 A very warm welcome to the first tutorial on this course, digital signal processing and its application. I am Aniket Sadashiv, I am an undergraduate student in the Department of Electrical Engineering IIT Bombay. I will be taking this tutorial in which we will go through certain interesting problems of the weeks 4, 5 and 6 of this course. So without further ado, let's start. So at first we will look at an example of using Parseval's theorem to evaluate certain integrals, namely the integrals which involve the sinc function. For example, currently we will try to evaluate the integration of sin x by x whole square dx over the entire real line using Parseval's theorem. So let us quickly recall what Parseval's theorem says. So it basically relates the two inner products in time to their inner products in the transform domain or the frequency domain. Let me call them x hat of omega and y hat of omega or if we convert the inner products to integral form, we can write it like y hat of omega d omega. Now we will use this Parseval's theorem to obtain the value of this integral. Now here I have used x, but so now let me take my x of t equal to y of t as sin of t by t and we know that the inverse Fourier transform of sin of t by t is this rectangle function with bounds as minus 1 and 1 and the amplitude as pi and this is in the omega domain or the transform domain. Now you can see that integrating this rectangle function is way more easier than integrating this function and that is what we are going to use. So integral of sin t by t the whole square dt over the entire real line is nothing but integral of minus 1 to 1 this pi times 1 because the rectangle function is 1 in minus 1 to 1 times pi times 1 because we have two functions x t and y t. So both will give 1, 1 rectangles and a 1 by 2 pi constant outside followed by d omega. So this is nothing but 1 by 2 pi into pi square into 1 minus minus 1 which is equal to pi by 2 times 2 which is equal to pi. So the value of this integral is equal to pi. Now we will move on to the next question. So this is week 5's first problem. So in this we have been given a stable and a causal system with impulse response as h of n and a rational system h of z. So and we are given that h of z has a pole at z equal to half and it also has a zero somewhere on the unit circle and about the other poles we have been given no information and we have to select the true statements amongst all these. So from the question we can conclude that the system will have all the poles inside the unit circle because the system is stable and causal. Also we can directly select that option a is correct why because we have been given that there is one pole at 1, 0 on the unit circle. So for that 0 on the unit circle we can say that h of e to the power j omega will be 0. Now we have to conclude something related to the duration of h of n like if h of n is finite or infinite which we will see how to deal with. Now before that a quick recap on poles and zeros at infinity like what do we mean by a pole or zero at infinity. So this is essentially related to the degree of numerator or denominator in case of a rational function. So a rational function will have both a numerator polynomial let me denote it by n of z and a denominator polynomial. Now if the degree of numerator n of z is greater than degree of d of z for example we have something like let me use z, z plus 5, z square plus 5 z plus 1 by z plus 1. Here we can see that degree of numerator is 2 and degree of denominator is 1. So for such a case we can say that there will be a pole at infinity why? So let me consider a simple example. So this is the case where you will have a pole at infinity. So for example let us take the simplest possible case something of the form z minus 4. So one thing which is obvious is it has a zero at 4. Now if we look at the behavior of this z minus 4 as z tends to infinity we can say that this function also blows up. And also we know that when we encounter a pole the value of function blows up. So in a way this z as it tends to infinity it behaves like a pole. So we say that it has a pole at infinity. Similarly if the degree of numerator is less than degree of denominator we will have a zero at infinity. So again we will consider a similar example 1 by z minus 4. Now we can clearly see that it has a pole at 4 and what happens as z tends to infinity? At z tends to infinity this thing, the entire thing, the denominator increases and the entire thing goes to zero. So we say that this thing as z tends to infinity behaves like a zero. So we say that it has a zero at infinity. Now there is this claim that if the sequence is a finite duration sequence then the region of convergence is the entire z plane except we include the possibility of z equal to zero or z equal to infinity. So let us say we have a finite duration sequence and let us say it is between two natural numbers or integers n2 and n1. Now we can have multiple cases. So one case is your n1 is greater than zero, n1 is strictly positive. So for xn if we want to write the z transform we can write it like summation x of n z to the power minus n. Now we know that n1 is greater than zero. So that means that x of n will have some finite value at n1. So let us consider that point for now. We will have other summation terms x of n1 z of minus n1 plus yeah. So we have this. Now what will happen as z tends to zero? So for this thing you can see that as z tends to zero this thing blows up it goes to infinity. So we can say that z equal to zero cannot be in the ROC of this case. So for z equal to zero we cannot include it in the region of convergence of this finite duration sequence in the case when n is greater than zero. So z equal to zero is a problematic point for this sequence. Now let us consider the case where n2 is less than zero. So we have something like this or it can be negative also but there is some value at n2 that the sequence has. Now again if I write the z transform and I only consider the point at n2 so x at n2 it will have z to the power minus n2. Now I know that n2 is less than