Lecture: 30

So, in the last class we were discussing about the various properties of Laplace transform. So, in today's class we will discuss about the inverse Laplace transform. So, let us start the discussion of today's lecture. Before going into the inverse Laplace transform, let us understand what we discussed in the last class related to the properties of the Laplace transform. First we discussed about linearity, where if the function is like a1 f1 t plus a2 f2 t, the Laplace transform of this would be a1 f1 s plus a2 f2 s. Scaling property if the function is f of a t, the Laplace transform of that function would be 1 upon a f s by a. So, means simply you will replace s by s by a. Now in case of time shift, if the function is like f t minus a and multiplied by unit function unit step function that is time shifted by a, then your Laplace transform would be e to the power minus a s into f s that is the Laplace transform of function f t. Now, then we discussed about frequency shift. So, it means that if we have a function like e to the power minus a t into f t, the Laplace transform of that would be f into f of s plus a. That means, that wherever we have s in case of the Laplace transform of f t, we will replace it by s plus a. Then we discussed about the time differentiation means that the derivative of f, if you have like df by dt, the Laplace transform of df by dt will be given by s f s minus f the value of function at 0. So, we can generalize also by saying that the n-th derivative of the function, if you have then you want to get the Laplace transform of n-th derivative of f. You can simply say that the Laplace transform of this function would be s to the power n f s minus s to the power n minus 1 f 0 that is the value of function at 0 minus s to the power n minus 2 first derivative of function and its value at 0 and so on. So, this will give you the Laplace transform for n-th derivative of a function. Then we discussed about the time integration, if you have a function the integral of that like 0 to t f t dt and you are asked to find the Laplace transform of the integral, its Laplace transform would be 1 upon s f s, where f s is the Laplace transform of function f t. Now, frequency differentiation, if you have function like t to the power n f t. So, its Laplace transform would be minus 1 to the power n d to the power n f s by d s n. So, particularly this function and this the Laplace transform in case of frequency shift and frequency differentiation will be used most frequently when we will do the inverse Laplace transform. So, all those the properties which we have discussed, we have to keep in mind because these will be used frequently in the calculation of inverse Laplace transform. Now, next the time periodicity, we discussed that if the function can be given by periodic function, the periodic series like f 1 t plus f 1 t minus t that means, time shifted by time constant t multiplied by unit step and so on. So, you can represent a periodic function as a combination of various functions and the first function is the value of this periodic function at first time period and then others will be like compilation of the first function with time shift conditions. So, this is time shifted by t, then time shifted by 2 t and so on. Now, if you are asked to find out the Laplace transform of f t, you will say f s is nothing but f 1 s upon e to the power minus t s, where f 1 s is the Laplace transform of first period of the function. We also discussed initial value and final value theorems. We calculated that if we want to find out the initial value of a function, the initial value can be directly given by the by finding the limit of s f s where limit s tends to infinity. Similarly, in case of final value theorem, we discussed that the value of function at infinity will be nothing but limit s tends to 0 for the function s f s. Now, these we have discussed in the previous class. So, let us start the discussion about the inverse Laplace transform. In this case, we are given the value of f s and now we need to transform it back to the time domain and obtain the value of f t. So, f s if you have f s in the general form like f s has numerator function that is N s divided by D s, where N s is the numerator polynomial and D s is denominator polynomial. Now, roots of N s equal to 0 are called the zeros of transfer function f s and roots of D s equal to 0 polynomial are called poles of function ah transfer function f s. Now, we have to keep in mind that f s is Laplace transform of one function which cannot necessarily be a transfer function of the function f. So, this we have to keep in mind, so that we are not confused with the transfer function and the Laplace transform. Now, to solve the ah to find out the ah the value of function f, we use partial fraction ah expansion method to break the function f s. So, when you break the function f s, you will break into simpler terms, so that you can easily find the inverse Laplace transform of f s. So, what are the steps are involved in the ah computation of inverse Laplace transform? We use two steps, one is we will first decompose the function f s into simple terms using partial fraction expansion, and then we find the inverse Laplace transform of each term which we have found in