# Lec 21 | MIT 6.450 6.450 Principles of Digital Communications I, Fall 2006

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MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: We’ve just started

talking about wireless communication. We spent a lot of time talking

about how do you communicate, essentially on wire lines, where

essentially the only problem is white

Gaussian noise. Namely, you transmit a signal,

noise gets added to it, you receive the sum of the

transmitted signal plus noise and all of the going from

base band to pass band. All of that stuff is all messy

analytically, but essentially all that’s happening is that

you’re going from — you take a signal, you move up

to pass band, you add noise, you move back to base band,

in which case you get the original transmitted signal

plus the noise moved back to — move back — moved to base band. So what we wound up

with there was a relatively simple situation. Wireless communication

is not so simple. As you all know, if you use a

cell phone, and I’m sure most of you do, because of something

called fading — what we want to understand is

where this fading comes from, how it arises and some of the

things you might do about it. One of the things you should

recognize right at the beginning — and we won’t

spend much time talking about it — is if you happen to be trying

to use a cellular phone and there’s a big wall, which is

perfectly reflecting right in front of you and the base

station you’re trying to communicate with is on the

other side of that wall, you’re not going

to get through. In other words, there are some

situations where no matter how you design a cellular phone,

you just can’t get any communication. It’s part of life. What we want to do, however, is

to make sure that when you can get communication,

you will get it as well as possible. You will prolong the period, but

you’re still communicating while the channel’s getting

worse and worse — and part of the way to do that

is to understand what it is in the physical mechanisms that’s

making the problem difficult. OK, so to start out with this,

we’ll start out kind of easy. We’ll assume an input, which is

just a cosine of 2 pi ft at some ficed antenna — and this is radiating outwards

and the electric field — anywhere in free space, if you

go a long distance away. If you go a very short distance

from the antenna, if you study electromagnetism,

you know that all sorts of crazy things are going on. When you get very far away,

there’s something called the far field and essentially what

happens is that all of these disturbances close in just sort

of disappear and very far away what’s happening is that

the field strength — and this is true for the

magnetic field also, which is just in — if the electric field is

this way, the magnetic field is that way. They’re both propogating

outwards and they’re going down as 1 over r, just like

when we deal with linear systems, we never talk about

voltage and current anymore. This is why we use a square

root of minus 1 at this point — because after you learn about

voltage and current, you don’t have to talk about

them anymore. You just deal with one of them

for the function that exists someplace and the other

one just follows along from the impedance. Same thing happens for

electric fields and magnetic fields. You don’t have to bother about

both of them, unless you’re really trying to solve Maxwell’s

field equations, which is a very challenging

endeavor. I admire any of you

who can do this. I used to be able to do it many,

many years ago and I’ve given up on it because I decided

that’s for younger people than me. This far field has a very

simple kind of behavior, because what happens is it has

to go down as 1 over r. How do you know it has to

go down as 1 over r? Namely, the distance away from

this radiating antenna. If you look at a sphere, which

is very, very far away from the radiating antenna, all you

find is this field that’s radiating outwards and you look

at how much energy is radiating outwards. The energy which is radiating

outwards, there’s no place to lose it because we’re

transmitting just an open space now — or at least that’s

what we’re imagining at this point. So we’re transmitting

out in open space. Energy doesn’t get beaten up

any place so it just keeps going out until it disappears

out of the outer edges of the known universe. It travels out there with

the speed of light. That’s what Maxwell’s laws

say if you solve them. Since this sphere has an area

which is proportional to r squared, the only thing that can

be happening in this wave, which is propogating outward

spherically, is it has to be going down as 1 over r, because

that’s the only way for the power to balance out. You can’t be creating power out

in this free space and you can’t be losing power. Whatever you’re sending is just

radiating outwards, so we have to have this 1 over r

dependence any time you’re dealing with free space. The other thing that’s happening

is we have an antenna pattern. The antenna pattern is a

function of two angles. We’ll think of theta as being

the angle around this way and psi as being the

angle this way. If you like to think

of angles in some other way, be my guest. The only thing is, when we’re

radiating outwards through this sphere, if the antenna is

directional, it’s going to be radiating more in some

directions than it is in other directions and this simply

takes that into account. I’m not going to pay any

attention to this at all. It just exists and it’s there. For people who design antennas,

that’s an important issue: How do you make

antennas which are directional, which will shine

their energy in one direction rather than another? We’re not going to go

into that at all. We’re just representing the

fact that it’s there. This is just a factor which

says, how much loss do you have in the antenna and how much

of the power that you’re radiating goes in each one of

these directions and how does it depend on the frequency that

you’re transmitting at? Antennas are sometimes

designed to be rather frequency dependent,

so they’re tuned to a certain frequency. They work very well at that

frequency and on other frequencies, they

just go to pot. The other part of this equation

is that if I send a signal, as it radiates outward,

it’s going to be radiating outward at

the speed of light. Therefore, whatever I receive is

going to be delayed by the distance that I am away divided

by the speed of light. So this equation is something

you don’t really have to know any electromagnetics

to derive. If you want to find out what

this term is, yes, you’d need some electromagnetics. All this is saying is the power

has to be going down as 1 over r and you have a

propogation delay, which has to be going as r divided

by the speed of light. Now, if we look at what

happens when we put a receiving antenna out at some

distance r away from the transmitting antenna and in this

direction theta and psi, that receiving antenna is

going to distort this electromagnetic wave locally

around the receiving antenna, but it’s not going to distort

the whole thing. In other words, its power

is radiating outwards. It doesn’t know anything about

this little receiving antenna until it gets close to it, then

the electromagnetic wave gets distorted somewhat. The only thing that’s happening

