 Welcome to the sixth session in the second module of the course signals and systems. Recall that we had looked at what happens when we give a sinusoid into a stable linear shift invariant system. We assume the impulse response of that system to be real, real valued. Now, you see the output did not make any sense in the sense that there was no convenient way to calculate what that output was. Let me do that little bit again. So, what I said was we could take a linear shift invariant system. Let us call it script S. This is linear shift invariant and stable. The impulse response h t is real, real valued. We gave the input x t equal to 2a cos omega t plus phi and we queried what the output was. The output was y t given by convolution equal to minus to plus infinity h lambda 2a cos omega times t minus lambda plus phi d lambda as we said not easy to evaluate. Now, let us think of that same sinusoid as broken into two rotating complex numbers. So, we have 2a cos omega t plus phi is a e raised to the power j omega t plus phi plus a e raised to the power minus j omega t plus phi and let us only apply this to S and query what the output is. And we will see there is something very interesting here indeed when we keep just one phasor here namely a e raised to the power j omega t plus phi the output is minus to plus infinity h lambda a e raised to the power j omega t minus lambda plus phi d lambda and this can easily be split as follows. So, if we get an integral minus infinity to plus infinity h lambda a e raised to the power j omega t plus phi times e raised to the power minus j omega lambda. In fact, you see this is essentially a consequence of the fact that we have a change of phase here essentially this is a change of phase and the change of phase or a time shift is reflected as a multiplying constant. And therefore, the output becomes minus infinity to plus infinity times the original wave form. So, in fact, you know the beauty is that you have essentially the original wave form coming into the picture once again or the input and this is independent of lambda can come out. So, that gives us a e raised to the power j omega t plus lambda time plus phi times minus infinity to plus infinity h lambda e raised to the power minus j omega lambda d lambda and notice that this quantity essentially depends only on omega and h. So, the beauty is we notice something very special here when we give one rotating complex number to this linear shift invariant system. All that happens is that rotating complex number is multiplied by a constant. Now, note here look at this expression here this expression is a constant independent of time and the constant depends on the system namely it is impulse response and of course, the frequency or the angular velocity of rotation omega. Now, because the system is stable you know if you look at it the stability of s guarantees this quantity integral minus to plus infinity h lambda e raised to the power minus t omega lambda is finite bounded if you might call it that bounded as a function of omega well that is very easy to see if I take its magnitude which I shall now write here. So, if I take its magnitude the magnitude is less than or equal to the integral of the magnitude of the integrand and if you look at this expression it has to be finite because of stability due to the stability of s. So, you know mod h lambda e raised to the power minus t omega lambda is the same as mod h lambda you say look at the expression here this quantity here mod h lambda e raised to the power minus t omega lambda is the same as mod h lambda and we know that the stability of the system implies and is implied by the absolute integrability of the impulse response and therefore, we are assured that this constant by which that rotating complex number is multiplied is necessarily finite bounded and bounded for all omega. In fact, the beauty is that the bound is the same as the absolute integral of the impulse response you could bound you have a concrete bound which you can associate with it and now let us see what happens to the other rotating complex number. So, here we had a rotating complex number rotating with an angular velocity capital omega let us see what happens to the other one the other phasor is multiplied by integral minus infinity to plus infinity e raised to the power minus j into minus omega lambda times h lambda d lambda and you notice that this is essentially the complex conjugate of the previous constant. It is a complex conjugate because h is real let me now formally show that complex conjugate of minus infinity to plus infinity h lambda e raised to the power minus t omega lambda d lambda is simply you know you can take the complex conjugation inside the integral sign because h is real this little change can be made here and now it is very clear to us that both the rotating complex number with angular velocity omega and minus omega have essentially the same multiplying constant except that one is the complex conjugate of the other. So, if you focus on one what happens to the other is known now we will use this in the next session to come out with the idea of what we call the frequency response of a stable linear shift invariant system. Thank you.