 Hello, welcome all. Welcome to video lecture on Z-Transform, myself S. N. Chamath Gouda from Walchand Institute of Technology, Solopu. At the end of this session, students will be able to determine Z-Transform of a given signals. Today in this lecture, we are going to consider two sided infinite discrete time sequence. So, let us consider an example here. So, x of n is equal to 2 to the power n u of n minus 3 to the power n u of minus n minus 1. Now, find find Z-Transform and ROC of the given signal. So, let us solve this one. We know that x of z is equal to summation n equal to minus infinity to plus infinity x of n into z to the power minus n, where x of n is equal to 2 to the power n u of n minus 3 to the power n u of minus n minus 1. By substituting x of n is equal to 2 to the power n u of n minus 3 to the power n u of minus n minus 1. In the above expression, x of z can be written as x of z is equal to summation n equal to minus infinity to plus infinity 2 to the power n u of n minus 3 to the power n u of minus n minus 1 into z to the power minus n, where the first term represents the right sided sequence and second term represents the left sided sequence. Therefore, the summation can be split it into two parts. Hence, x of z is equal to summation n equal to minus infinity to plus infinity 2 to the power n u of n into z to the power minus n minus summation n equal to minus infinity to plus infinity 3 to the power n u of minus n minus 1 into z to the power minus n here. Because of the unit step signal in both the terms, the limits of the summation are changed here. We know that unit step signal u of n is defined as u of n is equal to 1 for n greater than or equal to 0 and u of n is equal to 0 for n less than 0. Similarly, u of minus n minus 1 is a modified version of unit step signal u of n and this one is defined as u of minus n minus 1 is equal to 0 for n greater than or equal to 0 and it is equal to 1 for n less than 0. So, we are going to use these two definitions in our example here. Therefore, the x of z because of these two unit step signals x of z will be written as n equal to 0 to infinity 2 to the power n z to the power minus n minus summation n equal to minus infinity to minus 1 3 to the power n z to the power minus n here. Because of these two unit step signals, the limits have been changed in both the terms here. In the first term, it is changed into n equal to 0 to infinity whereas, in the second term it is changed into n equal to minus infinity to minus 1 here. Now, to understand how the limits have been affected because of the unit step signals, let us consider the graphical representation of the signal here. So, here summation are written as n equal to 0 to infinity 2 to the power n z inverse minus summation n equal to minus infinity to minus 1 3 to the power n z to the power minus 1. Now, to understand how the limits of these two summations have changed because of the unit step signals defined above here. Here you can see that u of n is equal to 1 for n greater than or equal to 0 and it is equal to 0 for n less than 0 and u of minus n minus 1 is defined as it is equal to 0 for n greater than or equal to 0 and it is equal to 1 for n less than 0. So, in the graphical representation, you can see that the first graph represents the 2 to the power n which is a growing exponential signal and in the second graph it represents the unit step signal as per the definition on the left side the values of u of n is equal to 0 whereas, for n equal to 0 to plus infinity the values are equal to 1 here. So, when we perform the multiplication between these two signals, we get the resulting signal that is which is equal to 2 to the power n u of n. So, on the left side 2 to the power n is multiplied with the 0 therefore, the resulting signal will be equal to 0 whereas, for n equal to 0 to plus infinity that is on the right side we can see that 2 to the power n is multiplied with the 1 therefore, you will get 2 to the power n here. Therefore, in this representation you can see that the signal exists from n equal to 0 to plus infinity hence the limits of the summation are written as n equal to 0 to infinity. Similarly, on the second on the right side you can see that another set of graphical representations are given here. The first graph represents 3 to the power n which is again a growing exponential signal and the second graph represents the modified version of unit step signal u of n which is defined as u of minus n minus 1 is equal to 0 for n greater than or equal to 0 and it is equal to 1 for n less than 0. According to that in the graph you can see that on the left side from n equal to minus infinity to minus 1 the values of u of minus n minus 1 is equal to 1 whereas, for n equal to 0 to plus infinity the values of u of minus n minus 1 is equal to 0. So, when we perform the multiplication between 3 to the power n u of minus n minus 1 we get the resulting signal has shown in this graph here. So, here you can see that for n equal to 0 to plus infinity the signal 3 to the power n u of minus n minus 1 is equal to 0. Whereas, on the left side from n equal to minus infinity to minus 1 