 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 operations on continuous time signal and the numericals related to this basic operation. At the end of this session, student will be able to apply the basic operations on continuous time signal. So, these are the contents. In this, we will see the numericals which are related to the time shifting operations, numericals related to time reversal and time scaling operations. So, before moving ahead, please recall the time shifting property. So, it is and the name suggests the shifting of the signal in a time. So, in case of time shifting property, the mathematical it is represented as y of t is equal to x of t minus t. So, here suppose the x of t signal, if the capital T value here if it is the negative, then this will lead to the shifting this signal towards the left side means advance the signal. If the capital T is a positive value means if t minus capital T and t is a positive, then which will leads to the shifting this signal towards the right side means delays this signal. So, this is the time shifting property. Then the second is a time reversal property. So, please recall the time reversal property. So, this time reversal of the signal is obtained by folding the signal about t is equal to 0. So, this mathematical it is represented as y of t is equal to x of minus t. If the x of t input, then the time reversal version of that signal is a y of t which is equal to x of minus t. So, suppose the x 1 of t the signal is given, x time reversal signal x 1 of minus t it is nothing but reflection about this t is equal to 0 or reflection about this y axis. So, this is the mirror image of this input signal x 1 of t. Then recall the time scaling property it is obtained by replacing t by alpha t in that signal x of t. So, mathematically it is represented as y of t is equal to x of alpha t. Now, let us see suppose this is the x of t input signal if you apply the time scaling property and if the alpha is greater than 1. So, it will lead to the compression of this signal. So, y of t is equal to x of 2 t suppose then it will lead to the compression of the signal by factor 2. And if the alpha value is in between 0 and 1. So, it will lead to the expansion of the signal and here x of half of t means it will lead to the expansion of the signal by factor this 2. Now, let us see the examples which are related to these basic operations of city signals. Sketch the signal y of t is equal to x of 2 minus t by 3. So, in this and here the x of t signal is given. Now, let us see the steps to solve this signal. The first step we have to apply the time shifting property on this x of t first. So, time shifting property in that case we required x of 2 plus t first. So, we will shift this signal towards the left side by 2 units. Then the second step we are going to apply the time reversing property on the step 1 signal output. So, if you apply the time reversing property then this x of 2 plus t signal becomes x of 2 minus t. And the steps 3 we are going to apply the time scaling property on the second steps output. So, here we are expanding the signal by factor 3. So, which is nothing but x of 2 minus t by 3. So, let us see one by one all steps. So, this is the x of t. Apply the time shifting property towards the left side by 2 units this gives the x of 2 plus t. So, we are shifting this signal x of t towards the left side by 2 unit. So, here the point A in x of t is at t is equal to minus 1. Now, in x of t plus 1 t plus 2 the point A will shift towards at t is equal to minus 3. Similarly, point B in a x of t is at t is equal to 0. So, if we shift this signal towards the left side then the point B is at t is equal to minus 2. Then the point C in a x of t is at t is equal to 1. So, if we shift this signal towards the left side then the x of t plus 2 the points C will be shift at t is equal to minus 1. Similarly, point D in a x of t is at t is equal to 2 and if we shift this signal towards the left side then the output at that point the D point will be at t is equal to 0 and E point in the new signal E point is at t is equal to 1. Now, the next is a we need this minus t here. So, for that we have to apply the time reversing property to get the 2 minus t. So, fold this signal about t is equal to 0. So, take the mirror image of the signal about this t is equal to 0. So, this is the time reversing property and then we have to apply the time scaling to get 2 x of 2 minus t by 3. So, here the time scaling factor is a 1 by 3. So, this signal is get expanded by factor 3 means at the initial the point is at a point is at 3. So, 3 into 3 this a point will be at t is equal to 9. Similarly, all points are get expanded by the factor 3. So, this is the required output resultant sketch. We can verify our answer. Suppose the point a in x of t is at t is equal to minus 1 then in x of 2 minus t by 3 at 2 minus t by 3 which is equal to minus 1 hence t is equal to 9. So, you can check whether your point a point is at 9. Similarly, for b you can verify your answer in x of t b point is at t is equal to 0 and in the x of 2 minus t by 3 this point will be at t is equal to 6. The point c in x of t is at t is equal to 1 then in x of 2 minus t by 3 it will be at t is equal to 3. Similarly, point d will be at t is equal to 0 and point e will be at t is equal to minus 3. So, in this way you can verify your answer. Then let us see the second example sketch the signal y of t is equal to x of 3 t minus 5 x of t is given. So, the step one we have to apply the time shifting towards the right side here we want x of t minus 5. So, shift this signal towards the right side by 5 units. Then the second step here the time we have to apply the time scaling to get this x of 3 t minus 5 and the scaling factor is the 3. So, let us see these two steps this is the x of t apply the time shifting by 5 towards the right side then we get the x of t minus 5. So, this x of t signal will shift to towards the right side then at this point the a point will be at 4 plus 5 means the a point will be at 9. Then the b point will be at 2 plus 5 this b point will be at t is equal to 7 like that all the signals are get shifted towards the right side by 5 units. So, you can verify this the point a in x of t is at t is equal to 4 then in this shifted signal the point a will be at t is equal to 9 the point b in a x of t is at t is equal to 2. So, in the shifted signal the point b is at t is equal to 7 then point c in a x of t is at t is equal to 0 then in the shifted signal the point c is at t is equal to 5 point d in x of t this signal is at t is equal to minus 2 in the shifted signal the point d is at t is equal to 3. So, you can verify your answer then the second step here we need x of 3 t minus 5. So, we have to apply the time scaling for the first step signal. So, here we have to apply the compression by factor 3. So, if we apply the compression by factor 3 means the point a at t is equal to 9 here this is get compressed means 9 by 3 it will comes at t is equal to 3. Similarly, point b here is at t is equal to 7 if we do the 7 by 3 then that value is for point b. So, in this way this signal is compressed here by factor 3. Now, you can verify this the point a in a x of t is at t is equal to 4 in x of 3 t minus 5 at 3 t minus 5 is equal to 4 hence t is equal to 3. So, point a is at 3 is equal to 3 correct then the point b in a x of t is at t is equal to 2 then in x of t 3 t minus 5 at 3 t minus 5 is equal to 2 hence the t is equal to 2.5. So, point b will be at 2.5. Similarly, you can verify for point c point c in a x of t is at t is equal to 0, but in x of 3 t minus 5 3 t minus 5 is equal to 0. So, the t value we get is it is equal to 1.6. So, point c will be at 1.6 t is equal to 1.6 then the point d in the x of t it is at t is equal to minus 2. So, in x of 3 t minus 5 at 3 t is minus 5 is equal to minus 2 hence t is equal to 1. So, the point t will be here in modified signals the point t d will be at t is equal to 1. So, this is our output. So, we can verify in this way whether our output is correct or not. These are the references. Thank you.