zero. So this negative thing will be positive minus n2 will be positive. Now again we can see that as z tends to infinity since this thing is positive it will again blow up. So we cannot include z tends to infinity in the region of convergence. So that is why if we have a finite duration sequence the ROC can be the entire z plane except z equal to zero or z equal to infinity as we saw in these two cases. Now another interesting fact is when is the ROC the entire z plane. You can see that when n1 is greater than zero we cannot have z equal to zero and when n2 is less than zero we cannot have z equal to infinity but we want to include both of these into our ROC. So the only point at which the function can take some positive value or some value in general is at z equal to zero and it cannot take any other value on any other points. So basically the only function or the set of functions for which the region of convergence is the entire z plane is the class of these functions where k is some constant. So for only this function since it takes value only at zero we can say that the region of convergence is entire z plane and for all other finite duration sequences either you will have problems at z equal to zero or z equal to infinity or both. Now let's move back to the problem. So now what we can directly conclude that h of n cannot be a finite duration because our system has a pole at z equal to half. So since it has a pole at z equal to half but we clearly saw that for a sequence to be a finite duration it can only have a pole at z equal to zero or infinity. So this pole at z equal to half make ensures that the sequence is not a finite duration so it has to be of infinite duration. So options c has to be correct. Now let us move on to the next part. So before that again we will quickly go through what happens when we convolve two causal sequences. So suppose xn and yn are two causal sequences that is x of n is equal to y of n is equal to zero for n is less than zero. Now what I am claiming is that the convolution of two causal sequences results in a causal sequence and we can quickly see that if we just write the definition of convolution xn convolved with yn is equal to this summation. This is the definition of convolution. Now we have been given that n is less than zero. So we have n is less than zero and we can also use the property that x of n the value of sequence x for negative indexes k will be zero. So we can run this summation from k equal to zero to infinity x of k y of n minus k because for negative k x of k will be zero so that will not contribute anything to the summation. Now when n is strictly less than zero look at this term when n is strictly less than zero and k is greater than or equal to zero. We can see that this entire thing n minus k this thing will be strictly less than zero. So since the argument of this thing is negative and we know that y of n is equal to zero for negative indexes so the entire convolution will be zero for n less than zero. So basically the convolution of two causal sequence when you take that you end up with a causal sequence again. So we have been given that g of n is equal to n times h of n convolved with h of n. So since h of n was given to be causal g of n will also be causal. Now if we quickly move back to the question we want to conclude about the stability of the system g. So before that let us look at how we are going to obtain the z transform of g. So we know that h of n convolved with h of n using the properties of z transform is nothing but h of z square because convolution in time domain results into product in the z domain. So g of n is n times this function h of n convolved with h of n. Now we know that if we multiply a sequence by n and if we take the z transform then it becomes this. So this is basically the multiplication of n in time domain results in this operation in the z domain. So we can quickly obtain the z transform of g. So let me denote it by capital G of z which is equal to minus z times derivative of h of z square which is equal to minus z into 2 h of z times derivative of. So this will be the z transform of g. Now since we know that h of z is stable all the poles will lie, let me change colors, all the poles will lie inside the unit circle or in another way we know that outside the unit circle the system will converge or the z transform will exist. Now what about the derivative of h of z? We need to focus on this term. So this term is stable so it will not cause any issues and again this only contributes a zero so this will again not affect stability because it is not contributing poles. So the only problematic term left now is derivative of h of z. So we need to deal with this. Now we know that the region of convergence of the derivative of a function for a rational function. Now here we will use the fact that h of z is rational. We will see why what happens when h of z is irrational. We cannot conclude that the region of convergence remains the same. But in case of rational systems the region of convergence of the ROC of h of z is the same as the region of convergence of h of z except inclusion or except the at the point z equal to 0. Apart from that the region of convergence remains the same for both and anyway z equal to 0 is not going to affect our stability because even if there are some issues at z equal to 0 we have a it will still be inside the unit circle. So it will not affect the stability as long as all the poles are inside the unit circle. So yeah so we can claim that g of z is stable because all these three terms will have poles inside the unit circle basically they all converge outside the unit circle. The region of convergence is greater than the unit circle so g of z is stable and hence we can mark option d as correct, option d is also correct. Now we can move on, okay there is one more thing. So I use this fact that the region of convergence is the same after differentiation. Now I can only use this fact when my system is a rational function. For a rational function this thing holds, this claim is correct. It's not exactly the same we need