yourself. We are going to evaluate only the real part. So, we are going to get 6 minus 6 into 11 which is equal to 1 minus 1 over sorry I this is that omega equal to square root of 11. This is what we have evaluated at omega equal to 0 j of j omega equal to that is easier at omega equal to 0 we get this is equal to 1 by 6. So, these are the only two places omega is equal to plus minus square root of 11 and omega equal to 0 are the only three places where we have real axis intersection. One of these cases is for the positive real axis. The other two intersections happen to be at the same point that is expected because we expect that it is symmetric with respect to the real axis and hence if one part of the plot for omega positive if it intersects at some point for omega negative also it will intersect at the same point. So, this particular point we have now inferred is equal to minus 1 by 60. We knew that this will be closer to the origin than 1 by 6 because the Bode magnitude plot is only decreasing because of only poles. So, to have only poles means it is going to be a low pass filter. So, it starts at 1 by 6 and goes on continuously decreasing to have any poles to have only real poles I should correct myself. If one has only real poles and no pole at the origin that is a situation where the magnitude can only keep decreasing. So, the next plot from wherever it starts will keep coming closer and closer to the origin. So, these are the points where it has intersected the negative real axis. So, what does this minus 1 by 60? What is the significance of this? That is what we will see now. So, if we have, so now let us by a small manipulation put the k up there. As I said it is only the loop gain that matters. If we have k here instead of 1 what we already concluded is that for unity feedback which is nothing but k equal to 1, k is equal to 1 in either this configuration where k is in feedback path or k is here in the feed forward path. In both these cases notice that the loop gain is the same and it is only the loop gain that matters as far as stability is concerned. The transfer function from R to Y is of course affected whether k is in feed forward path or feedback path but the closed loop stability conclusions do not depend on where inside the loop these k and g are located. As long as they are all in the loop they all give the same closed loop stability conclusion. So, coming back to this figure for unity feedback for k equal to 1 we have already concluded that the Nyquist plot does not encircle the point minus 1. This is minus 1, this is 1 by 6, this we have said minus 1 by 60 even though I had already told that it has to be to the right of the point minus 1. We have in fact calculated the point precisely it is at minus 1 by 60. So, we know that for unity feedback equal to 1 n is equal to 0, p is also equal to 0 because g is stable open loop stable g is equal to 1 over s plus 1, s plus 2, s plus 3. So, all the poles are in the left half complex plane. So, the loop gain g and hence k g also both have p equal to 0. Hence, now we plug this all into the formula n is equal to p minus z this gives us z equal to 0. So, for unity feedback that is for k equal to 1 we have concluded that the closed loop is also stable in addition to the open loop. But now we go on multiplying this k for k equal to 2, 3, 4, 5. Notice that if we have done so much effort in plotting the Nyquist plot of g plotting the Nyquist plot of 5 times of g just means that we multiply this Nyquist plot by 5 and it blows up. To multiply by any positive number means just scaling and making this Nyquist plot larger to multiply the Nyquist plot by minus 1 does not mean we reverse this like this. But what it means is we rotate this by 180 degrees to multiply by minus 1 does not mean that we swap this like this. We mirror image this with respect to the image axis but to multiply by minus 1 in fact means to rotate this whole plot by 180 degrees either plus 180 or minus 180 that is immaterial. That indeed makes a difference though you see the whether it is rotating a point by in a clockwise anti-clockwise orientation that is going to get reversed if we mirror image this plot. But whether it is rotating clockwise anti-clockwise that does not change, orientation of the plot does not change by rotating this by rotating the entire plot by 180 degrees. So please note these are extremely different things. When you multiply by a minus 1 then keep note to rotate the plot by 180 degrees rather than to mirror image the plot about the image axis. So now what we are able to say is for k larger than 1 or for k positive for larger than 1 this plot only becomes larger and larger and it will not encircle the point minus 1 until you multiply by k equal to up to 60. For k less than 60 n is still equal to 0 and by whatever k you multiply of course the number of unstable poles of kg do not change by multiplying by a constant k. So p continues to be 0 and hence z equal to 0. But what happens when you put k larger than 60 for k larger than 60 if you multiply this Nyquist plot by k larger than 60 then this this Nyquist plot would have become large enough so that this point of intersection on the negative real axis comes to the left of the point minus 1. So let me take an example. Let me take say k is equal to 65. Let us take an example and for the ease of understanding let us explicitly plot it for this situation. Nyquist plot of g of s equal to 65 over s plus 1 s plus 2 s plus 3. Are we going to do all the calculations again? Of course not.  