Lecture: 29

[Music] [Music] namaskar so in last class we were discussing about the few major properties of the laplace transform so in this class also we will continue our journey on ah discussing the few important properties of the laplace transform so let's see what we discussed in the previous class related to properties of the laplace transform we discussed about linearity so in that we demonstrated that the if the laplace transform of f1 t and f2 t are f1 s and f2 s so their linear combination like a1 f1 t plus a2 f2 t if you take the laplace transform of the linear term you will get the laplace transform like a 1 f 1 s plus a 2 f 2 s similarly for scaling the laplace transform of f of a t will be nothing but equal to 1 by a into the laplace transform of f t where you will replace s by s by a so this would be laplace transform would be nothing but the function of s by a now in case of time shift if your signal is shifting by sometime say a the laplace transform of that function would be e to the power minus a s into fs so these three we discussed in the last class now let us talk about the frequency shift now if f s is the laplace transform of f t then laplace of e to the power minus a t multiplied by f t would be if you take that function inside the integral so this is basically the standard definition of our laplace transform that is integration between 0 to infinity function f t into e to the power minus s t d t so instead of f t we will write e to the power minus 18 into f t so this will become f t into e to the power minus s plus a t d t so which is nothing but the laplace transform having s being replaced by s plus a so you can say that in case of frequency shift the laplace transform of e to the power minus a t f t can be obtained from the laplace transform of f t by replacing every s with s plus a so this particular property is called as frequency shift or frequency translation now let us take an example suppose the cos omega t is a function then its laplace transform would be s upon s square plus omega square similarly for sine omega t the laplace transform is omega upon s square plus omega square now if we use the shift property and we need to obtain the laplace transform of the damped sinusoid that is damped sine and cosine functions means we are multiplying cos omega t and sine omega t with e to the power minus 80 then its laplace transform can be given by simply replacing s that is there in the laplace transform of cos omega t will replace s by s plus a so in numerator we will replace s by s plus a in denominator it will become s plus a whole square plus omega square similarly in case of e to the power minus a t sin omega t it will become omega upon s plus a whole square plus omega square now let's talk about the time differentiation if f s is the laplace transform of f t then we need to find out the laplace transform of its derivative that is the laplace transform of d f by dt so we use the standard definition of laplace transform the integral from zero minus to infinity d f by d t e to the power minus s t d t now how you will solve this particular type of integral we will use the integration by parts method what does this method says suppose there are two functions u and v so integral of u v d x can be given as u into integral of v d x minus integral of derivative of u in bracket it is the integral of v dx and multiplied by the derivative of u and the integral of v d x and then the combined function will be having the integral with respect to x again so this is the property of integration by parts so the same property we will use for finding out the integral of this particular equation so here we will assume u is nothing but e to the power minus s t and v is tf by dt so then laplace of df by dt can be given as so you will take first the derivative of u will take out that is e to the power minus s t is out and then v dx integral that will simply become f t and the integration will be calculated between 0 minus 2 infinity then the second one you will integrate between 0 to infinity u dash that is d u by d t is given as minus s into e to the power minus s t and then again integration of v d x integration of v d x would be nothing but function f t and then the integration point that is dt so [Music] now if you see this particular expression when the value is infinity this term will become 0 so the whole function will be 0 so at infinity the value will be 0 for 0 the e to the power minus s t will become 1 so at 0 the value of function at 0 will be the value plus because minus s you can take it out take it out because s is constant when you are differentiate integrating with respect to t so this will become plus s and then within the integration what term you are left with is f t e to the power minus s t d t which is nothing but the laplace transform of function f t so finally what we get from here is the laplace transform of df by dt is nothing but sfs minus the value of function at 0 that is just before 0 so now similarly you can take the second derivative of f t also so what you will get in case of the laplace transform of d 2 f by d t 2 you will get s square f s minus s the value of function at 0 minus derivative of f at t is equal to 0 so this is what you get for second derivative laplace of the second derivative of the function if you continue in this manner you can create more generic laplace transform that is for nth derivative of f t which you can write as laplace of d 2 n f by d t n is nothing but equal to s to the power n into f s minus s to the power n minus 1 the value of function at 0 minus s to the power n minus 2 the value of derivative that is first derivative at 0 and so on up to the value of derivative n minus 1 a derivative at 0. so this is your generic expression for finding out the laplace transform of d2 and dnf by dt n now let's talk about the time integration now if fs is the laplace transform of f t then laplace transform of its integral so let us take the function that is now integral of f t d t so when you use the standard definition of laplace transform this will be the integration between 0 to infinity then function function is here 0 to t integral f x d x so multiplied by e to the power minus s t d t now you will use again the same property that is integration by parts let us assume u is nothing but 0 to t of x dx and v is e to the power minus s t so when you use and you use the integration by parts property you will get the function that is 0 to tfx dx into 1 minus s e to the power minus st that is the integration of e to the power minus st and then the value between 0 to infinity minus 0 to infinity minus 1 by s e to the power minus st into ft dt now if you see carefully