the first step. Now, there are three possible forms of the trans ah the Laplace transform. Now, let us try to understand how we can apply the above two steps in finding out the inverse Laplace transform. So, first form is when you have simple poles, that means, if you have a transfer ah Laplace transform f s which is ah represented as ah numerator polynomial divided by denominator, and where you can have the denominator as shown in the equation like if you can ah represent the denominator polynomial in the form of product of factors that is s plus p 1 into s plus p 2 into s plus ah p 3 and so on, then you can use the simple pole approach to find out the inverse Laplace transform. Now, what are the p 1 p 2 and p n? So, these are the simple poles of the Laplace transform function. Now, these poles are distinct. Now, if we assume that the degree of n s is less than the degree of d s, that means, the the highest degree of s in denominator is greater than the highest degree of s in numerator, we can apply the partial fraction expansion method to decompose the Laplace transform which was ah shown in the previous equation. So, this is the Laplace transform, and we want to decompose it. How will decompose? We can represent this particular function as decomposed term like k 1 upon s plus p 1 plus k 2 upon s plus p 2 and so on, where expansion coefficients k 1 k 2 to k n are called the residues of Laplace transform. Now, there are many ways through which you can find these values of expansion coefficients. The one the most popular technique is residue method. Now, in this what we will do? We will multiply both sides by s plus p 1. So, if you see the expression for ah Laplace transform which we decompose into various terms ah, so that you can solve it. The first term has the denominator s plus p 1. So, what we will do? First we will multiply both side by s plus p 1. So, when you multiply this will become s plus p 1 into f s equal to you will get a constant term k 1 plus the other terms where s plus p 1 would be in numerator. Now, if you set the value of s is equal to minus p 1 at both sides, the right side terms other than the k 1 will become 0. So, you can simply say that the value of k 1 is nothing, but s plus p 1 into f s the whole product of ah this would be calculated at s is equal to minus p 1. So, when you solve you will get simply the value of k 1. So, now, in more general form you can simply write for any factor k i ah you can write s plus p i into f s and the whole product will be evaluated at s is equal to minus p i. So, this will give you the value of constant term k i. Now, this particular ah property is called the Heaviside theorem. Now, once the value of k i's are known we can easily find out the inverse Laplace transform for f s. How? Because now we know that the it is ah decomposed into this form. Now, if you see this this is nothing, but the Laplace transform of k 1 e to the power minus p t p 1 t. So, this is nothing, but frequency shift property. So, if you recollect the frequency shift property you can simply find out the value of inverse Laplace transform of first term. So, this will become k 1 into e to the power minus p 1 t and similarly for other terms also you can find out the inverse Laplace transform. So, finally, when you club all of them you will get the inverse Laplace transform of the function ah f s that means, the inverse Laplace transform of f s is f t which can be given by this expression. Now, second case is that when you have repeated poles means suppose if the Laplace transform f s has n repeated poles at s is equal to minus p, then what you can write? You can write simply f s is nothing, but k n upon s plus p to the power n plus k n minus 1 upon s to the power s plus p to the power n minus 1 and so on. And the last f 1 s is the remaining part of f s which does not have pole at s is equal to minus p. So, this is the remainder of the ah the Laplace transform which is not represented in this particular form. Now, as we did in the previous case, in this case also we can find out the value of k n. So, k n would be given at as if you multiply s plus p to the power n at both sides and equate it to ah the the and put the value of s is equal to minus p, you will simply get the value of k n. Now, how to find the value of k n minus 1? For finding out the value of k n minus 1, we have to multiply each term by s plus p to the power n as we did in this case. So, if you multiply by s plus p to the power n, you will get k n plus k n minus 1 into s plus p and so on. Now, in this case you will have k n as a constant term, but we need to find out k n minus 1 because k n we have already calculated. So, what we can do? We can ah get rid of k n if you differentiate the whole expression. So, what we have to do? We have to first find multiply it by s plus p to the power n and then differentiate the whole expression to get rid of k n. So, finally, you can put the value of s is equal to minus p in the final expression and you will simply get the value of k n minus 1. So, what you can ah simply represent k n minus 1 is nothing, but the d by ds of s plus p to the power n f s and the whole expression will be evaluated at s is equal to minus p. Similarly, for others also you can calculate for k n minus 2 it will be 