All we are going to do is multiply it by 65. So we have got that it starts from 65 by 6 and slowly comes inside. We do not know where the point minus 1 is but notice that this point is equal to now minus 65 by 60. Nyquist plot is useless without the orientations especially now that the minus 1 point is getting encircled. How did we conclude that the minus 1 point is getting encircled? This negative real axis intersection has been calculated as 65 times minus 1 by 60 which is nothing but minus 65 by 60 which is slightly less than minus 1 which concludes that the point minus 1 is inside here. So now we can count how many times it is encircling this point minus 1. Where is the point minus 1? It is here. So it has encircled clockwise. So n is equal to 2. It has encircled the point minus 1 twice but clockwise. That if it is clockwise then we are supposed to take this number 2 as negative minus 2. p continues to be 0 and hence now we are going to put n is equal to p minus z. So we have got z equal to 2. So one of the best things of Nyquist criteria is that everything fits well like magic. For no Nyquist plot will you ever get z negative? It is not possible to say that the number of unstable right half, the number of unstable close to poles is minus 2. We can have negative number for n. For p and z we have to have positive numbers only. So if you have p is equal to 0 and if you have instability then the encirclement of the point minus 1 has to be clockwise. All these things are so well rigid. They are also interlinked that one part ensures the other part so well. So I am trying to say that you will never encounter a situation where z has become negative. So what it means is there are two unstable poles for k larger than 60. This is the conclusion for any larger value of k also. So for k greater than 60, n equal to minus 2, p continues to be 0 of course hence z equal to 2. We have two unstable poles of the closed loop for any value of k greater than 60. This is how one can use Nyquist plot for also finding the precise range of k for which we have closed loop stability just like one can use Ruth Hurwitz criteria also. So now let us come back to our saturation nonlinearity and see what this says for the saturation nonlinearity. So we have a constant gain k here and k is equal to let us say 65. This is y. We have 2, we have k here. We are going to analyze two systems closely. Here also r equal to 0. So now here we know that for k is equal to 65 we have closed loop stability even when r equal to 0 to have instability means that if the initial conditions are not equal to 0 for very small perturbations we will have oscillations building into the system, oscillations that slowly grow until things become unstable. So this is how y looks as a function of time y of t. If it starts from nonzero initial conditions then we have instability, oscillations that slowly grow. In other words, they cannot be sustained. These oscillations are slowly going to grow and become very large. Why? Because two poles are in the open right half complex plane for k is equal to 65. But when these oscillations are very small, when y is very small then we expect that it is inside the range plus minus 1. So they will continue to grow inside this range. As long as the output is inside this range it will grow. But the output cannot, after all y is limited what comes out here cannot be greater than plus minus 1. And it is getting amplified by only 65. Do we really expect that oscillations will become unbounded in this example? Here because of this saturation in the feedback path the oscillations cannot become unbounded.  No, they cannot die down to, as soon as the oscillations are smaller than this range as long as y is less than plus minus 1 there will always be a tendency to grow because if y is 4, 4 minus 1 less than y of t less than y less than plus 1. As long as y is inside this range the input here, the y input to this particular saturation nonlinearity block looks at this block as a linear system. And hence what comes out will only want to grow as it goes like this. Why would it want to grow? Of course this is just gain 1. For y inside this range this is just gain 1. But why would it want to grow? Because in that case it is like this system where you have 1 in the feedback path. And for k equal to 65 we know that this is unstable. So it will want to grow because of the dynamics of k and g together and the feedback aspect of this configuration. It will want to grow. But once it grows beyond plus or minus 1 it is getting chopped, it is getting saturated. So we cannot have a situation like this here. So we have to have oscillations, we have concluded and oscillations are sustained. What is the meaning of sustained?  Why they do not die to 0? Let us just recapitulate. It doesn't die to 0 because as soon as the range of y, the oscillations here come smaller than plus minus 1, this block behaves like a linear system with unity gain in the feedback path. That time it is like this system and we know that this is unstable. So the oscillations can't come smaller than plus minus 1. The amplitude will be greater than 1, cannot become unbounded. Why it cannot become unbounded? Because we know that there is a saturation feedback path. At this point the gain can never exceed the value of the signal here, can't exceed 1 in magnitude and this is at most 65, this is at most 65 magnification, r equal to 0 in both the cases, external input is 0. We are looking at oscillations of this autonomous system. So that is why we know that this system can't have unbounded trajectories either. It cannot become, oscillations can't become unbounded. In other words, it's sustained. This is an example of an oscillator where we have sustained oscillations but of course this is not a linear system. We need more techniques to find out the amplitude and frequency of the signal of the oscillations here. Of course at least the frequency is not hard to find, at least