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Welcome to lecture 3. In the last class, we
were looking at the equivalence between full
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rate sampling and time-interleaved sampling.
The motivation to consider timeinterleaved
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sampling is those applications where implementing
a full rate sample and
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hold might just not be possible. So, just
like if you are not able to work hard enough
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if
you hire more people to do the same job. Here
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the idea is to use 2 sample and holds
sampling at half the rate and working in parallel.
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And somehow you stitch the outputs
together in order to get an output sequence
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which is what the full rate sample and hold
would have given you understand. And. So,
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I have shown here this is the equivalent to
the full rate sample and hold right. And this
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shown below is a mathematical equivalent of
time-interleaved sampling.
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Please note the words mathematically equivalent
physically it will not be implemented
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like this. What makes more sense to implement
in practice is that the top sample and
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hold channel will be sampling at all even
multiples of the sampling. Sampling clock
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period and the bottom channel will be sampling
at a lot multiples of the sampling period
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which is T. To analyze it does not make a
difference if we have a mathematical
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equivalence which is why I simplified things
by sampling both of them at the same
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instant of time. However, 1 is artificially
been advanced by a time T.
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And if we put this whole contraction inside
a box the system inside the green box should
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in principle be not distinguishable from a
full rate sample and hold right. And during
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the
last class we wrote down the expressions for
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spectra at various places in the system what
are important are. Let us get some more intuition
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on this by actually plotting the
spectrum last time we had equations for the
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various spectra. Let us now plot and get
some intuition; let us just quickly remind
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ourselves what is the maximum bandwidth of
the continuous time input signal x of T which
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will allow us perfect reconstruction.
Status is.
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I mean please note also that we in the last
class we denoted 1 by T by by f s correct.
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So,
let us quickly remind ourselves again what
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is the maximum bandwidth if x of T which
will allow us to enable perfect reconstruction?
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F s by 2.
F s by 2 and in the interleaved sampling system,
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what are we sampling each channel at.?
No, please note this signal x of T can have
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a maximum bandwidth of f s by 2. And in
each in these in this interleaved channel
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each of these channels is sampling at what
rate?
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F s by 2 .
It is also sampling at F s by 2, you understand.
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So, the signal. So, as far as the each of
these channels is concerned the input signal
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bandwidth is f s by 2 and the sampling rate
is also f s by two. So, what do you think
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happens in each of these individual channels.
Aliasing.
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They will be aliasing in each of these individual
channels; however, from end to end we
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saw that the system is equivalent to a sampling
system operating at f s. So, if there is
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aliasing happening here and aliasing happening
in the lower channel. But when you add
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the 2 if it is equivalent to something where
there is no aliasing what must be happening?
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Transforming .
What must be happening is somewhere along
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the line the.
Alias .
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Alias components are.
Cancelled.
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Cancelled; so, let us simply the you know
take a look at this spectra and and plot them
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again and plot them this time last time we
saw this through the mathâ€™s. Let us actually
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draw pictures and convince ourselves that
that is indeed happen. So, can somebody
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please remind me what we call this to be consistent
with notation used last time what did
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we call this point?
C.
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So this was called point c all right what
about this point?
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J.
G for god.
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So, we called this g and what did we I guess
we must have called this a then and then b
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c
d e f and g.
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H.
And h is the final output, am I right?
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Yes.
So, I am not going to go over all the math
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again at c the spectrum must be of the form
1
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by 2 T sum over all k x of f s by 2 pi times
omega minus.
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Pi k pi k.
Pi k and at g we saw last time that the spectrum
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is 1 by 2 T sigma over all k x of f s by 2
pi times omega minus pi k times.
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E to the power.
E to the power j.
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Final minus j.
Minus j.
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K pi k pi.
K pi does it make sense? This are results
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that we derived the last time along.
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So, now let us sketch these how do you let
us say this was the continuous time signal
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and
let us assume that the continuous time signal
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is occupies the entire f s by 2. So, if this
signal was ideally sampled at f s then they
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would be no aliasing correct. However, how
does this look like?
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It is shifted horizontally.
So, let us remind ourselves again if this
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signal was sampled at f s, what would the
discrete time spectrum look like?
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Opening.
