 Hello, everyone. Welcome to this session. I am Deepali Vardhakar working as an assistant professor at WIT-Solapur. In this session, we are going to discuss the basic operation on DT signals and the numericals related to the basic operations on DT signals. At the end of this session, student will be able to apply the basic operation on discrete time signals. So, these are the contents. We will discuss the numericals which are related to time shifting, time reversal and time scaling operation. So, before moving ahead, please recall the time shifting operation on DT signal. So, in the time shifting operation, the mathematical it is represented as y of n is equal to x of n minus k. Suppose this x of n signal is given, here if the k is a positive value, then the shifting delays this signal means the signal will shift towards the right side. So, if there is a x of n minus k and if the k is a positive, then the signal will shift towards the right side and if the k is a minus value, so minus minus becomes plus. So, if there is a n plus k such type of values are there, then in that case the signal will shift towards the left side means advances the signal. Also recall the time reversal property, it is obtained by folding the signal about n is equal to 0. So, time reversal of the signal is obtained by folding the signal or by taking the mirror image of the signal about n is equal to 0. Then also recall the time scaling property of DT signal, in the time scaling property the n is replaced by alpha n. So, y of n is equal to x of alpha n. So, suppose x of n is given, if the alpha value is greater than 1, then that signal is a compressed and if the alpha value is in between 0 and 1, then that signal is get expanded. So, here the if alpha value is greater than 1 means x of 2 n there, then that signal is compressed. So, each n is equal to that signal is a compressed. So, let us see one example sketch the signal y of n is equal to x of 4 minus n. So, here the x of n is given. So, let us see the first step we have to apply that is the time shifting here. So, if you apply the time shifting to this x of n then to obtain this x of n plus 4, we have to shift this signal towards the left side. So, it will it is equal to x of 4 plus n, but we required x of 4 minus n then in the second step we have to perform the time reversing operation here we will get directly x of 4 minus n. So, this will the second step. Let us see the one by one steps. So, this is the x of n given in the sequential form we can represent this x of n like this. So, here this arrow represent the origin then to get the x of n plus 4 we have to shift this signal towards the left side by 4 units. In x of n plus 4 the signal is shifted towards the left side by 4 unit. So, the origin is add this 1 by 2 we can verify this signal for n is equal to minus 1 x of n plus 4 is equal to nothing, but x of 3 which is equal to 1. So, then for n is equal to 0 x of n plus 4 means 0 plus 4 which is equal to x of 4. So, x of 4 here is a half which is equal to half here. So, for n is equal to 0 the modified signal here the value is a half. Then for n is equal to 1 x of n plus 4 means 1 plus 4 which is equal to x of 5. So, x of 5 signal here is a 0. So, it will be 0. Then for n is equal to minus 2 x of minus 2 plus 4 which is equal to x of 2 and x of 2 signal what is here x of 2 it is nothing, but its amplitude is 1. For n is equal to minus 3 x of minus 3 plus 4 which is equal to x of 1 which is equal to 1. Then n is equal to minus 4 x of n plus 4 means minus 4 plus 4 x of 0 and x of 0 here x of 0 value is a 1. For n is equal to minus 5 x of minus 5 plus 4 it is equal to x of minus 1 and x of minus 1 here is a 1. So, here at n is equal to minus 5 it is 1. Now at n is equal to minus 6 x of minus 6 plus 4 which is equal to x of minus 2. Now x of minus 2 what is here x of minus 2 here 0. So, at n is equal to minus 6 this value sample value it is 0. So, you can verify in this way. Now let us see the second step. So, to obtain the x of 4 minus n we have to apply the time reversing on this x of 4 plus n. So, apply the time reversing property here means fold the signal about the n is equal to 0 axis. So, we get x of 4 minus n. So, this is the actually resultant sketch. So, here the x of 4 minus n that is the minus n plus 4 this signal is represented in a sequential form. So, origin is at this point means half. Then the second example sketch the signal y of n is equal to x of n into u of 2 minus n. Here the x of n signal is given then to obtain this y of n signal first we have to apply the time shifting property on this u of n. So, we have to obtain first the u of 2 minus n for that we have to apply the time shifting property. So, u of n plus 2 which is equal to u of 2 plus n. So, to get this we have to shift this signal towards the left side by 2 units. Now, the second step to obtain the u of 2 minus n we have to apply the time reversing property here. So, it will give the u of 2 minus n. Now, to get this y of n we have to multiply this u of 2 minus n by 2 x of n. So, we have to perform here signal multiplication x of n into u of 2 minus n. So, let us see all these steps one by one. So, this u of we know that this u of n signal then to get the u of 2 n plus 2 we have to shift this signal towards the left side by 2 units. So, this signal is shifted towards the left side by 2 units. This is the sequential representation of this original signal u of n. Then if we shift this signal towards the left side by 2 units then the origin will come at this point. So, similarly in the previous example you can verify like this. The second step to get u of 2 minus n we have to apply the time reversing property here. So, hold this signal about the n is equal to 0 and see the reflection of that signal. So, this is the u of 2 minus n. So, u of 2 minus n if you hold that signal about the n is equal to 0 then we will get the sequential representation of this signal like this. Then apply the multiplication to get x of n into u of 2 minus n. So, here this is the given signal x of n multiply the x of n by n to the this u of 2 minus n. Then at this point we get this resultant sketch here at the minus 2 the amplitude is 1 at the minus 2 n is equal to minus 2 here the amplitude is 0. So, 0 into 1 amplitude is 0. Then at n is equal to minus 1 here the amplitude is 1 at n is equal to minus 1 here amplitude is 1. So, 1 into 1 output is 1. Similarly at n is equal to 0 1 and 2 here the amplitude is 1 amplitude is 1. So, 1 into 1 here the amplitude is 1 for respective signal at n is equal to 3 here the amplitude is 0 and n is equal to 3 here amplitude is 1. So, 0 into 1 0. So, at n is equal to 3 the amplitude will be 0. So, this is the resultant sketch for this discrete time signal. So, these are the references. Thank you.