approximately but exact values are a very complicated task. We will use describing function method to find out the amplitude in particular. For this example, for k equals 65, 70, for different values, we would like to find out the values of the amplitude. This is an example where describing function method will help. So, what exactly is a describing function? We saw some motivation for describing function for finding out, some technique is required for finding out the gain and the frequency when we have a non-linear element in the loop, in the feedback loop. So, one should think of the describing function as a gain. It is a function of, it is a gain. We are going to denote it by eta, a gain that depends on the frequency and amplitude. So, for every block, linear or non-linear, we prefer calling the input as Y, output as U only because in our applications we will have this non-linear element in the feedback path. For example, it could be a saturation non-linearity which we have seen or it could be a dead zone or it could be a dead zone like this. So, corresponding to every block, we want to associate a notion of gain of this operator. And we are going to keep in mind that this gain can depend on the frequency omega, it can also depend on the amplitude. What are these amplitude and frequency? This is when the input Y is equal to the signal Y of t equal to A sin omega t. Now, we need a little more development before we reach the stage where we define the describing function as a gain. It is a complex gain. It has a real part and an imaginary part, but the motivation of this is completely linear time invariant systems. Why?  Hence, for non-linear systems also, we are going to associate a gain of this system, a gain of this operator and this gain of the operator we are going to call as eta. This is called eta, Greek alphabet for eta and this gain of system for input equal to A sin omega t. For this particular input, we are going to associate a gain of the system and this gain, we are going to allow it to vary as a function of omega, the frequency of the sinusoidal input. We can also allow it to depend on the amplitude of the input signal. So, what is the motivation for this? Because the motivation for linear time invariant systems behave like this. One can check that if G of s is a transfer function of a system of transfer function of an LTI system. Let us also assume that it is stable. Then, when you give input A sin omega t to the system G of s, what comes out is output equal to some transients. What is the meaning of transients? It goes to 0, stable, all poles in open left half complex plane. No poles on the imaginary axis also. Transients going to 0 plus steady state part. There is a steady state part that we are going to write now. This steady state part turns out to be nothing but A G of j omega sin omega t. This is the part we will focus on for the rest of this lecture and also the next lecture. When you give an input A sin omega t, this A sin omega t might have suddenly been switched on. That is why we might have some transients. But because all the poles of the transfer function are in the open left half complex plane, there is nothing on the imaginary axis, there is nothing in the right half complex plane. Hence, all these transients, because of the sudden switching on, go to 0. But eventually, there is a steady state part. This steady state part is, of course, it has sin omega t. This is what we like to think of as the forcing function, the forced part and the transients, the natural response and the forced response. We are speaking about the forced response. The forced response has exactly the same frequency sin omega t. If A was the amplitude, there is A here also. But there is a complex number associated to, complex number that is called multiplied to sin omega t. Of course, this is primarily notation.  How can the output become complex? I mean, how can it have an imaginary part? We measure signals that are functions of time, they are all real part. At any time instant, the value of function of t on the oscilloscope we measure, we do not measure complex signals. This is a notation. G of j omega is indeed a complex number and this complex number has some meaning. The real part and imaginary part have some important meaning, but the way it is written here, it is just notation. So, what is the meaning of this notation then, if it is just notation? So, if G of j omega is equal to some real part plus j times imaginary part. We can break its real part and imaginary part. There is some transfer function G of s. We evaluate it at a particular value of omega. When we evaluate G at some complex number, there is a very rare chance that it has become a real number. It could become a real number. For example, if the necklace plot for this omega intersects the real axis. Except for that rare case, this G of j omega has a real part and imaginary part both. Imaginary part by itself is some real number. It has got multiplied to j. Now, it has become purely imaginary and now you have added another real number. So, now together it has become a complex number. So, when we say G of j omega times sin omega t, what we mean by this is real part of G of j omega times sin omega t plus imaginary part of G of j omega times cos omega t. To associate the imaginary part with sin omega t and this j, this j when it applies to sin omega t, we will think of it as cos omega t. This is a very important fact. What is that important fact? You take a linear time invariant system with transfer function G. You give it A sin omega t. What comes out has a transient part that goes to 0 because G is stable. Stable by here we mean no poles on the imaginary axis, no poles in the right half plane and there is a forced part. This