What do we do to get from continuous time
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to discrete time? 2 pi multiples of 2 pi and
this is the continuous time signal first you
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make copies of a.
Copies and paste at what intervals.
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2 pi 2 pi.
At f s right. So, if we had sampled the original
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signal at f s this is what you would have
got. And then we replace the x axis with we
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scale the x axis we multiply this by you
know 1 over T and we scale the x axis. So,
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what does f s by 2 become pi pi. F s by 2
becomes pi this becomes minus pi and. So,
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this becomes 3 pi 5 pi and so on. What is
this
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in the discrete time domain? This is what
we would have got this is omega minus 2 pi
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k
the stuff that we already know. But what is
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this expression? What is the difference I
mean what is the difference between this expression
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and this expression?
Superior and 2 pi.
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One is repeated at.
Pi.
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2 pi where as the other one is repeated at.
Pi.
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At pi; so, how do you think this expression
will look like?
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Margin t
You will have.
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Aliasing aliasing.
Aliasing terms now repeated at.
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Pi pi.
You will have images repeated at pi. So, you
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will have right. So, this corresponds to
the
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spectrum at sorry at c, is that clear? Now,
I need to find this or draw the spectrum at
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g,
but before we go there please note that indeed
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we see aliasing in the top channel right the
spectrum at c itself is.
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Alias.
Alias; so, this makes sense because the signal
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bandwidth is f s by 2 and we are sampling
at f s by 2. Now, let us draw the spectrum
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at g and to do that I will copy
and paste, what
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is the only difference between the spectrum
at c and the spectrum at g?
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Ones.
The ones which you have shifted by.
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Pi.
Pi are now inverted in psi. So, instead of
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having I am sorry.
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Does it make sense here also we note that
that is aliasing correct, because the sum
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of the
spectra is is the black guy plus the.
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Red guys.
Red guys so, there is aliasing in both channels.
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However, when you add c and g together
what is happening into the step and repeat
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happening at odd multiples of pi get cancelled.
Whereas, those at the even multiples of pi
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reinforce each other and what we get
eventually
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is this. And we also understand that the strength
is 1 by 2 T the strength is also
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1 by 2 t. So, when you add these 2 the strength
will be at 1 proportional to 1 by T. And
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you get the resultant h is a spectrum like
this which is what we expected in the first
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place
given that this whole system is equivalent
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to a system which samples at rate f s it is
clear.
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Now, the next thing is to figure out what
happens when there are non idealities here.
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So, let me copy this again as I said a practical
realization of this approach would
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basically have 2 sample and holds 1 sampling
at all odd multiples
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of T and 1 sampling at
all even multiples of t. Then you have these
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2 sequences and you interpolate up by a
factor of 2 by inserting zeroes. Then move
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one of the sequences by 1 sample and add the
2 the several practical non idealities which
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come in which we never accounted for in the
derivation for 1 as I mentioned that day the
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signal is passing through different channels.
And each channel can add offset in other words
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the output here for example, is not
simply the sampled output plus some offset
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which we denote by v offset 1 and v offset
2
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right. So, the first non-ideality we consider
is offset now let us try and understand
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intuitively what we should expect for the
output sequence if each of these channels
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had
different offsets. Or before you go to different
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offsets may be we tried figure out what
happens when there are the same offset. So,
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when both of them have the same offset
what do you think will happen to the output?
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Corresponding same.
If all the channels have the same offset then
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the output will simply be.
Will be ideal.
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The ideal output plus.
Offset.
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The offset that each channel is add correct
now let us try and figure out what happens
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when the 2 channels have different offsets
what do you think will happen?
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Minus c cancellation.
How do you think the output will I mean let
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us try and understand this by putting 0 input.
If you put in a 0 input to the system ideally
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what should you see at the output?
T.
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Ideally you we should see 0. If both channels
have the same offset, what would you see?
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You would see a constant d c value of v offset.
Now, if both channels had different
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offsets v o offset 1 and v offset 2, what
would you see?
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P s a 1 offset.
So.
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Offset.
Very good. So, in all even samples you would
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see.
Offset.
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1.
1 all right and all odd samples you will see.
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Offset.
2 mind you the offset stays constant. So,
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every alternate samples value will be the.
Same.