because of this receiving antenna is that

there’s some added antenna pattern due to the receiving

antenna — namely, some added attenuation which multiplies

by the attentuation in the source antenna. You have both the source antenna

pattern, the receiving antenna pattern. We put the two together. We call that alpha and we don’t

bother about it anymore except recognizing that

it might change with frequency also. Then we have this propagation

delay between transmitting antenna and receiving antenna If

you’re transmitting in free space from one antenna to

another antenna, this is what happens, with some arbitrary

pattern here for the two antennas, which depends on

whether they radiate spherically or whether

they radiate in some sort of direction. That’s the received wave form. That said, if you look at it,

at the received field — this is supposed to be

valid for any — oh, sorry. Yes, that would help. This equation is supposed to be

valid for any frequency we want to transmit at, at

any time, and if we’re transmitting two signals, both

together, the response is going to be the sum of the

response to one signal and a response to the other signal. In other words, Maxwell’s laws

are linear and therefore the response that you get when you

solve Maxwell’s laws — I’ve never solved Maxwell’s

laws for anything this complicated and probably

none of you had either. You might have, but anyway, they

are linear and therefore, in fact, what’s going on is that

this gives you a system function which says what the

response is to any given input that you might want

to transmit. It says the response — at frequency f to a sinusoidal

input, which is what we were assuming before. We were assuming that we

transmitted cosine 2 pi ft. The system function is just this

antenna pattern times e to the minus 2 pi i times f

times the distance away divided by c divided by r. Namely, this takes into account

the propogation delay. The only thing it doesn’t

take into account is what the input is. The received field is then the

real part of the system function times what we already

assume that we were going to be transmitting. Namely, the real part —

e to the 2 pi i ft. So at this point, what we’re

doing is taking into account the fact that the solution to

Maxwell’s equations is going to be linear and therefore we

can just add up what happens for each frequency of input. What you notice from looking at

that is that when we have a fixed transmitting antenna, a

fixed receiving antenna and free space between them, we are

right back to the problem that we started with. Namely, white Gaussian noise on

a channel, because nothing is varying with time. There isn’t any fading. Nothing interesting

is going on. This is sort of like the case

of microwave towers. Microwave towers are set up and

they have nice directional antennas, nice horns which are

blowing at each other. Nothing changes except every

once in awhile is a rainstorm or something and the

communication goes to pot. Usually, you’re just

transmitting as if it was white Gaussian noise and you

usually view microwave towers sending microwave as being

almost equivalent to wire line communication. So there’s nothing very

interesting there. Now if the receiving antenna

starts to move — at this point, we are

transmitting from a fixed sending antenna. We have a receiving antenna,

which for example, is in a car where somebody’s running along,

driving a car with their feet and talking on

two cellular phones at the same time — and the car’s driving along

at 100 miles an hour and something’s going to

happen soon, but it hasn’t happened yet. At this point what we’re

interested in is not the response of some fixed place r,

but what we’re interested in is the electromagnetic

field in the absence of receiver at this point,

which is moving. We’re interested in the

electromagnetic field at a point r 0 plus v2, where v is

the velocity of this vehicle. We’ll assume for the time being

that the vehicle is going directly away from the

transmitting antenna. If it’s going at some angle,

it just changes these equations a little bit. And the electric field there,

before we put the car in the car changes all of the field

patterns, but just changes it in a local way again. Just like when we put the

receiving end antenna in a fixed location, it changed all

the local field equation, but it didn’t change anything

globally. Again, what we have

when we put in the receiving antenna — is now the electric field

at a point r, which is varying with time. There’s a time dependence

with this now. 1 over r 0 plus v.t. times the real part of this

antenna pattern — which we’ll assume remains fixed — times e

to the 2 pi if times t minus r 0 plus vt over c. This is for a velocity away

from the antenna. That’s just what this same

equation says, but we can now interpret this nicely if we take

the vt over c and combine it with the ft here and then we

get something which looks like this antenna

pattern again — e to the 2 pi i f times 1

minus v over c times t. This v over c here is

just this term here. It’s coming down to there. Nothing mysterious has happened

here, but what you see is this well known

phenomenon called Doppler shift. If you throw some screaming

person over a cliff, what you’ll hear coming back to you

is a scream in velocity much smaller than the actual scream

that the person is actually transmitting to you. You’re all familiar with this. You’re familiar with having

planes fly overhead and when you hear them coming towards

you, you hear a higher pitched sound. When it passes by you — this

is a nicer example than throwing somebody over

a cliff, obviously — and you then hear a lower

frequency sound as the plane starts moving away from you. This Doppler shift is a well

known phenomena as far as sound is concerned. The same phenomena exists with

electromagnetic radiation. And there’s nothing more

to it than just this — it’s just that as you are

transmitting from here to a point which is moving away, it

keeps taking longer for the electromagnetic wave to get out

to there than it takes to get here, so if you look at the

wave fronts going along, the peak of the wave as it

travels along, the peak of the wave takes a little longer to

get out here than it took to get here, which means that

from the viewpoint of the receiver, it looks like the

receive frequency is much smaller than it was when

it was actually being transmitted. We get this thing called

the Doppler shift. Just to get some idea of the

magnitude of this Doppler shift, what you’re interested in

is the speed of the vehicle divided by the speed of light. That’s the relative

change in the frequency that you observe. The situation here is quite

different than it is in sound. Sound travels rather slowly. Light travels pretty fast and