is also steady state part. At steady state, this A sin omega t will have an effect only on the forced part. The forced part is equal to real part of G of j omega times sin omega t plus imaginary part of G of j omega times cos omega t. So, one can check this by first principles. By substituting this into the differential equation, every transfer function after all is nothing but a differential equation between this and this. In the differential equation for output also, you substitute sin omega t and cos omega t. The coefficient that come out, I forgot to factor A here. If the amplitude is larger, the output also will have amplitude exactly larger, but how much part comes with sin omega t? How much part comes with cos omega t? That is determined by the real part and imaginary part of G of j omega. Now, you see there is no j here. Now, the output is a real signal. What happened to the imaginary part of G of j omega as far as our earlier notation is concerned? That has got translated to cos omega t amplitude. So, how do you, how does one verify this? We can take a very simple example and also verify this. Suppose, the transfer function of some system is 1 over s plus 1, input is A sin omega t, output is question mark. So, this particular, let us say input is, let us come back to the more traditional notation of using input U, output Y. So, the meaning of this transfer function is nothing but d by dt of Y plus Y equal to U. If this is the differential equation of a system, then we associate this transfer function to the system. Now, U is equal to A sin omega t. So, we will write U is equal to A sin omega t here. Now, for Y, of course, there is a e to the power minus t term, which is homogeneous for every initial condition there is a part, but that is going to die to 0. So, we are interested in the forced response part. So, we will write that this is some b times sin omega t plus c times cos omega t. We are going to find out values of b and c. So, we will substitute this into the differential equation here and when we substitute we get derivative of sin omega t is nothing but cos omega t.  So, this part here is the derivative of Y and this part here is nothing but Y. So, this is equal to A sin omega t. Now, sin omega t will equate to the coefficient will equate and cos omega t, there is nothing cos omega t here. So, sin omega t in cos, this in other words we say that sin omega t and cos omega t are independent functions of time. If this has to be satisfied for all values of time, then we can equate b omega plus c equal to 0 and A is equal to b minus c omega. So, we are going to solve these two equations simultaneously for obtaining values of b and c and by solving this we will get the values of b and c. We have two unknowns b and c and two equations in b and c. We will use this to find the values of b and c. So, what do this translate to? Let us write this into a matrix b c equal to one of them is of course A, other one is 0. So, the coefficients of b and c is 1 minus omega and the upper equation is omega 1. So, in order to get b c all we have to do is b c equal to inverse of this matrix applied to 0 A. So, the inverse of this matrix is obtained by just minus 1, minus 1 applied to 0 A divided by the determinant of this matrix.  So, it is 1 over minus 1 over omega square plus 1. So, this is how we can get b and c. So, b and c is equal to this nothing but is second column of this.  This is the value of b and c. So, what we can do is we can check whether, so there is an A missing this time t. We can check whether these are indeed the real parts of G of j omega. So, G of j omega from the theory that I said, we have got output y of t equal to A times A times real part of G of j omega is the coefficient of sin omega t plus A times imaginary part of G of j omega times cos omega t.  After checking that, we would have ratified this formula that we wrote for linear time invariant systems, which is what we will use as motivation for defining the describing function of a non-linear system. So, G of j omega is equal to 1 over j omega plus 1. This is equal to 1 minus j omega divided by omega square plus 1. This is exactly equal to 1 over omega square plus 1 plus j times. So, this is what we also obtained here. We have obtained that the real part is nothing but B, B over A. Notice that there is a factor. The part that comes as sin omega t of course, has a factor A that comes because the input has A sin omega t. If you scale the input by a constant, of course, the outputs will get scaled by the same constant both for sin omega t and cos omega t. But how much portion has gone to sin omega t? How much portion has gone to cos omega t? For linear time invariant systems, that fraction does not depend on A, a very important point to note that linear time invariant systems, it is only the relative partitioning. Relative, I do not mean to say that they add up to 1, that is not the case. But A does not play a role in these constants, the real part of G of j omega and imaginary part of G of j omega. In other words, the magnification also has some contribution coming into cos omega t. But scaling by A does not change these relative contributions. So, real part of G of j omega indeed is equal to 1 over omega square plus 1, which is what we got as B, B except for this A factor. And the part coming with cos omega t was exactly minus omega over omega square plus 1, which is nothing but the imaginary part. This is what we will like to generalize to describing functions also, to for non-linear systems also using this notion of disrupting function, where we allow dependence on omega and also A for non-linear systems, that is very much possible quite expectedly. We will see that for the case of saturation non-linearity.