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Same correct. So, if you look at this sequence,
what can you decompose this as can you
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see something in the sequence can you comment
on the spectral content of the sequence?
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This is a discrete time sequence.
Two different figure it seems different same
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period.
You please note that this is a discrete time
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sequence by definition it only exists as a
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sequence. So, there is no I mean the again
the. So, may at the a and that additional
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additional 2 1 at what frequency? Frequency
of time difference is t. So, 1 by T is please
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note that this is a discrete time sequence.
So, it is there is there is no you know there
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is
no question of you know frequencies f s you
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understand that was continuous time after
sampling you only have a sequence. So, whatever
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frequency you come up with must be
within the range 0 to pi correct. So, now,
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can you look at this sequence in general it
is
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very difficult to look at a sequence, And
tell I mean say what its spectral content
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is unless
you have a natural Fourier transform at your
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eyes you understand.
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But for this special simple sequence, we should
be able to look at it and say. And this is
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1
channel offset; this is the other channels
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offset
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and this is v offset 1; this is v offset 2;
this
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is clear.
Sir these are the amplitudes or this is the
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signal I mean v offset 1 of n.
V offset is a constant which is added to every
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sample in that channel correct see if the
input was grounded what would you expect ideally
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at the output?
0.
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All zeroes, now what is happening is that
each channel is adding a constant, but fixed
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offset I mean its I am sorry its its constant
offset to all samples. So, so instead of being
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instead of the output of this channel on the
top being 0 0 0 it is v offset 1 v offset
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1 v
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offset one. But these are only the even samples
this channel on the other hand also adds
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offset, but its offset is not the same as
the offset added by the first channel. And
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this can
happen in practice you have to mismatch in
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the circuits that I use to implement these
systems. So, let us not worry at this point
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about where these offsets are coming from
and
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why they are different we just take them for
granted that this is what would happen in
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practice. Now, this side the lower channel
also adds offsets which is v offset 2 v offset
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2
v offset 2 and so on. And since we are interleaving
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this, these 2 sequences it follows that
the output of the composite system.
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When the input is 0 will be a sequence like
this is this clear to everyone. Now, all that
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I
am saying is that we can think of this sequence
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as the sum of 2 sequences 1 which is.
V offset.
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The average
drawn in the blue squares correct plus another
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sequence which is what how
does the other sequence look different. The
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difference here will denote that by triangles
here what is the difference in the second
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sample.
Negative.
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Negative and these 2 will have the same magnitude
correct because by definition I have
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removed the average. So, this is the average
which is v offset 1 plus v offset 2 by 2.
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And
the amplitude of this sequence which as we
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can see is periodic with the period of 2
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samples is the peak to peak amplitude is v
offset 1 minus v offset 2. Since the discrete
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time tone is periodic with 2 samples the in
the frequency domain it must be at.
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Sinusoidal.
A sinusoidal to omega equal to pi is this
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clear. So, now, if we trace this back to the
continuous time domain for ease an understanding.
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We can think of this tone at omega
equal to pi as coming from a continuous time
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sinusoid operating at. It is like if we see
a
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tone in the discrete time domain at a frequency
omega equal to pi that must be coming
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from a. We would normally think that that
would be coming from the continuous time
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input at what frequency.
F s by 2.
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Correct, because if we put this whole thing
in a black box. And we look at the output
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sequence and the output sequence seems to
show d c plus a tone at omega equal to pi.
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And as far as we are concerned this is coming
from a sample and hold operating at a rate
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of f s which basically means that the input
must be coming from must have a tone at d
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c.
As well as f s by 2 in another words offset
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in each of these channels will make us believe
that the continuous time input has components
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at f s by 2. As well as d c while the real
continuous time input is 0 so; obviously,
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this is a some kind of error correct causing
us to
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get confused about the nature of the input
in continuous time. Now, if the input was
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not 0
what do you think will happen?
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So, this causes offset and a tone at omega
equal to pi which corresponds to f in equal
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to f
s by 2. Now, let us discuss what happens when
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apart from the offsets there is a regular
input coming into the time-interleaved sampling
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system what do you think will happen?
