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Practice English Speaking&Listening with: Lecture - 11 Bandpass Signal Representation (Part - 2)

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In the last class, we started bandpass signal representation. And, in this class, we will

also continue the same topic. And, we will start bandpass signal and more or less

bandpass system representation in this class. So, in the last class, what we have seen is

that a bandpass signal can be represented as a lowpass signal with someafter doing

some operation.

So, what we have seen is that, if we have a bandpass signal with this kind of spectrum

any spectrum; this is f c; then, we can first

suppress the negative side by multiplying by

two times u of f, that is, the unit step function. And then, we will get again a bandpass

one-sided spectrum. The resulting signal will have this spectrum; negative side will have

nothing. Then, we can doWe can shift this spectrum here; then, we will get a signal

with this kind of spectrum. So, this will be a lowpass signal and this is called the

lowpass equivalent signal of this bandpass signal.

So, we have seen the relationship between this

signal and this signal; and, in particular, we have seen how to construct the bandpass

signal from the lowpass equivalent signal and how to construct the lowpass equivalent

signal form the bandpass signal. The bandpass signal is a real signal; but,

the lowpass equivalent signal will be complex signal, because there may not be any equivalence;

it may not be any symmetry in one side of the bandpass signal. Whatever is there

in the positive side, will be also there in the negative side with some phase will be

negative, but magnitude will be same. But, in

one side itself, there may not be any symmetry. So, when we take only that part and

bring it to 0 centre, then we may not have any symmetry. So, the resulting lowpass

equivalent signal may be complex. In general, it is complex. So, we have also seen how

to construct this signal, what is the real part and imaginary part in terms of this signal,

how to construct this signal. And we have also seen that, for a communication

system, though we will be transmitting the bandpass signal through the channel ultimately,

the whole system in the transmitter as well as the receiver may not be designed

for that particular band, which is lying in very high frequency. So, what we can do is

that, we can design the signal as a lowpass equivalent signal. We can the design the lowpass

equivalent signal first, because that is the base band signal. That is the lowpass

signal with some band width. So, designing that

signal is easier because processing required to design that signal, to generate that signal

is easier withIt requires low cost electronics. And then, we can do up conversion to

get the bandpass signal for that and then transmit that. The receiver on the other hand,

we can first take thereceive the bandpass

signal, then do down conversion to take the lowpass equivalent signal, and then do required

processing to detect the message. So, the main significant processing at the transmitter

as well as the receiver may be done in lowpasson the lowpass signal in the low

frequency.

So, with that in mind, we saw that, we can represent a communication system this way.

This is the diagrammessage in the transmitter; first, this lowpass equivalent signal of

the transmitted signal is designed. So, this is design of some lowpass signal. So, this

is quite easy. And then, we do up conversion

to get the bandpass signal from the lowpass equivalent signal and then transmit it through

the channel. The receiver first does down conversion on the received signal to get the

lowpass equivalent signal and then does all the processing on this lowpass equivalent

signal to detect the message. So, if we represent this whole blockchannel including

up conversion and down conversion in the transmitter and receiver respectively,

we can consider this as a single channel; and

then, this equivalent channel takes a lowpass signal as input and gives a lowpass signal

as output. So, this channel is now a lowpass channel and it is

We can consider from now on that, the channel is always lowpass; but, but the channel is

now complex, because the lowpass signal is complex; lowpass signal, that is, input

transmitted is complex and the output is also complex. So, the channel is complex

baseband channel. So, we said that, from now on, we will considered such a channel and

consider the task of detection of message from the received lowpass equivalent signal

and how to design the lowpass equivalent also. But, there is one thing we have not yet

done, that is, we have said that, this is the equivalent lowpass channel. Now, if this

channel is given; say this channel has impulse response h t, which is bandpass; then, how

to get the equivalent lowpass channel. We are given the bandpass spectrumfrequency

response of the channel, which is bandpass. Now, from there we are going to actually

consider the lowpass equivalent channel. What is the impulse response of that lowpass

equivalent channel then? So, we are going to investigate about that in this class.

So, before doing that, we will consider also another topic and another small question

about the bandpass signals itself; and, that is, how does a bandpass signal really look?

If you plot a bandpass signal, how does it look?

