Let us continue our discussion on Demodulation of FM signals or Angle Modulated
Let me very quickly revise for you, what we discussed last time, basically for
demodulation of angle modulation signals. We need particularly FM signals; we need a
device which produces a response proportional to the frequency deviation of the input
So, if you remember, we read ideal of a characteristic that we would like from, so called
frequency discriminator will be something like this, it produce an output voltage, which
is linearly dependent on the input frequency deviation. Now, such a device, we call a
frequency discriminator and sometimes also called a slope detector. Because, basically
we are following of this device, basically we are producing an output, a detector output,
which is proportional to the slope of this line here. The approximate serialization of
this, we said could be done by having a differentiator,
I am just quickly reviewing, what we did last time.
The differentiator followed by an envelope detector and we discuss, how this would
approximately produce, what we want or in fact more or less, what we want.
The only thing to discuss was how to realize these differentiators and we discuss two
realizations of this differentiator, one using a small time delay that is using a time delay
and other using a bank pass filter that is why we do it?
So, let me quickly come to the bank pass filter, we can use a bank pass filter, which
produces a frequency sensitive response around the carrier frequency by choosing the
center frequency of this bank pass filter to be away from the actual carrier frequency.
So, we call it staggered unit that is the bank
pass filter is tune to a frequency, staggered from
the actual carrier frequency. The only difficulty with this approach is,
the linear range, the linear frequency sensitivity that we desire from such a filter is rather
small. And the second difficulty is, it does not
produce a 0 response and a carrier frequency accepts, which is what you like to have.
And as to take care of these things, we suggested an alternative, which is what we are
going to discuss now, things likely more detail and that is used bank pass filters.
Each of which is tuned to a frequency, away from the carrier frequency of interest, one
is tune to a frequency larger than the carrier
frequency and the other is tune to a frequency smaller than a carrier frequency. And the
envelope detectors, following this will produce an output, which is proportional to the output
of these filters. So, this magnitude response that I have plotted here is also in some sense
is representation of the magnitude of the voltage, that will be produced in the output
of the envelope detector. So, if you have an FM signal as the input,
which will varies on frequency along this axis
that as a function of time, the output voltage, the output of the envelope detector will
vary accordance to this curve. Because, the response of this circuit will be in accordance
to this curve, incidentally where I am making this treatment that is the slight
approximation involved in this treatment.
I do not know, whether you appreciate that fact or not, but I come back to this point
in minute. Similarly, this other twin circuit
will produce an output voltage, proportional to
this curve. Now, what is a approximation involved here, the approximation involved here
is as follows, your input signal is we are really not mean, what is called a steady state
analysis, we are not really make making a steady state argument here.
When, we talk about the output of a circuit as given by the input spectrum, multiplied
by the transfer function that is the steady state
statement. As the string, from a translucent behavior, but here we are slightly, we are
working actually in a translucent domain. If
you really think about it, is our input signal is constantly varying in frequency where,
we are saying that the output will be proportional
to whatever is the corresponding magnitude of that time.
Really, speaking your input is constantly changing it is never in a steady state in
that sense. So, even if a slight approximation
in the argument that we are giving here, but this
is usually valid for the case of FM signals. So, because of variations, it is slow enough
and we can assume that this steady state accrues; steady state is approximation is valid.
So, as an input signal frequency varies, the output of this circuit will vary accordance
to this curve.
So, please try think about this point that I making this is this kind of analysis is
called quasi steady state analysis, because we are
not ever in a true steady state, when with the
FM signal. Because, here input signal frequency is constantly varying, which is different
from input signal, which simultaneously has a large number of frequency components,
but here we are saying that at different instance. It is suppose to have a different instantaneous
frequency, we are not looking at it in spectral terms, we are looking at it in terms
of it is behavior at a specific point in time. And as that, is treating a frequency changes,
we are saying immediate to the response or situations, that is the point, that we need
and that is really this approximation, that I am
talking about the Quasi steady state approximation. So, nevertheless if you assume that this approximation
is valid, we can see that if we take a difference between the two outputs, the
two bank pass filter followed by envelope detector outputs, we will have this kind of
s curve. And now, if you restrict your input signal to have your peak frequency deviation
to lie in this linear range. If you this point,
goes only up to our input signal has a frequency deviation only up to this point or on the
other side up to this point. Then our input signal will always be in the
linear range of this characteristic and not only
that we would have produced a 0 response and the carrier frequency as desired. So, this
is a very good approximation to what we actually want to do, any questions on this.