What do you think the output sequence will
211
00:29:26,320 --> 00:29:40,130
be this superposition the input and the
offset? Very good. So, if the input was not
212
00:29:40,130 --> 00:29:47,940
grounded, but was a regular input that you
would normally use then the output sequence
213
00:29:47,940 --> 00:29:55,360
apart from containing the ideal output
sequence x of k T would also contain this
214
00:29:55,360 --> 00:30:05,200
extra sequence. Because this simply adds to
what whatever was suppose to come out ideally
215
00:30:05,200 --> 00:30:11,820
correct. So, what is the conclusion now?
Therefore, what is the conclusion?
216
00:30:11,820 --> 00:30:19,080
Super position.
217
00:30:19,080 --> 00:30:31,410
Yes. So, basically if you have offsets in
each of these channels the output spectrum
218
00:30:31,410 --> 00:30:37,690
will
be change of will be the ideal spectrum plus
219
00:30:37,690 --> 00:30:41,540
d c offset plus a tone at.
Pi.
220
00:30:41,540 --> 00:30:54,000
Omega equal to pi. Now, let us think what
would happen if instead of having a 2 way
221
00:30:54,000 --> 00:31:00,290
sample and hold. We had a 4 way sample and
hold what do you think will happen to
222
00:31:00,290 --> 00:31:07,670
offsets?
Pi by 2 pi by 2 I think.
223
00:31:07,670 --> 00:31:11,770
Pardon.
At pi by 2.
224
00:31:11,770 --> 00:31:15,110
Is it only at pi by 2.
Pi by 2 pi pi by 2 pi.
225
00:31:15,110 --> 00:31:24,540
Very good. So, if you have a 4 channel sample
and hold system you will see tones due to
226
00:31:24,540 --> 00:31:28,500
offset at.
Pi by 2.
227
00:31:28,500 --> 00:31:42,350
Pi by 4 2 times pi by 4 3 times pi by 4 and
actually I am sorry no if what would I say
228
00:31:42,350 --> 00:31:51,520
we
assumed what was this? So, I am sorry if we
229
00:31:51,520 --> 00:31:58,230
have a 2 channel sample and hold then we
see tones at.
230
00:31:58,230 --> 00:32:05,900
2 times pi.
At at pi. So, in other words its 2 pi by 2
231
00:32:05,900 --> 00:32:10,040
if we have 4 channels we would see.
4.
232
00:32:10,040 --> 00:32:12,460
Tones at.
2 pi.
233
00:32:12,460 --> 00:32:18,310
2 pi by 4.
2 pi and 3.
234
00:32:18,310 --> 00:32:24,510
And 2 pi by 2 and as well as d c you understand.
So, in other words if you translate this
235
00:32:24,510 --> 00:32:31,650
into the continuous time domain a 2 channel
system would make it look as if the input
236
00:32:31,650 --> 00:32:41,530
contained d c. As well as a tone at f s by
2 if you have a 4 channel system it will look
237
00:32:41,530 --> 00:32:48,650
a
like d c f s by 4.
238
00:32:48,650 --> 00:32:52,790
F s by.
F s by 2 and 3 f s by 4 which is the same
239
00:32:52,790 --> 00:32:59,160
as f s by four. So, in general if you have
an n
240
00:32:59,160 --> 00:33:06,970
channel sampling system you will see tones
at f s by.
241
00:33:06,970 --> 00:33:12,100
N
N and its multiples and that makes intuitive
242
00:33:12,100 --> 00:33:22,040
sense because every n-th sample is the same,
because the first channel processes the I
243
00:33:22,040 --> 00:33:27,860
mean the first sample its again exercised
only
244
00:33:27,860 --> 00:33:36,230
after n after all after you gone through all
the other channels. So, you come back every
245
00:33:36,230 --> 00:33:43,200
nth sample is the same right and the samples
from 0 to n minus 1 are all different because
246
00:33:43,200 --> 00:33:49,730
they are processed through different channels.
So, you can always expand this as a
247
00:33:49,730 --> 00:33:58,000
Fourier series with the fundamental period
equal to n samples which corresponds to f
248
00:33:58,000 --> 00:34:11,579
s
by n does it make sense. So, in other words
249
00:34:11,579 --> 00:34:20,800
we already see that using time-interleaving
has its problems means nothing comes for free.