you can see that the 3 to the power n u of minus n minus 1 is equal to 3 to the power n because 3 to the power n is been multiplied with u of minus n minus 1 that is 1 3 to the power n is multiplied with 1. Therefore, you will get the resulting signal 3 to the power n here. So, that is why the limits of the first summation has been changed to n equal to 0 to infinity and the limits of the second summation has been changed to n equal to minus infinity to minus 1 here. Therefore, x of z is written as x of z is equal to summation n equal to 0 to infinity 2 into z inverse to the power n summation n equal to minus infinity to minus 1 3 into z inverse to the power n here. Now, in earlier 2 example we have used the infinite summation identity that is summation n equal to 0 to infinity alpha to the power n the infinite summation identity that is summation n equal to 0 to infinity alpha to the power n is equal to 1 divided by 1 minus alpha. This infinite summation identity is valid for mod alpha less than 1 year. So, by comparing x of z with the infinite summation identity we can see that the first summation is identical with this infinite summation identity whereas, the second summation is not identical with the infinite summation identity. So, therefore, we need to first convert this second summation the limits of the summation identical to this infinite summation identity so that we can use this infinite summation identity to simplify the x of z further here. So, first we will change that limits of that summation here. So, to change that one so we are going to put k is equal to minus n k is equal to minus n in second summation here. So, k equal to minus n means minus plus n will be equal to minus k and similarly n equal to minus infinity minus n will be equal to plus infinity and minus n will be replaced by k. So, k is equal to plus infinity similarly n equal to minus 1 so minus n is equal to 1 here so minus n is replaced by k so k is equal to plus 1. So, these are the k values which we are going to use in the second summation here not in the first summation. Therefore, x of z can be written as summation n equal to 0 to infinity 2 z inverse to the power n minus summation k equal to infinity to 1 3 z inverse to the power minus k here. So, further this one written as n equal to 0 to infinity 2 z inverse to the power n minus summation the lower limits are written below the summation and upper limits are written above the summation. Therefore, k is equal to 1 to infinity and this is written as 3 inverse z to the power k here, but still the limits of the second summation are not identical with the infinite summation identity here. Therefore, we are going to use the concept that is a to the power 0 equal to 1 in this example here. Therefore, x of z is equal to summation n equal to 0 to infinity 2 z inverse to the power n plus 3 inverse z to the power 0 minus 3 inverse z to the power 0 minus summation k equal to 1 2 infinity 3 inverse z to the power k here. Now, we are going to merge these two terms here and this will be further written as x of z is equal to summation n equal to 0 to infinity 2 z inverse to the power n plus 3 inverse z to the power 0 is nothing but 1 minus summation we are going to merge these two terms. Therefore, k is equal to 0 to infinity 3 inverse z to the power k. Now, we can use this infinite summation identity in both the summations here to simplify x of z. So, in our example in the first term in our example in the first term in the first summation alpha is equal to 2 z inverse whereas, in the second summation alpha is equal to 3 inverse z. So, therefore, x of z can be written as x of z is equal to 1 divided by 1 minus 2 z inverse. So, x of z is equal to 1 divided by 1 minus 2 z inverse plus 1 minus 1 divided by 1 minus 3 inverse z here. Now, in the first term we multiply both numerator and denominator by z. So, therefore, it will be written as z divided by z minus 2 z inverse z plus 1 minus again in the in this term we are going to multiply in this term we are going to multiply numerator and denominator by 3 here 3 divided by 3 minus 3 inverse z into 3. So, 3 and 3 inverse will get cancelled here z and z inverse will get cancelled. So, this will be written as z divided by z minus 2 plus 1 minus 3 divided by 3 minus z. So, further x of z is simplified as z divided by z minus 2 plus 3 minus z divided by z minus 2 plus 3 minus z minus 3 divided by 3 minus z. So, 3 and minus 3 will get cancelled. So, this will be written as z divided by z minus 2 plus minus z divided by 3 minus z. Now, this second term is multiplied both numerator and denominator by minus 1. So, further this will be written as z divided by z minus 2 plus z divided by z minus 3 and this summation this x of z is valid if mod 2 z inverse is less than 1 and mod 3 inverse z is less than 1 here. So, this can be further written as mod z is between mod 2 and mod 3. Now, let us plot the region of convergence on the z plane here. So, region of convergence is as shown in the z plane shaded part represents the region of convergence and this region of convergence is between the two circles of radius 2 and radius 3. Thank you.