We have seen that, you can represent it as a lowpass equivalent signal; but, in practices,

how does the bandpass signal itself look? So, we are going to first see that. So, usually

the bandpass signal will look like something like this. It will have some envelope

say something like this. And, of course, the negative side also will be there. Then

So, this is how usually a bandpass signal will

look. So, this sinusoid, which is inside this envelope, has aboutthe frequency is about

f c. And, if weNow, this envelopethis

wave formthis is called the envelope of the

signal. Now, suppose we have s l t, we know that,

the bandpass signal s t can be represented in

terms of a lowpass equivalent signal, that is, s l t. Then, we can write s l t is a complex

signal. So, we can write it as somethingsome magnitude times e to the power j theta t;

theta t is a phase and a t is the magnitude. So, now, if we write that way, then we have

this representation of the lowpass equivalent signal. Now, what is a t? Here a t is the

magnitudex square t plus y square t. If you remember from the last class, x t is

the real part of s l t and y t is the complex imaginary

part of s l t. So, the magnitude of s l t is root

over x square t plus y square t; and, theta t is the phase of s l t. So, it is tan inverse

y t by x t.

Now, once we have that, what is s t? s t is the real part of s l t e to the power j 2

pi f c t. This we have seen in the last class that,

the bandpass signal can be expressed in terms of

the lowpass equivalent signal this way. And, that is real part ofTake this

representation of s l t and put it here. So, we will have a t times e to power j common

then, 2 pi f c t plus theta of t. Now, what

is the real part of this? This is a real quantity, a

real number, a real function. So, what is the real part of this? Cos of this. So, the

real part of the whole thing is a t cos 2pi f c t plus

theta t. Now, here is the envelope; a t is the

envelope. And, this is the phase. Now, you can see that, this signal if you

plot; this will really have this kind of shape. Cos

omega c t; 2pi f c is the angular frequencyomega c. So, omega c t plus theta t; cos

omega c t is the kind of the carrier signal, that is, this sinusoidal; and then, it will

have theta t phase shift; it will have some phase

shift, which will change with time. And then, a t is a lowpass signal; as you can see, it

is a real lowpass signal, because it is the real

part; it is the magnitude of a lowpass signal. So, it will be positive real signal. So, it

will be this kind of signal. So, this will be really

the envelope of this whole signal, because this is a very high frequency signal, which

is varying very fast. And then, you have the a

t times thatwill actually change the magnitude slowly; a t is a lowpass signal.

So, it will have some slow variation compared to

the high frequency carrier signal. So, it will

vary the magnitude of the sinusoidal signal. So, it will act as an envelope for the carrier

signal. So, here this is the envelope a t. This is the envelope a t. And, phase theta

t is actually wouldIt will change here. So,

we have seen what is envelope and phase of a

bandpass signal. And, envelope and phaseEnvelope specially has a very pictorial

pictorial representation. Envelope is basically actually the envelope of this signal as you

can see. This signal is the envelope. Now, we haveSo far, we have seen two things:

one isWe have seen two representations of bandpass signal. One is

as a lowpass equivalent signal. So, it will be

complex lowpass signal, which will represent the bandpass signal in a way. So, we can

get the bandpass signal from the lowpass equivalent signal; and, lowpass equivalent

signal from the bandpass signal. And, we have also seen that, another representation,

where s t is represented in terms of an envelope and a phase with some carrier frequency.

So, here you get an idea of how the signal will look like. So, it will basically have

that a t envelope. But, we need to also askwhen

we do this conversions from the complexfrom the bandpass signal to the lowpass equivalent

complex signal, how does the energy change? What is the relationship between the

energy of the baseband lowpass equivalent signal and the real bandpass signal? When

we do this conversion, what happens to the energy? So, that we will now see.

So, let usSo, this is we are going to find out the energy of a bandpass

signal in terms of the lowpass equivalent signal. So, the

energy is defined this way. This isNow, we

takeWe know that, the bandpass equivalent signal is this in terms of the lowpass

equivalent signal. If the low pass equivalent signal is s l t, then the bandpass signal

is real part of s l t e to power j 2pi f c t. Now,

how have we got this? This is basically here; you

can see that, from the previous page, we have this. So, real part of s l t e to the power

j 2pi f c t; that is real part. So, that is

the s t. So, we have here real part of this; basically,

this partthis thing. And, this is actually expressed in terms of envelope and phase here.