These are the outputs of two filters, followed by an, I will draw a diagrams to it soon;
basically we want an output which is proportional to the magnitude of this response. And
mind you, what are we looking at, we are looking at the frequency discriminator, which
is consisting of a really speaking an ideal differentiator, followed by an envelope
detector. So, we now instead of an ideal differentiator,
I have this kind of a circuit followed by an
envelope detector. So, what will a envelope detector produce, it will produce an output,
this magnitude output, envelope this magnitude is proportional to this and suppose the
input signal frequency is this much, the output on the bank pass filter will have the bank
pass modulated signal, whose amplitude is this much. And the envelope detector will
simply be producing an output proportional to the instantaneous amplitude, the
modulation will go. So, basically that is what we are talking
about, we look at a circuit this how, basically that differentiator followed by envelope detector
is being replaced with two filters of this kind. Each followed by a corresponding envelope
detector and the outputs of these two are subtracted from each other. So, basically
the system that we are depicting here is something like this.
You have bank pass filter 1, corresponding to H 1 f, so the bank pass filter 1,
corresponds to this H 1 f, this is followed by envelope detector 1. Similarly, you have
a bank pass filter 2, followed by the corresponding
envelope detector and we want the outputs of the 2 to be subtracted from each
other. Why are we using, I will second question first, why we using an en envelope
detector, what was that we are trying to do, we are trying to convert our input FM signal
into an AM signal. Because, there the variations in the frequency,
due to the message signal are being converted to variations in the amplitude using
the message signal. That was idea of using the slope detector in which we have a differentiator,
followed by envelope detector, so please consider, what we discuss last time,
this is something that we must understand.
Well, here was your differentiator and this is what it was doing, it was converting this
FM signal or angle modulated signal into an amplitude modulated signal. The only thing
is we are replacing this; we are trying to realize this differentiator by different means,
let me complete the answer. This differentiator,
it was suppose to have an ideal frequency response of g omega is being approximated
or being approximately realized by combination of bank pass filters.
Basically, I am replacing this differentiator or basic replacing this kind of a circuit
with another equivalent circuit, which will do
the same thing. How, is it doing the same thing,
basically what we want is an output, we are not interested in this part. This is the carrier
containing the model same modulation, we are not interested in this part, we are only
interested in is taking this amplitude. So, what will this bank pass filter do instead
of differentiator, I have a bank pass filter, which will once again produce an output, whose
amplitude will depend on the instantaneous frequency. That is a passive
state argument, I am giving that at any given time instant, you have a certain frequency
present the input signal and the bank pass filter will respond, according to the input
frequency. So, bank pass filter 1, for example suppose
the institute frequency somewhere here, will produce an output, which is a sinusoid again
at that frequency, sinusoid at this frequency, and let us see modulated sinusoid. But, the
amplitude of their output will be the
proportional of this and as the input frequency varies, produce an output proportional to
this, but all the time, we are talking of the amplitude.
But, this signal is always still present, the carrier is always still there, the amplitude
of the carrier at the output of the bank pass
filter is varying in accordance with the filter response. And now, we want to remove this
carrier itself, how do we remove the carrier, through the envelope detector that is the
basic idea. So, hope it answers, both you your
question as well as your question. Quasi steady state is argument is being invoked,
because we are assuming that as the input signal varies in frequency, instantaneous
as the output magnitude changes in accordance with the response of this filter.
That is a quasi steady state approximation that
I talked about, so these are the two arguments that you need to understand. So, hopefully
that we can realize these characteristics to the combination of two banks pass filters,
each followed by a corresponding envelope detector.
Because, then you will get an output, which is proportional to this, in the first filter,
in that first branch, proportional to this in
the second branch, then you subtract the 2 outputs, you will get an output voltage. So,
that we use a black pen here, think this as the
output voltage, which is proportional to the which is given by these characteristics versus
input frequency or frequency deviation or instant, I think best thing is to call it
instantaneous frequency at the input. Though, it is no were impulses here, how does this
question come to your mind Sir, I would saying that H 1 f and H 2 f are
They are the classs functions of the two bank pass filters; they are the frequency
domain requisitions looking for bank a pass filter
that is if you feed sinusoid at different frequencies, how your response did varies.
So, the frequency response of the entire system becomes something like this, I think
some of you have probably not understood what I am trying to say.
So, that is the different question, let us try to understand, what we are doing, whether
it is doing a job that we want it to do or not.
