 A warm welcome to the 12th session in the 4th module of signals and systems. We are all eager to learn how to deal with the most frequently encountered impulse responses. Let us review what kind of impulse responses are most frequently encountered in natural engineering systems, the poly-X response. Let us write it down again. A polynomial plus exponential or polynomial multiplied by exponential. What is the general form? The general form would be something like I am showing the continuous time form. Summation L going from 0 to m a L, t raised to the power L into e raised to the power minus alpha t times u t. Now, there are several variations here. One is that alpha is called the exponential parameter which can be real or complex. Now, if it is complex then it has a real part and imaginary part or if you are talking about the other way of describing you could have a magnitude and a phase. Here in the context of continuous time, we will think of it as real part and imaginary part. So, alpha is alpha r plus j alpha i. Now, you know the alpha r part is alright, but the alpha i part would give you a complex part in the expression. So, if the overall impulse response is real, this is just one of the terms remember. If the overall impulse response is real then alpha must come together with alpha complex conjugate. So, in other words, if you have one term of this form, you must have another term of the form which I am going to write now. So, essentially what we are saying is the complex conjugate exponential parameter is also needed. Also, here we have taken what is called a right sided function or a right sided signal. It could also be left sided. An example of a left sided signal would be as follows or a left sided poly x term. So, this u of minus t here is non-zero to the left of a certain point. Here of course, it is t equal to 0 here and therefore left sided. In general, we could have the following form. So, essentially a right sided signal is 1 which is non-zero after a certain point on the t axis. So, there is the other way of saying it is you can find the point t equal to t 0 say or that could just be t equal to 0 because you are free to fix the 0 where you like. You know you can fix the 0 where you like, but you cannot do it if there are multiple poly x terms because they may not have the same point from which they start. So, the statement that you can fix your 0 where you like is meaningful to the extent that you are dealing with just one term. Anyway, so a right sided signal is 1 which is non-zero after a certain point on t or the other way of saying it is it is 0 all before a certain point and a left sided signal is 1 which is non-zero before a certain point in t and completely 0 after. So, you could have right sided or left sided signals. Unfortunately, you cannot have signals at last all over time. We saw that if we had e raised the power of 2 t for all t it has no Laplace transform in the general sense of Laplace transforms using function you are not using impulses and things like that. So, if one wants to deal with functions in the s plane or in the z plane then exponentials lasting all over time have no Laplace transform or z transform. So, you need either left sided or right sided signals. Now, we already know how to invert if I encounter something like 1 by s minus 2 or 1 minus or 1 by s minus alpha you know this kind of thing we have been doing extensively. What would be the Laplace transform if I multiply by a polynomial we have multiplied by a polynomial of degree 0 a constant multiplied by 1 by s minus 2 or 1 by s minus alpha is no problem. What happens when I want to multiply by t or t squared for that we need to now establish property 5 the next property of the Laplace transform in the z transform let us do that. So, we need to establish what is called the property of differentiation in other words let x of t have the Laplace transform capital x of s with the region of convergence r x the question we are asking is what happens if I differentiate in the s domain. So, what is it that has a Laplace transform d x s d s with almost the same r o c we will explain what we mean by almost the same except some isolated contours perhaps. Let us write down x s in terms of x t and of course, there is an r x associated we will keep that in the background. Let us differentiate both sides with respect to s. Now, this is the derivative with respect to s if this function is differentiable we hope that we can operate only on that part of the integrand which is a function of s. So, I can take the derivative into the integral sign let me multiply both sides by minus 1 and what do I have here I have this is essentially the Laplace transform of t times x t as you can see. So, now, if you compare this is the Laplace transform of t times x t. So, now we can go back and answer the question that we raised just a few minutes ago firstly I put a minus sign here and then I answer this question by saying t x t here multiplication by t amounts to taking the derivative with respect to s and then negating. Let us take an example let us take e a is the power minus 2 t u t for a change not plus 2 t we know Laplace transform its Laplace transform is 1 by s plus 2 with the region of convergence real part of s greater than minus 2. What is the Laplace transform of t times this? The answer would be the negative of the derivative of this expression with essentially the same r o c essentially real part of s greater than minus 2 here we do not need to worry what is this derivative and then we need to multiply by minus 1. So, minus t d s this becomes plus 1. So, very simple that means the Laplace transform of this is 1 by s plus 2 squared. So, now we are quite competent to deal with poly x terms. In fact, now at least we are well posed to find out the Laplace transform for poly x term. Let us put down a procedure let us take the example that we took some time ago summation l going from 0 to m a l t raise the power l e raise the power minus alpha t u t. Now the Laplace transform of this is the sum of the Laplace transforms of the individual terms. Now each of them can be obtained by using the differentiation property repeatedly. So, e raise the power minus alpha t u t let us begin with that. So, e raise the power minus alpha t u t has a Laplace transform given by 1 by s plus alpha with the real part of s plus alpha greater than 0 or real part of s greater than minus real part of alpha or rather well it would be 1 by s plus alpha and yeah for example, if it were e raise the power minus 2 t then you would have real part of s greater than minus real part of alpha that is right minus real part of alpha this is all right. Now the same region of convergence essentially holds and you keep differentiating repeatedly with respect to s every time multiplying by minus 1 and you get additional multiplications by t and of course, multiplication by a constant is no problem the Laplace transform is linear. The Laplace transform is additive and homogeneous there is no problem. So, now we are well equipped to study the Laplace transform of a poly x term. So, at least we are now competent to deal with a large number of engineering systems which have such impulse responses. All the typical networks electrical networks or mechanical systems using masses and frictional parts and springs would typically have these kinds of responses. So, it is very clear that we now know how to recognize what happens when you have a square or a cube when you have something like 1 by s plus 2 squared we know where it comes from it comes from a poly x term and we can recognize it. So, we can now go back and ask the question how do I invert based on this knowledge? We shall see that in the next session. Thank you.