Now, if we take the square of that, which is required; so, here this will be squared.

If we take the square of this, what we have is cos

square this. Now, what is cos square theta? What is

We can express cos square theta as half 1 plus cos 2 theta. So, using that, now, we

have this term. Square of this is a square t. So,

squares square t is a square t half of 1 plus

cos two times this, that is, 4pi f c t plus 2 theta t. So, now, if we takeintegrate

separately these two parts as we need here; so, this inside quantity is basically this.

This quantity is this. Now, we need to integrate

these two partshalf a square t and half a

square t times cos of this. So, once we do that, we actually have this. Integrate half

a square t; a square t is nothing but s l t

square, because a t is the magnitude of s l t. So, we

have this part and then half inside s l t square. This is magnitude square times cos

of 4 pi f c t, that is, 2 times this angle; this plus

2 times theta t dt. So, once we have, this can be written approximately

as half… – only the first part. Why? Because what is this signal? What are we integrating

in the second integration? What we are integrating there is that, if you see

thetry to draw the wave form of this inside function, it will also be a bandpass signal,

because this a carrier signal with some 2 times

theta t; that is, the frequency is 2 f c and the phase is 2 theta t; and, the envelope

is s l t mod squaremod s l t square. So, it will

look like s l tmod s l t square is the envelope. And then, this is the signal. So,

this is a very high frequency2 times f c is

usually a very high frequency. And, this is a baseband real signal. So, this is the

envelope. And, we are integrating this signal. Now, onceWhen we are integrating this

signal, we can see that, since 2 times f c is

very high, there are manyThis oscillation is very high frequency. So, if you now take

one cycle at a time, what is the integration? One cycle at a time? If we take only one

cycle, because this oscillation is very high frequency, what you can expect is that, in

the one period of this oscillation, this envelope

is not changing much, because this is a low frequency baseband signal; mod s l t square

is a baseband signal; it is lowpass signal. So,

if you magnify and draw only thisone period here, you will find it is a signal

of this type; that the magnitude is notThis is

same as this. So, here magnitude is not varying much. The envelope is almost fixed here, because

this is a very small part of this envelope and envelope is a lowpass signal.

So, this integration of this signal in this interval will be 0, because it is a sinusoid

of period 1one period integration of a

sinusoid. So, now, if you continue doing that, in this

small interval it is 0integration is 0.

Similarly, in this integralsmall interval, it is 0. This way we keep doing it. The whole

signal will give you almost 0. It is almost because the envelope is not really constant

here; it is changing its radiuschanging very slightly. So, this integration will not

be exactly 0, but almost 0. The error will be

negligible. So, we can say that, this integration is only this part, because this part will

give you almost 0. So, this is almost half times

energy of s l t. So, what we have seen is that, the energy of the bandpass signal is

almost same as the energy of the lowpass equivalent

signalhalf of that. So, this is something interesting, because now, we know that, we

canIf we treat the lowpass equivalent signal; if we know its energy, then we know

that, the bandpass signals energy is proportional to that. So, we can keep that

in mind. And, this will simplify lot of our analysis. So, we can treat now the lowpass

equivalent signal and assume the bandpass equivalent signal to have the energy half

times the energy of the lowpass equivalent signal.

Now, we will go into the representation of the linear bandpass system. So, in the last

class we saw that, this is an equivalent diagram of a communication system, where this

whole thing is treated as a channel and this is now a lowpass complex channel. But, if

we are given the impulse response of this channel,

then what is the impulse response of this channel? Now, we will see that relation now.

So, to do that development, we need to analyze, represent a bandpass system also

just like we represented bandpass signal in terms of lowpass system.

So, suppose the impulse response of the bandpass system is h t. And, it has frequency

response H f. Then, we know several things. One is that, h t is real and h t is bandpass.

Now, h t is real; that means that, H of f is H star of minus f. This is a conjugate

symmetry. The magnitude is same at f and minus f; and, phase at f is negative of phase

of phase at minus f. So, this is the conjugate symmetry. That is the value at f is the

conjugate of the value at minus f. Now, just like we did for the bandpass signals, let

us say this is theSuppose this is the H f.