As to why, we go in to many other things is something that I think at this point, it is
difficult to explain, because there will be many
such possibilities, which need to see, why they will not work, that will not work. Why,
what you want is an output proportional to the instantaneous frequency, what we are
getting from the top system, if please try to understand if this point is clear.
What we are getting from the top system, the top branch here, if you look are this output,
the input to this envelope detector is an amplitude modulated signal. However, the
amplitude of this is proportional to the input signal empty, only if amplitude lies in
certain range. Otherwise this is amplitude modulation, but there is non-linear
modulation, as input frequency varies, the output amplitude varies, according to this
at this point.
But, you still have a modulated carrier present here; this is the important point to
understand here, there were modulated carriers, which contain only frequency
modulation. Here, we have a modulated carrier, which contains amplitude modulation,
but the amplitude modulation is proportional to the message signal mt. Only in this
interval and I mean suppose your input frequency varies beyond this, will contain some
kind of a non linear amplitude modulation. That is why, we do not want to go beyond this,
that is why we want to keep make sure, that your input signal frequency lies in this
feature that means you need to show your peak frequency deviation is less than f 1
minus fc on this side and fc minus f 2 on the
other side. But, important point is you get an amplitude modulated signal here and all
you want to do is check this out, detect out, amplitude of this output.
You have a modulated carrier; you want to get an output proportional to the
instantaneous amplitude of the carrier. So, what should we do, use an envelope detector
that is all, envelope demodulator, similarly here, you set for the 2, you get what you
want, if you appreciate this, other questions will disappear. So, this is the thing to really
appreciate, think about it, I think it will be clear, if you really think a little more
about it yourself. But, if you have some specific question
at this point, I am ready to stop for a couple of seconds and see what those doubts
and questions are.
2 delta f is approximately remember the constant formula is 2 delta f plus twice of f. So,
approximately this is also proportional to bandwidth, but our argument is not based on
of the bandwidth of the FM signal our argument
is based on the peak instantaneous frequency deviation.
I dint get that point, can you explain again
Oh, yes, yes the each individual filter will be a small linear range
But, I mean just look at this is precisely, why we are doing all this, it is because that
each individual filter has a small linear range
and using two staggered filters and the difference output reduces the characteristic
like this. So, the range of this is let us say
somewhere here, the range of this is somewhere here, but together the range becomes
this. So, if I choose by frequencies f 1 and f 2
in accordance with my delta f, I have done my
job, I have increase my linear range. Suppose, I have a certain peak frequency deviation
to work with as I say 75 kilo hertz, so what should I do, how should I design these two
filters? I should choose the center frequency of f 1 to be 75 kilo hertz, away from fc
similarly f 2, I should select 75 kilo hertz away from fc on the other side.
So, we have choose the frequencies f 1 and f 2 to take care of that peak frequency
deviation that I will have on either side of fc, I would have got a linear range that
No, all the frequencies are passing through both the filters, but they are producing non
linear amplitude responses, by subtracting the 2, I am making sure that over the entire
range of interest, we get a linear response. Because, the filter is a filter which will
respond to the input in accordance with it is characteristic, whatever frequencies are
present in the input are being processed by both of them. But, this response to the input
in a particular way, this response the input will be different way.
And when, two are combined, this is the net way in which they are together effectively
seemed to be responding to the input. So, I hope with these questions and answers some
of these things are clear, but we settle in it to interest that and think about it to
make it absolutely clear to yourself.
Actually, phase response is of no consequence, because we are only looking at the
amplitude of the output, we are only looking at how the
instantaneous amplitude varies as a functional type. As the input signal amplitude
changes, we were like the output signal amplitude to vary with the input moderating
signal amplitude linear, that is what we want. So, first we have we converted the amplitude
variations in frequency variations that is what frequency modulation is all about and
now through the artifice of these two bank pass filters and converting those frequency
variations back to amplitude variations. And then, detecting the amplitude variations through
the envelope demodulator, think about it.
We look at the mathematics of it also very soon, first I watch it with the physics effect,
now how do we realize this in practice, well we can realize it exactly a the way I have
And finally, if we if we realize it like this, what you like to do is to have some kind of
a some kind of a circuit, which takes a difference
between the 2 output amplitudes and how can we do that, we can use a different
amplifiers are something like that. So, this techniques for combining the e d
outputs, envelope detector outputs, we can combine them by using a different difference
amplifier. But, there is a more popular passive alternative, the passive circuit is
which will realize, this in one whole and not
need a different amplifier, like to just draw that very quickly.