1 upon 2 factorial d 2 s by ds 2 ah the double ah the second order derivative of s plus p to the power n f s and evaluated at s is equal to minus p. So, in general for mth term you can simply say that it is ah k n minus m. So, you are finding out the value of constant term k n minus 1 k n minus m and it will be given by 1 upon m factorial and then mth derivative of s plus p to the power n f s and then the whole expression will be evaluated at s is equal to minus p. So, now, you have all the constants in your hand. Now, if you remember the inverse Laplace transform of 1 plus ah 1 upon 1 plus a to the power n is divided by t to the power n minus 1 e to the power minus a t upon n minus 1 factorial. So, that is nothing, but with the help of frequency ah differentiation ah property you can find out the ah value of inverse Laplace transform of 1 upon s plus a to the power n. Now, using this you can find out the inverse Laplace transform of ah the ah transfer function given as f s. So, if you compile the individual terms for first one you can simply write k 1 into e to the power minus p t for second you can write k 2 into t e to the power minus p t and for third term similarly you can write k 3 upon factorial 2 into t square e to the power minus p t and so on. So, for the kth term you will for the nth term you will write k n upon n minus 1 factorial into t to the power n minus 1 e to the power minus p t plus f 1 t that is the inverse Laplace transform of remainder of the function that is f 1 s. So, with this way you can find out the inverse Laplace transform of function f t. Now, we have a third case where we have complex poles. Now, pair of complex poles is simple if it is not repeated and if it is repeated then complex poles have double or multiple poles. Now, simple complex poles may be handled the same way we handle the simple real poles, but since we have a complex algebra involved the result may be a little bit cumbersome. So, we will use an alternative approach to find out the inverse Laplace transform in case we have complex poles. So, the approach is the method which is known as completing the square. Now, what does it mean? This is the idea to express the each complex pole pair in the denominator as a complete square such as s plus alpha square plus beta square. So, whatever we have in the denominator we can represent them as ah set of complete square plus remainder of the term which are there in the denominator. Then we can easily find the inverse Laplace transform of the function f s. What does it mean like ah how we will solve it? Let us take one simple example. So, that you can understand how we can proceed for this kind of condition if ah as n s and d s always have real coefficients. So, the complex roots of the polynomial with real coefficients will always occur in conjugate pairs. So, we will keep this in mind and try to solve the ah try to find out the inverse Laplace transform of f s. Suppose Laplace transform f s is ah given to you and you want to ah find out its inverse Laplace transform. So, the f s function you can decompose into two forms two parts. First is like if you can represent the section where you have complex poles as a 1 s plus a 2 divided by s square plus a s plus b plus f 1 s where you do not have any complex poles. So, f 1 s is the remaining part of f s which does not have pair of complex poles. Now, what you have to do next in the denominator you have term called s square plus a s plus b you need to complete the square in the denominator. So, we add few ah terms in the polynomial and ah subtract few of them in the polynomial and rearrange in such a way that it looks like s square plus 2 alpha s plus alpha s square plus beta square. So, if you can rearrange the polynomial in such a way you can simply say that this particular polynomial can be represented as s plus alpha s square plus beta square. Similarly, in the numerator also you can arrange the polynomial in such a way that it looks like a 1 into s plus alpha plus b 1 beta. So, when you rearrange numerator as well as denominator what you get you can simply represent ah the function f s as a 1 s plus alpha upon s plus alpha square plus beta square plus b 1 beta upon s plus alpha square plus beta square. So, this is what you got in the numerator and in the denominator we have s plus alpha square plus beta square. So, if you rearrange this term and this term in the expression you can say that this term is a 1 l s plus alpha upon s plus alpha square plus beta square plus b 1 beta upon s plus alpha square plus beta square and then rest of the terms which does not have complex pole. Now, if you see the first term if you remember when we ah when we were discussing about the Laplace transform ah we discussed that if we have ah the if you want to find out the Laplace transform of cos omega t it becomes s upon s square plus omega square. So, using that expression for cos omega t that is the you can find out the inverse Laplace transform because if you see this expression this is nothing, but the frequency shift property plus the cos beta t term. So, what you can simply write you can simply write ah the inverse Laplace transform of this would be a 1 into e to the power minus alpha t because here you have s plus alpha. You will use frequency