We are drawing only the magnitude. Then, just like we did for the bandpass signals, we say

that, H plus f isTake only the positive side. So, basically, H f if f is greater than

0; and, 0 if f less than 0. So, this you can also

write as u f times H f. Now, here for bandpass signals, we actually

had two times this here. But, here we are not

doing that for some reason; we will see later. Now, again similarly, this is missing as

scaling; so, which we will neglect for time being and later we will explain why we have

done this. And again, just like bandpass signals, we will shift this. So, this is basically

only one side H plus f. Now, H l of f is H plus f plus f c. So, you basically shift it

here. So, this spectrum is now. This f c is brought

to 0. Now, in terms of H l f; now, we can write H l ofSo, what is H l of f minus

f c; it is basically H plus of f. So, H plus of f can

beSo, we can write thisH of f for f greater than 0 and 0 for f less than 0.

So, we can write that, H of fSo, H of f is H l f

minus f c. So, H l f minus f c is basically H plus f;

that is this. So, if you add this with the negative part of it, which we have removed.

So, that part we can get by taking H l star minus

f minus f c. So, what did you have here? H f

We want to get this signal from this. This is H l of f minus f c. So, how to get this

signal? First keep this signal and then add this part.

How do we get this part from this signal? It

is basically first invert it. How do you do that? Take this positive side into negative

side and negative side into positive side. We simply

take the minus of that function. So, H l; replace f by minus f. So, that is H l of minus

f minus f c. But, this is not really the negative flipped version of this signal. This

has the phase negative of this. So, we have take to conjugate of this. The magnitude we

want same; but, the phase we want negative. So, we need to have H l star of that function

flipped function. So, flipped function is H

l minus f minus f c. Then, we need to take the conjugate of that. So, that is H l star

of minus f minus f c. So, once we add these two,

we get this function. This part is basically the positive part; this part is the negative

part. So, from here we can also write that, h t is

h l t e to the power j 2 pi f c t plus h l star t e to power minus j 2 pi f c t. This

is basically the inverse Fourier transforms of these two.

So, h l t is the inverse Fourier of H f t. H f

Hcapital H l of f. So, then we can use the well known properties of Fourier transforms

to see that; really the Fourier transform of this; inverse Fourier transform of this

is this. Similarly, here.

Now, this can also be written as real2 times real part of h l t e to power j 2 pi

f c t. Now, this is because, this part is the conjugate

of this part. So, we can take. Only the real parts will remain; imaginary parts will actually

cancel. So, this is what we have. The actual impulse response is expressed in terms

of the lowpass equivalent of the impulse response. This is what we also did in the

case of representation of bandpass signals. And,

we have done the same thing again for bandpass systems. Now, we will see the

relationship. Why have we done these two things? We have represented bandpass signals

in terms of lowpass equivalent signals; lowpass equivalent signal of the same signal.

Then, we have represented bandpass system in terms of lowpass equivalent of that

system. NowSo, we knew that, we have a bandpass signal; we can give that as input

to the bandpass system. Then, we will get some output. That is also a bandpass signal.

Then, we have a lowpass equivalent system of that; lowpass equivalent signal of this

signal. Then, if we give this as the input to this signal, what do we get as the output?

Would we really get the output as a lowpass equivalent signal of this output? That is

what we what to see.

And, this is a very interesting idea. We will use that also later. So, what we are saying

is that, we have a bandpass signal s t. A bandpass

system h t. We give this as input to this.

And, we get some output say r t. We receive this. From here we have lowpass equivalent;

lowpass equivalent of this signal is s l t. And, we have also seen how to get lowpass

equivalent system of this system, that is, h l t. It is very similar to this. We have

seen just now. Now, we give this input; what would we

get as this output? That is the question. Shortly, we will see that, we will actually

get the output, which is the lowpass equivalent of this output. So, that is something very

interesting. So, this is also lowpass equivalent

lowpass equivalent. Now, let us see that. So, this we are seeing

response of a bandpass system to bandpass signal; response of this bandpass system to

bandpass signal. We want to express this whole operation in terms of lowpass signals

and lowpass systems; that is what we want to see. So, r t is the output for s t input.

So, r t. Now, we express in terms of the lowpass equivalent signal. Now, we do notWe have

not yet seen that, the output of this system is actually the lowpass equivalent signal.

