We now have all that we need to model a channel.So what I'd like to do now is talk about the channel that we'llbe using in this lab, and particular in the online labsthat you'll be working with to get a better physical understandingabout the limitations of communication systems.And for our system, the step response willbe modeled by this formula over here.And so this formula over here is what we call the exponential step response.It includes this exponential component over here.And this exponential step response can model two types of different phenomena.One is changes in amplitude through the parameter k.And so the parameter k is describing what the final value of the responseis.And so typically, the value of a here will be something between zero and one.So when I take a to the power, let's say a is equal to 0.5,when I raise it to a power, what happens is that a to that powerkeeps on getting smaller and smaller.For example, if a was equal to 0.5, I raise it to the first power,I get 0.5 or 1/2, raise it to the second power,I get 1/4, third power, 1/8, 1/16, 1/32 and so on.So this term, as n gets bigger, keeps on getting smaller and smaller.So when n goes out and gets very, very large, goes out to infinity,this thing disappears and I just get one. So k then is the final value.And so here, you can see that this thing does indeed increase and approacha final value of k, in this case here, I have k is equal to one.The other property of this step response is that it is equal to zerofor all values of n less than zero.And the way I can enforce that is just taking this formula hereand multiplying it by u of n, which is the unit step.Remember, the unit step is equal to one for all values of n bigger than zero.And so if this is just one, it's not going to change anything.Everything's going to be dictated by what's happening here.But for n less than zero, u of n is zero.And what's going to happen is that that's justgoing to force everything over here to be zero, because zero times anything is justgoing to give us zero.And the final thing that this thing will dois it will enable us to model the blurring of the transitionsthat we'll see when we look at the actual channel responses.That's the fact that even though the unit step goes up instantaneouslyfrom zero to one, this one takes some time to reach its final value.In fact, because of the exponential, it will take infinite timeto reach that because this will always have some small value.Might be very, very small, but it will always have some small value.But the blurring of the transitions here is modeled by this a parameter.In particular, the value of that parameter awill determine how quickly this moves up.And it turns out that the larger that value of a is,or the closer that value of a is to one-- because the largest value of awill be one -- the closer that value is to one,the slower this transition is going to be.The easiest way to see that is just to look at this first value over here.If we look at this first value over here, in this case, n is equal to zero.When n is equal to zero, I remove this term here.I just get one minus a.And so if I have one minus a, this value here is one minus a,I know that if a, let's say, is 0.9, then one minus 0.9 is 0.1.But if a is some value that's small, let's say very close to zero,then what I'm going to get is, let's say, 0.1.I'm going to get one minus 0.1, or 0.9.And that means a very quick increase up to here.So let's take a look at a few different examples of different parameterssetting for this exponential step response.So these are all step responses that are generatedaccording to this formula over here.The only thing I've done is change the values of k and a.So this is the thing that we saw before where I have the value of kis equal to one, and the value of a is equal to 0.8.So remember, this value here was one minus 0.8, or around 0.2.Now what I'm going to do is I'm going to take k and multiply and set kto be 0.5, but leave a to be the same.And because k multiplies this whole thing, what's going to happenis that all the values here are just going to be multiplied by 0.5.So instead of going from zero to one, it's going to go up to 0.5.And this value here will also be dropped down by a half.And so instead of being 0.2 over here, it will be 0.1.But you can see that roughly speaking, it'sstill increasing at about the same rate, although to a lower value here.But the amount of time it takes to reach somewhere close to that valuelooks to be about the same.If I compare these two waveforms over here,here I've left the k value the same.So the final value of these two waveforms is exactly the same.But what I've done is I've changed the value of a.And so when I've changed the value of a, I've made it smaller in this case.And remember, when a is smaller, it corresponds to faster transitions.Remember I told you that the value of this first sample is one minus a.And so if I have a is equal to 0.5, then one minus 0.5is 0.5, or 1/2, versus one minus 0.8, or 0.2.So you can see that already at the first sample, it's moved up by much more.And so it's going to approach this final value of one much quicker than this.And so now given the fact that we understand this step response,then we can use the fact that the channel is linear and time invariantto predict the response to any kind of input.And what I'll do next is show you the responseto different bits where I've chosen different numbers of samples per bitto represent that bit.So this is a single bit waveform.It goes from zero to one, holds that one for a certain number of samplesper bit, in this case, 32, and then goes back down to zero.So this is just one single bit.And this is the response of the channel using those same parametersthat I started out this section of the lecture with.And so remember, how do we compute the response?Well, the input to the channel, remember,is this single bit waveform, which I canmodel as the sum of a unit step that brings me up at time zeroand brings me back down at time 32 over here.So this is evaluated n minus SPB, in this case, it would be 32.So that's a delay of 32 over here.And this is shown in the blue waveforms that I've shown you here,and if I