Subject: δ(ω/ T) = T.δ(ω)
I'm sorry that this email turned out long. My apologies that I
couldn't figure out my exact question. So only managed to disperse all the
information I could that hopefully revolves around it.
To start with, I wish to understand the difference between the unit impulse
function (or dirac-delta function δ(t) ) and the unit sample sequence (or
discrete-time impulse, not a function though, δ(n) ). For the same, a
brief prelude follows:
When x(t) becomes x(nT), I thought it was sampling. But no. I realized that it
was just an impulse-train modulation, whereby x(nT) is still an analog SIGNAL
(not a sequence); just that x(t) was sifted by an impulse train comprising the
dirac-delta impulse function.
So x(nT) is x(t) found at intervals of time T, and not found (or zero) at all
other values of t -- still continuous and analog. So when we plot x(nT), it is
still against t on the abscissa, telling me that I'm still in the analog
domain: the signal x(nT) is a continuous signal in 't', though it once
appeared to me that it looks like a sample sequence x[n], which is not
continuous. So the difference.
Transpiring the frequency domain, t -> Ω (i.e. post-impulse train
modulation, t found at nT), I realize --
is not the same as t -> f (pre- impulse
train modulation), f = 1/ t.
When we talk of Ω, we also talk of Fs = 1/T. And though its called
sampling frequency, we're not yet in the discrete-time domain.
Moving from the impulse train modulation -> discrete-time domain:-
- x(nT) becomes x[n], i.e. a time normalization by a factor of T, and
correspondingly
- In the frequency domain, Ω gets scaled/ normalized by an inverse factor
1/T or Fs = 1/T.
So Ω/Fs = ΩT = ω. Then may I conclude, that:
{ Ω : t :: ω : n }
I don't know if I should call Ω an analog frequency and ω a
discrete-time frequency. If yes, then what I should I call f, where f = 1/t
before we knew time as t = nT !
Actually, it is Fs which we call the sampling frequency expressed as Hertz,
while Ω is expressed as radians/ sec.
For e.g. Ωs = 2π/ T. And in general Ω = ω/ T. So maybe I
shouldn't even term it 'frequency'.
In the transition from the impulse-train modulated signal -> discrete-time
sequence, may I comment on what we know as the impulse:
if { δ(Ω) or δ(ω/ T) : Area of the dirac-delta impulse
function }, then
{ δ(ω) : the sample value or the height of the impulse sequence,
and not the area }
Then w.r.t the equation in the subject (as the title) of this email, how can we
equate δ(ω/ T) with δ(ω)! I mean, the author states that
scaling the independent variable, also scales its AREA. Here I presume that both
Ω and ω are independent variables.
First of all δ(ω) is not an area. So scaling it by T and then
assigning to δ(ω/ T) which is actually an area doesn't make any
sense to me. Notably, the author says δ(ω/ T) = T.δ(ω)
Only a dirac-delta impulse FUNCTION is ascribed an area. A unit sample impulse
sequence is not a function, but just a floating value in height, because:
when we scale Ω by T, i.e. ω/ T * T, we're loosing information
about time in toto, i.e. when we reach the abscissa scaled in terms of
ω, we only know the value of the sample in frequency domain, not its
position in time, i.e. we've lost even its spacing because space-time being
a continuum.
I dunno. Maybe using the sample value δ(ω) and the value of ω,
we can calculate x[n] in the time domain,
(where ω = ωo +/- nωs; ωo = ΩoT = 2πFoT, Fo
being the fundamental frequency of the signal x(t); ωs = ΩsT =
2πFsT = 2π).
And by the way, I'm thinking that { δ(ω) : frequency domain ::
δ(n) : time domain }. Moreover, as we poked Ω, can ω be even
termed 'frequency'! I mean, if we call ω discrete-time
frequency, then what should I call -- some variable = 1/n in dealing with
ω, just as we know Fs = 1/T in dealing with Ω. Just that we've
dropped information about T before we're left purely with n. I mean, does
it imply that sampling implies dropping time info in moving from nT -> n or
ω/ T -> ω, having nothing to do with impulse-time modulation.
Drawing attention to pg. 174, the author seems to state that δ(ω/ T)
= π and δ(ω) = π/ T (from the equation δ(ω/ T) =
T.δ(ω)), which doesn't make sense to me at all, because the the
graphs on the page clearly show that δ(ω) = π and
δ(ω/ T) = δ(Ω) = π/ T. I mean:
ref. pg. 173 says x(t) = cos(4000πT), so in the frequency domain we have
X(jΩ) = πδ(Ω - 4000π) + πδ(Ω + 4000π).
So that's what I was referring to when I wrote δ(ω) = π
above.
And by the way, I forgot: "The area of an impulse is either '0' or
'1'; I mean the dirac-delta impulse function δ(t). Just alike,
the same applies even to the unit sample impulse δ(n), i.e. it is either
'0' or '1'. Then what about { δ(Ω) : δ(t)
:: δ(ω) : δ(n) }. Do the same properties also apply to
δ(Ω) and δ(ω)??"
If yes, then why not keep the area of the impulse seperate from
'π', in X(jΩ) = 'π' δ(Ω - 4000π)
+ 'π' δ(Ω + 4000π). To me, the π is like A in the
function x(t) = A cos(Ωot). It has nothing to do with the area of the
impulse. I can understand if π gets multiplied with δ(ω), because
in this case, it is not an area that is getting scaled, but just the value of
the sample, which is '1', being multiplied by π which modifies the
height of the sample. Well, you can go ahead and multiply π with
δ(Ω) too, but don't say that the area is now scaled, because
here its a dirac-delta function involved, whose area is ALWAYS 1. Multiplying it
by π doesn't change its area.
And coming back to the same question, "How can you equate an area with a
sample". I mean, is the author trying to say that multiplying a unit sample
impulse with T mutates it into a dirac-delta impulse?? That is ridiculous. Once
you've lost information about T in moving from nT -> n, you can never go
back. You just know the value of the samples, but you don't know the rate
at which they arrive. Which is like saying, "I know that the second sample
proceeds the first, but I don't know at what interval."
Or should I think that though I've lost T, the bastion of intervals is
compensated by ω! Because as said earlier,
ω = ωo +/- nωs. Or for what matter, Ω = Ωo +/-
kΩs.
So with a quantity δ(Ω - (Ω)), i.e. the points at which t = nT
(in the frequency domain ofcourse), we've
δ(Ω - Ωo +/- kΩs) = δ(ω/T - 2πFo +/-
k.2π/T), because ΩT = ω; and when Ω = Ωs, ω =
2π. So Ωs = 2π/ T.
So when in moving from Ω -> ω, the above quantity becomes
δ(ω - 2πFoT +/- k.2π). Which to me, as I see it, implies that
we've really not lost any information about time, because it is retained in
the quantity 2πFoT, where it does not get cancelled. Because the fundamental
frequency is independent of Ω or ω. It was in existence even before
they came into picture.
I mean, if the information about the interval is lost, there is no way that we
can recover the original signal x(t).
Sorry. I know that I'm only able to envisage. But I'm unable to vivify
the exact problem that lurks behind this email. If somebody could please help me
identify the same.
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