So, here is a based on this arguments, some kind of a circuit, which will work as a FM
discriminator, the input modulating signal is coming in and through a suitable center
tap rf transformer, it is fed to 2 circuit is
which are operating in parallel. Each of them has a
bank pass filter, which I am representing with approximately by a parallel LC circuit.
This can be more elaborate than a simple parallel LC circuit, but this is the most common
way of doing it and this is followed by an envelope detector.
So, this is one circuit and the envelop detector, if you remember contains a diode,
followed by an RC circuit is in parallel. So, at any given time instant, suppose you
have a certain signal here, FM signal coming in,
is as a positive envelope and a negative envelope, so at any given time instant, the
polarities will be like this. So, I have parallel similar circuit in the lower branch and a
diode detector also present here and how I do is
simply combine them like this, had the 2 outputs like this.
So, as you can see the output between these two points, this point and this center point
here, may be proportional to the positive envelope that is coming here and the output
between these two points will be proportional to the negative envelope that is coming
here. Because, at any given time either this diode conducts, this diode will conduct from
in the positive half cycles; this diode will conduct through the negative half cycles with
respect to the input. Of course, it has to be positive half cycle
for the diode, but the posit positive half cycle
of this diode will be the negative half cycle of this diode. So, this essentially this is
the top circuit will be sensitive to the positive
envelope and the bottom circuit will be sensitive to the negative envelope. And if
you add the 2 outputs effectively, you would a
subtracted what you wanted to do, you have carried out the subtraction that you really
wanted to carry out. Because, you are adding a positive thing with
the corresponding negative thing, of course the two envelopes are symmetrical, we assume
that the two envelopes, after the bank pass filter are symmetrical. So, here is the
FM signal coming in here, is at this point at
these two points you have the two amplitude modulated signals, one AM signal is here
and the other AM signal is here. This diode detector is positive to the positive
envelope, this diode detector is positive to
the negative envelope of the same AM signal, because the 2 AM signals are identical.
And therefore, if you simply add the 2 outputs, you are able to reduce, so if I call this
e plus t corresponding to the positive envelope.
And this has e minus t corresponding negative envelope, y sub d t the detector
output would be e plus t plus e minus t, that e
minus t is minus of e plus t. If you have carried out the difference between
wanted to carry out, of course these two LC circuit is are tune to two different frequencies,
they are not tune into the same frequency. What, we are saying is e minus
t will be the negative of e plus t , so if I add
the 2 is like subtracting the same thing with, I should call them differently, I will call
this e 1and I call this e 2, I think that is a
problem, otherwise this becomes identically 0.
So, this is e 1, this is e 2, e 1, e 2 thank you for that pointing out that mistake, because
the two envelopes are slightly different, I mean
the envelopes that we are looking at through the diode detector are different, because
you have different tune frequencies here. This is
tune to frequency f 1, this is tune to frequency f 2 and the responses are different, that
is the whole point.
So, this is e 1 plus t, this is e 2 minus t, so the point that I meet earlier that the
envelopes identical has to be modified. Envelopes are
different, because the two bank pass filters are different and for that wrong statement
and what you get here is this difference output that you want it.
So, in another words let me just complete this discussion, output of your upper end
is proportional to that, we just summarized what
I have discussed, the amplitude of this is the important thing to appreciated. The output
of the upper and lf detector here is going to be proportional to the magnitude of H 1
f, where H 1 f is the frequency response of this LC circuit.
Similarly, output of the lower envelope detector will be proportional to magnitude of H 2
f and the detector output yd t will be proportional to the difference between these two
and this is what you want to do. This again the question that you phrased some time ago
cannot be first do the subtraction and then envelope detection that is the answer to
question no, but this is what you want to do. You do not want to do envelope of H 1
f minus H 2 f magnitude of H 1 f minus H 2 f.
We want the two magnitudes to be subtracted with each other that are the important point
to appreciate it. Such device that, I have just indicated is known as the balanced
discriminator, just like the balanced modulator that we discuss some time ago, so this is
the balanced discriminator. Basically, what we are saying the term balanced is use to
indicate that the response to the carrier is balanced out and the respond and for the
final output to carrier frequency itself is 0.
The two circuit is will have identical response as the carrier frequency and the two, for
the response of the carrier frequency, the final detected output due to the carrier
frequency will be 0.When, an input signal is at a carrier frequency, then the output
will be at that instant 0, because the two circuit
is will behave identically at that time, so that
is the balanced discriminator. Now, there are several variations of this
thing and they are number of variations of this
practical implementation, which improve upon this basic principle and I will since we
will not have time to discuss all those in the class. I will just mention the names of
a few other discriminators, which improve on this
basic idea further. So, I like you to read about them on your own.
