 Hello everyone, I am K. R. Biradhar, Assistant Professor, Department of Electronic Telecommunication Engineering, Palchand Institute of Technology, Sallapur. Welcome to video lecture on event and odd signals. Let us start with the learning outcomes first. At the end of this session, students will be able to identify event and odd signals, find event and odd component of the signal. These are the some of the contents, event signal, odd signal, properties of event and odd signals, event and odd component references, event signal. A signal x of t or x of n is referred to as an event signal if it is identical to its time reversed counterpart that is with its reflection about the origin. In continuous time, a signal is even if it satisfies the condition x of minus t is equal to x of t for all t. Similarly, for a discrete time signal, e is even if it satisfies the condition x of minus n is equal to x of n. Event signals are symmetric about the vertical axis or time origin, x of t is an event signal. If you take x of minus t, time reversal of this signal or mirror image of this signal both must be same. Let us see here. So, this is my x of minus t. If these two signals are same, then it is an given signal is an event signal. So, this is an example for discrete time event signal. Then odd signal, a signal is referred to as odd if it satisfies the condition x of t is equal to minus of x of minus t for all t for continuous time signal, x of n is equal to minus of x of minus n for discrete time signal. Then odd signal must necessarily be 0 at t equal to 0 or n equal to 0. Odd signals are anti symmetric about the time origin. The given x of t is an odd signal. Now, you find the x of minus t that is time reversal of this signal or mirror image of this signal. Next, you find the minus of x of minus t which is the amplitude scaling by minus 1. That means, the signal which is present above this x axis will shift towards below this x axis. The signal which is present below this x axis will shift the above x axis. The first x of t and x of minus of x of minus t are same therefore, it is an odd signal. Discrete time example for odd signal let us see here x of n this is minus of x of minus n these two must be same. Now see step by step my x of minus n will be whatever the signal which are present at n equal to minus 1, minus 2, minus 3 should shift at n equal to 1, 2, 3 respectively. Similarly, the signals which are present at n equal to 1, 2, 3 should shift at n equal to minus 1, minus 2, minus 3 respectively. This is a x of minus n or mirror image of this signal x of n. Next amplitude scale by minus 1 the signal which are present at the below this x axis will shift to above this x axis and signal which are present at above the x axis will shift to below the x axis. This is x of n this is x of minus minus of x of minus n these two are same therefore, it is an odd signal. Then properties of even and odd signal first one is product of an even signal and an odd signal results into odd signal. Second one the product of two even signals results into even signal and two odd signal results into even signal. Third one the summation of an odd signal x of n for all n is always 0. And proofs we shall see now product of an even and odd signal consider x 1 of n be an odd signal and x 2 of n be an even signal. So, therefore, x 1 of minus n is equal to minus x 1 of n this is equation number 1. Similarly, x 2 of minus n is equal to x 2 of n this is equation number 2 because this is an even signal this is an odd signal. Let y of n is equal to x 1 of n multiplied by x 2 of n equation number 3 replace n by minus 1 in the above expression therefore, y of minus n is equal to x 1 of minus n into x 2 of minus n equation number 4 substituting equation 1 and 2 in 4 therefore, y of minus n is equal to x 1 of minus n is odd signal it minus of x 1 of n. Similarly, this if I consider even x 2 of minus n becomes x 2 of n this is equal to minus of x 1 of n multiplied by x 2 of n this is equal to minus of y of n therefore, y of minus n is equal to minus y of n therefore, it is product of even and odd signal results into odd signal. Proof number 2 product of 2 even and 2 odd signals let y of n is equal to x 1 of n multiplied by x 2 of n replace n by minus n y of minus n is equal to x 1 of minus n multiplied by x 2 of minus n let x 1 of n and x 2 of n be even signals. So, y of minus n is equal to if I substitute that is even signals it becomes x 1 of n into x 2 of n therefore, y of minus n is equal to y of n product of 2 even signals results into even signal. Similarly, if x 1 of n x 2 of odd signals then y of minus n is equal to. So, x 1 of minus n becomes minus x 1 of n x 2 of minus n becomes minus x 2 of n because it is an odd signal. So, if you product of these 2 becomes equal to y 1 of n multiplied by sorry x 1 of n multiplied by x 2 of n which is equal to y of n their product is x 1 of n multiplied by x 2 of n which is equal to y of n. The product of 2 odd signal results into an even signal summation of an odd signal consider x of n is an odd signal y of n is equal to summation equal to minus infinity to infinity x of n if I divide the limits from minus infinity to infinity. So, if I change a sign here n equal to n equal to minus 1 2 plus minus infinity now becomes equal to 1 to infinity. If I change a summation sign here the n sign also changes here summation equal to 1 to infinity x of minus n plus x of 0 minus summation equal to 1 to infinity x of n. We know if it is an odd signal x of minus n equal to minus x of n this becomes minus of summation n equal to 1 to infinity x of n plus x of 0 plus summation equal to 1 to infinity x of n. So, there is a minus sign there is a plus sign these 2 get cancels only x of 0 remains which is equal to 0 as x of n is 0 at n equal to 0 for any odd signal even and odd components. The x of t can be expressed as even and odd components x of t is equal to x e of t plus x o of t this equation number 1. Replace t by minus t in the above expression x of minus t is equal to x e of minus t plus x o of minus t x of minus t is equal to if it is an even signal x of minus t is equal to x of t. Therefore, x e of minus t becomes x e of t and x of o of minus t becomes minus x of t this is equation number 2 adding equation 1 and 2. So, x of t plus x of minus t is equal to 2 x e of t the even component is given by x e of t is equal to 1 by 2 into x of t plus x of minus t subtracting equation 2 from 1 x of t minus x of minus t is equal to 2 x o of t. Therefore, the odd component is given by x o of t is equal to 1 by 2 into x of t minus x of minus t. Similarly, for discrete time signal the even and odd components are x e of n is equal to 1 by 2 into x of n plus x of minus n x o of n is equal to 1 by 2 into x of n minus x of minus n. Draw the waveform and identify the even and odd signals for the following signal first one is y is equal to cos x second y equal to sin x. Pause the video for a moment and think and write your answer. This is a cosine wave or cos of x it is an even signal. Similarly, this is a sin wave we can see here this is an odd signal. Find the even and odd components of x of t for the given signal x of t. We know that the even and odd components is given by x e of t is equal to 1 by 2 into x of t plus x of minus t x o of t is equal to 1 by 2 into x of t minus x of minus t. If I want to find even component we need to have x of minus t also convert that is x from x of t to x of minus t that is x of minus t is the mirror image of this signal. Take x of t and x of minus t find x of t plus x of minus t. So, from minus 2 to minus 1 this is 0 and this is minus 1 you are going to get minus 1 from minus 1 to plus 1 this is 1 this is also 1 1 plus 1 equal to 2 from plus 1 to 2 this is minus 1 this is 0 you are going to get minus 1. Next is x e of t is equal to 1 by 2 this much you divide it by this signal by 2 you are going to get this signal. Similarly, x of t minus x of minus t subtract x of minus t from x of t and subtract x of minus 2 to minus 1 this is 0 and this is minus 1 0 minus of minus 1 equal to plus 1 from minus 1 to plus 1 this is 1 this is also 1 it becomes 0 from 1 to 2 x of t is minus 1 x of minus t is 0 that is minus 1 minus 0 equal to minus 1. If you divide it by this signal by 2 you are going to get x of t these are the references I have referred to prepare the above PPT. Thank you.