this particular expression what you will see when the term t tends to infinity let us talk about the first expression which first term which you have in this expression so if t tends to infinity then this value will tends to zero and again when it is when time t is equal to 0 this integration will become integration between 0 to 0 f x dx which will again be 0 so it means that this particular term that is first term of the expression is 0 between both of the limits so we can ignore this particular term so finally what we left with is the integration of the laplace of integration between zero to t f t d t is nothing but one by s zero to infinity e to the power minus s t f t d t and this is nothing but the laplace of f t so what we get finally that the laplace of the integration between 0 to t f t dt is nothing but 1 by s f s now let us talk about the frequency differentiation now if f s is the laplace transform of f t so that is by its standard definition you can represent f s is nothing but 0 to infinity f t e to the power minus s t d t now if you take the derivative of function f s with respect to s what you will get so if you differentiate both sides the df s by ds will become here if you differentiate it will become 0 to infinity minus t because now you are differentiating with respect to s so it will become minus t f t e to the power minus s t d t so which is nothing but the laplace transform of the function minus t f t so what you can write that laplace transform of t f t is nothing but equal to minus of the differentiation of laplace transform f s with respect to s so more generic you can write like if you are taking the laplace transform of t to the power n f t its laplace transform would be minus 1 to the power n d n f s by d s n so this expression will give you the the frequency differentiation property now let's talk about the time periodicity if a function f t is a periodic function which is shown in the figure it is periodic with period t so it can be represented as the sum of time shifted functions how let's see in the next slide so if you have this particular function you can divide this the periodic function into three parts so one is say f one t which have a period between zero to t second is f two t which has a period between t two two t and third is having the period between two t to three t now if you are compiling the function ft you will say that this is a summation of f1t plus f2t plus ft3 or if if it is continuing then so on so what you will write you can simply write in terms of the step functions so you will simply write f1t as f1t but f2t you can write t minus t that is the time shift function that function f1 is shifted by time t into unit step function t minus t similarly for f3 you can write f 1 t minus 2 t because it is a time shift function with time equal to 2 t so you can write f 3 also in terms of f f1 and so on so for f1t what is f1t f1t is nothing but the function ft with limited in the period that is limited by u t minus u t minus t so now if you are asked to find out the laplace transform of the periodic function that is f s what you will write you will write the laplace transform of function f1 t that is f1s then we will use the time shift property and we will write the laplace transform of the second express second term that will become f one s e to the power minus t s for third term it will become f one s e to the power minus two t s and so on so what will happen you will finally get one expression which is f one s and within bracket you will have you will get one plus e to the power minus t s plus e to the power minus two t s and so on so now if you sum up the series that is 1 plus x square plus x cube and so on the value of the sum of this particular series is 1 upon 1 minus x provided the mod of x is less than 1 so now if you use this particular property you can simply write the value of fs that is the the laplace transform of a periodic function what we can write we can write fs is nothing but f one s upon one of one minus e to the power minus t s so what we can observe from this we can observe that f one s is the laplace transform of f one t that means it is the laplace transform of function defined over its first period only now what we get is the f 1 s if you want to find the value of fs that is the laplace transform of complete periodic function so what we have to do we have to multiply with the value of 1 upon 1 minus e to the power minus t s with f 1 s then we will get the laplace transform of complete periodic function or we can say that if we divide the f one s by one minus e to the power minus t s will get the value of laplace transform of a periodic function now let's talk about two important properties that is basically will say as ah initial value and final value theorems so these two theorems are very important in finding out the initial and final values of function so that means this allow us to find the value of function at time t is equal to 0 and the value of function at infinity and we can find out these two values directly with the help of laplace transform of ft that is fs how will get let us try to understand now let us take the differentiation property of the function that is the laplace of d f by d t so what we write we write laplace of d u by d t is nothing but s f s minus the value of function at 0. now if we let s tends to infinity then the integrand in the above equation will vanish because of the damping exponential factor so if you put the value s as infinity this will become 0 so the whole of the expression ah to the right of laplace of d a by d t will become 0 so when it will become 0 you can simply write it as limit s tends to infinity s of f s minus the value of function at 0 equal to 0 or you can write because this value will be a constant value so you can simply write limit of s tends to infinity s f s is the value of function at time t is equal to 0 so this particular property is known as the initial value theorem now the in case of final value theorem let us take the same property of the laplace transform for the df by dt so laplace of d f by d t is nothing but s f s minus f 0 so that is nothing but the value of the integration between 0 to infinity d a by d t e to the power minus s t d t now if s tends to zero what we can write limit s tends to zero s of s minus f zero equal to zero to infinity df by dt e to the power 0 t so this will become now 1 so that means you can simply write the this particular expression is nothing but integration between 0 to infinity df or you can write the value of function at infinity minus value of function at 0. now you have value of function at 0 at both sides those both will cancel out so finally