250
00:34:20,800 --> 00:34:26,270
So, we thought we would make our life
easy by using 2 slow sample and holds and
251
00:34:26,270 --> 00:34:34,800
stitching their outputs together. And we
already see that because of offset there is.
252
00:34:34,800 --> 00:34:43,060
Artifacts in the sampled output which would
not have been there if the sampling rate was
253
00:34:43,060 --> 00:34:52,950
f s I mean was it was a true full rate sample
and hold, is that clear? Now, the next non,
254
00:34:52,950 --> 00:34:55,580
any questions?
255
00:34:55,580 --> 00:35:31,640
The next non-ideality is gain error. So, let
me again copy this and paste now just like
256
00:35:31,640 --> 00:35:37,920
each channel introduces its own offset. Because
these 2 channels are built using different
257
00:35:37,920 --> 00:35:44,360
circuitry though they are nominally identical
it will turn out and there will be a small.
258
00:35:44,360 --> 00:35:47,330
Gain.
Gain error in other words the gains of all
259
00:35:47,330 --> 00:35:54,440
the channels will not be the same you
understand. So, let us now consider gain error
260
00:35:54,440 --> 00:36:13,850
separately in a in the. In other words the
nominal gain is 1 plus alpha and the nominal
261
00:36:13,850 --> 00:36:24,550
gain is 1 plus beta in the lower channel
ideally alpha equal to beta equal to alpha
262
00:36:24,550 --> 00:36:32,730
equal to beta equal to 0. Both are very very
small numbers in practice alpha will not be
263
00:36:32,730 --> 00:36:40,510
equal to beta again. Because of random
mismatch between both the channels, what do
264
00:36:40,510 --> 00:36:44,730
you think before we get into the math? Let
us try and understand intuitively what we
265
00:36:44,730 --> 00:36:53,550
should expect? What do you think will happen
if alpha was equal to beta? What do you think
266
00:36:53,550 --> 00:37:06,960
will happen?
Scaled.
267
00:37:06,960 --> 00:37:11,130
If alpha is equal to beta there is no real
problem in the sense that it will appear as
268
00:37:11,130 --> 00:37:15,880
if the
entire sequence of the output is in has a
269
00:37:15,880 --> 00:37:20,880
gain error of 1 plus alpha I mean has an error
of
270
00:37:20,880 --> 00:37:27,110
gain error of alpha correc. Because it is
like the even samples getting multiplied by
271
00:37:27,110 --> 00:37:29,900
a
constant 1 plus alpha and the odd samples
272
00:37:29,900 --> 00:37:34,200
also getting multiplied by a constant 1 plus
alpha. So, when you stitch the 2 sequences
273
00:37:34,200 --> 00:37:36,720
together it is pretty much like the original
274
00:37:36,720 --> 00:37:43,790
sequence except that all numbers have been
multiplied by this by this extra factor 1
275
00:37:43,790 --> 00:37:48,620
plus
alpha. So, there is no real problem now, what
276
00:37:48,620 --> 00:38:00,870
do you think will happen when alpha is not
equal to beta. There is normal term.
277
00:38:00,870 --> 00:38:05,690
Performance.
Very good. So, let us recall that what were
278
00:38:05,690 --> 00:38:34,470
we depending on. So, if
279
00:38:34,470 --> 00:38:40,450
the channels had
identical gains then we recall that we are
280
00:38:40,450 --> 00:38:48,270
adding the spectra at. And we understand that
each of the sub channels is not sampling at
281
00:38:48,270 --> 00:38:49,270
nyquist.
282
00:38:49,270 --> 00:38:52,300
So, they will be aliasing in each of these
sub-channels; however, when we stitch the
283
00:38:52,300 --> 00:39:02,530
signals of the sequences together magically
the aliases are getting canceled off correct.
284
00:39:02,530 --> 00:39:11,680
Now, if there was a gain error in both these
channels in other words this if you recall
285
00:39:11,680 --> 00:39:14,530
g is
the output of the channel on top h is the
286
00:39:14,530 --> 00:39:18,240
output at the spectrum at the output of the
lower
287
00:39:18,240 --> 00:39:26,690
channel. So, if the gains were different this
is 1 by getting multiplied by 1 by 2 T into
288
00:39:26,690 --> 00:39:29,580
1
plus alpha. Whereas, the lower channel is
289
00:39:29,580 --> 00:39:40,740
getting multiplied by 1 by 2 T times 1 plus
beta, is this clear? Because all that we have
290
00:39:40,740 --> 00:39:51,250
done we have already computed the spectra
at this point and this point correct.