Some of these are phase shift discriminator and there is a ratio detector
and there is a foster seelyFM detector or discriminator.
So, I like you to read more about them on your
own, find out the suitable book in which they are available to read about them. Now, all
of them would be discussed in your book I think, may be at this stage, I should put
the entire query that we have discussed on a slightly
formals fitting. By considering precisely, how this discriminator
works by developing a model for these discriminators, theoretical models. And then,
looking at the output of response, I thing once we have that, then many of the questions
that have been raised will disappear hopefully.
So, let us look at the mathematical theory of slope detection, basically that is what
we are discussed so far, in developing a mathematical
theory, we will work with slightly idealized filters. Basically to elaborate
further as to what I mean by this idealization, in
the two bank pass filters that we just discussed, we are really using only the linear
portion, we are interesting in using the linear portion of the two bank pass filters.
So, in a way remember, these bank pass filters are motivated by what, what do we really
want, you want a differentiator, to be approximately and that differentiator characteristic
to be available around the carrier frequency. So, the g omega characteristics that we
want, we want it to shift them to the frequency of interest the carrier frequency. So,
really speaking let us look at the filter one of them, let us say H 1 f.
So, ideally we wanted filter characteristics like that, let me call it H 1 f upon j, from
a single slope detector, this is the kind of
thing I want. And since, I am plotting remember, I want a g omega kind of thing here, the differentiation
and that is what I am plotting H 1 f by j, so what will I get on the other side,
it will be this. So, a start from minus fc minus
B T by 2 to minus fc plus B T by 2, where B T is your transmission bandwidth of a FM
signal. So, this is an idealized filter that who have
served my purpose, corresponding to one of the two bank pass filters, basically by using
any of the two bank pass filters, I am trying to realize this kind of a characteristic.
At the moment, let us work with the case of a
single bank pass filter, this idealization is for the case, when we are using the single
bank pass filter.
Of course, the actual bank pass filter will have a complete response, but really speaking
the portion of the bank pass filter is pass, which I am effectively trying to utilize is
the linear portion and I am idealizing that as
if I have filter with these characteristics. Basically therefore, we need to analyze the
behavior of this filter followed by envelope detector.
If you understand this, you understand that, would you appreciate this model, I mean you
go along with this model, because we need we do not need to look at the complete
behavior of the bank pass filter. You are not going to use the characteristics of the
bank pass filter, beyond the peak and we are representing
the rest of it by linear approximation of this kind. That is what the ideally important,
if you about to use a single bank pass filter, how this changes, how the picture
changes, when you are using two bank pass filters, will take up subsequently, yes please
Precisely, this is all I am saying is this is the approximation that I would get, if
about to use a single bank pass filter case system,
which we discuss earlier to get the 0 response at
fc; obviously, I have to use the two bank pass filter system, which I will come to later.
Let us, discuss first the single bank pass filter or single slope detector output, so
you have a single slope demodulator.
Let us represent this, the characteristics of this filter by it is complex envelope,
what will be the spectrum of it is complex envelope,
what will that look like, let me do this brought down to 0, this will be between minus B T
by 2 to plus B T by 2, sub t by 2. And let me
say, the slope here is2 pi, some constant a, this is just an arbitrarily selected, this
to be2 pi a, the a is a constant and the slope here
would be 4 pi a, because the amplitude here would be double, when you consider the complex
envelope. So, this is a plot of H1 tilde f upon j, I
am going to use complex envelope representation of both the FM signal as well as this filter
to study the response at the output of this filter.
So, suppose I was to mathematically represent this picture, what we are saying this
complex envelope representation in a frequency domain of H 1 f and a H 1 f. I will call it
the slope circuit and no longer call it the bank pass filter. It is a slope circuit, which
request particularly you raise the bank pass filter, which I am denoting by H1 tilde f
which such that this is equal to just please look at this is equal to j 4 pi a into f plus
b by B T by 2.
