 So, welcome to the seventh session. Now, here we take up from where we left in the previous session namely what happens when we give a rotating complex number or a phasor to a stable linear shift invariant system. Let us quickly recapitulate. So, we said that I have this stable linear shift invariant system S impulse response h t which is real well that makes sense because you know most physical system would have a real impulse response. When we give it a e raised to the power j omega t plus phi what we get is a e raised to the power j omega t plus phi multiplied by and we recognize that this is essentially a constant independent of time. It is a constant that depends on h and depends on omega, but otherwise it is independent of t that is what I meant by constant. And now if you replace omega by minus omega in fact if you complex conjugate this let me write on the same picture, but in a different colour. So, if I complex conjugate this what I would get is again the same thing e raised to the power j omega t plus phi complex conjugate, but multiplied by complex conjugate of this. So, it is just a change of constant and those constants are complex conjugates of one another. Let us give this constant a name now. Let us give this constant a name. So, let us call this constant as capital H as a function of omega. So, you know we are saying two things we are saying that it depends on h and it depends on omega. We are saying both together and this is a constant independent of time. Now, we give this constant a name we call this constant the frequency response of this linear shift invariance table system evaluated at the frequency omega. The frequency response of s evaluated at omega. Now, frequency response essentially refers to the angular frequency of the sinusoid. And when I say frequency response I have a frequency response at omega I have a corresponding frequency response at minus omega. These are complex conjugates. So, let me write that down mathematically. Mathematically H omega is H minus omega complex conjugate and therefore, in total I could write down when 2 a cos omega t plus phi goes into the system script s. What comes out is H omega times a e raised to the power j omega t plus phi plus h minus omega times a e raised to the power minus j omega t plus phi and now you can aggregate them. This can be written as mod H omega a times e raised to the power j omega t plus phi plus angle H omega. You see you recall that we can write H in polar form. Similarly, plus mod H of minus omega well you know you could write minus j omega t plus phi plus angle H of minus omega. So, of course you know I need to make a little change here I could write e raised to the power separately that might be more convenient. So, this is ok, but I could make a little change I could write this as a e raised to the power minus j omega t plus phi into e raised to the power j angle H of minus omega. And now I make an explicit relation. These are the same and these are negatives and therefore, in total I have the following expression and I can aggregate them. Now once again notice something interesting. These are complex conjugates and you can add them together and take only the real part left twice the real part. So, I have and this gives a very beautiful interpretation. What has happened? We have seen the sinusoid go into the linear shift invariance stable system it has come out as a sinusoid of the same angular frequency, but it has undergone a change of amplitude and a change of phase. And in fact that constant which we referred to all this while gives us an insight into this change of amplitude and the change of phase. Let us see what happened. So, here we are. We had this sinusoid which went in and what came out is this and as you can see this is the change of amplitude and this is the change of phase and except for this change of amplitude and change of phase has no other change fundamental change in the nature of the input. A sinusoid has gone in and come out as a sinusoid of the same frequency the change of amplitude and change of phase are predictable by looking at what happens to any one of those rotating complex numbers. You can look at the rotating complex number rotating with an angular velocity omega or you could look at the rotating complex number rotating with an angular velocity minus omega. The amplitude changes of both of those are the same equal to mod h omega. The phase changes are opposite and you could look at any one of those phase changes and interpret the phase change that the sinusoid sees. In fact, one concrete way to do this is look at the phase are rotating in the counter clockwise direction. Whatever change of amplitude undergoes is also the change of amplitude undergone by the sine wave. Whatever change of phase it undergoes is also the change of phase that the sine wave undergoes. So, now we have a beautiful situation. I can conveniently evaluate the frequency response of this linear shift invariant stable system and use that frequency response to understand what that system would do when a sinusoid or a sine wave is given to it. And of course, we know that a sine wave will result in a sine wave of the same frequency except for a change of amplitude and phase and we know what that amplitude change and phase change are. Sinusoids have a special relationship, have a special place in the context of linear shift invariant systems. In fact, now suppose I had multiple sinusoids. Suppose I had 2 sinusoids of 2 different frequencies. Let me put the problem down before you and let me answer it in the next session. The problem is as follows given to you to think about. Find the output. I have the same linear shift invariant stable system. Impulse response h t. The input is now a sum of 2 sinusoids of different frequencies. 2a1 cos omega 1 t plus phi 1 plus 2a2 cos omega 2t plus phi 2. The question for you is, find the output. What is the output here? And the hint is very simple. Remember the system is linear. So if I have a sum of 2 inputs, you could find the output individually to each of those inputs and calculate the net output. I leave you with this question and we will meet again in the next session. Thank you.