We do not yet know this. We are trying to see

that this is true. So, we know that, r t is a real part of r l t e to the power j 2 pi

f c t. Similarly, s t is real part of s l t e to

the power j 2 pi f c t. And, again h t is 2 times real

part ofRemember there was a difference of scalar in the representation of bandpass

systems. So, now, we can also write these things in terms of in the frequency domain.

What do we have? In the frequency domain, R of f, that is, the spectrum of this received

signal is S of f times H of f. That is the spectrum of theThe frequency response

of this system times the spectrum of this input signal.

Now, this can be written as half. What is the expression of s f in terms of the lowpass

equivalent signal of s f. So, s f in terms of s l of f. s f in terms of s l f. We have

seen just now for systems; it will same thing for signals

also. We have seen in the last class; but, it

will have only a difference of scalar. Just now, what did you see for systems. We saw

that, H f in terms of H l f is this. Similarly, for signals also, we will have the same thing,

but there will be half here. So, we have half S l f minus f c plus S l star minus

f minus f c times H f. H f just now we saw that, H f is

H l f minus f c plus H l star minus f minus f c.

So, this is we have just replaced S f in terms of S l f and H f in terms of H l f.

Now, remember that, s l of t and h l of tthe inverse Fourier transform of the S

l f and H l f are narrow band signals. So, we can say

that S l of f minus f c is 0 for f less than 0.

Why? Because S l f is narrow band. It is lowpass narrow band signal. So, it is frequency

response of this Fourier transform is like this. It lies here. Something like say this

S l f. So, what is S l f minus f c? It is the shift

here. So, S l f minus f c. This is S l f. And, this

is S l f minus f c. This will be something like this. So, this is 0 on the negative side.

So, this is what it is. Similarly, H l f minus

f c is 0 for f less than 0. Now, let us multiply these terms. This is

0 and this is 0 for the negative side. So, this is 0

on the negative side. And, what about S l star of minus f minus f c. It is the flipped

version of this and then conjugate. So, flipped version of this will have positive site 0.

So, if you take this product; this is at less than 0. This is 0. This is 0 at f greater

than 0. So, the product will have 0 everywhere. We

have one function, which is 0 at negative side; it has something like this spectrum.

And, this part has positive side 0. So, let us say

it is like this. Then, the product of these two will be 0, because this is 0 at negative

side; this is 0 at positive side. Everywhere the

product will be 0. So, only productSo, similarly, this time, this also will be 0.

So, only product, which will be nonzero is this

times this.

So, we can say that R f. So, R f is half and these two also. Here this both of them have

some positive side components. So, these two times these two. These cross terms will be

zerosplus S l star minus f minus f c times H l star minus f minus f c. Now, this is R

f. Now, what is this R l f? You can see that,

this is R l of f. We have replaced this as R l of

f is S l of f times H l of f; that means it is the product of these two frequency responses;

that means R l of f is the spectrum of the output signal here. So, that isSo, R l

f is the output spectrum of this; and then, R f is

the output spectrum of this. This input and this

system. SoAnd, we have got a relationship between them. The output spectrum here

and output spectrum here. And, that relation is that, R f is this in terms of R l f. So,

this spectrum in terms of this spectrum is expressed

here. Output spectrum here is expressed in terms of the output spectrum here.

SoBut, what is this reallythis thing, which is the output of the bandpass system

R f? This is nothing but the up converted version

of the R l signal. So, here we can see that, this R l f is really the lowpass equivalent

of R f, because it is expressed this way. So, R l

is really the lowpass equivalent signal spectrum of the R f. So, we have showed that, this

output will really be R lthe lowpass equivalent signal. So, how did you show it?