combine this blue waveform with this blue waveform,I get this nice blue waveform over here.The output of the channel then, for this particular input,is just the step response that we talked about-- that'swhat we plotted in the previous page.And now if we take this waveform over here,then by homogeneity and time invariance,I know that the response will be the same as this response, only delayedby 32 and multiplied by minus one.So instead of going up from up zero up to one, it goes from zero down to minus one.And then I know by additivity, that I can compute the responseto this input, which is the sum of this blue waveform plus that blue waveform,as the sum of this black waveform plus this black waveform over here.And what you can see is that for all of these points here where it's zero,I'm going to simply get the sum being just this waveform over here.And at this point here, it's pretty close to one.And if I add this one plus this negative waveformover here, that's going from zero down to minus one, then I'm going to go from oneand slowly go back down to somewhere close to zero, at which pointI'm going to be adding plus one and minus one.Now what we can look at is what happens if wechange the number of samples per bit, making the bit time smaller.And so we can do that by simply taking the bit time and changing it to 16.And the only thing that really changes isthat this waveform remains the same.This one changes.It goes to a faster transition down to here, sample number 16.And that's introduced by this unit step downwards at time 16 now.And so what happens is the input basically remains the same as before,just delayed and again, scaled by minus one.And then I get the response by adding this one to this one.And you can see that I'm going to get this difference mainly just over here.So if I compare this one over here and this one over here,the difference is really just in this latter part over here.Now, if I continue to decrease the number of samples per bitthat I'm holding the bit, then I'm transmitting information faster.And you can see that the analysis, is going to be the same.I'm going to start going up, but at some point, I'm going to start pulling down.And as I try to push that limit and try to transmit informationfaster and faster by decreasing the bit time,decreasing the number of samples per bit,you can see what's happening is that it starts going up,but before it has time to reach the final value, whathappens is that this downward going step kicks inand starts pulling the value back down because of that additivity.And so now you can see it starts to go up.But at this point, it comes down.And then the rise here is canceled by the downward trend over there.And if I keep on going, you can see that when this bit is really, really,really short, that I only go up for a little bit of time and immediatelyI start coming back down again.And you can see that if I have a very, very short sample per bit,then it just goes up only a little bit and thenstarts immediately coming down.And so from this then, we can use this and extend this analysisto more general inputs.And the way that we do that is simply by building upon what we already knowand extending it to the more complicated situation.So here, I show you a more complicated bit stream correspondingto encoding the bit sequence 1010001 and so on at five samples per bit.And we already know that we can express thisas a sum and difference of a bunch of unit steps.And then by exploiting homogeneity and time invariance,we can predict the output to each one of these individual unit steps.For example, the unit step leads to the step response here,if I delayed by five and multiply by minus one,I just take the step response delay by five and multiply by minus one.And I continue to do that for each of the components that you see over here.And then finally, I can exploit additivityto take all of these individual predictionsand combine them together so that I finallyget my prediction for the input to outputrelationship of that channel for this particular input, x of n.And so we can take almost any input and predictwhat its output is to this linear and time invariant portionof the channel using the methodology that I've talked about.And so if we want to get a complete description of whatthe channel actually looks like and what is received by the receiver,then we have to add in one more term.And so to think about this, think about the situation where you're at a partyand I'm trying to talk to you, but there are additional people at the partyand they are also talking.And so what is actually the signal that's received by you?It's not just the signal that's coming from my mouth,but it's also additional noise that's comingin from other people inside the room, and perhaps there'ssome background music.And so the actual signal that's received by your ears-- and youcan think about now in this example here rather thanbeing an antenna as being your ears-- the signal that's receivedis not just what I'm saying, the informationthat I'm trying to transmit to you, whichis going through some kind of channel and it's being distorted in some way.For example, by the room acoustics.But also the fact that there's additional signals that are justambient within the environment.For example, the party noise that we model over here.But with our model, what we're going to dois we're going to assume that those two effects are additive.Just like we assume that this effect of the individual components herecombining together was additive, we're goingto assume that the effect of this additional ambient noiseis also additive.And in that case, we can do the same kind of thingwhere we do the pointwise addition.In this case here, we have zero plus this component over here will give us this.This nice and clean transition over here, plus this over herewill give us this more noisy looking waveformover here, plus the constant bias.And we can just write an expression for that using this.And so that then gives us the complete descriptionof what's going to be seen by the receiver.And then the receivers job then is to take this and tryto figure out what the original bit waveform thatwas sent by the sender over here was.