[Music] welcome back so ah till now we were interested in understanding various quantities associated with the transfer function both in transfer in the continuous domain as well as in discrete time domain ah and its effect on the dynamical behavior ah when we said the dynamical behavior we were basically focusing on ah the rate of response of the speed of response one important characteristics of ah dynamical feature is the stability of the system ah till now we have not paid pretty much attention on deciding whether the system would be stable or not ah when subjected to
a given ah input to the system so we will focus on this particular topic in this ah lecture so what we have in front of us ah is ah system in state space domain ok so imagine that i have a autonomous system which is described in state space domain given as d x under bar by d t is equal to a double under bar x right so this is an nth order system and at order autonomous system and we spend a lot of time to understand ah how to assess the stability of the system are
apart from the dynamical features of the system and how would you ah decide on the stability of the system well you know that the general solution of the system is given as x under bar of t is equal to summation i is equal to 1 to capital n c i e to the power lambda i t times the corresponding eigen vector v i ok so what you basically do is you list out all the eigen values eigen values and what are the two major categories in which the eigen values may fall they may either be
real or they may be complex ok and depending upon whether they are all real or complex you can immediately see the eigen values and comment upon the stability so when the real eigen values ah such that you have all negative every single eigen value is negative then you definitely see that the system is stable which means every single state which the system can sample would be bounded no state will blow up to infinity this is the simple definition of stability and otherwise otherwise the system would be unstable even if there is one eigen value which
is positive then e to the power positive quantity would blow up to infinity as t tends to infinity fine and when you have ah complex eigen values then you wanted to look at the real parts and if the real parts all the real parts are less than zero let me emphasize all real parts are less than zero negative the system was stable and otherwise the system was unstable right so this was ah what you could infer from the state space domain analysis you looked at the eigen values every single negative eigen value or in case
of complex ah eigen values every single real part being negative is the assurance of stability and if even one of this is not satisfied the system is going to be unstable pretty straight forward now you have an input output type system where you would like to know the response of a system subject to a given input and then now you have to define stability so you need to define stability and we know that we need to do this analysis in ah the transform domain so let us see how do we do that so the first
thing which you need to do is you need to define what exactly is the meaning of stability ok so if in response to a bounded input so the first thing which you need to worry about is bounded input what is the meaning of boundary input there is a minimum and maximum which has been defined for the input ok so there is a minimum and maximum which is defined for the input ok so if in response to a bounded input the dynamic output of the system remains bounded which means the response also is bounded between maximum
and minimum then the system is said to be stable otherwise unstable so ah if we understand the meaning of boundary input obviously the meaning of bounded response would follow so let us try to understand the meaning of bounded input ah let us imagine a step input of magnitude a right so what happens here in this particular case you give your system so this is u of t and this is t this the function is defined such that you have the value 0 and a value a at t is equal to 0 which is sustained there
ok so this is 0 this is a so what is the input what are the bounds on the input minimum is zero and the maximum is a right so there there are bounds so therefore a step input follows the condition of bounded input a step input is a bounded input what about a sinusoidal input for example input ok so i have here t and i have here u of t is equal to a sine omega t what is going to happen in this case your function is going to look like this right so what is
the maximum that you have it is a this is the maximum and the minimum that you can have is minus a so again your input is bound between two limits the upper one being plus a the lower one being minus a so therefore sinusoidal input also is an example of a bounded input what about a ramp input so let me write here bounded bounded and i have a ramp input now on what is the functional form of a ramp input i have u of t is equal to a t and draw it versus t you
see here the input goes to infinity as t tends to infinity ok so therefore maximum is unbound there is no upper cap on the maximum and therefore you have an unbounded input so therefore you may not say anything or you may not be in a position to comment upon the stability of your system if you subject the system to a ramp input but you may subject the system to a step input and you know that the maximum that my input or my forcing function can have is a the minimum that it can have is 0.