This is the linear equation of a straight line, the equation of the straight line can
be written as j 4 pi a into f plus b sub t upon
2 in the range between mod of f line between b
sub t by 2 just look at this carefully. So, this is f equal to minus B T by 2, this becomes
0, at f of equal to plus B T by 2, it becomes
4 pi a into B T. So, therefore that is why, the slope is 4
pi a, this value will become 4 pi a into B T at f is
equal to B T by 2 and it is 0 outside this. So, this is the mathematical representation
of the complex envelope of the filter the slope
circuit, that I am using, what is the input that
you are feeding into it, the input that you are feeding into it is your FM signal. And
it is a sub c cosine of2 pi fct plus2 pi kf; I am
being consistent with my earlier notation that
does not matter. May be, I think the constant2 pi, which was
not there earlier that does not matter, what is
the complex envelope of this, pi inspection and if you assume, if you look the usual
assumptions that we make about the representation of aero band signals. So, we can the
complex envelope of this as ac into exponential of j2 pi k sub f integral m tau, d tau
between 0, minus infinity to t also 0. Let us say this has some spectrum S tilde
f, this is the time domain representation, let us
say the frequency domain representation is full transform is some function S tilde f.
We do not need to find it, because we will not
need it, we will come back to S tilde f will come back to time domain. So, to find the
output, what do we need to do multiply this S
tilde f with the H1 tilde f, that will give you the complex envelope of the output
spectrum, rather the spectrum of the complex envelope are the are the output.
So, the output of a slope circuit, therefore would be S 1 tilde f, let me call this output
spectrum on denoting the S 1 tilde f and that will be equal to half of H1 tilde half times
S tilde f. Remember that, the output is half
of the complex convolution of the two and the
convolution in the frequency domain will be replaced by multiplication. So, if I
substitute for H 1 tilde f that is given by j, which was 4 pia, so it will become j2 pia
into f plus b sub t upon 2 into S tilde f. For mod
of f lying between b sub t by 2 and will be 0
elsewhere. Now, I do not know S tilde f, so what I will
do is, I will take this equation back to time domain, take the inverse transform of this,
what I will get, S1 tilde t, which is the complex envelope of the output. So, how convenient
it becomes, we use a complex
envelope representation everything becomes a simple, this is equal to how can you tell
me, what the corresponding time domain equation would be.
The first equation j2 pi a f times S tilde f, this is the frequency domain relationship,
what is the corresponding time domain use Fourier
transform properties that is a times ds tilde f upon dt derivative of st, s tilde f, a is
common, so I will keep it like this. I want me the
second one this is the constant2 pi a into B T by 2 everything is a constant here, so
I left with plus B T by 2, maybe I will make j2 pi.
So, this should be this two factor will go and then will be j there into pi a, a is outside
have made a mistake of it. So, let me rewrite, this is a times d s tilde f upon dt plus j
pi b sub t into s tilde f, that is the corresponding
time limit equation. So, I have expressed the
output in terms of the input, so what do we find the output is a sum of two things, the
input itself plus it is derivative. So, this shows, that this circuit, the slope circuit
that I have started with thus carry out the differentiation
that I want you to carry out, but in addition, it also produces this additive component.
So, let us look at this again, let us substitute for s tilde t, remember what is your s tilde
t, s tilde t is this complex envelope, and so
let me substitute for this in the last equation. So,
I will need to use this as well as it is derivative what will be it is derivative of this will
become j2 pi k f into ac into mt. The derivative of the exponent which will be j2 pi k of
mt times ac times exponential of the same argument and s tilde t is of course this.
So, I need to use both this, in fact do that, it is a straightforward operation, I am left
with the output s 1 tilde t is given by I am just
writing the final expression, you can get it
yourself the way I have explain it a few minutes ago. It will term out to be j pi b sub t a
times a sub c into 1 plus 2 k sub f upon b sub t into mt into exponential of j2 pi k
sub f integral m tau d tau, this is what you get
when you substitute for s tilde t in this expression.
So, I substitute for s tilde t, find out the derivative, combine these two terms, simplify
it keep that this expression, this was your s
tilde t part, which comes outside as a common factor and this is what you get as the other
factor and you are looking in a angle up of the
output. How is the real envelope, relative to the complex envelope, just the magnitude
If you want, you can first go to the bank pass signal, by combining it with e to the
power j2 pi f not t and taking the real part of
that and then taking the envelope in the usual way,
but there is no need to do all that. The actual envelope of the signal s 1 t would be
nothing but the modulus of the complex envelope, this is something that we already
know and what that is going to be equal to what is the modulus of this; it is piB T a,
ac into 1 plus 2 kf upon B T into mt.
So, that is what you get, you get an output, that the envelope of the output is proportional
to ac signal plus some dc component same as before that we discussed. So, this is the
analysis of the single slope circuit, based discriminator will complete this analysis.
Next time, when you ask to consider the double
slope circuit that just takes
a minute. Thank you very much.