Hey guys she bear 1,000 here. I want to show you something. I got it's not really a yard cell find or a Flea market fine, but kind of this has been at the shop where I work for I don't know over a year and a half now and The owner which is my boss. I asked him what he wanted for today and he said You know it It is a husky Yeah, it's a husky, so I'm thinking that's what tractor supply I'm believing. I'm not sure what year or anything like that Yeah, that says TSC industry is incorporated. I believe that to be tractor supply. I Don't know anything about it. Don't know if it works But Let's plug it in and check it out guys. All right. I'm gonna put you in the stand here. I hope for a tomorrow I will have an unboxing for you Something else that I have ordered that supposed to be in tomorrow So Get you down here All right Now let's get get a plug in and see if it works All right Over here Let's plug her in. Well, it's running and pumping. It is very quiet Which is cool, but I'm either thinking about Let's let it run for a minute. See what if it builds up pressure. I don't know what that knocking noise is I'm not sure But if it runs alright either that Or that Is going to get my two horsepower bridge and strap but first I gotta figure out what this knock is Used to be building up pressure you probably See that 25 psi It is a hundred psi air compressor So I'm gonna pause you guys for a second. It's my kid bothering me. I'm gonna pause you guys for a second Somewhere It is pumping air you have to weld that if I decide to use it But I'm gonna pause you guys for a second and I'm gonna let it run Let me check the oil and see where it goes from here. So I'll be right back with you We're up to So it should be shutting off here in a minute providing the pressure switch works It's been about Four minutes But that is a very quiet running air compressor Of course, I'm gonna have to put a new a new cord on it. It's been taped up and chewed up Regulator sitting on about 90 right now pushing about 98 pounder it does have a Vile on the bottom where you can drain the water out, but a lot of people won't do that And of course, it's air you get water and it's condensation and if you don't grain it out every so often then it sets for years This thing's been sent in for ten years.  That's what I'm waiting on tomorrow There's the little filter on about a hundred and two now Regulator still on about 90 But it's a hundred PSI so It could have a turned up a little bit, maybe Of course it might not be shutting off because of that leak, but you would think it build up to a hundred pound Should shut off on switch kind of holding steady at a hundred That might be why it's not shutting off At a hundred and one hundred and two But this very very quiet, so I like that that's cool. Probably what it is.  I know all that thing I'm just going to go ahead and unplug it right now turn it up here a little say The leak is right there a little pinhole right there. No biggie no leak there, but Anyway guys There's my little to play with fix up. All right guys So I'm going to sign off for now. I just wanted to show that with you and Keep an eye out for more videos Saturday night, I don't know if you'll see it Saturday, but you probably see it Sunday I've got another little surprise that a little challenge. I've been seeing on YouTube thought I'd try it So ordered what I needed offline Online what anyone call it and got it. So I'm gonna do that Saturday night for you All right guys, so that being said remember check out monkey 1000 check her channel out.  So it's pretty cool not crap craft She'd like to do some craft stuff for around the house and things so Very entertaining very interesting. So check her out and as always guys. Thanks for watching I'll see you soon and Remember she better the myth man legend. I'm going for now. Bye. Bye guys. Take care


Okay, I'm having a minor dilemma here, which is that I'm working on something kind of huge, like a life-changing project. And, you know, I have a lot of imaginative things going on, a lot of creative, unorthodox processes and stuff. Basically, I'm at the point now where I need to lock myself in my artist's studio, or my workshop, or whatever I have. And a lot of you guys will understand this, I think, because a lot of you guys are artists, or creative people, or general crafty people of some kind.  So I need to do that. So what do I do about this video documentary? Because this is a big part, this is the culmination of a lot of things that have been leading up to this for years. And I know I'm not going to be able to document it the way I've been doing it, which is, you know, I video everything as I'm going, and put the videos up as I'm going, and then discuss it as I'm going, and I'm not going to be able to do that. So what I'm going to do is I'm going to bring the camera with me, I'm going to get video of whatever I'm doing, and I'm just going to dump it onto my computer, and at some point in the future, I will open that giant folder of like 50 gigabytes of video I've gotten for this project. And then I'll put it all together and, you know, put all the videos online. So there's going to be a break between now and then, and it's going to be a long enough break that I feel like I needed to say something about it. So you're not just like, Jamie, where'd you go? Well, I just need to lock myself in for a while and really focus on this. And so like I said, a lot of you guys will understand that. You'll just be like, oh yeah, I totally get what you're saying. Do it. No problem. But I also know that there are a lot of people who aren't artists or imaginative type creative people or whatever, and you won't understand the need for an artist to lock themselves in. Why do you have to do that? Why do you just do what you've been doing? What's the big deal?  