therefore, you need a really rapidly speeding vehicle to make

this be any appreciable fraction of one. So it looks like this is

a very small effect. The trouble is, what you are

multiplying this effect by is the carrier frequency,

which can be up in the gigahertz range. To look at it another way, we

are looking at situations where the wavelength is small

fractions of the meter. What this equation says is that

any time this receiving vehicle moves by one wavelength,

namely a small fraction of a meter, the crest

of this wave goes from maximum down to minimum back up

to maximum again. In other words, in a quarter

wavelength, it will go from maximum down to zero. What you are observing at this

carrier frequency is very, very different and it keeps

changing rather rapidly. All these other terms are

just junk, of course. This f times r 0 over c — all that is is just a fixed

phase difference, so we don’t care about that. This is just some fixed term. This quantity here is changing

with t also. If we’re thinking of a distance

away, it’s several kilometers and we’re thinking of

the amount of time for this to become an appreciable

fraction of one. For this to becoming an

appreciable fraction of this, that’s a pretty long time. The amount of time for this to

change appreciably is seconds or minutes. The amount of time for this to

go through a wavelength change is milliseconds. So despite the fact that you

see this sitting in an important place down there,

this is not important. Everything that goes on as far

as fading is concerned is tied up with that term there. This is important. So we now have a system

which is linear. We still have the linear field

equation, but it’s not time invariant anymore. It’s changing with time. The response is changing

with time. You send an exponential,

what you received. If you have a linear time

invariant system, when you send an exponential of frequency

f, you receive an exponential with frequency f. The only thing that a linear

time invariant system can do is change the phase of that

signal and change the amplitude of it. Can’t do anything more

complicated than that. That’s why we love to study it;

because it’s so simple. Now we have something more. We have a system that can also

change the frequency of what’s getting received. This small change down here — and of course, if obstacles get

in the way or something then there’s this huge shadowing

difference and all those important things,

but you can’t do anything about that. You can do something about

this, which is why we’re focusing on that. Let’s go to the next example. Incidentally, that example

is no problem at all for communication. I’ll show you why

in a little bit. You can get around that problem

very, very easily and I’ll show you why. This is a problem you can’t

get around so easily. Here we have a vehicle, which

is travelling, say, at 60 kilometers an hour. Person’s talking on his two

cell phones, has his eyes closed because something

surprising is happening, there’s this big reflecting wall

right in front of him and he doesn’t see it at all, so

he’s speeding into this wall. We’re going to analyze

this problem right before he hits the wall. We have two paths here. We have one path which is the

path from here out to the vehicle, which has a

length, r of t — this is the distance away from

the sending antenna to the receiving antenna. We have another path which has

a length d, then it gets reflected and this distances

is d minus r of t. The total length — and you’re adding up this length

with this length — is 2d minus r of t. The reason is — do I have

an extra picture there? No, I didn’t make my extra

pretty picture. The reason is that one way to

deal with electromagnetic radiation is when you see a

wall, the thing that happens is that you get a reflection

which is coming back this way. The reflection has a strength

which is equal to the radiation that you would get if

there weren’t any wall — uhhuh except, of course, that

you’ve changed directions. In other words, this wall has

generated a new plane wave which is going backwards, which

is just enough to make the electric field strength on

this wall equal to zero, because we’re assuming a

perfectly reflecting wall. You can satisfy Maxwell’s

equations by having this incoming electromagnetic wave. You would like to have an

outgoing electromagnetic wave, but you can’t do that because

there’s no way for the wave to get through it. The only way you can do it is to

generate a new wave, which is moving backwards, which

cancels out the incoming wave right at this point. We really have a path here of

length 2d minus r of t and as a result of that, the electric

wave has two components. One is the component we were

dealing with before where there’s this Doppler shift

because this is moving away from the sending antenna. The other term — in fact, we are moving closer

to the wall and the distance in this path is getting shorter

and shorter as doom approaches. Here we have a positive

Doppler shift. Here we have a negative Doppler

shift and we have these junk terms

in both places. One is fr 0 over c. One is 2d minus rc over c. Here we have r0 plus vt. Here we have 2d minus

r0 zero minus v.t.. As we said before, this term and

this term are not changing very rapidly. It makes it a little easier to

analyze this if we say, let’s suppose that this is

equal to this. In other words, we’d like to

look at this right before the car strikes the wall. That also is where this

approximation is best because for those of you who have

studied electromagnetism, you know that if a plane wave

impinges on a wall, funny things happen. If you look very far away from

the wall, you will find this electromagnetic wave, which

looks like a plane wave if the wall is distant from

the source. So this electromagnetic wave

coming in, there’s this wall in here and what happens to the

electromagnetic radiation is outside of the wall that’s

going to go out past the wall — and because of Maxwell’s

equations, it just sort of gathers together beyond

the wall and it sort of comes together. What you find is a disturbance,

which is just around the wall. Far away from the wall, you get

the same electromagnetic radiation that you had before

and close to the wall you have this disturbance. If you look at the

situation — here it is. If you look at what happens here

and the wall is not big enough, if the wall is very

small, this reflection is really going to look like what

happens when you have an electromagnetic wave hitting

the wall and the wall then re-radiates an electromagnetic

wave, which very far away, this wall just looks like

a point source. What you have is instead of a

1 over r, 1 over 2r minus 1 over 2d minus r attenuation,

you have a 1 over d attenuation multiplied by a 1

over d minus r attenuation. If you didn’t get all

of that, fine. Doesn’t make any difference. The point that I’m trying to

make is that this analysis is really limited to the case where

there the wall is rather small, where the wall is very

large, because otherwise you won’t have just this plane

wave radiation effect. What we wind up with these two

terms instead of one term. If I throw away all of the phase

terms and I assume that the denominators are equal — I’m going through this for

some sort of reason — I wind up with two sinusoids:

e to the blah and e to blah plus. When I take the real part of the

sum of two sinusoids, and I look in all of the high school

books I can think of about elementary geometry and

playing around with sine waves, what I find is that this

collapses into 2 alpha times the sine of 2 pi

ftv over c times the sine of 2 pi ft. In other words, it collapses

into a sinusoidal term, which is the major part of this term,

ft, and the major part of this term. In other words, I can cancel

out the terms here that are the same as the terms here. When I cancel out those

same terms, that’s the term that comes out. When I look at the other terms,

I get an e to the minus vc over t and an e to the

plus vc over to t. When we look at e to the minus

vc over to t and e to the plus vc over t, even I know how

to deal with that. It looks like either a cosine

term or sine term, depending on whether the sines

in the same or the sines are different. What we have at that point is

this sinusoid is really a sinusoid at the carrier that

we’re transmitting at. This term here is really

something which expands and contracts slowly. So it’s a beat, which says that

if you’re transmitting from this source at this

receiver, what you’re hearing is something which contracts,

expands, then contracts, then expands again. There’s nothing you can do about

that problem either. There’s just no energy there

part of the time. This sine term here is running

along, sort of changing from maximum to minimum at a few

milliseconds time period. If you have this vehicle

traveling at 60 kilometers per hour — you just work out the numbers

there with the velocity of light and all of that stuff. You find that you really can’t

communicate over your cellphone in that situation,

because of these peak frequencies. It’s too fast. It’s too slow to ignore and ride

over it and it’s too fast to be able to get

all of your data transmitted before it happens. It sort of is a catastrophe. So that’s what happens because

of Doppler shift. You get a response which is

periodically fading at the Doppler frequency. This is called multipath

fading or fast fading. It’s called fast fading because

it happens so fast. It’s called multipath fading

because it happens because of multiple paths, which each

have lengths which are changing relative

to each other. Let’s go back and look at the

thing we had before, where we just had a moving antenna. Here you have a Doppler

shift also. Why doesn’t it bother you? Here you have this Doppler shift

and you’re transmitting, let’s say, a gigahertz and what

the receiver is getting is the gighertz minus — perhaps a kilohertz

or something. So why isn’t that a problem? If I demodulate at the carrier

frequency, I’m sort of in bad luck because I have a

signal then which is changing very rapidly. I have something which looks

like a time varying system. But what’s going to happen? If I use the same kind of

frequency recovery system that we talked about earlier, that

frequency recovery system has all the time in the world to

track that frequency which is one gigahertz minus

a kilohertz. It can track it perfectly, which

says so what happens is we start out with a signal. We move it up in frequency

by one gigahertz. The Doppler shift moves it

down by a kilohertz. We track that frequency, then

we move it down again by a gigahertz minus a kilohertz and

everything works fine and nobody even knows that there’s

any Doppler shift there. So the problem is not

Doppler shift. The problem is multiple Doppler

shifts which are at different speeds relative

to each other. That’s an important point and we

will come back to it as we move along. If you put all of those phases

that we neglected in the analysis that we just went

through, this is the equation that arises. I write this down not because

it’s important, but because the notes — this is

in lecture 20 — have an error. In the sine term, it fails to

put an i in, which should be there and this is the correct

term and that is off like ninety degrees. If you write this down, you will

then see what’s going on. Equation 7 in the notes has

an e to the 2 pi f.t. minus f.d. over c, which is not

the right thing. As we said, the fading is due

to Doppler spread between different paths. The single Doppler shift does

not bother us at all. If you have a vehicle which

is traveling away from the sending antenna and you have

some kind of reflector, which is not something you’re running

into, but which is a reflector above, a reflector

below or something like that, the thing which is going to

happen is that both of those paths then are going to have

roughly the same Doppler shift in them and if they both have

roughly the same Doppler shift, it’s not going to be this

kind of beat cancellation that we have here, which

is something you really can’t get rid of. The other thing that this points

out again is that this variation is going to be in

terms of minutes or seconds and anything you’re doing to

track the signal is going to be adequate for that until you

get to the point where there just isn’t enough energy anymore

and then of course you have to move to a different

base station or something else. Want go through one more example

of electromagnetism because it’s so surprising — at

least, it was surprising to me when I found this out. If you have a sending antenna

— think of this as a base station, which is high

up at about maybe 15 meters or something. It’s sending to some receive

antenna, which is at some height above the ground — usually quite a bit smaller. Suppose there is some more or