so therefore if my response also is bounded then the system is called to be stable ok fine so now lets see how we can understand the dynamical behavior with respect to the stability of the system the condition for stability is that a continuous system is stable if every pole associated with its transfer function is negative or has a negative real part so this means that i can simply look at the transfer function and say that i know that this system is going to be stable i know this system is going to be unstable the way
i used to look at the eigen values in the state space domain analysis and the moment i used to see that the eigen values are positive i have to say that the system is going to be unstable when i saw all the eigen values to be negative as to say that the system is going to be stable when the eigen values used to be complex numbers with all negative real parts then i used to know that ok the system will have oscillations but oscillations will eventually die out and when they had positive part ah i
used to know that there would be oscillations and the oscillations would ah increase in amplitude with time the oscillations would sustain if you have a you had a purely imaginary eigenvalue so similarly now what i can do is i can look at the transfer function the laplace domain transfer function in case of continuous systems and i would look into the poles of the system ok and if every pole associated with the transfer function is either negative or the real part is negative if it is a complex pole then the system will be stable let us
see if we can ah establish this so imagine that i have a p q order system ok so let us take the most general case so i have g p q of s and what would this look like this would look like e zero plus a one s plus a 2 s squared up to a q s to the power q this is the numerator having a polynomial of degree q divided by an polynomial of degree p so this would be 1 plus ah b s so b 1 s plus b 2 s square plus
up to b p s to the power p ok so this i am considering the most general case now if this be the case ah i would like to know the dynamical response and the input itself has to be bounded so therefore let us take the case of a step input ah one thing which i have forgotten is that i need to multiply it with the static gain k so this is my general transfer function so therefore what would be y of s bar would be transfer function multiplied by the laplace transform of the step
input so that would be a k 1 upon s multiplied by a 0 plus a one s plus a two s squared up to a q s to the power q divided by one plus b s b one s plus b two s squared up to b b s to the power b how will i do the laplace inverse inversion well i know that i will need to do a partial fraction so imagine that i can factorize the denominator i do not worry about the numerator i can i need to factorize the denominator and let
us say that the denominator can be factorized as follows so the numerator is simply a 0 plus a 1 s plus a 2 s squared up to a q s to the power q and i have factorized my denominator into p number of factors so lets say the factors are s minus ah one s minus r two up to s minus r p right what would be my next step i will need to do a partial fraction i will not worry about what exactly would be the form of the partial fraction but i know what
would be the functional form of the final finally partial final partial fraction which i get right i hope you would agree that the final partial fraction would look like this c 0 upon s plus c 1 upon s minus r 1 plus c 2 upon s minus r 2 up to plus c p upon s minus r e there would be some numbers associated with c 0 c 1 and so on ah i do not need to worry about that at this point of time because they are not going to affect the stability of the
system they would affect the dynamics but it will not affect the stability we will see why so whats going to happen to y of t so now i will do a laplace inversion so i will get y of t is equal to what a k multiplied by c 0 upon s and its laplace inverse is simply c 0 plus c 1 la plus inverse of this is not difficult e to the power r 1 t plus c two e to the power r two t up to c p e to the power r p t
fine now you see the first term on the right hand side is a constant function so its always going to be bound but at rest every term on the right hand side has e to the power r i t right so i have e to the power what is the nature of e to the power plus ah t this is what this is e to the power plus r t versus t it looks like this which means on bound and e to the power minus r t and in every case r is greater than zero
ok so it will power minus r t you will have e to the power minus r t again ah greater than zero versus t this will this is bound you can find the minimum x minimum maximum in that case so when ah eyes are real when r as a real stability dictates that ah ice must be less than zero for y of t to be bound if y of t has to be bound then r i have to be less than zero you can see the graph on the right hand side now the problem is
when r i belong to complex numbers when r i belong to complex numbers then what happens so i have e to the power ah i t and let me write e to the power rit as what e to the power the modulus ah e to the power i theta where theta is equal to tan inverse imaginary upon real right i can do this fine so then what is going to happen in this case ah even before this this happens what i can do is i can simply write e to the power ah i t is
equal to e to the power z one plus i z two times t right in this case i do not need to even convert it to ah the r theta form ah i simply convert it to r i is equal to z one plus i times z two so this is what e to the power z one e to the power minus i z two t t right and now it is very clear that this will give you oscillations this will give you oscillations to your system because those can be converted to sines and cosines
