 Today I'm going to discuss the topic called correlation of signals. Let us start with the learning outcomes first. At the end of this session, students will be able to identify similar activity between two signals. Find the auto correlation and cross correlation of the signals. These are the some of the contents. Start with the definition of correlation and its application types, implementation of auto and cross correlation and references. Let us start with the definition. Correlation is defined as a similarity between a pair of signals. Basically, it is used to compare two signals. The application of correlation is in radar system where the location of the target is measured by comparing the transmitted and reflected signals. You transmit a signal from the transmitter. It reaches to receiver and it hits back the signal and those two will be compared. Then you can identify the distance between the transmitter and receiver. This is one of the important application of this correlation. There are two types of correlation. One is auto correlation and second one is cross correlation. Let us discuss auto correlation first. Auto correlation of a sequence is correlation with itself. The auto correlation of sequence x of n is given by rho xx of l is equal to summation n equal to minus infinity infinity x of n x of n minus l. In this auto correlation, you are going to get only one sequence and you need to take the time reversal of that one. That means, you need to take the flipped version of the sequence x of n that is called the x of n minus l. Let us take an example to identify or find the correlation of signal x of n is equal to 1 2 3 4. We need to write the x of n as well as a x of n minus l in the diagram shown. So, x of n will be written in the first row and x of n minus will be written in the column in time reversal format. Then first element in the column will be multiplied with first element of the row and that is 4 into 1 is 4. Similarly, first element in the column will be multiplied with second element in the row and 4 into 2 is 8. Similarly, first element in the column will be multiplied with third element in the row 4 into 3 is 12. Similarly, 4 into 4 will be 16. This will be continued for element present in the column that is second element 3 will be multiplied with all the elements in the present in the row. Similarly, third element in the column will be multiplied with all the elements in the row we are going to write 2 4 6 8. Similarly, 1 that is last element in the column will be multiplied with all elements in the row that is 1 2 3 4. Then combine those things and sum it up as shown in the diagram. The first still single element we are going to get that is 4 second 8 and 3 will be combined 8 plus 3 is 11. Similarly, third group will be 12 plus 6 18 plus 2 that is 20 and next 30 and 20 11 and 4. So, this will continue. Once if you see the result R excess of L then 30 will be the central element or middle element. Then at 30 we are going to get the symmetrical sequence or symmetrical to the center of element of this sequence. Then now let us move with the cross correlation. The cross correlation of a signal and its impulse response is given by rho x y is equal to summation n equal to minus infinity to infinity x of n y of n minus l where l is equal to 0 comma plus or minus 1 comma plus or minus 2 etcetera. The computation of correlation takes the following steps or we need to follow these different steps to find the cross correlation of the sequence. First one is obtain the sequence y of n minus l that is nothing but the second sequence and its time reversal by shifting the sequence right by the time lag l. Multiply the second will be multiply the shifted sequence y of n minus l by x n and sum all the values to obtain rho x y of l. Third one repeats steps one and two for all values of the lag one. Let us see by taking one example how we are going to follow and find the cross correlation of the sequence. The question arises whether the procedure for finding the cross correlation is same as auto correlation? Fast for a moment and think whether the procedure which is used in the auto correlation is same as the cross correlation. The answer is yes. Let us consider an example to find the cross correlation of the sequence. Given sequences are x of n is equal to 1 2 1 1 and y of n is equal to 1 1 2 1. The shifted sequence y of n is y of n minus l that is 1 2 1 1 that is 1 1 2 1 will be time reversal we can get 1 2 1 1. Same procedure as that of auto correlation write the x of n in the first row and y of n minus 1 second sequence and its time reversal will be written in the column. Then multiply the first element in the column with all the elements present in the row and write as 1 2 1 1. Similarly second element in the column will be multiplied with the all the elements in the row that is 2 4 2 2. The third element in the column will be multiplied with all the elements present in the row at 1 2 1 1. Similarly last element in the column will be multiplied with all the elements in the row 1 2 1 1. Again group as shown in the diagram and submit we are going to get the cross correlation of the sequence. First element is 1 second element is 2 plus 2 4 third element will be 4 plus 2 6 fourth element is again 6 fifth element will be 1 plus on 2 and last will be 1. The cross correlation of the sequence is 1 4 6 6 5 2 1. This is not symmetrical to the element present in the center of this sequence. Let us take one more example to find the cross correlation of the sequence. But interesting thing is when I take the same sequence of x of n and y of n then you are going to get the result as same as the auto correlation sequence. Let us see here x of n is equal to 1 2 3 4 and y of n is equal to 1 2 3 4. The shifted sequence of y of n is y of n minus l which is 4 3 2 1. Then write the first sequence in the first row and time shifted sequence y of n minus will be written the first column. Then multiply the first element present in the column will be with all the elements in the row you write 4 8 12 16 second element will be multiplied with the all the elements present in the row and write 3 6 9 12 and third element will be multiplied with all the elements present in the row so on. Then group those elements and sum it up first will be 4 and second will be 8 plus 3 11 12 plus 6 18 plus 2 20 third will be 30 fourth will be 12 plus 6 18 plus 2 20 fifth will be 11 and last one will be 4. When you look for this answer for cross correlation of two sequences then interesting thing is it is symmetrical to the central element present in the output sequence that is 30 with respect to 30 you can see that is this is a symmetrical sequence. These are the some of the references I used thank you.