Let us summarize. We have seen now this. We have

seen that, this output is r t expressed in terms of the lowpass equivalent. We do not

know what is the output of this is really r l t;

we are going to verify that. So, we have taken represented r t in terms of r l t; h t in

terms of h l t; s t in terms of s l t; then, computed

the spectrum of this as product of these two spectrums. Then, taken their lowpassRepresented

those two spectrums in terms of

their lowpass equivalent spectrums this way and taken that, these two products. And,

these two productsthe product of the lowpassthese two things, we have got

as this. And then, we have assumed that, this lowpass

equivalent spectrum of this and lowpass equivalent the spectrum of this s l t and

spectrum of h l tthe product of them we have

assumed to be R l of tR l of hS l f and H l f, the product is R l f. Then,

we have seen here that, this R f, which is the output of

the bandpass system is really expressed in terms

of the product of S l f and H l f this way. And, from this representation, we see that,

R l f, which is the product of these two, is really

the lowpass equivalent of R f. So, R l f, which

is the product of these two means which is the spectrum hereoutput spectrum is really

the lowpass equivalent of this spectrum; that means the output of this is really the

lowpass equivalent of this output. So, we have verified that, if we have this

bandpass system and if we represent the signals and the systems in terms of their lowpass

equivalent signals and systems, then we really have this output of the lowpass equivalent

system. When it is excited by the lowpass equivalent signal of this input, and the output

will also be the lowpass equivalent of this output. This is what we have verified. So,

now, we can treat a bandpass system without any loss. We can treat a bandpass system in

terms of lowpass system, because whenever there is a bandpass system, it is difficult

to suppose implement such a system. We want to implement a bandpass system with some transfer

functions. Then, we can take the corresponding lowpass equivalent system and

design that. Then, suppose we want to give a bandpass signal as input to bandpass

system; we have a bandpass system; we want to implement that. And then, we want to give

a bandpass signal as input and then we want to find out the bandpass output.

Then, we can do that whole thing in terms of a lowpass system and lowpass signal. This

is exactly like we had drawn here. So, we want to implement h t and we want to find

out this output. So, what we can do is take the

lowpass equivalent signal. If it is difficult to

implement this system, then it is not possible to do this way. Then, we can take this. So,

do to this down conversion, do this down conversion also and implement its lowpass

equivalent system and find out the output. ThisThen, do up conversion. So, we will

get r l t here and then we can find out r t by doing up conversion. So, this is basically

down conversion; this is the lowpass equivalent system. So, instead of implementing this

system, we can do this way, because this will be cheaper, because this is a lowpass

system. So, we have seen in this class thatWhat we had seen in the last class; we have

actually refined that. Now, we had seen in the last class that, we

can represent this up conversion channel and down conversion as a single channel as lowpass

channel. Now, we know now that, there is a relationship between this channel and

this channel. If this channel has impulse response h t, this channel has the impulse

response h l of t, which is the lowpass equivalent of this impulse response. So, it

is something interesting. So, this is what we

have seen.

Now, we willThisAll these things you can study from the book by Proakis. I am

also teaching from this bookDigital Communication by Proakis. And, at this point,

after these two classes, you should be able to also solve this exerciseexercise 1

from Proakisfourth chapter; fourth chapter

of this bookexercise number 1; that is, the

exercise 4.1 from this book Proakis. It is basically an exercise on Hilbert transform.

It has these components. Please try this exercise

at home. And, I will solve some particular parts of this exercise in the next class.

So, it also asksIt actually asks to prove some

relationship on Hilbert transform.

So, for example, part a is if x t is same as x of minus t; that means it is symmetric

signalevent; also called event; that means the

positive side is exactly the mirror image of the

negative side. So, for example, if it is like this, then it is like this; then, the Hilbert

transform x hat t is minus x hat minus t. B is similar if x t is minus x of minus t,

then x hat t is x hat minus t. So, this can be just

use theFor proving these two, just use the

definition of Hilbert transform; the convolution of x t with 1 by pi t. That is the Hilbert

transform of x t. Use the definition and simple substitution of variable will lead you to

the solution of these problems. And then, there arec for example, is if

x t is cos omega naught t, then x hat t is sin

omega naught t. So, for solving this problem, you can use this. And then, solve this

problem in frequency domain. You know the spectrum of Hilbert transformerH f;

multiply with this. So, this has X f as half delta omega. So, if omega minus omega

naught plus delta omega plus omega naught. Then, multiply by H f and take the inverse

Fourier transform; you will easily get this. So, this is how you can solve this exercise.

And, there are some more parts not similar to this. But, you can show that, the energy

for example, f is actually to show that, the energy

of x t and its Hilbert transform are same. I

will not give a hint to this. This is actually easier. So, you should be able to do this

yourself without any hint. So, we will see you again in the next class.

The Description of Lecture - 11 Bandpass Signal Representation (Part - 2)