less partly reflecting plane, like a road. Here we have a vehicle which is

travelling along a road and we have ascending antenna, which

is also close to the road, which is sending the

signal so we have two paths, one which is the direct path

from sending antenna to receive antenna. The other is a reflecting path

which goes down here and comes back up again. The rather surprising thing

here, the thing that I couldn’t believe when I saw it,

because it contradicted all of my intuition, is that

when r gets quite big, the difference between these two

path lengths goes to 0. Is that surprising

to anybody else? Anybody awake enough

to be surprised? This difference in path length

here really goes down as 1 over r. That’s an easy geometric

problem to solve. You just write down what this

length is and the sum of these two lengths and you will find

when you do it that the difference is proportional

to 1 over r. What happens then is that as

r gets big, these two path lengths get closer and

closer together. As they get closer and closer

together, eventually they’re much closer than one wavelength

to each other. When they’re much closer than

one wavelength to each other, the thing that happens is that

we get a reflection in the electric field here, so. This field and this field

are going to be canceling each other. They’ll be canceling each other,

except for this phase difference which is proportional

to 1 over r, and the phase difference which is

proportional to 1 over r is the only thing that gives us

any power here at all. As we move further and further

away, that phase difference goes down as 1 over r, which

means what’s happening is the overall electric field that you

receive here, instead of going down as 1 over r, which

it would in free space, is going down as 1 over

r squared. Since it’s going down as 1 over

r squared, it means that the power that we’re receiving

is proportional to 1 over r 4th instead of 1

over r squared. The analysis of this is not

particularly important. Why it is that surfaces such as

macadam reflects so well is not particularly important

to us either. The point is that when you look

at all the problems of electromagnetic radiation, in

actual situations, you find some situations which behave

like this, you find some situations which behave

like this nice plane wave and free space. You find other things where in

fact the radiation goes down as 1 over r 6th, rather than 1

over r squared, so that we wind up with each path has its

own particular kind of attenuation in with path lengths

and it’s kind of hard to figure out what

all of those are. Then you go on and say that

things are even worse than this because sometimes you’re

communicating through a wall which is only partly absorbing

and if you’re transmitting through a wall which is partly

absorbing, then the attenuation is exponential

in the width of the wall. You have a lot of different

paths, all of which are very messy electromagnetic radiation

problems and all of which have attenuations which

range from 1 over r squared to 1 over r 6th, sometimes with an

exponential thrown in for good measure, which says that if

you really try to find the electromagnetic field at a

wireless cell phone, you’re in very deep trouble. It’s certainly not something

that your cellular phone is going to solve for you and it’s

certainly not something that you’re going to have

solved ahead of time and program into your cellular phone

or into the base station or anything else. The question is, what

do we do about this? One thing is, we’re not going to

study these electromagnetic phenomena any further because

it’s a losing game. If the thing that you’re

interested in is finding out where to place base stations,

all of this kind of analysis is useful and you can go much

further with it and you should go much further with it. If your interest is in, how do

you build third or fourth or fifth or 20th generation

wireless systems? All the electromagnetics that

you study is only going to give you gross ideas of what

kind of phenomena you have to deal with. We already have some idea of the

kind of phenomena we have to deal with. We have to deal with paths

which have different attenuations on them, which

have different propogation delays on them, and all of

these multiple paths are things that we have to somehow

deal with without analyzing them in detail. The question is, how

do we do this? Everything that we’ve done so

far is called rate tracing. In fact, even with all the

complexity that we’re dealing with now, we have highly

oversimplified it. Each of these paths that we have

is going to give rise to an attenuation. We’ll now call the attenuation

beta of j of t and a propogation delay, which we’ll

call passive j of t and the propogation delay is just what

we get by assuming that a plane wave is going from source

to destination and we add up the distance from

source to reflector to destination and that gives us

this propogation delay. These are going to vary with

time, but in our rate tracing approximation, we’ve assumed

that they’re independent of frequency. I originally assumed that the

antenna pattern was a function of frequency. We didn’t want to say

anything about that. So we have a total of

capital J paths. we put in an input of cosine 2

pi ft, what’s going to come out is an electromagnetic

radiation, which is a sum of these different attenuation

factors times e to the 2 pi ft minus this propogation delay. Everything you can do

with ray tracing is included in this formula. What you might be able to do

in a wireless system is you might, by looking at the

received wave form and knowing things about the transmitted

wave form, you might be able to figure out what these

attenuation factors are and what these propogation

delay factors are. Just like when we tried to do

frequency recovery, we can find out what the transmitted

frequency was. We can play the same sorts of

games here, but they’re harder and we’ll talk about

that later. If you ever hear the term rake

receiver, a rake receiver is a receiver that in fact measures

all this stuff and responds to it and we’ll talk about that

probably next Monday. If we want to look at that the

reflecting wall just as an example of what this formula

means: beta 1 of t, namely for the direct path. We have an attenuation, which

is the magnitude of the antenna patterns divided

by r0 plus vt. For the attenuation on the

return path from the wall, it’s the same alpha. Why is it the same alpha? Because we assumed it was the

