 we could also obtain y n point by point for each n. Here all that we need to do is to reinterpret the expression for convolution. So what we are saying is take note of the expression for convolution once again. Instead of now thinking of x of k as a constant and h of n minus k as a sequence, I told you the interpretation is important. We will think of both x of k and h of n minus k without the summation as sequences indexed by k. So x k is a sequence indexed by k that is easy to understand h n minus k is also a sequence now not indexed by n but indexed by k okay. So think of h of n so the difference in that approach is think of h of n minus k as a sequence in k not in n for fixed n you see because you are trying to calculate point by point and how do we obtain that sequence we need to interpret that. So towards interpreting that we will follow two steps we will first go from h k to h k plus n for a fixed n and obviously for a fixed n h of k plus n is essentially the in the sequence h of k shifted backward by n. So h of k plus n is h of k shifted backward by n and the next step is of course to replace k by minus k. So when we replace k by minus k what we are doing is to reflect about k equal to 0 replacing k by minus k essentially means switching every pair of points for example switching 1 and minus 1 switching 2 and minus 2 switching 3 and minus 3 keeping 0 where it is. So it really means you are making a mirror image of that sequence with the mirror placed at k equal to 0 remember the sequence that you are mirroring is h of k plus n alright. So we are mirroring h of k plus n and we can now visualize what to expect in fact I do not need to take a specific example we can write in general in general you know when you have at 0 something you might have h of 0 and so on is not it h of 1 h of 2 and so on this is the sequence here h of minus 1 and can go on behind if this is h k then h of n minus k is going to be obtained by shifting this backwards so maybe we will first write h of n plus k shifting this backwards so this appears at minus n and this can go on and then reflection now when you reflect minus n goes to plus n h of 1 which appears here after minus n will now appear before plus n. So what you do is move this to the point n and then switch or mirror all the points around that point n alright. So what we have is as follows h of n minus k is therefore at n put h of 0 at n minus 1 put h of 1 h of 2 and so on behind and here you have h of minus 1 h of minus 2 and this can go on this was not too difficult to see generally but we can of course fix our ideas by taking the same sequence that we did a minute ago for h n namely if we consider h n as we did there or you see of course here you can call it h of k if you like to be specific and h of n minus k is essentially at n put 1 and then therefore at n minus 1 you will also have a 1 simple. So essentially what we are saying is that this h n or h of n minus k more precisely has a 0 at the has 1 at the point n and 1 at the point n minus 1 and therefore if us if a system had the impulse response given by this h n if an LSI system have this h n as it is impulse response remember this tells me everything about the LSI system once I know this unit impulse response I know everything about the LSI system. So once I write this I have specified it completely if I have an xn what I am saying is y of n is essentially x of n times 1 plus 1 times of x of n minus 1. So the impulse response also gives me an interpretation of what the system does in fact more particularly a finite length impulse response immediately gives us an interpretation the moment you have a finite length impulse response a finite length impulse response means an impulse response where the number of non-zero samples is finite. So here for example the number of non-zero samples in the impulse response is only 2. So the beauty is that with this interpretation of the operation of convolution I also have an interpretation for how the impulse response describes the system for a finite length impulse response it is very straight forward it tells you that the output at any point becomes a linear combination of the input at that point and a few points around it. Now the key issue is whether you involve only points from the so-called past or also points from the so-called future I am saying so-called because past and future have a meaning when the independent variable is time not otherwise but if the independent variable is time it is very easy to see that if the impulse response has non-zero samples for negative n then you are going to involve so-called future samples and if the impulse response has non-zero samples only for non-negative n that is n equal to 0 and positive n then the output at a given point depends only on the input at that point and on points before it so to speak points from the past all right we will come to that a little later but let us use this expression let us use this idea to recalculate the output that we did a few minutes ago. So there we had x of k given by 261 n minus 1 is not it now you can visualize h of n minus k so I will just write it down on the side you see when I want to calculate so h of n minus k look like this 1 1 at n so now I can recap I can recalculate y n using this very easily you see I can visualize that when n is minus 1 I can visualize this sequence being brought here and the 1 comes in contact with 2 you multiply them point wise the 1 gets multiplied by 2 2 plus this one gets multiplied by a 0 so the output is 2 here on the other hand when n is 1 and sorry when n is 0 then this 