291
00:39:51,250 --> 00:39:56,640
Now, if there was a gain error in this channel
it is simply taking the spectrum at c for
292
00:39:56,640 --> 00:40:04,290
instance and multiplying it by 1 plus alpha.
In a similar fashion if there is a gain error
293
00:40:04,290 --> 00:40:08,480
in
the lower channel it is like to compute the
294
00:40:08,480 --> 00:40:11,070
spectrum at g we just take the old spectrum
295
00:40:11,070 --> 00:40:27,180
and multiply it by 1 plus beta instead correct.
So, since the picture is worth a thousand
296
00:40:27,180 --> 00:40:33,790
words what were we seeing earlier if alpha
was equal to beta was equal to 0. Then which
297
00:40:33,790 --> 00:40:38,940
of these components were which of these colors
were was getting cancelled out. The
298
00:40:38,940 --> 00:40:46,230
images in red were getting were cancelling
exactly where as the images in black were
299
00:40:46,230 --> 00:40:50,801
adding up. Now, what do you think will happen
if alpha is not equal to beta?
300
00:40:50,801 --> 00:40:56,350
Beta will cancel
There will be A red component in others words
301
00:40:56,350 --> 00:41:02,260
there will be an.
Alias component.
302
00:41:02,260 --> 00:41:47,300
Alias component. So, if I ((no audio 41:01
to 41:45)). So, the strength of the black
303
00:41:47,300 --> 00:41:53,400
images will be 1 by T plus 1 by 2 T times
alpha plus beta the red ones. However, will
304
00:41:53,400 --> 00:42:15,241
be
this is somewhat exaggerated, because the
305
00:42:15,241 --> 00:42:36,270
cancelation is not perfect any more does it
make sense. And what will be the height of
306
00:42:36,270 --> 00:42:49,410
these images it will be 1 by 2 T into alpha
minus beta clearly when alpha is equal to
307
00:42:49,410 --> 00:42:54,380
beta again the image is cancelled. And there
is
308
00:42:54,380 --> 00:43:00,430
simply a gain error correct, but when alpha
and beta are not equal we see that there are
309
00:43:00,430 --> 00:43:02,560
alias components.
310
00:43:02,560 --> 00:43:16,480
Now, specifically if the input consisted of
2 sinusoids or consists of a sinusoid what
311
00:43:16,480 --> 00:43:35,450
do
you think will happen
312
00:43:35,450 --> 00:43:48,460
if the input consisted of a sinusoid? The
discrete times spectrum
313
00:43:48,460 --> 00:43:56,810
would ideally have done something like this.
Now, what do you think will happen? I
314
00:43:56,810 --> 00:44:06,680
mean I do not really need to I mean and draw
the periodic extension, because I know its
315
00:44:06,680 --> 00:44:19,370
periodic between minus pi and pi. I want to
get rid of the scaling factor also just to
316
00:44:19,370 --> 00:44:24,340
make
the diagram a little clearer and what else
317
00:44:24,340 --> 00:44:37,580
what other tones would be present. So, there
would be something here and something there
318
00:44:37,580 --> 00:45:21,800
correct. So, what is I am sorry I made a.
So, if the input was continuous time and had
319
00:45:21,800 --> 00:45:33,230
a frequency of let me erase and redraw.
320
00:45:33,230 --> 00:45:45,860
It clearly I am going up just that region
between minus pi and pi what we would have
321
00:45:45,860 --> 00:46:03,790
got
in something like this. And the red part would
322
00:46:03,790 --> 00:46:17,730
have done something like this is minus pi;
this is pi. Now, what I am asking you is what
323
00:46:17,730 --> 00:46:26,040
would happen for the specific case of the
input being a tone at some frequency f in
324
00:46:26,040 --> 00:46:37,690
right in which case the input spectrum would.