same alpha to make things simple for ourselves

— divided by 2d minus r0 minus vt. If you look at these propogation

delay terms, the propogation delay terms are r0

plus vt divided by c and this gives us the Doppler shift that

we’re interested in here. We’re also going to have an

extra term here, which is really caused by the phase

change at the transmitting antenna and the phase change at

the receiving antenna and a phase change at a reflector,

if there’s any there. We have the same sort of term

at both of these places. The reason that I talk about

that is if you look at the electromagnetic wave that you

received for this reflecting wall problem that we’ve talked

about a good deal, the second term is there with a negative

sign rather than a plus sign. Here, everything is put

in with plus signs. You can create negative signs

by phase changes of pi. So the assumption is, we put

in a phase change of pi as part of this term here. So all of these terms can be

expressed in this general forum here. As we said before, you only

have two choices with a cellular system. You cannot solve the

electromagnetic field problem at the cell phone. The person using the cell phone

is not going to do it. The cell phone is not

going to do it. The base station is not going to

do it and you’re not going to store all those changes

because these radiations change remarkably within a

period of just small fractions of one meter. You have a coverage here, which

in fact, is at least area coverage of one kilometer

times one kilometer. Then the reflectors are going

to be moving also, so you can’t deal with them

very easily either. It’s a hopeless problem,

to try to solve the electromagnetic problem and

store it some place. Electromagnetism helps us to

limit the range and likelihood of choices, but it doesn’t

help in actual detection. So we’re now going to deal with

the kind of thing we just talked about, which is this sort

of general expression for electric field in terms of

attenuation factors and phase changes, as opposed to

anything which is wh much more detailed. We’re going to define a channel

system function as just this sum of these

attenuation terms times phase change terms. The reason that we’re doing this

is that if we put in an input — e to the 2 pi ift,

then what we get is this system function here — times x

of t, which is e to the 2 pi ift, so we get this

quantity here. Here’s the minus

2 pi if tau — that’s that term coming down

there and here is the e to the 2 pi ift coming down here. So all of this term and this

term are both included in this system response term. This is linear also, so we know

what the response is to an exponential that we put in. I’m cheating you a little

bit here by going from real to complex. And the notes do that a little

more carefully, but we ought to be used to that now. If I put in an input, x hat of

f e to the 2 pi ift df and integrate it. In other words, if I put in an

arbitrary input, which I represent in terms of its

Fourier transform, I now know what the response is to

every x-hat of f. It’s given by this

response here. So I just integrate over that

and I find that the response y of t to an arbitrary input now

is the integral of x-hat of f, h-hat of ft, e to the

2 pi ift df — namely, the same game

that we always play. Namely, that’s just the system

analysis way of looking at arbitrary systems in terms of

their Fourier transforms. So the output y of t is just

this integral here. Important point: When you look

at this, this says this is the same as any old linear time

invariant system. This is not a linear time

invariant system. If you try to take the Fourier

transform of this to get y-hat of f, you’re not going to get

x-hat of f times h-hat of f. Namely, this is not

equal to that. Why isn’t it equal to it? First reason is, try to take the

Fourier transform of this and see what you get over here

and when you deal with the fact that there’s a t in here,

you will find that there’s nothing you can do. You’ll find you’re stuck. So you can’t derive this

equation here. Next, argument is this quantity

here is not a function of t. This quantity here is

a function of t. You can’t have a quality between

something which is not a function of t and something

else which is a function of t. It just can’t happen. The final argument is, if you

look at what’s happening here, when you put in a single

frequency — when you put in x of t equals e to the 2 pi

ift, what comes out? In terms of this reflecting wall

example, the thing that came out was not one sinusoid,

but two sinusoids. One sinusoid a little bit above

the carrier, the other sinusoid a little bit

below the carrier. In other words, because of

Doppler shifts, when you put in a sinusoid, what comes out

is not a sinusoid, but a modulated sinusoid. It’s something spread out over

a region of frequencies. So one of your favorite tools

for dealing with linear time invariant systems is

no longer adequate. Doesn’t work. Just make a mark of that,

because every time you see a problem like this, everytime I

see it, the first thing I try to do is go through that most

familiar and most favorite form of linear system analysis,

which is the Fourier transform of convolution, is the

same as multiplication in the frequency domain. You cannot do that anymore. However, convolution still work,

so that’s the next thing we want to look at. So the thing that we have is

the output of the system is now going to be this integral

over frequency — x-hat of f times the frequency

response function for the linear but time varying

system, times e to the 2 pi ift. This is what we derived on the

last page, on the last slide. It is the thing which just

automatically happens here. If I let h of tau and t be the

inverse Fourier transform of h-hat of ft — and here what I’m doing is I’m

regarding t as a parameter. So this is a function of tau now

and this is a function of tau for a given t, I can take

the Fourier transform of this. Take the Fourier transform of

any old thing at all, so long as it’s L2. I will sort of half-pretend

it’s L2. We’ll worry about that later. So this is the Fourier

transform of this. So then, the thing that happens

is that y of t is going to be equal to this

quantity here, except in place of the system function, h and f

of t, for a particular value of t, I’m going to put in this

inverse Fourier transform. So it’ll be h of tau and

t, e to the minus 2 pi i of tau d tau. This integral here is just

h-hat of f and t. If I take this quantity here

and I move this term inside and I interchange orders

of integration — incidentally, when we’re dealing

with wireless, we’re going to forget about all of the

nice things that we know about L2 functions. There’s just too much new stuff

that’s going on here to worry about that. So what you want to do is just

take Fourier transforms like while, interchange orders of

integration, interchange everything you want to and

simply forgot about all the mathematical problems

that might arise. After you understand this, at

that point, go back and straighten out the mathematical

issues. This in fact is the way we deal

with any problem, or the way you should deal

with any problem. You don’t bring the mathematics

in unless it’s going to help you solve

the problem. You don’t bring it in to

frustrate yourselves. So the thing we’re going to do

now is to interchange these orders of integration. We’re going to integrate over

tau on the outside and f on the inside. So we’re going to bring the

function of tau outside. This is an integral in tau. The function of f is going to go

on the inside. we have e to the 2 pi ft — that’s this quantity here. We have this term there. When we look at this, we see

something very nice because this is in fact just the

Fourier transform of x of t minus tau. When we take that out, what we

get is the integral of x of t minus tau times h of

tau and t be tau. In other words, this is time

varying convolution. Nice, simple equation,

makes a lot of sense. It says you have this impulse

response here. h of tau and t now can be interpreted as the

response at time t to an impulse tau seconds earlier. If you have a system which is

changing very, very very, slowly, then this is essentially

just a function of tau and it’s the usual impulse

response that you’re familiar with — namely, this convolution

equation gives you the response at time t to an

impulse tau seconds early. Now we just have something which

says this is a linear time varying filter and in all

the cases, we’re interested in this linear time varying filter,

changes its impulse response very slowly

as time changes. Relatively fast change

with tau, relatively slow change with t. So this is very similar to

linear time invariant convolution. Channel behaves like a slowly

time varying filter and that’s the bottom line of this. For these ray tracing models we

were looking at, the system function is a sum of terms at

the sum of attenuations times phase change terms. If we take the inverse Fourier

transform of this — remember, we’re taking the

inverse Fourier transform on f and putting in a tau. So we’re taking this inverse