shift property and you will say that it is e to the power minus alpha t into the function f t. So, which is cos beta t here similarly for this case also you can write a 2 e to the power minus alpha t into sin beta t because here also you will see that ah this particular segment is nothing, but ah the Laplace transform of sin beta t plus the frequency shift because here you have s plus alpha. So, you will simply write a 2 e to the power minus alpha t sin beta t plus some function f 1 t which is the inverse Laplace transform of f 1 s. So, when we have complex poles you can rearrange the expressions in such a way that it can be arranged ah in a particular fashion then you can easily find out the inverse Laplace transform. So, whether the pole is simple repeated or complex the general approach that can always be used in finding the expression is the method of algebra. That means, that whatever did in algebra the same can be applied here also if you want to find out the the constant components in any case of the Laplace transform which we discussed till now. So, to understand these concepts let us understand with the help of one example. So, that you are more clear about how to proceed with finding out the inverse Laplace transform. So, let us take one example we have been given a Laplace transform of f t which is given as s square plus 12 upon s into s plus 2 into s plus 3. So, here you can see in the denominator you can clearly distinguish all 3 ah poles that is poles are at s is equal to 0 at s is equal to minus 2 at at s is equal to minus 3. So, this particular expression can be represented as a by s plus b upon s plus 2 plus c upon s plus 3. Now next task is that you need to find out the value of the unknown constants which are a b c. So, we can use the residue method which we discussed just now. So, in this case we will first multiply the ah Laplace transform with s and equate it for s is equal to 0 we will simply get which is 2. Similarly for finding out b we multiply the Laplace transform with s plus 2 and equate and find out the value of the complete function at s is equal to minus 2 and we get the value of b as minus 8. Similarly for c we multiply with ah the Laplace transform by s plus 3 and find out the value of complete expression at s is equal to minus 3 and when we simplify we get the value of c as 7. So, with this you can easily find out the inverse Laplace transform. Another method is that you can use simple algebraic method which you generally use for solving the ah general ah expressions. So, here we will apply those ah technique to find out the inverse Laplace transform. So, what we can do we can multiply both sides by s into s plus 2 into s plus 3. So, what we get? So, if you see the expression here s square plus 12 is there in the numerator and in the denominator we have ah s into s plus 2 into s plus 3. So, if you multiply ah the both sides by this what you get you will say that the s square plus 12 is nothing, but a into s plus 2 into s plus 3 plus b into s into s plus 3 plus c into s into s plus 2. So, if you see here if you ah multiply both sides by s into s plus 2 into s plus 3. So, what you will get? The denominators will cancel out and the numerator what you get is s square plus 12 is equal to a into s plus 2 into s plus 3 plus b into s into s plus 3 plus c into s into s plus 2. So, this is what we got from the ah multiplication. Now, what next you have to do? You have to just equate the coefficients of like powers of s. So, for constant terms if you equate the only constant terms you will simply get the value of a that is equal to 2, when you equate ah only the ah terms having s you can you will get the value of ah expression you will get the equation that is 3 b plus 2 c is equal to minus 10. And when you equate the terms of s square you will get another equation that is b plus c is equal to minus 1. Now, you have 2 equations and 2 unknowns you can solve and you will get the value again as a is equal to 2 b is equal to minus 8 and c is equal to 7. So, either you use this method or the previous method which we discussed that is the residue method you will get the same results. And finally, for the ah given Laplace transform f s you can ah get the inverse Laplace transform ah by simply decomposing you can easily say that the inverse Laplace transform of f s is this is 2 into unit step function minus 8 into e to the power minus 2 t plus 7 into e to the power minus 3 t provided the time t is greater than 0, because this is applicable for t greater than or equal to 0. So, now, you can ah easily understand that how you can proceed with finding out the inverse Laplace transform. So, with this we can close our ah today's session where we discussed about how to proceed with finding out the inverse Laplace transform. So, next week we will ah continue our ah utilization of Laplace transform in analyzing the various circuits. So, remember if we ah when we discuss the ah first order second order circuits, we saw that there were ah various ah second order and first order differential equations which are sometimes difficult to solve. So, we will use the Laplace transform techniques and see how we can easily solve those kind of equations when we ah see the second order or first order circuits. Thank you . .