For instance look would have 2 impulses here
325
00:46:37,690 --> 00:46:45,540
and if the input was at f in and the discrete
spectrum where would they occur at.
326
00:46:45,540 --> 00:46:48,390
f in .
F in by.
327
00:46:48,390 --> 00:46:49,640
F s.
2 pi.
328
00:46:49,640 --> 00:46:53,690
Times 2 pi correct.
Yes.
329
00:46:53,690 --> 00:47:05,190
And because these alias components do not
cancel they will be
330
00:47:05,190 --> 00:47:18,500
something there all right
and this is f in by f s times 2 pi what is
331
00:47:18,500 --> 00:47:20,670
this distance.
.
332
00:47:20,670 --> 00:47:32,880
Is also f in by f s times 2 pi correct. So,
ideally if we had just a sample and hold running
333
00:47:32,880 --> 00:47:38,360
at the full rate what would which of these
colors, would you see.
334
00:47:38,360 --> 00:47:40,490
Black.
Only the spectral components shown in black
335
00:47:40,490 --> 00:47:51,150
could appear now, because of timeinterleaving
and gain mismatch. We see that there are other
336
00:47:51,150 --> 00:47:55,530
artifacts in the spectrum and
they are at what frequency?
337
00:47:55,530 --> 00:47:58,250
High frequency.
At high frequency and what is that high frequency
338
00:47:58,250 --> 00:48:01,270
here?
F s by 2 and.
339
00:48:01,270 --> 00:48:04,520
F s by 2.
If this distance is f in by f s times 2 pi.
340
00:48:04,520 --> 00:48:17,210
It is pi minus f in by f s times 2 pi translated
into the continuous time domain it is like
341
00:48:17,210 --> 00:48:21,800
having an input tone at.
F s by 2.
342
00:48:21,800 --> 00:48:36,650
So, in continuous time it is like having f
s by 2 minus f. In minus f in or equivalently
343
00:48:36,650 --> 00:48:40,920
you
can say is f s f in f s by 2 minus plus f
344
00:48:40,920 --> 00:48:44,640
in, because whether you have a tone at f s
by 2
345
00:48:44,640 --> 00:48:59,770
minus f in or f s by 2 plus f in when you
sample it at at at f s. They will you understand
346
00:48:59,770 --> 00:49:04,890
intuitively why does this make sense is. So,
if you have a tone at f s by 2 minus f in
347
00:49:04,890 --> 00:49:06,991
what
can you tell me, why does the this intuitively
348
00:49:06,991 --> 00:49:13,110
make sense?
The sample f f s by 2 aliasing would happen
349
00:49:13,110 --> 00:49:14,380
at.
Yes.
350
00:49:14,380 --> 00:49:44,280
Now, is that aliasing part is not? If you
alias if you sample this f in f in tone.
351
00:49:44,280 --> 00:49:47,160
At f s.
Yes.
352
00:49:47,160 --> 00:50:00,790
We would aliasing would happen exactly at
f in by f s by 2 minus. No how is that I
353
00:50:00,790 --> 00:50:04,860
means the question I am trying to get at is
how is that if we would samples sinusoidal
354
00:50:04,860 --> 00:50:16,300
tone at a frequency f in with a single channel
sampling at the rate f s. We see only these
355
00:50:16,300 --> 00:50:20,500
2
tones in black; however, when you we use a
356
00:50:20,500 --> 00:50:25,200
2 channel sample and hold with gain
mismatch. We see additional tomes please note
357
00:50:25,200 --> 00:50:31,290
that the amplitude or the strength of these
tones is proportional to alpha minus beta
358
00:50:31,290 --> 00:50:37,360
correct. So, why does this make sense? No
in
359
00:50:37,360 --> 00:50:41,320
each of the channels you are actually things
I mean.
360
00:50:41,320 --> 00:50:42,320
Yes.
Aliasing is happening.
361
00:50:42,320 --> 00:50:44,720
Correct.
Because there is a gain mismatch.
362
00:50:44,720 --> 00:50:45,720
They are not cancelling.
Cancelling.
363
00:50:45,720 --> 00:50:50,450
All right I mean you know that is the math
in words, but is there more intuition to this
364
00:50:50,450 --> 00:50:51,450
then.