1 clashes with a 6 and this 1 clashes with a 2 and therefore you have 6 plus 2 that is 8 similarly when n is 1 you have this one colliding with this one and this one with the 6 you have 7 there and when n is 2 you have this one colliding with minus 1 so the product is minus 1 and this collides with 1 so you have 1 into minus 1 plus 1 into 1 that is 0 and finally when n is 3 you have 1 colliding with a 0 and 1 colliding with a minus 1 and therefore you have just a minus 1 there for n less than minus 1 it is very easy to see that these 2 ones collide only with 0s here and for n greater than 3 it is very easy to see again that these 2 ones collide only with 0s and therefore there is no non-zero sum at all no it is convenient to give another mnemonic to this operation and that mnemonic is to think of one sequence as being static and the other sequence as being dynamic here you might think of this sequence as the passengers on a platform and you might think of this sequence as the passengers inside the bogies of a train that moves and you might visualize that this train moves one step at a time and there is a handshake between the passengers on the platform and the passengers on the train at every point the output is interpretable as the net effect that the train feels due to this handshake right so it is a combination you know some passengers are stronger some passengers weaker you might think of that as the number put there of course some passengers make a handshake outwards and some passengers make a handshake inwards that is the positive and negative part of it so altogether at every point there is a net impact of all these handshakes and the train moves one step at a time and the net impact is interpreted as the product of the strength of the person inside with the strength of the person outside summed over all the passengers which does make a handshake and of course after some time you run out of passengers now this is easy to understand when you have a finite number of passengers train has the finite number of passengers and the platform also has a finite number of passengers but it is not too difficult for us to extend the idea to a context where the train has an infinite length and the number of passengers on the platform is also infinite it takes just a little more visualization to arrive at an interpretation for that context is that right so this is another convenient mnemonic to interpret convolution is that right convolution is a very fundamental operation in the study of linear shift invariant systems convolution is an operation in its own right what I mean by that is other than the fact that it occurs in the context of linear shift invariant systems in a very very meaningful way you can think of it as an operation between two sequences independent of the context what I mean by that is you could have taken two sequences multiplied them point by point to get an output sequence that is also an operation between two sequences you could have taken two sequences added them point by point and that gives me an operation between two sequences of course those are what are called point operations that means each point of the output involves only the particular point of the input and the particular particular point of the two sequences if you have sequences point by point then you are operating on in a point wise way so the output is a point wise function of the input and the impulse response or the output if you do not want to call it them input and impulse response if you have two sequences the output is a point wise function of the points of the two sequences. However, convolution is not a point wise function. In convolution, we must now see from this example that in principle, all the points of the input and all the points of the impulse response have come into the picture to create a point of the output. Again going back to the mnemonic of the train and the platform, the impact felt by the train at every move is dependent on all the passengers on the platform and all the passengers in the train. It is not passenger by passenger. Is that right? So, it is truly convolution is truly an operation between two sequences and not point wise at all. This must be emphasized and understood very clearly. So, are there any doubts before we proceed? It is a very important operation and we must be absolutely clear how this is done. Now I put before you an exercise to do and I also introduced the verb form of convolution. Convolution is a noun. It is a name of the operation. The verb form is convolved. So convolved xn which is 7 minus 4, 3, 6, 2 with hn given by 2 minus 3 minus 6, 0 and let me put another point there to make you work a little harder. So, I leave this for you as an exercise, obtain this convolution and we conclude this lecture with this exercise and just give a feel of what we are going to do the next time as a trailer. Next time we are going to take note of this relation between the input, the impulse response and the output and we are going to fulfill our first promise namely the whole reason why we want to look at linearity and shift invariance was that if I gave a complex exponential as an input I would have expected a complex exponential to emerge. But we said we need one more thing that is stability. And the next lecture we will begin by seeing what happens when we give a complex exponential as the input to an LSI system and then see whether we are happy with the system being only LSI or we need something more that would bring us to a few more properties that we desire of linear shift invariance system. Thank you.