Fourier transform for a particular t. We have a function

of f and of t. We take the inverse Fourier

transform with respect to the f and get a tau here. This then becomes this

quantity here. How do I interpret that? If I look at a single term

here, what is it? A single term here is just

an attentuation factor, a constant times a sinusoid. At a particular value of

t, this is just the constant here also. So for a particular t, all

I have is a sinusoid. What’s the inverse Fourier

transform of a sinusoid? I told you all along that it

doesn’t exist, but for the time being we will assume that

it’s what you learned early, that the inverse Foruier

transform of the sinusoid is an impulse. So here we are with

our impulse there. This says that the response at

time t to an impulse at tau is going to be zero unless tau

is equal to one of these propogation delay terms. In other words, I have a system

where I’m putting in an input and this input comes in. The response to the input is

a number of different path delays and at each path delay,

what I’m going to get out of the system is just a delayed

and attenuated version of what I put in. Namely, the system isn’t a

function of frequency at all. That’s what I get through

using ray tracing. I mean, it’s one of

the consequences of using ray tracing. So what I wind up with is a

system function which is a string of impulses and the

output then, the convolution of this with that — you’re probably better at

working with impulses than I am — and it’s y of t is just

the sum of these attenuation terms times x of t at these

various delays. So what we’re doing is we’re

putting in an arbitrary input. What’s coming out is that

attenuated input coming out at various different times,

due to these various different paths. I get various paths that are

delaying the input by different amounts and out

that delayed input comes at various times. This is a nice sanity check

because if you think about it, that’s exactly what ought

to come out of here. On the other hand, you ought to

wonder about this impulsive impulse response, because that

clearly doesn’t make any sense physically. So what’s going on here? The thing that’s going on is

that when we started, we said, if we’re putting in a narrow

band input, we don’t care about the frequency response

because the frequency response on these different paths cannot

change very quickly and therefore, we’re just going

to have a fixed frequency response term. Then we’ve worked with that

thing which is not a function of frequency and then finally

we get down here, where in fact, what we’re doing is

looking at the output. Due to a bunch of delayed input

terms, if this input term here — if x of t is in fact the

narrow band term — I guess the way to see

that as to look at — where do I look at it? I want to look at this

expression here. If I have a narrow

band or maybe — I guess this one

is better here. Let’s look at this expression. If my input is in fact narrow

band, it’s only going to be non zero over a small range

of frequencies. If it’s only non zero over a

small range of frequencies, I don’t care what this is, except

over that small range of frequencies. All of this gets filtered out. In other words, this filters

out this, opposite of the usual case. So I don’t care about what this

is at different frequency ranges and therefore, we simply

have the consequences of this, which is something that

you see in linear system theory all the time. We’ve sort of ruled out impulses

and sine waves because they don’t carry

information, but in terms of looking at things as

intermediate points and going through filters and things like

that, they’re perfectly fine So here, for simplicity, we

assume that these channel filters really do

not have any — don’t respond to frequency

changes, whereas in fact they do, and all we’re doing is

modelling them in certain frequency bands, which, when

we get all done, is what really gets rid of all our

problems, due to this sort of input because this input is

smooth now and therefore, the output is smooth also. The next thing I want to spend

a little bit of time on and we’ll come back to it next time

is, how do you deal with all of this at baseband? I should warn you here that the

notes don’t do a terribly good job of this. They have all of the results

that you need. They don’t seem to put

them in a very nice, well organized fashion. I’m not sure there is a

nice, well organized fashion to put them in. But anyway there’s a lot of

stuff going on when you try to move this down from pass

band down to baseband. I will try to change the notes a

little bit to make it clear, but I’m not sure that I can. The kind of system that we’re

looking at now is our usual QAM type system, which can

be generalized somewhat. We have a binary input

coming in. We have a baseband encoder. That baseband encoder is

creating baseband signals, which are being added together

to give us a baseband complex input to the channel. This is being frequency

modulated up to some function, x of t, which is just the real

part of ut times e to the 2 pi if of ct as usual. So we now have a real part of

this signal here, modulated up by the carrier frequency. This is going through what we’ll

now regard as a time varying channel filter. We talked a little bit about

channel filters before when we were talking about Nyquist

theory, because we said in general, you want to take your

input, you want to pass it through a pulse p of t. That goes through another

filter, which is some h of t and that goes into another

filter, which is at the receiver, which is q of t and

you want the product of t and you want the convolution of all

of those to satisfy the Nyquist criteria. So we’re back with that in

spades because now this is varying with t also. So this goes through

this channel filter, up at pass band. The channel filter up at pass

band, we’ve seen that one of the things that it can be viewed

as doing is putting Doppler shift into this input. So what comes in at some

frequency f is now going to be coming through here at some

slightly different frequency. We then add white noise to it. We then get y of t out, which

