 Hello. Good morning. I am Mr. V. D. Sathe, Assistant Professor, Department of Mechanical Engineering, Vulture Institute of Technology. Now today we are going to study the first unit of our transient system and in that transient system as usual we are going to see that what will you will be able to do after finishing this session that is called as learning outcome. So learning outcome for today's session is that you will be able to discuss the necessary condition for stability. Now the important thing is say before we start actual discussion why it is important to study the necessary condition then we will see in the part 2 sufficient condition then we see for other complications and say mathematical procedures for how to find out whether the system is stable unstable or say marginally stable. Now the simple question is what is the meaning of transient system? Transient system is a system in which whenever we give input the output changes for a short duration of time and that is called as a transient behavior and that behavior we are going to analyze. Now for our analysis what we are going to study is whenever we have been given a characteristic equation of a system that is a closed loop control system how to decide whether the system is stable or not. If it is stable whether it is permanently that is asymptotically stable or not or if it is unstable what are what are the reasons for its instability and if it is marginally stable how we can identify from the particular say equation whether it is under these three categories. Now my first question to is you already know you have studied the block diagram and you know that we can have canonical form that is known as a block diagram in which we have got one input then this is the summing point then we have forward gain that is g then this is the output which is taken take take off point and there is a feedback which is say h and I have here the negative feedback. So this is my summing point this is the input signal this is negative and this is the diagram okay. Now we do our analysis assuming that it is a negative feedback now sometimes question may answer why there is what if there is a positive feedback what to do see we are doing our calculations with negative feedback and if there is a positive feedback you have a choice what we can do you can just say make this plus you can make this plus and have in negative sign in the feedback loop. So with a small change in our gain function by changing its sign from suppose it is a plus 5 then we can make it minus 5 and minus 5 added it is equivalent to negative addition. So we have no problem whether we take our system as positive feedback or negative feedback so we will do our analysis assuming that there is a negative feedback. Now we know from our black diagram algebra that the closed loop transfer function for this has got say g h upon g h okay so we have got a numerator here and denominator is 1 plus g h if it is a negative feedback I am not talking about numerator because in the numerator we have do not have to worry because even if numerator is 0 0 is not an mathematical problem but when 0 comes in the denominator it creates a singularity singularity means a problem is there for singularity is we cannot define the function there function is not continuous at that particular point and if it is continuous within the proximity we have to go for a very fine calculus for that particular thing okay now the next thing is for this 1 plus g h what will happen this will be a polynomial in s this will be a polynomial in s so there is some polynomial there is some polynomial in s now what is the degree of polynomial that degree of polynomial may be it may be linear it may be quadratic third order fourth order nth order polynomial I can have now today we are restricting ourselves to the equation which is our say characteristic equation which we are getting in the form of a quadratic suppose it is a s square plus b s plus c now this is a quadratic equation now the main thing about our transient system stability is whenever I draw a diagram or a graphical representation on this axis I have real number on this axis I have got omega that is say negative number that is a imaginary number so any point lying on this axis this particular plane is on the left of s plane so this side is known as left of s plane because we are working in s plane now and this is the right of s plane so this is the right side now see there is a slight difference in our regular coordinate system and the system of coordinate system that we use for our say transient system analysis which again we use for our root locus analysis also now you can see here the particular say equation has got roots now for a quadratic you know what is the possibility for roots all 2 roots may be real all 2 roots may be imaginary 2 roots may be irrational numbers and out of these 2 real roots both may be same and out of these 2 real roots they may be positive or they may be negative or one is positive and one is negative so as far as our stability is concerned we are interested only in the left of s plane so this is a stable region so always remember this is my stable region so I want my equation that is the characteristic equation which I have obtained for a closed loop transfer function and its denominator equated to 0 I will get some polynomial in s now the question is how to identify whether my system is going to be stable or not so first thing is first criteria is it all the roots must lie on the left side so all roots must be present in this region okay now in this region means there are 2 possibilities either my root may be somewhere here on this axis that is the real axis which is a negative x axis or my root may be here see what how I am drawing the particular say roots when I say the roots are here on the left half of s plane when I show one root here second root here one root here second root here means whatever the roots I get whatever the roots I get they are symmetrically placed about the real axis because the complex roots I am getting in this plane as long as I am on this line it is real but when I shift either above or below I will get a complex number complex number has a form that is the simple form that you know which you have already studied in your previous classes which is a plus I b a plus I b is my complex number so when b is 0 it is a real number when a is greater than 0 it is to the right side if a is less than 0 it is to the left side of 0 is that not and if a 0 and b 0 it is a origin so now when my root is lying on this axis it is going to be on the negative real axis and if it root if the root lies on the complex plane if it is a quadratic both the roots must be complex conjugates because that is the necessary condition necessary condition that is the first part but the stability criteria requires certain things which are mathematical in nature it says that whatever the polynomial you have whatever the polynomial you have it must be complete polynomial now what is the meaning of complete polynomial suppose I write a polynomial 3 x cube plus 2 x square plus 3 is equal to 0 if I write this polynomial the problem is with this is that the x cube term is present x square term is present but term in x is absent so when the term in x is absent that system can never be a stable system means first condition necessary condition is that equation must be a complete polynomial and because it is complete polynomial that is a square plus b s plus c or I will take another example say it is a square plus 5 s say plus 6 is equal to 0 now if I see this polynomial and if I ask you factor this so you will factor this as s plus 3 and s plus 2 is equal to 0 so what are the values s is equal to minus 2 and s is equal to minus 3 so it will be minus 2 here and minus 3 here so essential condition is that all the terms must be present it must be a complete polynomial that is the first condition second important condition is that the coefficients of all the terms that is s square s and constant or s cube s square s and constant must have same sign must have same sign same sign means it can be plus plus plus or it can be minus minus minus because if it is minus minus minus it makes no difference I can get still by multiplying minus 1 the same answer but on the contrary suppose it is a square minus 5 s plus 6 is equal to 0 now my roots are s minus 2 into s minus 3 is equal to 0 and s is equal to 2 and s is equal to 3 so what is happening now I will get 2 roots here so the stability condition says that your roots must be on the left half of the s plane it should not be on the right half of the s plane that is the necessary condition that is the first part and if my polynomial is only up to say quadratic linear and quadratic this condition is also the sufficient condition means if my particular function is like this I will say that without any problem as it is quadratic as it is a complete quadratic and as the coefficients of all the terms have same sign not necessarily positive it is same sign then the system is going to be stable because all the roots will lie on the left half of s plane so this is our first introduction about the transient system in which we find out whether the system is stable or not if it is a quadratic we have studied that for quadratic equation to be stable necessary and sufficient condition is the same that is it must satisfy complete must there must be complete polynomial no term is missing and the coefficients of all the terms are having same sign so I think with this particular you have got the idea about the particular thing and you can now if you want to refer you can refer automatic control engineering by Revan by Bakshi and by Barapet. So thank you for your patience listening now in the second module we will see the further extension of this topic thank you.