Sir in each channel you are sampling at f
365
00:50:51,450 --> 00:50:55,780
s by 2. So, there is a multiplication of a
tone at f
366
00:50:55,780 --> 00:51:10,710
in and I mean some frequency components at
f s by 2 and multiples of it. So,
367
00:51:10,710 --> 00:51:14,790
multiplication in time domain is nothing but
in frequency domain you have the sounds
368
00:51:14,790 --> 00:51:15,790
and differences of those frequency components.
369
00:51:15,790 --> 00:51:16,790
So, here because f s by 2 and f in are there
we have sums and differences. Well, you kind
370
00:51:16,790 --> 00:51:23,420
of there, but let us see. So, if you think
of what is really happening right ideally
371
00:51:23,420 --> 00:51:33,090
we were
expecting. Let us say some sequence
372
00:51:33,090 --> 00:51:45,760
the crosses correspond to 1; 1 channel and
the
373
00:51:45,760 --> 00:51:50,020
circles correspond to the other. This is let
us say the ideal sequence expected out of
374
00:51:50,020 --> 00:51:53,940
the
whole system because of gain error what is
375
00:51:53,940 --> 00:52:01,130
happening every odd sample is getting
multiplied up by some constant which is different
376
00:52:01,130 --> 00:52:07,650
from unity. And every even sample is
getting multiplied by another constant which
377
00:52:07,650 --> 00:52:12,960
is not the same as the constant which was
multiplying the odd samples. So, you can think
378
00:52:12,960 --> 00:52:31,250
of this as taking this ideal sequence and
multiplying it by a sequence which is you
379
00:52:31,250 --> 00:52:46,460
understand what is this sequence 1 1 plus
alpha. This is the ideal output the actual
380
00:52:46,460 --> 00:52:52,840
output can be thought of is taking the ideal
output and multiplying it by a sequence where
381
00:52:52,840 --> 00:52:57,820
the even samples are.
1 plus alpha.
382
00:52:57,820 --> 00:53:00,530
Odd ones are.
And the odd samples are.
383
00:53:00,530 --> 00:53:07,650
1 plus beta.
1 plus beta correct and just now, we said
384
00:53:07,650 --> 00:53:11,080
that this kind of sequence what does the spectral
contain.
385
00:53:11,080 --> 00:53:14,720
C and c 1.
386
00:53:14,720 --> 00:53:16,620
In consist of d c.
And.
387
00:53:16,620 --> 00:53:23,200
A tone at.
A tone at at pi which in the continuous time
388
00:53:23,200 --> 00:53:24,740
domain maps to.
F s by 2.
389
00:53:24,740 --> 00:53:33,280
F s by 2. So, if you take a sequence and multiply
it by another sequence which has got
390
00:53:33,280 --> 00:53:40,310
both a constant term and a tone at pi what
would you expect for the output sequence.
391
00:53:40,310 --> 00:53:46,740
Sum sum and difference on.
You will see sum and.
392
00:53:46,740 --> 00:53:54,260
Difference of I mean a tone the tone at pi
will beat with the input tone and give you
393
00:53:54,260 --> 00:53:58,930
the
sum and difference. You understand that is
394
00:53:58,930 --> 00:54:13,590
why it makes sense that you see components
at I mean we see the original spectrum plus
395
00:54:13,590 --> 00:54:28,890
some components at pi minus f in by f s
times 2 pi. The intuition is that you multiply
396
00:54:28,890 --> 00:54:44,380
the ideal sequence with a sequence which is
got a constant value plus a tone at omega
397
00:54:44,380 --> 00:54:50,970
equal to pi which was you see other artifacts.
Now, taking this further to an n channel system,
398
00:54:50,970 --> 00:55:07,318
what do you think we will see? You will
have n n minus 1 components v c. So, at if
399
00:55:07,318 --> 00:55:15,130
n equal to 2 we have components around f s
by 2 correct if we have n equal to 4, you
400
00:55:15,130 --> 00:55:20,670
will have components around.
F s by 4.
401
00:55:20,670 --> 00:55:31,540
Right and its multiples you understand, is
this clear? So, we will stop here for today
402
00:55:31,540 --> 00:55:32,770
and
we will continue tomorrow.