and this will give you decay of growth depending upon z one so therefore you will have decay for z one less than zero and growth for z one greater than zero why because you have e to the power z one t same thing which is there on the figure on the on the right hand side ok so therefore its pretty clear that when you have a system whose poles are all negative or the if the poles are complex numbers all of the real parts are negative its only then that you can expect the system to
be stable ok so lets quickly ah look into the statement which we had a continuous system is stable if every pole associated with the transfer function has a negative ah real part or is itself negative ok we ah looked into the transfer function for a first order system and we got for the equation tau d y by d t plus y is equal to k times u from here we got g of s is equal to k upon tau s plus 1. what was the response of such a system subjected to ah y of t
and u of t as a function of t if this is a the response was bound bounded and this was the response right what was the pole r in this case was minus 1 over tau which is less than 0 i encourage you to find this out that for an equation of the form d y by d t minus y is equal to k u for which g of s would be k upon tau s minus 1 your r will become 1 over tau which is greater than 0 right so we had y of t
is equal to a k 1 minus e to the power minus t by tau as the solution if you follow the exact same procedure here you will find that y of t would now become e k minus 1 plus e to the power t by tau ok and what would this function look like this function would look like this this is y of t or u of t versus t this is e and what would the response look like you can plot the response looks like this so bounded and tau r was minus one over
tau less than zero unbounded r was one over tau which is greater than zero as simple as that so you can simply look at the transfer function and comment about the stability now can we extend this to discrete system let's let us look and write down the general discrete general discrete time pulse transfer function so g of z would be what g of z would be a 0 plus a 1 z inverse plus a 2 z inverse squared up to plus a q z inverse to the power q divided by 1 plus b 1 z
inverse plus b 2 z inverse square up to b p z inverse to the power minus b multiplied by k the static gain so this is the general pq order discrete time first transfer function will do the same analysis subject this to a step input what is going to happen y of z subject to a step input would be a k 1 upon 1 minus z inverse a 0 plus a 1 z inverse plus a 2 z inverse squared plus a q z to the power minus q divided by now let me factorize the denominator
if you remember ah the lecture in which we introduced z transforms we can factorize the denominator now into p number of factors as this 1 minus r 1 z inverse one minus r two z inverse up to plus ah multiplied by one minus r p z inverse this is the way we will do the factorization then what we will do same we will do a partial fraction so what is going to happen this will be a k multiplied by c 0 upon 1 minus z inverse plus c 1 upon 1 minus r 1 z inverse
plus c 2 upon 1 minus r 2 z inverse up to plus c p upon 1 minus r p z inverse and therefore y of n t the inverse z transform will give me what a k times c zero because one over z inverse is simply the z transform of ah step function unit step function plus c one we will take the inverse laplace and this would be e to the power n ln r one please refer to our previous lecture you will get this plus c c2 e to the power n ln r2 up
to plus c p e to the power n ln r p fine so the first term on the right hand side is bounded because c 0 is a constant now every term rest of the of the terms have e to the power n l n r i right ah in general r i may be complex so therefore let me write this as e to the power n ln the modulus of r e to the power i theta ok and therefore this would result in e to the power n e to the power i theta ah
simply i theta i theta e to the power l n modulus of r ok so now what we see here is that we have ah e which is raised to the power lnr and what is lnr ah r modulus is z one squared plus z two squared root over right so if i draw a plane in which i have z one the real part on x axis and z two the imaginary part on the y axis then i have a unit circle ok and then for e to the power ln r to be less than
1 or to be less than ah sorry for lnr not e to the power l for lnr for ln r to be less than 0 you need r to be less than 1 right the log of quantity less than 1 is negative so therefore in this region as long as you have z1 z2 in this region your system is stable otherwise the system is unstable and i can draw a corresponding plot so this is for discrete and what will be the corresponding plot for continuous you have z one you have z two all of this
has the real part which is less than zero so therefore you have system which is stable here so you can always draw the complex plane and locate the poles there for the case of continuous systems the poles lie on the second and third quadrant the system would be stable for the case of a continuous system you draw a circle of unit length and if the pole lies inside it or even on it you will have stability otherwise you will not have stability so ah this is not this is what ah what we have we had
to study we which we had planned on studying we started with ah the analysis of the systems in transform domain studied everything about the continuous domain analysis then we went to the discrete time domain analysis we studied the effect of the form of ah transfer functions on the dynamical features the speed of dynamics basically and today we also looked into the ah effect of that features of ah the transfer function on the stability of the system ah in the next lecture which would be the final lecture we will recapitulate everything which we studied in this
ah in this particular course ah and for today we will stop here we meet again for the last lecture tomorrow till then good bye
