 A very warm welcome to the second half of the tutorial one on the course digital signal processing and its application. In this tutorial again we will continue solving some more problems related to weeks 4, 5 and 6 of this course. So we will look at week 5 problem 10. It is very similar to week 5 problem 9. So here we have been given the sequence x of n as sin square of n by 2 and we have to again make certain claims related to the Z-transform of the function, the existence of Z-transform of the sin square function. So we can quickly go through when does the Z-transform exist or converge. So the Z-transform of a sequence exists or converges when the sequence is absolutely summable for that particular Z. So this is what we saw in the previous lecture. So if we want to check it for any Z let's say Z equal to 2 so I will plug in Z equal to 2 in place of Z here and I will evaluate the summation. If the summation comes out to be finite then the Z-transform exists and it converges and if not then we say that the Z-transform does not exist or the sum diverges. Now we will make use of this fact to solve this problem. So we need to check for two cases like what happens when it is on the unit circle and what happens when it is outside of the unit circle or what happens if it does not exist at all. So again sin square of n by 2 will be 1 for n equal to 0 will be 0 for n equal to 2 m just to remind sinc of x or sinc of n by 2 is equal to sin of n pi by 2 over n pi by 2. So when n is equal to 2 m sin of n pi by 2 will be sin of m pi and which is 0 so 0. And now when n is odd so I am considering this case when n is equal to 2 m minus 1 sinc square of n by 2 is equal to sin of 2 m minus 1 pi over 2 the whole square by n 2 m minus 1 pi over 2 the whole square. So which is equal to now we know that sin of 2 m minus 1 pi by 2 is minus 1 to the power n and when we will square it that the effect of that negation goes off. So we will be left with 1 so which is minus 1 to the power 2 n which is nothing but 1. So we can write it as 1 divided by 0.25 1 by 4 is 0.25 pi square 2 m minus 1 the whole square. So that is why we get the third term. Now again let us check what happened at z equal to 1. So we evaluate this absolute summation it is 1 plus 8 by pi square 2 m minus 1 the whole square. Now we know that summation 1 by n square n equal to 1 to infinity the Bezel's home it is equal to pi square over 6 right. So this summation clearly which is equal to pi square over 6 is greater than this summation I am considering only this half. So this thing if you write a few terms of the summation it is 1 by 1 plus 1 by 3 square plus 1 by 5 square and so on and if you expand this this is 1 by 1 plus 1 by 2 square I should write 1 by 1 square plus 1 by 3 square plus 1 by 4 square plus so on. So we can clearly see that this summation is greater than this summation. So if this thing is finite there is no means by which this thing could be infinite. So this thing is less than infinity in fact you can actually evaluate this summation as an exercise if you know how to evaluate this. We can also use properties of Fourier transform to evaluate these summations. So we can say that for z equal to 1 for mod z equal to 1 that is on the unit circle this summation is finite. So we can claim that the z transform exists on the unit circle. So clearly option 1 would be false because it says that it does not exist for all that we have found at least some set of z on which the z transform exists and that is the unit circle. So this option is true. Let's see if there is any other option which is true since we have been asked which of the following is or are true so they can be multiple correct options. Now for z not equal to 1 again we have a similar situation as in the ninth problem. So again we consider two cases first is when mod z is greater than 1 strictly greater than 1. So in those cases we can see that this term will blow up because when mod z is strictly greater than 1 minus n will be strictly greater than 0. So z to the power minus n is equal to z to the power some positive quantity let me call it k where k is greater than 0 and as k tends to infinity in this summation k tends to infinity that means as n tends to minus infinity this z to the power k will tend to infinity. So the summation will blow up and it won't converge. Now again we can look at this term sum in the second case when mod z is less than 1. Again this term will be the troublemaker so z to the power minus n now n is positive and so this will be something like 1 by z to the power n. Now here we will start facing problems as n tends to infinity again because as n tends to infinity z to the power n will tend to 0 which will imply that 1 by z to the power n tends to infinity so this will blow up. So for both these cases we see that our summation diverges. Now okay so both these options are false so the only correct option is option 2. This is false because it includes both z equal to 1 and z greater than 1 that is why it is false because it does not hold for this case. Now there is one more problem in which we have to consider a digital filter which has been given by y of n is equal to p of n minus k by 4 p of n minus 1 and y p of n is given by x of n minus 2k by 3 p of n minus 1 and now we need to conclude something regarding stability. We know that a system will be stable when the poles of the system will lie inside the unit circle. So basically we need to find out the transfer function that is y of z by x of z for this digital filter and we can then conclude something regarding the stability of the system depending on what poles the system has alright and it is not direct since y depends on p and p depends on x. So we need to first find the transfer function of p and x the system p and which includes p and x and then we will use it to find the transfer function of y and x. So let me take the let me call it equation 2 let me call it equation 1 let me take z transform on both sides of equation 2. So taking z transform on both sides of 2 we get p of z I am denoting capital p of z to denote the z transform this is equal to x of z minus 2k over 3 when we have a delayed function. So we multiply it by z inverse times p of z using this property that z the z transform of x of n minus k is z to the power minus k times x of z. So when we have a delay of 1 we will have z to the power minus 1 times p of z. Now let us collect all the terms with p of z on one side. So p of z times 1 plus 2k by 3 is equal to x of z this gives p of z to be equal to x of z divided by 1 plus 2k over 3. Let me call it as my third equation. So I have p of z in terms of x of z. Now let me take the z transform on both sides of the first equation taking z transform both sides of equation 1 I will have y of z to be equal to p of z minus k by 4 again delay of 1 unit. So z inverse times p of z collecting all the terms of p of z together gives p of z 1 minus k by 4 z inverse I made a mistake here I will have a z inverse term. So z inverse z inverse okay so it is because of this term. Now okay so now we have p of z but we need y z in terms of x of z. So we can use equation 3. So using equation 3 we can substitute p of z as x of z by 1 plus 2k by 3 z inverse. So y of z is equal to x of z into 1 minus k by 4 z inverse by 1 plus 2k by 3 z inverse. So this is our final transfer function this thing in the green box. Now we can see that it has a 0 at so you set the denominator 0. So z inverse k by 4 equal to 1 this implies at z equal to k by 4 we have a 0 and it has a pole at 1 plus 2k by 3 equal to 0 which implies 2k by 3 z inverse equal to 0 which implies z is equal to minus 3 by 2k. So we can see that it has a pole at z is equal to minus 3 by 2k okay I think I made a mistake is equal to 0 this implies z inverse 2k by 3 is equal to minus 1 this implies z is equal to minus 2k by 3 okay. So we have pole that z is equal to minus 2k by 3. Now so for the system to be stable the poles must lie inside the red circle. So all poles must lie inside the unit circle. So we have a pole at z is equal to minus 2k by 3. So this thing so the if it lies inside the unit circle that means mod of z should be less than 1. So this gives us mod of minus 2k by 3 should be less than equal to 1. So this gives k is less than equal to 3 by 2. So let us see if we have an option close to that okay. So option 2 option 3 satisfies this k less than equal to 3 by 2. So we can see that the system which is 1.5 the system will be stable when your k is less than 1.5 yeah. So with this we conclude the first tutorial session for this course digital signal processing and its applications and thank you for attending thank you.