what you will get the value of function at infinity is nothing but limit s tends to 0 s f s so this particular expression is called as a final value theorem so if you compare both of them here when s tends to infinity the value of function s f s will give you the initial value while in case of final value f infinity the limit will tends to zero and you will find the value of function s f s when limit s tends to zero so you can simply remember it by understanding that when s is tending to infinity means the value which you will get will be at origin and when l when s tends to 0 the value which you will get for the function is at infinity so these two are very important theorems but the final value theorem has its own limitation what are those limitations the final value theorem will be valid only when all poles of fs are located in the left upper of s plane what do you mean by poles let us take this particular function that is the laplace transform of e to the power minus 2 t sine 5 t so here if you are asked to find out the poles of f s what you will do you will simply take the denominator and equate it to 0 and you try to find out the roots of this particular expression so since it is a two second order equation you will get two roots of s so those two roots will be nothing but the poles of fs so what you need to do you have to observe whether the value of those roots are having any negative real part or not if poles are having negative real part then only you can apply the final value theorem in the function fs the only exception in this case is the case when f s has simple pole at s is equal to 0 means if this expression is having another component like 1 by s then you will have 1 pole at s is equal to 0 so that is the only exception in in finding out the final value theorem so these two things you have to always keep in mind when you are asked to find the value of final value of the function f t that is f infinity now let's take the example the function is f t e to the power minus 2 t sin 5 t its laplace transform would be 5 into s plus 2 whole square plus 5 square so the final value theorem if you apply the value of f at infinity will be 0 because limit s tends to 0 5 s divided by s square plus 4 s plus 29 will be 0 and this is what you can see from the function f t also because you have a exponentially decaying component with the function so final value will always be zero now let's take another example let us take the function is sine t so its laplace transform would be f s is equal to one upon s square plus one now if you apply final value theorem limit s tends to zero sfs you will get the value 0 but this is wrong why because function f t that is sin t will always oscillate between plus 1 and minus 1 and it will not have limit as t tends to infinity now the final value theorem cannot be used in this case because if you take the poles that is the roots of this particular second order equation you will see the poles as the value s equal to plus minus j which are not in the left half of s plane because these two poles doesn't have any negative real value so that means that in this case you cannot apply the final value theorem so you have to always keep in mind where you can apply the final value theorem where you cannot use the final value theorem so you can summarize that the final value theorem does not apply in finding the final values of sinusoidal functions because these functions are these functions are oscillatory in nature and does not have the final values now initial values and final value theorems shows the relationship between origin and the infinity in the time domain as well as in the s domain and these two theorems also serve as a useful check on laplace transform so now let us see few of the example so that we can understand what we discussed till now more clearly let us see there is function f t which is a combination of a unit impulse unit step and the exponential function so what you will use you will use linearity property here and find out the value of the laplace transform of f t so let us apply the linearity property the laplace of delta t will be 1 laplace of the 2 you can take out laplace of u t will be 1 by s and then the laplace of e to the power minus 2 t will be s 1 upon s plus 2 so now you know the laplace transform of all these three terms you will simplify then you will get the laplace transform of complete function f t that is s square plus s plus 4 divided by s into s plus 2 now let's take another example if function f t is t square sine 2 t multiplied by unit function u t so now what we know is that the laplace of a periodic function that is sine 2t is nothing but 2 upon s square plus 2 square that is omega upon s square plus omega square this is what we know about the laplace of sine function so this will become 2 upon s square plus 2 square now what you have to do you have to use the frequency differentiation because you have a t square term so what you have to do you have to double differentiate this particular x ah term that is 2 upon s square plus 4 so and then multiply by minus 1 whole square and then when you simplify you will get the value of the laplace transform of t square sine 2t and what we get is 12 s square minus 16 upon s square plus 4 whole q now next is if we are asked to find the laplace transform of a gate function so what we can see here this is a gate function which starts from 2 and completes at 3 so you can say that this particular function is is combination of two unit step function how you can say of this function gt is nothing but 10 because 10 is the amplitude multiplied by unit step function u delayed by two time periods so u t minus 2 so you can say there is one function like this and minus of u t minus 3 means there is another function which is like this so when you will say when you combine both of them minus of u t minus 3 you can write so what you will say that the laplace transform of u t is a combination of two unit step function one is delayed by two time period and second is delayed by three time period so what you can write now you can just take the laplace transform of both of the unity step function delayed by their respective time period so the laplace transform of gt will become gs that is nothing but 10 into e to the power minus 2 s by s minus e to the power minus 3 s by s when you simplify you will get the value of gs that is 10 by s e to the power minus 2s minus e to the power minus 3s so now with this we can close our today's session in this session we discuss about various properties of the laplace transform and the next class we will discuss about the calculation of inverse laplace transform thank you [Music] [Music] [Music] you