has now been shifted around and smudged in frequency

a little bit. We go through a frequency

demodulation. We get down frequency

to modulation by this carrier frequency. We get down to v of t, which

is now a baseband complex function again, which is

supposed to be the same as this except for the white

Gaussian noise and except for the fact we’ve gone through this

filtering operation here. Then we want to do base band

detection at this point. What we would like to do and

what will make life a little easier for ourselves because

we all got sick of this business of looking at filters

at baseband and also looking at filters at pass band, we

would like to be able to take some baseband equivalent

of this filter here. So what we’re going to do

is look at the baseband equivalent. The system function at baseband

corresponding to this system at carrier frequency

will just be this response moved down by s of t. In other words, when you take

what comes out of here, you multiply it, and you shift it

down in frequency by s sub c, what happens is that this gets

shifted down in frequency by s sub c and this channel filter

gets shifted down in frequency by f of c and therefore, what

happens is the effect of passing y of t through

a base band filter — h-hat of f plus fc and t or

0 for f, less than or equal to minus fc. So far, this is all kind of

straightforward and not too mysterious. So you wind up then in — and this is pure analogy to

what we did before — the output is then going to be

the integral of this Fourier transform of the input. The system function for the

baseband filter times e to the 2 pi ift df. Same equation as we had before,

but before we did it at pass band and now we’re

doing it at baseband . For the ray tracing model that

we looked at, this function here down at baseband is going

to be the same as it was at pass band, except in place of

the 2 pi i tau jft, we now have f plus the carrier

frequency times tau jft. So when we take the inverse

Fourier transform of this, we’re given parameter t. What we’re going to wind up with

is this quantity here. This is the same as we had

before, with the difference that now we’re stuck with this

carrier frequency times propogation delay, which

is occurring in time t. We still have the same delta

functions here and the output v of t is the same time variant convolution as we had before. I’m doing this much too

fast for you to follow this in real time. What I’m saying is, this is

exactly the same thing as we did before and the only thing

new that happens is now this carrier frequency term is coming

in here on this term. If we’re using the ray tracing

model then, v of t is equal to this quantity here. The reason I write this down is

that this shows you quite simply what’s going on in terms

of the Doppler shift, because these terms here, these

delay terms are changing with Doppler shift and now we’re

going to see exactly what they do to us. So we’re going to represent

the propogation delay on each path. That’s tau j of t, which is

tau j prime, which is the propogation delay of time zero

minus the Doppler shift times t divided by f. This is just Doppler shift,

which is a frequency term, and it’s increasing linearly. The propogation delay is

increasing linearly with t. The Doppler shift is a shift in

frequency and, as I think we saw before, you need a 1 over

f to compensate for this Doppler shift. Namely, this is in

terms of hertz. This is in terms of hertz. This is in terms of time. This is in terms of time. So you need this frequency here

to be dimensionally right at any rate. If we’re using narrow band

communication, then v of t is just going to be this difference

here, which is just the last equation with the

Doppler shift put into it. So all of this says that when

we’re now looking at things at base band instead of a pass

band, what has happened is that this term has been added. Before, we just had these delay

terms here and we had these attenuation terms. Now what’s happening is because

we are modulating up with carrier frequency s sub c,

we then get a Doppler shift that moves us down

a little bit. We’re then moving

down by s sub c. What we wind up with is

something which is off by that Doppler shift. If we only had a single

term, we wouldn’t be demodulating by s sub c. We’d be demodulating by

something which would get rid of this term for us. Then all we’d have is just

some arbitrary phase term here, because this

isn’t important. This is just a phase term. This is the term which

is important. The reason for going through

all of this — and I’m going to come back

to this next time — is that when you use frequency

recovery at the receiver and you always use frequency

recovery at the receiver, what’s going to happen is you’re

not going to shift down by s of c. You’re going to shift down s of

c, plus some average value of Doppler shift. If they have a bunch of terms,

each with different Doppler shifts in them, when you try to

measure how much Doppler — when you try to measure what

the received carrier is, you’re going to be frustrated

by all of these different Doppler shifts. You’re going to come up with

some average value in your frequency circuit, which means

that in place of this quantity, you will get something

which replaces each of these D sub j’s by not the

actual Doppler shift, but by how far this Doppler shift

is away from the main Doppler shift. So these terms here, which are

the things that make this system function in here change

with time, are in fact changing according to how far

away each of these Doppler shifts are from the main Doppler

shift, which means that the amount of time this

system function in here is going to remain stable depends

on how much the Doppler shifts vary from each other on

these different paths. In other words, the important

thing is the Doppler spread between the biggest Doppler

shifts and the smallest Doppler shifts. That Doppler spread is going to

determine how long you’ve got something that looks like a

linear time invariant system function which says that every

once in awhile, if you’re trying to measure what’s going

on at the channel, at intervals of time approximately

one over two times the Doppler spread, you’re

going to have to change those measurements. So whether you can make cellular

telephony work or not depends on whether you can make

measurements at a speed which is equal to one over two

times the Doppler spread. I’m going to do more about

that next time. I don’t expect you to

understand it now. Maybe after you read about it

and we talk about it more, because in fact that’s — if you look at this question of

Doppler spread and what its effect is on how long a system

looks like it’s time invariant, this is one of

the key parameters to understanding how any

kind of wireless system is going to work.

The title is wrong. I think it should be 2006 not 2106

@MIT OpenCourseWare: Is there any special lecture series especially on Wireless Communications/Communication Networks, Protocols and Standards and evolving technologies in these areas?