to still check at the behavior of the function at 0 but for most of the cases ROC will still remain the same after differentiation. Now what will happen in case of irrational function? You can see that this claim will not be true and I have referred a link below. You can go and check it out they have given an example in which the region of convergence of the derivative of the transform is different from the region of convergence of the function and for the sake of time we are skipping that but you can go over it for a quick read. Okay next we consider another problem in which we have to deal with the existence of the transform of the sinc function. So here I have defined sinc function as sin of n pi by 2 over n pi by 2 and we have to conclude something regarding the existence of the transform over all z over the unit circle if it's region of convergence greater than 1 or if it is less than 1, okay. So let's quickly go over it. So I have just defined the sinc function here. It is sin n pi by 2 over n pi by 2. It's equal to 1 when n is equal to 0 by definition. For all even integers, this 2 and 2 will get cancelled. So we'll have sin m pi on the numerator which is 0 and for all odd integers, we know that sin of 2m plus 1 by 2 pi is equal to minus 1 to the power m. So using that we get this expression, all right. Now the next thing that we need to recall is when does the z transform exist? Or here existence by existence we mean when does the z transform converge? So we are basically concerned with what is the region of convergence of the z transform of the function. Now in the theory lectures we have seen that this z transform of the function converges when our sequence at that point let's say here there should have been a z, this thing is absolutely summable. So for all such z for which this summation comes out to be finite or less than infinity, we can say that the z transform exists at that point. So we'll plug in different, let's say if I want to check if my z transform converges at z equal to 2. So I'll put in z equal to 2, I'll evaluate this summation. And on the basis of that, if it converges then I'll say that the z transform exists at that point or if it diverges then we'll say that the z transform does not converge. Now let us first look at the convergence at mod z equal to 1. So at mod z equal to 1, I'll quickly write the terms involved in the z transform. So summation n equal to minus infinity to infinity. And let me denote sinc of n by 2 as x of n, x of n into. Now I'm taking mod z equal to 1, so this thing will go off since mod z is equal to 1, x in. Now we know the value of functions at certain points. So let me break this summation for the cases where n is odd and n is even and n is 0, so for n is equal to 0 we have 1 plus for n belonging to even numbers we'll have 0. And for n is equal to odd, let's say we'll have terms of the form 0.5 times 2n minus 1. So here m's claiming that n is equal to 2m minus 1 and your m will range over all integers. So we'll get this summation. So the scripted z denotes the set of all integers. Now we can see that by symmetry in this term over the I'll have an absolute sign as well. Over the negative terms as well as positive terms, the value remains the same. For example, when consider the case where m is equal to 1 and let's say m is equal to 0. So when m is equal to 1, this term becomes minus 1 to the power m will go off because of this absolute sign, so we need not be concerned with this. This term will be 0.5 pi times 1 over 1 and for m is equal to 0, this term will be 1 by 0.5 times 2 into 0 minus 1. But this absolute sign will remove this minus 1, so this will again become 1 by 0.5 pi times 1. So by symmetry, we can see that these two terms will be like for every plus 1, for every positive odd integer, we'll also have a corresponding negative odd integer in this summation. So I can just sum over the positive integers and multiply my summation by 2. Also, I'm taking this 0.5 pi as 1 by 2, so I'll have a 4 by pi here. So this is 2 into 2, 2 for the removal of the negative half and 2 from this 0.5 pi and this is 1 by 2 m minus 1. So this is what my summation turns out to be. Now this thing, what I'm claiming is this thing will diverge. So this is nothing but 1 by 1 plus 1 by 3 plus 1 by 5 till infinity. And this thing diverges. This is something that we know by properties of sequences. So this thing is a divergent sequence. So clearly, since our sequence does not converge, so our Z-transform does not exist at Z equal to 1. So we can rule out the, so this option is ruled out. Since our Z-transform does not exist for Z equal to 1, so clearly it does not exist for all Z. And it clearly does not exist for Z equal to 1, as we saw right now. Now let us look at what happens for other Z, which is not equal to 1. Again, I can expand my Z-transform as this. So I can take the summation from minus infinity to 1 and from 0 to infinity. Now, there can be two cases. One is when mod of Z is greater than 1. So let me take this case first. So when mod of Z is greater than 1, you can see that this thing will diverge. Because n is negative, so this thing will be positive. So as this n tends to infinity, this term will diverge, the first term. Now, for the second case, when mod of Z is less than 1, you can see that this term will diverge. Because these will be terms of the form 1 by Z to the power n. And your modulus of Z is less than 1. So when your modulus of Z is less than 1, as you move towards n equal to 0, this thing will again tend to infinity. So because of that, this term will diverge because of this Z to the power n. So in either case, we can see that our summation diverges. So we can say that for Z not equal to 1, also we don't have. So this is also false, this is also false. I'm sorry, it says does not exist. So yes, so it does not exist on the unit circle. It does not exist for Z greater than equal to 1. It does not exist for Z less than 1. So option A is correct because it says does not exist. So yeah, so we'll look at some more problems in the next lecture. So stay tuned, thank you.