And maybe in the comments, if there are people who don't understand that, some of you other artists and creative people can chime in and kind of fill in the blanks that I'm missing here. But basically, when you're... I'm just going to use the term artist for just to cover. You do a lot of things that are unorthodox, that are outside of what everyone else is doing. So you're not following the instructions, you're not doing what other people are doing. You're coming up with ideas and trying them. Some of them work, some of them don't work.  And then you follow that for a while. You have this really loose process where there are a lot of variables all the time. And I think that's the big piece of it. When you're an artist, you have a lot of variables going on. Whereas the way we're generally taught to think in school, which I wish they wouldn't do so much, is to think more like a computer where things are constants. A plus B equals C. And that's how you get to the answer. So when you start doing things in these kind of more imaginative artistic kind of ways, you get a lot of questions about it. And a lot of the questions are fine. And a lot of them are like, wait, why aren't you just doing it like everyone else? This is tried and true. You shouldn't be doing that. You get a lot of you shouldn't be doing that. But then you also just get a lot of distracting questions. And they're not bad. It's not like the people are trying to cause any problem. They're just interested in the process. Like, what are you doing? What are you doing? And the problem is when you're doing these kind of imaginative open-ended things, a lot of times there isn't an answer to the questions. Someone will ask, why are you doing this in this way? And as the artist, you're like, because there are 20 different variables, and I used my intuition and some part of my brain that I can't explain well to take this cloud of intangible stuff and come up with some answer. And they're like, wait, but what was the information? What was the thing that got you there? And it's just like, it doesn't work that way. I can't answer that question in any reasonable... That's not the right question. And this is why humans can do things that computers can't. Because to a computer, if you have 20 variables, and each variable has a million different possibilities, the computer looks at that and says, wait, a million to the power of 20? That's too big a number, too many possibilities. However, in our brains, we have some kind of software or something that can take 20 different variables, each with a million possible combinations or a million options, and we can somehow process that into something and do it in a way that's not really explainable, other than to say, I'm using intuition, I'm connecting with the universe in some imaginative way that's hard to explain unless you've done it. So what artists do at times is they say, okay, I can't explain any of this to anyone, I just need to focus on it because processing all these variables takes a lot of energy and all the questions and all the distractions and everything is confusing and I really just have to have my mind open to all these things and when my mind is so open, any kind of input kind of throws it off. So that's basically why artists lock themselves in their studio and sometimes for months at a time and just won't talk to anyone or deal with anyone and it's not that they're trying to be rude or disrespect anyone, they're just saying, I need to put my mind in a state that is not conducive to other people in order to make some incredible thing happen. So that's where I am right now and like I said, a lot of you guys don't even need that explanation, you're just like, oh yeah, totally get it and hopefully that explains it to some of the other people. But anyway, all right, I've got this project and it is 100% my main focus, top priority right now, I gotta go do it. All right, and I promise to get some video that I'm just not going to be able to show for a while. Oh, I left a little bit out of my explanation of taking all these variables and somehow funneling them into an answer. I should also mention that often, I mean usually, the answer is also a variable, so it's in flux. So you take all these variables and it's like, okay, right now I have an idea of what to do and two and a half minutes later it's going to be different and it's just going to slide around and constantly change and tomorrow I'll be doing something that's not recognizable compared to what I was doing today and it's like everything is in flux. So there are no concrete answers to anything. Okay, I gotta go.